Methodology for estimating the intensity of failures of functional nodes integrated schemes

Baryshnikov A.V.

(FSUE Research Institute "Automation")

1. Introduction

The problem of predicting the reliability of radio-electronic equipment (REC) is relevant for almost all modern technical systems. Considering that REA includes electronic componentsThe task of developing techniques to evaluate the intensities of failures (IO) of these components. Often, the technical requirements for the reliability presented in the technical tasks (TK) on the development of the REA are in conflict with the requirements for the weights and dimensions of the REC, which does not allow the requirements of the TK by, for example, duplication.

For a number of types of REC, elevated reliability requirements are presented to controlling devices placed in one crystal with basic functional assemblies of equipment. For example, the addition scheme for module 2, providing control of the main and duplicating nodes of any blocks of equipment. Increased reliability requirements can also make any memory areas in which the information needed to perform the algorithm for the operation of the equipment is stored.

The proposed technique allows you to evaluate the IO of different functional areas of the microcircuit. In memory chips: operational storage devices (RAM), constant storage devices (ROM), reprogrammed storage devices (RPPU), these are the intensities of the failures of drives, decoders and control circuits. In the schemes of microcontrollers and microprocessors, the technique allows you to determine the memory regions, an arithmetic logical device, analog-digital and digital-analog converters, etc. In programmable logic integrated circuits (plis), and the main functional nodes, from which the FPGA consists of: a configurable logic unit, an input / output unit, memory area, JTAG, etc. The technique also allows you to determine the io of one output of the chip, one memory cell, and, in some cases, and the individual transistors.

2. Purpose and scope of the technique

The technique is intended to evaluate the operational io λ e of different functional assemblies of chips: microprocessors, microcontrollers, memory chips, programmable logical integrated circuits. In particular, inside the crystal areas of the memory, as well as the IO cells of the storage storage devices of foreign proceedings, including microprocessors, FPGA. Unfortunately, the lack of information about the enclosures does not allow to apply the methodology for domestic chips.

The IOs defined for this technique are the source data for calculating the reluctal characteristics during engineering studies of the equipment.

The technique contains an IO calculation algorithm, an algorithm for checking the obtained calculation results, examples of the calculation of the functional nodes of the microprocessor, memory schemes, programmable logic schemes.

3. Advisions of the methodology

The technique is based on the following assumptions:

Failures of the elements are independent;

Io chip is constant.

In addition, these assumptions will be shown the possibility of separation of Io chip on the enclosure and the intensity of the crystal failure.

4. Original data

1. Functional purpose chip: microprocessor, microcontroller, memory, plis, etc.

2. Technology of manufacturing chips: bipolar, CMOS.

3. Restaging the intensity of the microcircuit failures.

4. The block diagram of the microcircuit.

5.Type and the amount of memory stackers.

6. The number of body conclusions.

5.1. According to the known values \u200b\u200bof the Io chip, the IU of the body and crystal are determined.

5.2. According to the found value of the IE crystal, for the memory chip, based on its type and manufacturing technology, the drive of the drive, decrypt schemes, control circuits are calculated. The calculation is based on standard construction electrical schemesserving the drive.

5.3. For a microprocessor or a microcontroller using the calculation results obtained in the previous paragraph, the memory regions are determined. The difference between the Io crystal and the found values \u200b\u200bof the memory regions will be the value of the remaining part of the chip.

5.4. According to the known values \u200b\u200bof Io crystals for the FPGA family, their functional composition and the number of single-type nodes, a system of linear equations is compiled. Each of the equations of the system is compiled for one symptominator from the FPGA family. The right side of each of the equations of the system is the amount of the values \u200b\u200bof the values \u200b\u200bof the functional nodes of a certain type on their number. The left part of each of the equations of the system is the value of the Io crystal of a specific model of FPGA from the family.

Maximum amount The system equations are equal to the number of FPGS in the family.

The solution of the system of equations allows you to obtain the values \u200b\u200bof the FPGA functional nodes.

5.5. Based on the results of the calculation obtained in the previous paragraphs, the values \u200b\u200bof the individual memory cell, the output of the chip or transistor of a particular block diagram can be found, if the electrical principal node scheme is known.

5.6. Checking the results of the calculation for the memory chip is performed by comparing the IO value for another memory chip received standard method, With the value of the IO of this chip, calculated using the data obtained in paragraph 5.2 of this section.

5.7. Checking the results of the calculation for FPGA is made by calculating the Io crystal of one of the symptoms of the considered family of FPGA, which was not included in the equation system. The calculation is carried out using the values \u200b\u200bof the functional nodes obtained in paragraph 5.4 of this partition, and the comparison of the obtained value of the FPGA with the value of the IO, calculated using standard methods.

6. Analysis of the method for predicting the intensity of failures by microcircuits in terms of the possibility of separating the intensity of the chip failures in the amount of the intensities of the crystal failure and corps

Io crystal, hull and external conclusions Microcircuits are determined from the mathematical model for predicting Io foreign integrated circuits for each model of IP.

Let's analyze the terms of the mathematical model to calculate the operation

io λ. e digital and analog integrated circuits of foreign production:

λ e \u003d (with 1 π t + with 2 π e) π q π l, (1),

where: C 1 is the component of IO IP, depending on the degree of integration;

π T - coefficient taking into account the overheating of the crystal relatively ambient;

C 2 - the component of IO IP, depending on the type of body;

- π e is the coefficient, taking into account the rigidity of the operating conditions of the REA (group of operation group);

- π q is a coefficient that takes into account the quality of the manufacture of ERI;

- π l-cell, taking into account the workshop technological process Manufacturing ERI;

This expression is fair for chips made both by bipolar and MOS technology, and includes digital and analog schemes, programmable logical matrices and pls, memory chips, micropro-processors.

The mathematical model of the predicted IO integral chip, for the original source of which the standard of the US Department of Defense is taken is the sum of the two terms. The first term characterizes the failures determined by the degree of integration of the crystal and the electrical mode of the microcircuit (coefficients C 1, π T), the second term characterizes the failures associated with the type of hull, the number of body conclusions and operating conditions (C 2, - π E coefficients).

Such a separation is explained by the possibility of issuing the same chip in different types of buildings substantially differing in their reliability (resistance to vibrations, tightness, hygroscopicity, etc.). Denote the first term as an Io-defined crystal (ΛKR ), and the second is the case (Λkorp).

From (1) we get:

λKR \u003d C 1 π t π q π l, λkorp \u003d c 2 π e π q π l (2)

Then the Io of one conclusion of the chip is equal to:

λ 1 \u003d λkorp / n output \u003d c 2 π e π q π l / n output,

where n output is the number of conclusions in the housing of the integrated circuit.

We will find the ratio of the case of the case to the operational Io chip:

λkorp / λ. e \u003d C 2 π E И q π L / (C 1 π T + C 2 π E) π q π L \u003d C 2 π E / (C 1 π T + C 2 π E) (3)

We analyze this expression from the point of view of the influence of the type of body, the number of conclusions, the overheating of the crystal due to the power dissipated in the crystal, the stiffness of the operating conditions.

6.1. Effect of hardness of operating conditions

Dividing the numerator and denominator of expression (3) on the ratio π E we get:

λkorp / λ. e \u003d C 2 / (with 1 π t / π E + C 2) (4)

Analysis of the expression (4) shows that the percentage of the case of the case and the operational Io chip depends on the operation group: the more tougher conditions of operation of the equipment (more value of the coefficient π E), the greater the fraction of the failure falls on the body failures (the denominator in equation 4 decreases) and AttitudeΛkorp / λE to strive for 1.

6.2. Influence of the type of body and the number of housing conclusions

Selecting the numerator and denominator of the expression (3) on the C 2 ratio:

λkorp / λ. e \u003d π E / (with 1 π t / s 2 + π e) (5)

Analysis of the expression (5) shows that the percentage ratio of the case of the body and the operational Io chip depends on the ratio of coefficients with 1 and C 2, i.e. From the ratio of the degree of integration of chip and body parameters: more quantity Elements in the chip (more coefficient C 1), the smaller share of failures falls on the case failures (ratioλkorp / λ. e strive for zero) and the more the amount of conclusions in the case, the greater the weight gain the failures of the case (attitudeλkorp / λ. u to strive for 1).

