Stabilizer on the operating amplifier circuit. PWM knob on the operating amplifier. Distant voltage stabilizers chips

Fig. eight. The main circuit of inclusion of the KR142EN regulator1

The reference voltage at the output 5 of the chip is about 2 V, and the voltage divider, removed from the support stabilion, is introduced into the microcircuit. Due to this, when constructing stabilizers with output voltages from 3 to 30 V, the same inclusion scheme with an external output voltage divider is used. Additionally, we note that the chip KR142EN1.2 has free conclusions not only inverting (conclusion 3), but also non-converting (conclusion 4) Amplifier inputs, which simplifies the stabilizer of the denial voltage from this IS. This is the main difference between the KRN2ESH chip, 2 from the 142En1.2 microcircuit of the earlier release.

External transistor V. T1. - This is an emitter repeater to increase the load current to 1 ... 2 A. If no more than 50 mA current is required, then the transistor should be excluded using the output 8 Chip instead of the emitter output of the transistor V T1.

As part of the chip, there is a transistor that protects the output cascade from current overload. Toko-restrictive resistance resistor R4. Choose from the calculation of the voltage drop on it 0.66 V when the emergency current flow. Without a snitter repeater v T1. The resistor should be installed R4. Resistance 10 ohms.

To create a "falling" characteristic of the overload current limit, a divider is connected. R2R3 and make calculated on the following dependencies:

Example, I max \u003d 0.6 A (set); I K3 - 0.2 A (choose at least 1 / s i max); U BE \u003d 0.66 V; U out \u003d 12 V (set); a \u003d 0.11 (by calculation); R3 \u003d 10 com (typical value); R2. \u003d 1.24 koi; R4. \u003d 3.7 Ohm.

In addition, there is an output in the microcircuit 14 To control the stabilizer. If you submit to this input a single TTL-level + (2.5 ... 5) B, then the output voltage of the stabilizer will fall to zero. So that the reverse current in the presence of a capacitive load did not destroy the output transistor, the diode V was installed D1.

Capacitor C1. With a capacity of 3.3 ... 10 MK suppresses the noise of Stabilon, but it is not necessary. Capacitor C2. (with a capacity of up to 0.1 MK) - the element of the frequency correction; permissible instead to connect the output 13 With "Earth" wire through a serial RC-chain 360 Ohms (maximum) and 560 PF (minimum).

On the basis of chip KR142ESH.2 (Fig. 8), you can create stabilizers of negative stresses (Fig. 9).

Figure 9. Stabilization of negative tension

At the same time Stabilitron V D1 shifts the voltage level on the output 8 relative to the input voltage. Basic transistor current V T1. It should not exceed the maximum allowable current of Stabilon, otherwise the composite transistor should be applied.

The wide possibilities of chip CR142EN1,2 allow you to create a relay voltage stabilizers based on them, the example of which is given in Fig. 10.

Fig. 10. Relay voltage stabilizer

In such a stabilizer, the reference voltage, as in the stabilizer according to the diagram. 8, installed by the divider R4R5, And the amplitude of the pulsations of the output voltage on the load is given by the auxiliary divider R2R3 and equal & U \u003d u B. x-R4IR3. The frequency of self-oscillation is determined from the same considerations as for the stabilizer according to the scheme in Fig. 7. It should only be borne in mind that the load current cannot be changed widely, usually no more than twice from the nominal value. The advantage of relay stabilizers is the high efficiency.

It is necessary to consider another class of stabilizers - current stabilizers, converting voltage in the current, regardless of the change in the load resistance. From such stabilizers that allow to ground the load, we note the stabilizer according to the scheme in Fig. eleven.

Fig. eleven. Current Stabilizer on OU

Stabilizer I load current u. \u003d U. B-X. .lrl. Interestingly, if the voltage U. BX Serve On the inverting input, only the current direction will change without changing its value.

More powerful sources of current involve connecting to AU amplifying transistors. In fig. 12 Dana Current source scheme, and in fig. 13 - current receiver scheme.

Fig. 12. Precision diagram of the current source; Input voltage - negative

Figure 13. Scheme of precision current removal; Input voltage - positive

In both devices, the current is determined by the calculation in the same way as in the previous version of the stabilizer. This current, the more accurate depends only on the voltage of the U Vx and the nominal resistor R1, The smaller the input current of OU and the smaller the control current of the first (after the OU) of the transistor, which is selected therefore fields. Load current can reach 100 mA.

Scheme of a simple powerful current source for charger Shown in Fig. fourteen.