6.3. The effect of power dissipated in a crystal

From the expression (3) it can be seen that with an increase of π T (the coefficient reflecting the overheating of the crystal due to the power dissipated in the crystal), the value of the valve equation increases, and, consequently, the fraction of failures per case decreases and crystal failures acquire greater relative weight.

Output:

Analysis of the change value of the relationship λkorp / λ. e (equation 3) depending on the type of body, the amount of conclusions, the overheating of the crystal due to the power dissipated in the crystal, and the rigidity of the operating conditions showed that the first term in equation (1) characterizes the operational Io of the crystal, the second - the operational IU of the body and equation (2) may be Used to estimate the operational io directly semiconductor crystal, housing and IO of the body conclusions. The value of the operational Io crystal can be used as the source material for evaluating the IO of the microcircuit functional nodes.

7. Calculation of the intensity of failures of the memory cell of storage devices that are part of the memory microcircuit, microprocessors and microcontrollers.

To determine the IO, which occurs on the bits of semiconductor information, consider their composition. The semiconductor memory of any type includes :

1) Drive

2) Framing scheme:

o Address part (lowercase and column decoders)

o Number part (write and read and read amplifiers)

o Local control unit - carries out coordination of all nodes in storage, recording modes, regeneration (dynamic memory) and erasing information (RPPU).

7.1. Estimation of the number of transistors in various areas of the memory.

Consider each component of IO Ply. The general value of the IO memory for chips of different types with different volume of the drive can be determined using. The enclosures and crystal are calculated in accordance with section 5 of this work.

Unfortunately, in technical materials on foreign memory chips there is no total number of elements included in the microcircuit, and only the information capacity of the drive is driven. Given the fact that each type of memory contains standard blocks, we estimate the number of elements included in the memory chip based on the drive. To do this, consider the circuitry of the construction of each block of the memory.

7.1.1. Drive RAM

The electrical concepts of storage rates of RAM, made according to TTLS, ESL, MOS and CMOS technologies are given. Table 1 shows the number of transistors from which one memory cell (1 bit of information of the RAM) is constructed.

Table 1. Number of transistors in one memory cell memory

Type of RAM	Manufacturing technology
		TTLSH	ESL	MOP	CMOS
Static	Amount of elements				4, 5, 6
Dynamic

7.1.2. ROM and PPZ drives

In bipolar ROM and PPZ, the storage element is implemented on the basis of diode and transistor structures. They are performed in the form of emitter repeaters onn - p - n and p - n - p transistors, transitions collector-base, emitter base, Schottky diodes. As a storage element in the schemes manufactured by MOS and CMOS technologies are usedp and N. - Channel transistors. The storage element consists of 1 transistor or diode. The total number of transistors in the ROM or PPZ accumulator is equal to information container Bis.

7.1.3. RPZU drive

The information recorded in the RPPU is stored from several to decades. Therefore, RPPU is often called non-volatile memory. The basis of the mechanism of registration

the mining and storage of information is the processes of accumulation of charge when recording, saving it when reading and when the power is turned off in special MOS transistors. Storage elements RPPU are built, as a rule, on two transistors.

Thus, the number of transistors in the RPPU drive is equal to the information container of the RPPU multiplied by 2.

7.1.4. Address part

The address part of the memory is based on decoders (decoders). They allow you to determineN. -Regable input binary number by producing a single value of a binary variable on one of the device outputs. To build integrated circuits, it is customary to use linear decoders or a combination of linear and rectangular decoders. Linear decoder hasN inputs and 2 n logic schemes "and". We find the number of transistors necessary to build such decoders in the CMOS basis (as most commonly used to create an BIS). Table 2 shows the number of transistors necessary for the construction of decoders to a different number of inputs.

Table 2. Number of transistors required for constructing decoders

Number of Inputs	Address inverters		Schemes "and"		The total number of transistors in the de-cipher 2 * n * 2 n + 2 * n
	Number of Inverters	Number of Transistors	Number of cHEM.	Number of transistors 2 * n * 2 n
				4*4=16	16+4=20
				6*8=48	48+6=54
				8*16=128	128+8=136
				10*32 = 320	320+10 = 330
				64*12 = 768	768+12 = 780
				128*14=1792	1792+14=1806
				256*16=4096	4096+16=4112
				512*18=9216	9216+18=9234
			1024	1024*20=20480	20480+20=20500

For linear decoders, the discharge of the decoded number does not exceed 8-10. Therefore, with an increase in the number of words in the memory, more than 1k use the modular principle of building a memory.

7.1.5. Number part

(Record and read amplifiers)

These schemes are designed to convert the levels of readable signals in the output levels of the logical elements of a particular type and an increase in load capacity. As a rule, they are performed according to the open collector (bipolar) or three states (CMOS). Each output schemes may consist of several (two or three) inverters. The maximum number of transistors in these schemes at the maximum bit of microprocessor 32 is no more than 200.

7.1.6. Local control unit

The local control unit, depending on the type of memory, can include lowercase and column buffer registers, address multiplexers, regeneration control units in dynamic memory, information erasing circuits.

7.1.7. Estimation of the number of transistors in various areas

The quantitative ratio of RAM transistors included in the drive, decoder and local control unit is approximately equal to: 100: 10: 1, which is 89%, 10% and 1%, respectively. The number of transistors in the RAM drive cell, ROM, PPZ, RPPU is given in Table 1. Using the data of this table, the percentage ratios of the elements included in various areas of RAM, as well as assuming that the number of elements in the decoder and the local control unit for the same The volume of the drive different types The memory remains approximately constant, you can estimate the ratio of transistors included in the drive, decoder and a block of local control of different types of memory. Table 3 shows the results of such an assessment.

Table 3 Quantitative ratio of transistors in different functional areas

	Quantitative relationship of elements of various areas
	Storage device	Decoder	Local control unit
ROM, PPZ.

Thus, knowing the volume of the drive and Io crystal zoom, you can find the drive of the drive, the address part, the numeric part, the local control unit, as well as the IU of the memory cells and transistors included in the framing schemes.

8. Calculation of the intensity of failures of functional nodes of microprocessors and microcontrollers

The section shows an algorithm for calculating the functional nodes of microcircuit of microprocessors and microcontrollers. The technique is applicable to microprocessors and microcontrollers with a bit of no more than 32 bits.

8.1. Source data for calculating failure intensity

Below are the initial data necessary to calculate the io microprocessors, microcontrollers and parts of their electrical circuits. Under part of the electrical scheme, we will understand as functionally completed microprocessor nodes (microcontroller), namely, different types Memoys (RAM, ROM, PPZU, RPPU, ADC, DAC, etc.) and individual valves or even transistors.

Initial data

The discharge of a microprocessor or a microcontroller;

Microcircuit manufacturing technology;

View and organization inside crystal memory;

Information capacity of the memory;

Power consumption;

Heat resistance Crystal - body or crystal - Environment;

Microcircuit type type;

The number of body conclusions;

Increased working temperature Environment.

The level of quality of manufacture.

8.2. Algorithm for calculating the intensity of the failure of a microprocessor (microcontroller) and microprocessor functional nodes (microcontroller)

1. Width the operational Io of the microprocessor or microcontroller (λE MP) using the source data using one of the automated calculation programs: "ASRN", "Asonyika-K" or using the Military Handbook 217F standard.

Note: Further, all calculations and comments will be brought from the point of view of the use of ACR, because Methodologies for the use and content of programs, "Asonyk-K" and the "Military Handbook 217F" standard have a lot in common.

2. Determine the value of the IO of the microprocessor (λ E RAM, λ e-ROM, PPZU, λ E RPPU), assuming that each memory is a separate chip in its housing.

λ e zo \u003d λ RAM + Λkorp,

λ e-ROM, PPZ \u003d λ ROM, PPZA + Λkorp,

λ e rpza \u003d λ rpze + λkorp,

where λ e is the operational values \u200b\u200bof the IO of different types of memory, λkorp, - the enclosures for each type: λ RAM, λ ROM, PPZ, Λ RPZ - IO RAM, ROM, RPZ, RPPU without taking into account the housing, respectively.

Search for source data to calculate the operational values \u200b\u200bof AO of different types of memory technical information (Data Sheet) and catalogs of integrated circuits. In this literature, it is necessary to find a memory, the type of which (RAM, ROM, PPZU, RPPU), the volume of the drive, the organization and manufacturing technology coincide or close to the microprocessor (microcontroller). Found technical characteristics of memory microcircuits are used in ACRs to calculate the operational Io chip. The power consumed by the memory is selected based on the electrical mode of the microprocessor (microcontroller).

3. Determine the values \u200b\u200bof the IO inside the crystal areas of the microprocessor (microcontroller), the memory and alu excluding the housing: λKR MP, λ RAM, λ ROM, PPZE, λ RPZU ,. Λ Al.

Io inside the crystal areas of the microprocessor, RAM, ROM, PPZE, RPPU is determined from the relation: λKR \u003d C 1 π T π Q π L.

Io Alu and parts of the crystal without memory schemes is determined from the expression:

. λ Alu \u003d ΛKR MP - Λ RAM - λ ROM, PPZE - λ RPPU

The values \u200b\u200bof the other functionally completed parts of the microprocessor (microcontroller) are similar.

4. Determine the drives inside the crystal memory: λ H RAM, λ n ROM, PPZE, λ n RPPU.

Based on the data of Table 3, it is possible to express the percentage ratio of the number of transistors in different functional areas of the memory, assuming that the total number of transistors in the memory is 100%. Table 4 shows this percentage ratio of transistors included in the crystal memory of different types.

Based on the percentage ratio of the number of transistors included in the different functional areas and the found value of the Io inside the crystal part, the functional nodes are determined.

Table 4. Percent Transistor Relationship

	The quantitative ratio of transistors of functional regions ZU (%)
	Storage device	Decoder	Local control unit
ROM, PPZ.

λ n RAM \u003d 0.89 * λ RAM;

λ n ROM, PPZ \u003d 0.607 * λ ROM, PPZA;

λ n RPPS \u003d 0.75 * λ RPPU,

where: λ n RAM, λ n ROM, PPZ, λ n RPPU - Io drives of RAM, ROM, PPZ, RPPU, respectively, but.

8.3. Calculation of the intensity of failures of functional nodes: decoders, address part, control circuits.

Using data on the ratio of the number of transistors in each part of the memory (Table 4), you can find the intensities of the failures of the decoders, the address part and the control circuits. Knowing the number of transistors in each part of the memory can be found intensity of group failures or individual transistors of the memory.

9. Calculation of the intensity of failures of functionally completed memory microcircuits

The section shows an algorithm for calculating the functionally completed nodes of memory chips of storage devices. The methodology is applicable for memory chips given to ACR.

9.1. Source data for calculating failure intensity

Below are the source data required to calculate the functionally completed memory microcircuits. Under the functionally finished memory chips, we will understand the drive, address part, control circuit. The technique also allows you to calculate the Io parts of functional nodes, individual valves, transistors.

Initial data

Type of memory: RAM, ROM, PPZ, RPZU;

Information capacity of the memory;

Organization of RAM;

Manufacturing technology;

Power consumption;

Microcircuit type type;

The number of body conclusions;

Heat resistance Crystal - body or crystal - Environment;

Group of operation of the equipment;

Increased working temperature of the environment;

The level of quality of manufacture.

9.2. Algorithm for calculating the intensity of failures of memory schemes and functionally completed memory schemes

1, determine the operational IU of the memory chip (λE P) using the source data using one of the automated calculation programs: "AsRN", "Asonika-K" or using the Military Handbook 217F standard.

2. Determine the values \u200b\u200bof the IO of the crystal of the memory without the case of λCR.

ΛKR \u003d C 1 π T π Q π L.

3. Calculation of the IU of the drive inside the crystal memory and the functional components of functional nodes in accordance with section 8.2.

10. Calculation of the intensity of failures of functionally completed nodes of programmable logical integrated circuits and basic matrix crystals

Each FPGA family consists of a set of original microcircuits of the same architecture. The architecture of the crystal is based on the use of identical functional nodes of several types. The chips of different symptoms within the family differ from each other by the type of housing and the number of function nodes of each type: a configurable logic unit, an input / output unit, memory, JTAG and the like.

It should be noted that, in addition to configurable logical blocks and input / output units, each FPGA contains a matrix of keys forming links between elements of FPGA. Given the fact that these areas are distributed evenly by crystal, except for the input / output blocks that are placed on the periphery, we can assume that the keys matrix is \u200b\u200bpart of the configurable logical blocks and input / output units.

To calculate the values \u200b\u200bof the intensities of the failures of function nodes, it is necessary to make a system of linear equations. The system of equations is compiled for each FPGA family.

Each of the equations of the system is equality, in the left part of which the value of the Io crystal is recorded for a particular symptomine of the chip from the selected family. The right side is the amount of the amount of the number of functional nodes N category I on the IO of these nodes λni.

Below are given general form Such a system of equations.

λ e a \u003d a 1 λ 1 + a 2 λ 2 + ... + a n λ n

λ e b \u003d b 1 λ 1 + b 2 λ 2 + ... + b n λ n

……………………………

λ e k \u003d k 1 λ 1 + k 2 λ 2 + ... + k n λ n

where

λ e a, λ e b, ... λ e k - operational Io microcircuits of the FPGA family (microcircuits A, B, ... K, respectively),

a 1, a 2, ..., a n - the number of functional nodes 1, 2, ... n category in the chip A, respectively

b 1, b 2, ..., b n - the number of functional nodes of category 1, 2, ... n, in the chip in, respectively,

k 1, k 2, ..., k n - the number of functional nodes of category 1, 2, ... n, in the microcircuit to, respectively,

λ 1, λ 2, ..., λ n is the functional nodes of category 1, 2, ... n, respectively.

The values \u200b\u200bof the operational Io microcircuits λ e a, λ e b, ... λ e k are calculated according to ASR, the number and type of functional nodes are given in the technical documentation on Plis (Data Sheet or in the domestic periodical).

The values \u200b\u200bof the functional nodes of the FPGA family λ 1, λ 2, ..., λ n are from the solution of the equation system.

11. Check the results of the calculation

Checking the results of the calculation for the memory microcircuit is made by calculating the Io crystal of another memory chip using the obtained value of the memory cell of the memory cell and the comparison of the obtained value of the Io crystal with the value of the IO, calculated using standard methods (ASR, Asonyika, etc.).

Checking the results of the calculation for FPGA is calculated by the calculation of the FPGA crystal of another called from the same family using the found values \u200b\u200bof the FPG functional nodes and the comparison of the obtained values \u200b\u200bof the FPGA with the value of IO, calculated using standard methods (ASRN, asionics, etc.) .

12. An example of calculating the intensities of failures of FPGA functional nodes and checking the results of the calculation

12.1. Calculation of IO of Functional Nodes and Plis Cases

The calculation of the IO was carried out on the example of the FPGA of the SPARTAN family developed by Xilinx.

The Spartan family consists of 5 PLIT Tyobominations, which include the matrix of configurable logic blocks, the input / output blocks, the logic of the boundary scanning (JTAG).

Flags included in the Spartan family are distinguished by the number of logical valves, the number of configurable logical blocks, the number of input / output blocks, types of housing and the number of body conclusions.

Below is the calculation of the IO of the configurable logical blocks, the input / output units, JTAG for FPGA XSS 05XL, CSS 10XL, CSS 20XL.

To verify the results obtained, the operational Io FPGA XSS 30XL is calculated. Operational Io PLIT XSS 30XL is calculated using the values \u200b\u200bof the functional nodes of PLIT XSS 05XL, CSS 10XL, CSS 20XL. The obtained value of the FPGA XSS 30XL is compared with the value of the IO, calculated using the ACR. Also, to verify the results obtained, the values \u200b\u200bof the AO of one output for different packs of FPGA are compared.