Fig. fourteen. High power current source

Here R4. - Toko-measuring wire resistor. Nominal value of load current I n. \u003d DU / R4 \u003d 5 And set. Approximately with the middle position of the resistor engine R1. When charging automotive rechargeable battery Voltage U Vh\u003e 18 V without taking into account the ripples of the straightened alternating voltage. In such a device, it is necessary to use OU with an input voltage range up to a positive supply voltage. OU K553UD2, K153UD2, K153UD6, as well as KR140UD18, possess such capabilities.

Literature

Bokuniev A. A. Relay stabilizers of constant voltage - M: Energy, 1978, 88 p.

Rutxvsky J. Integral Operations Amplifiers. - M.: Mir, 1978, 323 p.

Xorolas P, Hill W. Art scheme engineering, t. 1. - m.; World, - 1986, 598 p.

Spencer p inexpensive power supply with zero ripples. - Electronics, 1973, No. 23, from 62.

Shilo V. l Linear integrated schemes. - M. Cov. Radio, 1979, 368 p.

The advantages of the PWM regulators using operational amplifiers are so that it is possible to apply almost any OU (in a typical inclusion scheme, of course).

The level of output efficient voltage is regulated by changing the voltage level on the non-converting input of the OMA, which allows the use of a scheme as composite part Different regulators of voltage and current, as well as schemes with smooth ignition and incandescent bulbs.
Scheme Easy in repetition, does not contain rare elements and with good elements starts to work immediately, without configuration. The power field transistor is selected by the current current, but to reduce thermal dispel power, it is desirable to use transistors designed for high current, because They have the smallest resistance in the open state.
Radiator area for field Transistor Fully determined by the choice of its type and current current. If the scheme is used to control the voltage in the onboard networks + 24V, to prevent breakdown of the shutter of the field transistor, between the transistor collectorVT1 and VT2 shutter it should be enabled to resist the resistance of 1 K, and the resistorR6. credit any suitable stabilion by 15 V, the remaining elements of the scheme do not change.

In all previously considered schemes as a power field transistor usedn - Channel transistors as the most common and best characteristics.

If you want to adjust the voltage on the load, one of the conclusions of which is connected to the "mass", then schemes are used in whichn - the channel field transistor is connected by the flow to + power source, and the load circuit turns on the load.

To ensure the possibility of full opening of the field transistor, the control circuit must contain a voltage increase node in the shutter control circuits to 27 - 30 W, as is done in specialized chipsU 6 080B ... u6084b, L9610, L9611 , then there will be a voltage of at least 15 values \u200b\u200bbetween the shutter and the source if the load current does not exceed 10a, it is possible to use power fieldp. - channel transistors whose assortment is much already due to technological reasons. The type of transistor changes in the diagramVT1. , and adjusting characteristicsR7 Changes to the opposite. If the first circuit has an increase in the voltage of the control (the engine of the variable resistor moves to "+" the power source) causes a decrease in the output voltage at the load, then the second diagram is inversely inverse. If the specific scheme requires inversely from the initial dependence of the output voltage from the input, then in the schemes it is necessary to change the structure of the transistorsVT1, Ie transistor VT1 In the first diagram you need to connect asVT1. The second scheme and vice versa.

The stability of the supply voltage is a prerequisite proper work Many electronic devices. To stabilize the constant voltage on the load during fluctuations in the network voltage and changing the current consumed between the rectifier with the filter and the load (consumer) put stabilizers of a constant voltage.

The output voltage of the stabilizer depends on both the input voltage of the stabilizer and the load current (output current):

Find a full differential change in voltage when changing and:

We divide the right and left parts on, as well as multiply and split the first term in the right part on, and the second term on.

Introducing the designation and passing to end increments, we have

Here is the coefficient of stabilization equal to the ratio of increments of input and output voltages in relative units;

Internal (output) stabilizer resistance.

Stabilizers are divided into parametric and compensatory.

The parametric stabilizer is based on the use of an element with nonlinear characteristic, for example, semiconductor stabilion (see § 1.3). The stress on the stabilone on the reversible electrical breakdown site is almost constant with a significant change in the reverse current through the device.

The diagram of the parametric stabilizer is shown in Fig. 5.10, a.

Fig. 5.10. Parametric stabilizer (A), its substitution scheme for increments (b) and external characteristics of the rectifier with a stabilizer (curve 2) and without stabilizer (curve) (B)

The input voltage of the stabilizer should be greater than the stabilization stabilization stabilization. To limit the current through the stabilion, a ballast resistor is installed. The output voltage is removed from Stabilon. Part of the input voltage is lost on the resistor, the remaining part is applied to the load:

Consider that we get

The greatest current through Stabilodron proceeds

The smallest current through Stabilod processes

When providing conditions - currents of stabilion, limiting the stabilization section, the stress voltage is stable and equal. From.