12.1.1. Calculation of the intensities of failures of function nodes FPGA XSS 05XL, CSS 10XL, CSS 20XL

In accordance with the above calculation algorithm to calculate the FPG functional nodes, it is necessary:

Make up the list and values \u200b\u200bof the source data for FPGA XSS 05XL, CSS 10XL, CSS 20XL, CSS 30XL;

Calculate Operational Io PlisXs 05xl, hss 10xl, hss 20xl, hss 30xl (calculation is carried out by using source data);

Make a system of linear equations for crystals FPG CXS 05XL, CSS 10XL, CSS 20XL;

Find the solution of a system of linear equations (unknown in the system of equations are functional nodes: configurable logical blocks, output blocks of the output, border scanning logs);

Compare the values \u200b\u200bof the FPG XS 30XL crystal, obtained in the previous paragraph, with the value of the Io crystal obtained using the ACR;

Compare the output values \u200b\u200bfor different enclosures;

Formulate the conclusion about the validity of the calculations;

When obtaining a satisfactory coincidence of failure intensities (from 10% to 20%), stop calculations;

With a large discrepanization of the results of the calculation, the initial data correction is corrected.

In accordance with The source data for calculating the operational Io FPGS are: manufacturing technology, the number of valves, power consumption, the cooler overheating temperature relative to the environment, body type, the number of body conclusions, thermal resistance of the Crystal Corps, the level of production quality, the operation of the equipment in which Plce is applied .

All source data, in addition to the power consumption, the temperature of the crystal overheating and the instrumentation group, are given in. Power consumption can be found either in the technical literature, or by calculating or measuring on the board. The overheating temperature of the environment relative to the environment is like a product of power consumed and thermal resistance Crystal-housing. The instrument of operation of the equipment is given in technical conditions on the equipment.

The initial data for calculating the operational intensity of references of PLIT XSS 05XL, CSS 10XL, CSS 20XL, XSS 30XL is shown in Table 5.

Table 5. Original data

Source	Tyobominal Plis
	Xs 05xl	Xs 10xl	Xs 20xl.	Xs 30xl
Technology manufacturing
Maximum number of logs clear valves
Number of con-appear logic. Blocks, N CLB
The number of IS-used inputs / outputs, N B / out
Type of shell	VQFP.	TQFP.	Pqfp.	Pqfp.
Number of body conclusions
Heat Safety Crys-Tall - Case, 0 C / W
Question level of manufacture	Commercial
Operation group apparatus

To determine the overheating temperature of the crystal relative to the ambient temperature, it is necessary to find power consumed for each chip.

In most CMOS integrated circuits, almost all dissipated power is dynamic and is determined by the charge and discharge of internal and external load tanks. Each conclusion in the microcircuit dispels the power according to its capacity, which is constant for each type of output, and the frequency at which each output switches may differ from the clock frequency of the microcircuit. General dynamic power is the amount of power dissipated on each output. Thus, for calculating power, you need to know the number of elements used in Plis. In for the Spartan family, the values \u200b\u200bof the current consumption current of the input / output (12mA) with a load of 50 PF, the supply voltage of 3.3 and the maximum frequency of FPGA 80 MHz. Assuming that the power consumption is determined by the number of switching blocks input / output (as the most powerful energy consumers), and due to the lack of experimental data on consumption power, we estimate the power consumed by each FPGA, given that 50% of the input / output blocks are simultaneously switched Some fixed frequency (when calculating the frequency was selected 5 times lower than the maximum).

Table 6 shows the values \u200b\u200bof the power consumed of the pliz and the overheating temperature of the crystals relative to the microcircuit body.

Table 6. Power consumed byplla

	Xs 05xl	Xs 10xl	Xs 20xl.	Xs 30xl
Consumed power, W
Crystal overheating temperature, 0 C

Calculate the values \u200b\u200bof the coefficients in equation (1):

λ e \u003d (with 1 π t + with 2 π e) π q π l

The coefficients π T, C 2, π e, π q, π l are calculated on the ASR. Coefficients with 1 find using approximation of the values \u200b\u200bof the coefficient C 1 given to the ACR for PLITS of varying degrees of integration.

The values \u200b\u200bof the coefficient C 1 for FPGA are given in Table 7.

Table 7. Coefficient values \u200b\u200bwith 1

Number of valves in Plis	Coefficient values \u200b\u200bwith 1
Up to 500.	0,00085
From 501 to 1000.	0,0017
From 2001 to 5000	0,0034
From 5001 to 20,000	0,0068

Then for the maximum number of valves FPS CSS 05XL, XSS 10XL, CSS 20XL, XSS 30XL We obtain the value of the coefficient C 1, 0.0034, 0.0048, 0.0068, 0.0078, respectively.

Values \u200b\u200bof coefficients π T, C 2, π E, π Q, π L, values \u200b\u200bof Io crystals and enclosures, as well as operational values \u200b\u200bof Io chipXSS 05XL, CSS 10XL, CSS 20XL, CSS 30XL shown in Table 8.

Table 8. Operational values \u200b\u200bof IO Plis

Designation and name of coefficients	Values \u200b\u200bof coefficients
	Xs 05xl	Xs 10xl	Xs 20xl.	Xs 30xl
π T.	0,231	0,225	0,231	0,222
With 2	0,04	0,06	0,089	0,104
π E.
π Q.
π L.
The intensity of Cree Stall failures,λKR \u003d. C 1 π t π q π l * 10 6 1 / hour	0,0007854	0,0011	0,00157	0,0018
Corusa failure intensityλkorp \u003d C 2 π E π Q π L * 10 6 1 / hour			0,445	0,52
Operational failureλE * 10 6 1 / h	0,2007854	0,3011	0,44657	0,5218

Find the values \u200b\u200bof the configurable logical blocks λ CLB, blocks / output blocks λ w / out and border scanning logicλ jtag for pls xss 05xl, xss 10xl, hss 20xl . To do this, make a system of linear equations:* S 05 XL - Io crystal, number of configurable logical blocks, number of input / output blocks for FPGA XSS 05XL, respectively;

λKR XC S 10 XL, N CLB XC S 10 XL, N B / You xs S 10 XL - Io crystal, number of configurable logical blocks, number of blocks input / output for PLIT XSS 10XL, respectively;

ΛKR XC S 20 XL, N CLB XC S 20 XL, N B / you xs S 20 XL - Io crystal, number of configurable logical blocks, number of input / output blocks for PLIT XSS 20XL, respectively.

Substituting into the system of equations Io crystals, the number of configurable logic blocks and input / output units, we obtain: 0.00157 * 10 -6 \u003d 400 * λ CLB + 160 * λ И х / out + λ jtag

The system of three linear equations with three unknowns has a single solution:

λ CLB \u003d 5.16 * 10 -13 1 / h;λ w / out \u003d 7.58 * 10 -12 1 / h; λ. Jtag \u003d 1,498 * 10 -10 1 / hour.

12.1.2. Check the results of the calculation

To verify the solution obtained, calculate the IO crystal of FPGSXs s 30 xl λKR xs s 30 xl using found valuesλ CLB, λ И х / out, λ jtag.

By analogy with system equationsλkr xs s 30 xl 1 is equal to:

ΛKR XS s 30 xl 1 \u003d λ bbb \u200b\u200b* n CLB xs s 30 xl + λ w / out * n q / you xs s 30 xl + λ jtag \u003d

576* 5,16*10 -13 + 192*7,58*10 -12 + 1,498 * 10 -10 \u003d 0.0019 * 10 -6 1 / hour.

The value of the IO crystal obtained using AsRN is equal to (Table 9): 0.0018* 10 -6. The percentage of these values \u200b\u200bis: (ΛKR XC S 30 XL 1 - ΛKR XC S 30 XL) * 100% / λKR XC S 30 XL 1 ≈ 5%.

AO of one output obtained by the division of Io on the number of conclusions in the housings for FPGA xsS 05 xl, xs s 10 xl, xs s 20 xl, xs s 20 xl , equal to 0.002 * 10 -6, 0.00208 * 10 -6, 0.0021 * 10 -6, 0.0021 * 10 -6, respectively, i.e. differ no more than 5%.

The difference in the values \u200b\u200bof the accounting of about 5% is determined, probably adopted when calculating the approximate values \u200b\u200bof the dispersion capacity, and, as a result, inaccurate values \u200b\u200bof the coefficientsπ T, as well as the presence of unaccounted elements of FPGS, information about which is missing in the documentation.

The Appendix contains a block of calculation scheme and testing the intensities of the failures of the functional regions of FPGS.

13. Conclusions

1. It is recommended by the method of evaluating the functional nodes of integrated circuits.

2. It allows you to count:

a) for memory schemes - Io storage storage devices, memory cells, decoders, control circuits;

b) for microprocessors and microcontrollers - io storage devices, registers, ADCs, DAC and based on their basis of functional blocks;

c) for programmable logical integrated circuits - IO, which are included in them blocks of different functional purposes - configurable logical blocks, input / output units, memory cells, JTAG and based on their functional blocks.

3. The method of checking the calculated values \u200b\u200bof IO functional nodes.

4. Application of the test methodology, calculated values \u200b\u200bof the functional assemblies of integrated circuits, showed the adequacy of the proposed approach to evaluate the IO.

application

Blockey scheme for calculating the intensity failures of FPGA functional nodes

Literature

PORTER D.C, FINKE W.A. REABILITY CHARACTERIZATION AN PREDICTION OF IC. Pads-TR-70, P.232.

Military Handbook 217F. "REABILITY PREDICTION OF ELECTRONIC EQUIPMENT". Department of Defense, Washington, DC 20301.

“Automated system The calculation of reliability ", developed by the 22CNII of the Ministry of Defense of the Russian Federation with the participation of the Renia" ELECTRONSTANDART "and JSC" Standorelektro ", 2006

"Semiconductor storage devices and their use", V.P.andreyev, V.V. Baranov, N.V. Bekin et al.; Edited by Gordonov. M. Radio and Communication. 1981.-344.

Development prospects computer equipment: B. 11 kN: Ref. manual / edited by Yu.M.Smirnova. Kn. 7: "Semiconductor storage devices", A.B.Akinfiyev, V.I. Mironetsev, GD Sofia, V.V. Titsrkin. - M.: Higher. shk. 1989. - 160 p.: Il.

"Schemery of bis permanent storage devices", O.A.Petrosyan, I.Y.Kozyr, L.A. Koldov, Yu.I.Schetinin. - m.; Radio and communication, 1987, 304 p.

"Reliability of operational storage devices", computer, Leningrad, Energoisdat, 1987, 168 p.

TIIER, T.75, IP.9, 1987.

Xilinx. The Programmable Logic. Date Book 2008http: www.xilinx.com.

"Sector of Electronic Components", Russia-2002g - M.: Publishing House "Dodeca-XXI", 2002.

DS00049R-Page 61  2001 MICROCHIP TECHNOLOGY INC.

TMS320VC5416 Fixed-Point Digital Signal Processor, Data Manual, Literature Number SPRS095K.

CD-ROM firms Integrated Device Technology.

Holtec Semiconductor CD-ROM.

test

3. Calculation of failure intensity

Calculate the intensity of failures for the specified values \u200b\u200bT and T

The control subsystem includes k series connected blocks (Fig.3.1).

Figure 3.1 - Electronic Block Connection Scheme

The failure intensity is calculated by formula (3.1).

where is the statistical probability of device failure on the interval (T, T + DT)

P (T) -Beverness trouble-free work devices;

DT \u003d 3 · 103 hours. The observation interval adopted earlier in the work;

I define the statistical probability of the device failure at a given interval (12.5 · 103h) from table (2.1) and I find the intensity of failures;

Provided that the intensity of failures does not change during the entire service life of the object, i.e. L \u003d const, then the operation to failure is distributed across the exponential law and the probability of trouble-free operation of the block in this case is determined by formula (3.2)

And the average blocking of the block to failure is determined by formula (3.3)

The intensity of failures of the LP (T) subsystem formed from K-sequentially included blocks, finding the formula (3.4)

Since all blocks have the same failure system, I define according to formula (3.5)

The probability of trouble-free operation of the subsystem determines according to formula (3.6)

The average time for the failure of the subsystem define similarly by formula (3.3)

The results of the calculation of the dependences of the probabilities of the trouble-free operation of one block and the subsystem from the operation I will none in Table 3.2

Table 3.2.

I build a graph of dependencies and

Figure 3.1 - Dependence Schedule and.

For any distribution of operational failure, the probability of trouble-free operation of a subsystem consisting of k-series connected blocks is associated with probabilities of the trouble-free operation of these blocks by the formula (3.7)

If the blocks are equally reliable, then the probability of trouble-free operation of the subsystem define by formula (3.8)

I calculate the likelihood of the trouble-free operation of the subsystem during an occasionally equal to formulas (3.6) and (3.8) and compare the results:

The results of the calculation on both formulas are the same.

To resolve practical tasks on the organization of road traffic, recommendations may be used to select the values \u200b\u200bof the emergency coefficients shown in Table 2.2 ...

Analysis of the road safety of the Vaninsky District of the Khabarovsk Territory

To calculate the average annual daily intensity, the transition coefficients from AUC 42 - 87 / / are used. The calculation is made by the formula: (2.3) where: traffic intensity per hour ...

The reliability of non-standard aircraft products

The trouble-free operation of the air conditioning system of the aircraft

The distance between the extreme sections on the constructed time diagrams determines the scope of which the value of which is divided into L of intervals and cross sections of the diagram corresponding to the interval boundaries are carried out ...

To assess the real loading of the crossroad by transport to use the absolute value of intensity incorrectly, since it does not take into account the composition of transport flows (TP) ...

Modeling of the Transport Flow of Grinchilds and Greenberg

Construction of the main chart according to the main equation of transport flow: n \u003d k V, (4.1) where N is the intensity of the transport flow, the auth. / h; k - density, auth. / km; V - speed, km / h. With a known pricker and vcClik from formula (4.1), we express: kccik \u003d NcIKl / vc, (4 ...

Organization of safety in road transport

The intensity of the mixed stream movement is determined by the formula: where the IIJ is an incoming transport stream according to the i-th direction of the Jth component,% K is the percentage of the type of transport included in the estimated stream ...

Road traffic

The intensity of the vehicle movement in the direction in the above NDR units is determined by the formula: (1) where Ni is a given intensity of the movement according to the i-th direction, aut / h; i is the movement direction number; RL, RG ...

Basics of the theory of reliability and diagnostics

The failure intensity (L), thousand km-1, - the conditional density of the probability of the failure of the failure of the current collector L -13U, determined for the time under consideration, provided that before that moment the failure did not appear ...

Evaluation of the reliability of the turning and rope machine brand 1K62 CJSC "Av'ar Airlines"

The failure tree or the tree of accidents is a complex graphic structure underlying verbally - graphic Fashion Analysis of the emergence of the accident from sequences and combinations of faults and failures of the elements of the system ...

Crossroads ul. Leithesin - ul. Revolution

The calculation of the intensity is carried out separately for pedestrian and transport streams, for each direction of movement. At the specified UDS plot, it is necessary to calculate the number of vehicles (TC) and pedestrians passing through the intersection ...

Payment optimal number Mechanization on the airport cargo courtyard

The intensity of the flowing stream of type I from the departure warehouse per side:, [pallet / min], where - the maximum amount of shipments in the watch "Peak", the day "Peak", the month of "peak", t / h; - Coefficient Taking into account the long and heavy loads (0.85--0 ...

Repair of devices of electrical centralization of arrows management on railway

Arrows together with electrical drives on them are essential electrical centralization nodes. The failure of the arrows can minimize the reliability of any centralization system and lead to the most severe consequences ...

Electric locomotive control circuit diagnostics system

Improving the organization maintenance Cargo wagons

The initial data for the calculation is shown in Table 2.1 Table 2.1 - the number of compounds that followed the Pinsk Zhabinka area and the number of cars as part of the month indicator 1 2 3 4 5 6 7 8 9 10 11 12 4.5 4,6 5,1 5, 5 5.8 4.8 4.7 4.1 3 ...

Part 1.