With an increase, the current is growing, the voltage drop is increasing. With an increase in the load resistance, the load current decreases, the current of the current is growing through the stabilion, the voltage drops on and on the load remain unchanged.

To find, we construct a stabilizer substitution scheme. 5.10, and for increments. The nonlinear element operates on the stabilization site, where its resistance to the variable Goku is the parameter of the instrument. The stabilizer substitution scheme is shown in Fig. . From the substitution scheme we get

Considering that in the stabilizer, we have

To find, as well as when calculating the parameters of amplifiers (see § 2.3), we use the theorem on the equivalent generator and put, then the resistance at the output of the stabilizer

Expressions (5.16), (5.17) show that the parameters of the stabilizer are determined by the parameters of the semiconductor stabilion used (or other instrument). Usually for parametric stabilizers not more than 20-40, and lies in the range from several Ohm to several cells.

In some cases, such indicators are insufficient, then apply compensation stabilizers. In fig. 5.11 is one of the simplest compensation stabilizer schemes, in which the load is connected to the input voltage source through the regulating nonlinear element, Transistor V. On the transistor database through the OE, the OS signal is supplied. There are voltages from a high-resistance resistant and reference (reference) voltage to the OU entrance.

Fig. 5.11. Simple scheme compensatory stabilizer with OU

Consider the work of the stabilizer. Suppose that the voltage increased, after it increases, and at the same time, a positive voltage increment is supplied to the inverting input of the OU, and at the OU output there is a negative voltage increment. The transistor of the transistor's control emitter transition is applied the difference in basic and emitter voltages. In the mode under consideration, the transistor current V decreases and the voltage of the lines is reduced almost to the initial value. Similarly, the change in listed by increasing or decreasing will be worked out: it will change, the corresponding sign will occur, the transistor current will change. Very high, since in the process of operation, the mode of operation of the stabilion is practically not changed and the current is stable through it.

Compensation voltage stabilizers are available in the form of an IC, which include an adjusting nonlinear element, transistor V, OU and chains that bind the load with its input.

In fig. 5.10, the external characteristic of the power supply with the stabilizer is shown, its working plot is limited to current values

Scheme:

Voltage stabilizer on operating amplifiers (OU) is sometimes not started, i.e. It does not enter the stabilization mode when power is turned on, and the voltage at its output remains almost equal to zero. After replacing the chip, the stabilizer begins to work normally. Check replaced by OU indicates that it is absolutely correct. When re-installing this OU into a working stabilizer, the above phenomenon is repeated - the stabilizer is not started again. The above is the scheme of one of the typical stabilizers in which such a phenomenon was observed.

After a number of experiments was established. that its reason is the offset voltage by UCM of the operating amplifier, shown below conditionally as a source of constant voltage:

The input resistance of the operating amplifier depicts the RVX resistor. The voltage of the mixing OU, as is known, can be any polarity. Suppose it turned out to be as shown in the figure. Then at the first moment after switching on the output voltage of the stabilizer, and therefore, the voltage between the OMU inputs is zero, and the negative pole of the UCM source is connected directly to the unconvertible input of OU. The voltage at its output decreases and with a sufficiently large value of the CCH (for K1UT531B, for example, it can reach 7.5 mV) due to the large voltage gain coefficient, the OU output cascade is strongly saturated, the output voltage is only tenths of Volta. . This voltage is not enough to open the regulatory transistor of the stabilizer and therefore it does not start. If it turns out that after replacing the chip in the newly installed OE, the displacement voltage value is not too large or its polarity is reversed in Fig. 2a stabilizer will start normally.

Get rid of the need for the time-consuming selection of an OU instance for each specific stabilizer different ways. One of them, for example, is to use to start a voltage divider stabilizer with a separation diode (Fig. 2b). The voltage on the R2 resistor should satisfy the following inequalities:

where:
URH.min and UVK.Max - the minimum and maximum input voltage of the stabilizer;
UD is the maximum voltage drop on the diode V1;
UCM.Max is the maximum voltage of the displacement of the OU;
U3 Nom - voltage at the inlet 3 OU (see Fig. 1) at nominal stabilizer mode.

When the stabilizer is connected to the power source, the positive voltage from the R2 resistor (Fig. 2. b) via the VI diode is supplied to the unconforming input of OU. The output voltage of OU at the same time sharply increases and the regulator transistor of the stabilizer opens.

After the output of the stabilizer to the nominal mode, the diode VI closes and disconnects the voltage divider from the EU entrance. For the most full elimination The influence of the launching value to the work of the diode stabilizer should choose a silicon, with a small reverse current.