Introduction
The development of modern equipment is characterized by a significant increase in its complexity. The complication causes an increase in the guarantee of timeliness and correctness of solving problems.
The problem of reliability originated in the 50s when the process of rapid complication of systems began, and new objects began to be put into effect. At this time, the first publications that define the concepts and definitions relating to reliability were created [1] and was created by the method of assessing and calculating the reliability of devices by probabilistic-statistical methods.
Investigation of the behavior of equipment (object) during operation and evaluating its quality determines its reliability. The term "exploitation" comes from the French word "Exploitation", which means receiving the benefit or benefit from anything.
Reliability - the object property perform the specified functions, keeping the values \u200b\u200bof the installed operational performance in the specified limits.
To quantify the reliability of the object and for operation planning, special characteristics are used - reliability indicators. They allow us to evaluate the reliability of the object or its elements in various conditions and at different stages of operation.
In more detail, the reliability indicators can be found in GOST 16503-70 - "Industrial products. Nomenclature and characteristics of the main indicators of reliability.", GOST 18322-73 - "Systems of maintenance and repair of technology. Terms and definitions.", GOST 13377-75 - "Reliability in the technique. Terms and definitions."

Definitions
Reliability - Property [Next - (s)] object [hereinafter - (OB)] Perform the required functions, maintaining its operational performance during a specified period of time.
Reliability is a comprehensive-in, combining the concept of working capacity, reliability, durability, maintainability and safety.
Performance - is a state of about which it is capable of performing its functions.
Undetyability - In order to maintain its performance for a certain time. An event that disrupts the performance of OB is called refusal. The self-configuration is called a failure.
Durability - It is necessary to maintain its performance before the ultimate state, when its exploitation becomes impossible for technical, economic reasons, safety conditions or the need for overhaul.
Maintainability - Determines the adaptability of the warning and detection of faults and failures and eliminate them through repairs and maintenance.
Saveability - It is necessary to continuously maintain its performance during and after storage and maintenance.

Main indicators reliability
The main quality indicators of reliability is the likelihood of trouble-free operation, the intensity of failures and the average operation to failure.
Probability of trouble-free work P (T) represents the likelihood that within the specified period of time T., the refusal does not arise. This indicator is determined by the ratio of the number of elements about, trouble-freely worked out until the moment of time T. To the total number of elements about workable at the initial moment.
Failure intensity l (T) - this is the number of failures N (T) Elements of the time per unit, assigned to the average number of elements Nt. About workable by the time D.t.:
l (T) \u003d N (T) / (NT * D T) where
D. t. - A specified time cut.
for example: 1000 elements about worked 500 hours. During this time, 2 elements were denied. Hence l (t) \u003d n (t) / (nt * d t) \u003d 2 / (1000 * 500) \u003d 4 * 10 -6 1 / h, i.e. In 1 hour, it may refuse the 4th element from a million.
Indicators of the intensity of failures of components are taken on the basis of reference data [1, 6, 8]. For example, the intensity of failures is given. l (T) Some elements.

	Name of element	Failure intensity, * 10 -5, 1 / h
	Resistors
	Condencators
	Transformers
	Inductivity coils
	Switching devices
	Combination soldering
	Wires, cables
	Electric motors	Reliability of how the systems is characterized by the failure stream L.numerically equal to the sum of the intensity of failures of individual devices: L \u003d ÅL i The formula calculates the flow of failures and individual devices about consisting, in turn, from various nodes and elements characterized by its failure intensity. Formula is valid for calculating the flow failure stream from N. Elements in the case when the refusal of any of them leads to the failure of the entire system as a whole. Such a connection of elements is called logically sequential or main. In addition, there is a logically parallel connection of the elements when the output of their structure does not lead to the failure of the system as a whole. Constability of probability of trouble-free work P (T) and failure stream L. Determined: P (T) \u003d EXP (- D T) , it's obvious that 0AND 0< P (t )<1 and p (0) \u003d 1,but p (¥) \u003d 0 Medium operation before refusal To. - This is a mathematical expectation of the work of OB to the first refusal: To \u003d 1 / L \u003d 1 / (Ål i) , or, hence: L \u003d 1 / To The time of trouble-free operation is equal to the inverse value of the failure intensity. for example : Element technology provides the average failure intensity. l i \u003d 1 * 10 -5 1 / h . When used in about N \u003d 1 * 10 4 Elementary details Total failure intensity l. oh \u003d. N * L i \u003d 10 -1 1 / h . Then the average time of trouble-free work on To \u003d 1 / l o \u003d 10 h. If you fulfill OB based on 4 large integrated circuits (bis), then the average time of trouble-free work will increase in N / 4 \u003d 2500 times and will be 25,000 hours or 34 months or about 3 years. Calculation of reliability Formulas allow you to calculate the reliability of it, if the initial data is known - the composition of the on, mode and conditions of its operation, the failure intensity of its component (elements). However, with practical calculations of reliability there are difficulties due to the lack of reliable data on the intensity of failures for the nomenclature of elements, nodes and devices about. Output from this position gives the use of the coefficient method. The incidence of the coefficient method is that when calculating the reliability, it is not used absolute values \u200b\u200bof failure intensity l I., and the reliability coefficient ki.Binding l I. With the intensity of failures l B. any basic element: ki \u003d l i / l b Reliability coefficient ki. It practically does not depend on the operating conditions and for this element is a constant, and the difference in operating conditions ku. Considered by appropriate changes l B.. A resistor is chosen as a basic element in theory and practice. Reliability indicators of components are taken on the basis of reference data [1, 6, 8]. For example, the reliability coefficients are shown ki. Some elements. In tab. 3 shows the operating conditions coefficients ku. Works for some types of equipment. Effect on the reliability of elements of the main destabilizing factors - electrical loads, ambient temperature - is taken into account by the introduction of correction coefficients A.. In tab. 4 shows the coefficients of conditions A. Works for some types of items. Accounting for the influence of other factors - dustiness, humidity, etc. - It is performed by correction of the intensity of failures of the base element using correction coefficients. Resulting coefficient of reliability of elements OB, taking into account correction factors: ki "\u003d a1 * a2 * a3 * a4 * ki * ku, Where Ku. - nominal value of the operating conditions Ki. - Nominal value Reliability factor A1. - The coefficient considering the effect of electrical load by U, I or P A2. - The coefficient considering the effect of the temperature of the medium A3. - the coefficient of lowering the load from the nominal via U, I or P A4. - the use ratio of this element, to work on the general Operating conditions COefficient of conditions Laboratory conditions Stationary equipment: Indoors Outdoor Mobile equipment: Shipping Automotive Trained Name of the element and its parameters Load coefficient Resistors: By tension By power Condencators By tension By reactive power Direct current Over the opposite tension By transition temperature By current collector By voltage. Collector Emitter. For power dissipated The procedure for calculation is as follows: 1. Determine the quantitative values \u200b\u200bof the parameters characterizing the normal operation of the software. 2. Make up the element principal scheme for the determining connection of the elements when they perform the specified function. Auxiliary elements used in the execution of the OR function are not taken into account. 3. The initial data is defined to calculate reliability: type, quantity, nominal data elements mode of operation, temperature medium and other parameters the use ratio of elements operating Coefficient System the basic element is determined l B. and the intensity of failures l B." according to the formula: ki "\u003d a 1 * a 2 * a 3 * a 4 * ki * ku The coefficient of reliability is determined 4. The main indicators of reliability indicators are determined, with a logically sequential (main) connection of elements, nodes and devices: probability of trouble-free work: P (t) \u003d exp (- L b * to *) where Ni - the number of identical elements in n - the total number of elements in about the main connection working on failure: To \u003d 1 / (L b *) If in the scheme about there are areas with parallel connection of the elements, then the calculation of reliability indicators is made separately for these elements, and then for about as a whole. 5. Found Reliability Indicators are compared with the required. If you do not comply, then measures are taken to improve reliability about (). 6. Means for improving reliability about are: - The introduction of redundancy that happens: intarelement - Application of more reliable elements structural - Reservation - General or Separate Example of calculation: Calculate the main performance indicators for the fan on an asynchronous electric motor. The scheme is given on. For starting m closure QF, and then SB1. KM1 gets food, works and with its contacts km2 connects to the power source, and the auxiliary contact is shuting SB1. For shutdown M serves SB2. The protection of M uses Fa and the thermal relay KK1 with KK2. The fan operates in a closed room at T \u003d 50 C in a long mode. To calculate the coefficient method using the reliability coefficients of the circuit component. We accept the intensity of the bounce of the base element l B \u003d 3 * 10 -8. Based on the concept of the scheme and its analysis, we will make a basic scheme for calculating reliability (). In the calculation scheme, components are included, the refusal of which leads to the complete failure of the device. Initial data we reduce in. Basic element, 1 / h l B. 3*10 -8 Coef. Operating conditions Failure intensity l b ' l b * ku \u003d 7.5 * 10 -8 Work time, h Element concept Element of the calculation scheme Number of elements Coef. reliability Coef. Load Coef. Electric load Coef. Temperature Coef. Loads for power Coef. Use Coef's work. A. Coef. reliability S (ni * ki ') Working to refusal, h 1 / [L b '* s (ni * ki')] \u003d 3523.7 Probability e [- L b '* to * s (ni * ki')] \u003d 0.24 According to the calculation, it is possible to draw conclusions: 1. Working before the device failure: to \u003d 3524 h. 2. The probability of trouble-free operation: P (T) \u003d 0.24. The likelihood that within the specified time of operation t in the specified working conditions will not fail. Private case of reliability calculation. 1. The object (hereinafter OB) consists of N blocks connected in series (). The probability of trouble-free operation of each block p. Find the probability of trouble-free operation P of the system as a whole. Decision: P \u003d p n 2. OB consists of N blocks connected in parallel (). The probability of trouble-free operation of each block p. Find the probability of trouble-free operation P of the system as a whole. Decision: P \u003d 1- (1- P) 2 3. OB consists of n blocks connected in parallel (). The probability of trouble-free operation of each block p. The probability of trouble-free operation of the switch (P) P1. Find the probability of trouble-free operation P of the system as a whole. Decision: P \u003d 1- (1-p) * (1-P1 * P) 4. OB consists of N blocks (), with the probability of trouble-free operation of each block P. In order to increase the reliability of the ongoing duplication, still the same blocks. Find the probability of trouble-free operation of the system: with duplication of each PA block, with duplication of the entire PB system. Decision: PA \u003d N pb \u003d 2 5. OB consists of n blocks (see Fig. 10). With a good c probability of trouble-free operation U1 \u003d P1, U2 \u003d P2. If the probability of trouble-free operation u1 \u003d p1 ", u2 \u003d p2" is defective. The probability of trouble-free operation C \u003d PS. Find the probability of trouble-free operation P of the system as a whole. Decision: P \u003d PS * + (1- PS) * 9. OB consists of 2 nodes U1 and U2. The probability of trouble-free operation for T nodes: U1 P1 \u003d 0.8, U2 P2 \u003d 0.9. After time, T is not unfair. Find the probability that: - H1 - faulty U1 node - H2 - faulty U2 node - H3 - faulty nodes U1 and U2 Solution: Obviously, there was H0 when both nodes are working. Event A \u003d H1 + H2 + H3 A priori (initial) probabilities: - P (H1) \u003d (1-P1) * P2=(1-0.8)*0.9=0.2*0.9=0.18 - P (H2) \u003d (1-P2) * P1=(1-0.9)*0.8=0.1*0.8=0.08 - P (H3) \u003d (1-P1) * (1-P2)=(1-0.8)*0.9=0.2*0.1=0.02 - A \u003d i \u003d 1 å 3 * p (hi) \u003d p (h1) + p (h2) + p (h3)=0.18+0.08+0.02=0.28 Aposterion (finite) probabilities: - P (H1 / A) \u003d P (H1) /a\u003d0.18/0.28\u003d0.643 - P (H2 / A) \u003d P (H2) /a\u003d0.08/0.28\u003d0.286 - P (H3 / A) \u003d P (H3) /a\u003d 02/02/0.28\u003d0.071 10. OB consists of M blocks of type U1 and N blocks of type U2. The probability of trouble-free operation for T each block U1 \u003d P1, each block U2 \u003d P2. For work, it is enough that any 2-A block of type U1 work worked during T and at the same time any 2-A block of type U2. Find the probability of trouble-free work. Solution: Event A (trouble-free work OB) There is a product of 2 events: - A1 - (at least 2 of M blocks of type U1 work) - A2 - (at least 2 of N blocks of type U2 work) The number X1 of working unfortunately blocks of type U1 is a random variable, distributed by a binomial law with parameters M, P1. Event A1 is that X1 will take a value of at least 2, so: P (A1.) \u003d P (x1\u003e 2) \u003d 1-p (x1<2)=1-P(X1=0)-P(X1=1)=1- (G1 M + M * G2 M-1 * P1), where g1 \u003d 1-p1 similarly : P (A2) \u003d 1- (G2 N + N * G2 N-1 * P2), where G2 \u003d 1-P2 The probability of trouble-free work about: R.\u003d P (a) \u003d p (a1) * p (a2) \u003d * , where G1 \u003d 1-P1, G2 \u003d 1-P2 11. OB consists of 3 knots (). In the U1 N1 node of elements with the failure intensity L1. In the U2 node of the elements with the failure intensity L2. In the U3 node of the elements with the failure intensity L2, because U2 and U3 duplicate each other. U1 fails if it refused at least 2 elements. U2 or U3, because Duplicate, fail if they refused at least one element. Oh fails if refused u1 or u2 and u3 together. The probability of trouble-free operation of each element p. Find the likelihood that during T, it does not fail. The probabilities of failure U 2 and U 3 are equal: R2 \u003d 1- (1-P2) N2 R3 \u003d 1- (1-P3) N3 The probability of failure in total about: R \u003d R1 + (1-R1) * R2 * R3 Literature: Malinsky V.D. and others. Tests of radio equipment, "Energy", 1965 GOST 16503-70 - "Industrial products. Nomenclature and characteristics of basic reliability indicators." Shirokov A.M. Reliability of radio-electronic devices, M, Higher School, 1972 GOST 18322-73 - "Systems of maintenance and repair of technology. Terms and definitions." GOST 13377-75 - "Reliability in the technique. Terms and definitions." Kozlov B.A., Ushakov I.A. Handbook on calculating the reliability of radio electronics and automation equipment, m, owls. Radio, 1975 Perprote A.I., Storchak M.A. Reliability issues of REA, M, Sov. Radio, 1976 Levin B.R. The theory of reliability of radio engineering systems, m, owls. Radio, 1978 GOST 16593-79 - "Electric drives. Terms and definitions". I. Bragin 08.2003

Intensity of failures It is called the ratio of the number of non-samples of the equipment per unit of time to the average number of samples, working in a given period of time, provided that the refused samples are not restored and not replaced with good.

This characteristic is indicated. According to the definition

where n (t) is the number of refused samples in the time interval from to; - time interval, - the average number of working samples in the interval; N i is the number of working samples at the beginning of the interval, n i +1 is the number of working samples at the end of the interval.

Expression (1.20) is a statistical definition of failure intensity. For a probabilistic representation of this characteristic, we establish the relationship between the failure intensity, the probability of trouble-free operation and the frequency of failures.

Substitute an expression for N (T) from formulas (1.11) and (1.12) to the expression (1.20). Then we get:

.

Considering the expression (1.3) and the fact that N cf \u003d n 0 - n (t), we will find:

.

Aspiring to zero and moving to the limit, we get:

. (1.21)

Integrating expression (1.21), we get:

Since, on the basis of expression (1.21) we will get:

. (1.24)

Expressions (1.22) - (1.24) establish the relationship between the probability of trouble-free operation, the frequency of failures and the intensity of failures.

The expression (1.23) may be a probabilistic definition of failure intensity.

The intensity of failures as a quantitative characteristic of reliability has a number of advantages. It is a function of time and allows you to visually establish the characteristic areas of the equipment. This can afford to significantly increase the reliability of the equipment. Indeed, if the time of work is known (T 1) and the end of the work time (T 2), then it is reasonable to set the time of training the equipment before it starts

predation and its resource before repair. This allows you to reduce the number of failures during operation, i.e. It leads ultimately to increase the reliability of the equipment.

The failure intensity as a quantitative characteristic of reliability has the same disadvantage as the frequency of failures: it allows you to simply simply characterize the reliability of the equipment only before the first failure. Therefore, it is a convenient characteristic of the reliability of systems of one-time use and, in particular, the simplest elements.

According to a known characteristic, the rest of the quantitative characteristics of reliability are most simply defined.

The specified properties of the failure intensity allow it to be considered the main quantitative characteristic of the reliability of the simplest elements of electronics.

There are probabilistic (mathematical) and statistical indicators of reliability. Mathematical indicators of reliability are derived from theoretical functions of the distribution of probabilities of failures. Statistical reliability indicators are determined by experimental ways when testing objects based on the statistical data operation of the equipment.

Reliability is a function of many factors, most of which are random. It is clear that there is a large number of criteria to assess the reliability of the object.

Reliability criterion is a sign that is assessed by the reliability of the object.

Criteria and reliability characteristics are probabilistic, since the factors affecting the object are random and require a statistical assessment.

Quantitative reliability characteristics can be:
probability of trouble-free work;
the average time of trouble-free operation;
failure intensity;
failure frequency;
Various reliability factors.

1. The probability of trouble-free work

Serves as one of the main indicators when calculating reliability.
The probability of trouble-free operation of the object is called the likelihood that it will save its parameters within the specified limits for a certain period of time under certain conditions of operation.

In the future, we believe that the exploitation of the object occurs continuously, the duration of operation of the object is expressed in units of time T and operation started at time t \u003d 0.
Denote by P (T) the probability of trouble-free operation of the object on the length of time. The probability considered as the function of the upper border of the segment of the time is also called the reliability function.
Probabilistic rating: P (T) \u003d 1 - Q (T), where Q (T) is the probability of refusal.

From the schedule it is obvious that:
1. P (T) - non-gaining function of time;
2. 0 ≤ p (t) ≤ 1;
3. p (0) \u003d 1; P (∞) \u003d 0.

In practice, sometimes a more convenient characteristic is the likelihood of faulty object work or a possibility of refusal:
Q (T) \u003d 1 - P (T).
Statistical characteristic of the probability of failures: Q * (t) \u003d n (t) / n

2. Frequency failure

The frequency of failures is called the ratio of the number of refused objects to their total number before the start of the test, provided that the failed objects are not repaired and are not replaced with new ones, i.e.

a * (t) \u003d n (t) / (nΔt)
where A * (T) is the frequency of failures;
N (T) - the number of facility facilities in the time interval from T - T / 2 to T + T / 2;
ΔT - time interval;
N - the number of objects involved in the test.

The frequency of failures is the density of the distribution of the work time to its failure. Probabilistic definition of failure frequency A (T) \u003d -P (T) or A (T) \u003d Q (T).

Thus, between the frequency of failures, the probability of trouble-free operation and the probability of failures in any law of distribution of the failure time there is a unambiguous dependence: Q (T) \u003d ∫ A (T) DT.

Failure is interpreted in the theory of reliability as a random event. The theory is based on statistical interpretation of the probability. Elements and systems formed are considered as mass objects belonging to one general aggregate and working in statistically homogeneous conditions. When they speak about the object, in essence, at the bottom of the taking object from the general aggregate, a representative sample from this totality, and often the entire general population.

For mass facilities, a statistical estimate of the probability of trouble-free operation P (T) can be obtained by processing the test results for the reliability of sufficiently large samples. The method for calculating the evaluation depends on the test plan.

Let the sample tests from N objects were carried out without replacement and recovery to the failure of the last object. Destinators the duration of time before the failure of each of the objects T 1, ..., T n. Then statistical assessment:

P * (T) \u003d 1 - 1 / N Ση (T-T)

where η is the singular function of Hevisida.

For the probability of trouble-free operation on a certain segment, the estimate is convenient for P * (T) \u003d / N,
where N (T) is the number of objects that refused to T.

The frequency of failures determined by the replacement of non-product referenced products is sometimes called the average frequency of failures and is denoted ω (t).

3. Intensity of failures

The intensity of failures λ (t) is called the ratio of the number of failed objects per unit of time to the average number of objects operating in this period of time, provided that the failed objects are not restored and not replaced with good: λ (t) \u003d n (t) /
where n cf \u003d / 2 is the average number of objects that operated in the time range ΔT;
N i is the number of products that operated at the beginning of the ΔT interval;
N i + 1 is the number of objects that operated at the end of the time interval Δt.

Resource tests and observations over large samples of objects show that in most cases the intensity of failures varies in time non-monotonic.

From the curve of the relationship of failures on time, it can be seen that the entire period of operation of the object can be conditionally divided into 3 periods.
I - th Permanent period.

Performance failures are usually the result of the presence of defects and defective elements, the reliability of which is significantly lower than the required level. With an increase in the number of elements in the product, even with the most strict control, it is not possible to completely exclude the possibility of entering elements that have certain hidden defects. In addition, errors in assembling and installation can also be denied to refuses during this period, as well as the insufficient yield of the facility by the service personnel.

The physical nature of such failures is random in nature and differs from sudden failures of the normal period of operation, the fact that refusals may not take place at elevated, but with minor loads ("burning defective elements").
A decrease in the value of the intensity of the object failure in general, with the constant value of this parameter for each of the elements separately, is precisely explained by the "burning" of weak links and their replacement is the most reliable. The cooler curve in this area, the better: less defective elements will remain in the product in a short time.

To increase the reliability of the object, given the possibility of permanent failures, you need:
conduct more strict rejection of elements;
carry out tests of the object on modes close to operational and use only the elements that have passed tests during the assembly;
improve the quality of assembly and installation.

The average battery time is determined when testing. For particularly important cases, it is necessary to increase the period of development several times compared with the average.

II - Time - Normal Operation
This period is characterized by the fact that the earmarket failures have already ended, and the failures associated with wear have not yet come. This period is characterized by extremely sudden failures of normal elements, the work on the refusal of which is very large.

The preservation of the level of failures at this stage is characterized by the fact that the refusing element is replaced by the same, with the same probability of failure, and not the best, as it happened during the accuracy.

Rejection and pre-running in the elements that are replaced by the refusal, has even greater importance for this stage.
The designer has the greatest possibilities in solving this task. Often changing the design or facilitating the operation modes of only one or two elements provides a sharp increase in the reliability of the entire object. The second way is to improve the quality of production and even cleanliness of production and operation.

III - 1 - wear
The period of normal operation ends when wear failures begin to occur. Comes the third period in the life of the product - the wear period.

The likelihood of failures due to wear with approach to the service life increases.

From a probabilistic point of view, the system failure in this period of time ΔT \u003d T 2 - T 1 is defined as the probability of failure:

∫a (T) \u003d Q 2 (T) - Q 1 (T)

The intensity of failures is the conditional likelihood that in the period of time ΔT a refusal will occur, provided that it has not occurred λ (T) \u003d / [ΔTP (T)]
Λ (T) \u003d LIM / [Δtp (t)] \u003d / \u003d q "(t) / p (t) \u003d -p" (t) / p (t)
Since a (t) \u003d -p "(t), then λ (t) \u003d a (t) / p (t).

These expressions establish the relationship between the probability of trouble-free operation, the frequency and the intensity of failures. If a (t) is a non-gaining function, then the ratio is true:
ω (t) ≥ λ (t) ≥ A (T).

4. Average time of trouble-free work

Average time of trouble-free work is called mathematical waiting time of trouble-free operation.

Probabilistic definition: the average time of trouble-free operation is equal to the area under the curve of the probability of trouble-free operation.

Statistical definition: T * \u003d Σθ I / N 0
where θ i is the time of operation of the i-th object to failure;
N 0 - the initial number of objects.

Obviously, the parameter T * cannot fully and satisfactorily characterize the reliability of long-term systems, since it is a characteristic of reliability only to the first failure. Therefore, the reliability of long-term systems is characterized by an average time between two adjacent failures or on the failure of T CP:
t cf \u003d σθ i / n \u003d 1 / ω (t),
where n is the number of failures for T;
θ i - the operating time of the object between (I - 1) -M and I-M failures.

Drawing on failure is the average value of the time between adjacent failures, subject to the restoration of the failed element.