Practical check confirmed the effectiveness of the application of the described chain - the stabilizer with it was started correctly with any values \u200b\u200band polarity of the UCM voltage. While without it, sometimes the inclusion of the stabilizer did not occur. The effects of the starting chain on the indicators of the stabilizer (the coefficient of stabilization is more than. 6000, the output resistance of 8 MΩ) was not observed.

As you know, a stable current is required to power the LEDs. A device that can feed the LEDs to a stable current is called the LED driver. This article is devoted to the manufacture of such a driver using the operational amplifier.

So, the main idea is to stabilize the voltage drop on the resistor of the known nominal value (in our case - R 3) included in the circuit sequentially with the load (LED). Since the resistor is turned on consistently with the LED, then the same current flows through them. If this resistor is selected in such a way that it practically does not heat up, then it will be resistant. Thus, stabilizing the voltage drop on it, we stabilize the current through it and, respectively, through the LED.

And here is the operational amplifier? Yes, despite the fact that one of his wonderful properties is that the OU tends to such a state when the difference of stresses at its inputs is zero. And he does this by changing its output voltage. If the difference U 1 -U 2 is positive - the output voltage will increase, and if negative is to decrease.

Imagine that our scheme is in a certain equilibrium state, when the voltage at the OU output is valid. At the same time through the load and the resistor flows the current I n. If for such reasons the current in the chain will increase (for example, if the resistance of the LED decreases under the action of heating), then this will cause an increase in the voltage drop on the resistor R 3 and, accordingly, an increase in voltage in the inverting input OU. A negative voltage difference will appear between the inputs of OU (error), striving to compensate for which the operator will reduce the output voltage. It will do it until the voltages at its inputs are equal, i.e. So far, the voltage drop on the R 3 resistor will not be equal to the voltage on the unconforming input of the OU.

Thus, the whole task is reduced to stabilize the voltage on the non-invisor entrance of the OU. If the entire scheme is powered by a stable voltage U p, then for this, a fairly simple divider (as in Scheme 1). Once the divider is connected to a stable voltage, the yield of the divider will also be stable.

Calculations: For calculations, choose a real example: Let we want to save two superwear backlight LEDs cell phone Nokia from voltage UP \u003d 12V (excellent flashlight in the car). We need to get a current through each LED 20 mA and at the same time we have a kindled motherboard Dual operational amplifier LM833. With such a current, our LEDs are shine much brighter than in the phone, but burn and not going to, significant heating begins somewhere closer to 30 mA. We will conduct the calculation for one channel of the operator, because For the second, it is absolutely similar.

voltage on unconvertising input: U 1 \u003d U P * R 2 / (R 1 + R 2)

voltage in inverting input: U 2 \u003d i H * R 3

from the conditions of equality of voltages in a state of equilibrium:

U 1 \u003d U 2 \u003d\u003e I H \u003d U P * R 2 / R 3 * 1 / (R 1 + R 2)

How to choose the nominations of the elements? First, the expression for u 1 is valid only if the input current of the operating amplifier \u003d 0. That is, for the perfect operational amplifier. So that you can not take into account the input current of the real OU, the current through the divider should be at least 100 times more than the input current OU. The magnitude of the input current can be viewed in the datasheet, usually for modern OSH, it can be from dozens of picoamper to hundreds of nanosper (for our case input Bias Current max \u003d 1 μA). That is, the current through the divider must be at least 100..200 μA. Secondly, on the one hand, the greater R 3 - the more our scheme is sensitive to changing the current, but on the other hand, the increase in R 3 reduces the efficiency of the circuit, since the resistor dispels the power proportional to the resistance. We will proceed from the fact that we do not want the voltage drop on the resistor more than 1B.

So, let R 1 \u003d 47kom, then, taking into account the fact that u 1 \u003d u 2 \u003d 1B, from the expression for u 1, we obtain R 2 \u003d R 1 / (U P / U 1 -1) \u003d 4,272 -\u003e from the standard row We choose a resistor by 4.3 com. From the expression for u 2 we find R 3 \u003d U 2 / I H \u003d 50 -\u003e select a resistor for 47 ohms. Check the current through the divider: i d \u003d u n / (R 1 + R 2) \u003d 234 μA, which is quite suitable. The power dissipated on R 3: P \u003d I H 2 * R 3 \u003d 18.8 MW, which is also quite acceptable. For comparison, the most conventional MLT-0,125 resistors are designed for 125 MW.

As already noted, the scheme described above is designed for stable power U p. What to do if the nutrition is not stable. Most. simple decision It is replacing the resistance of R 2 divider to Stabilod. What is important to consider in this case?

First, it is important that Stabilong can work in the entire supply voltage range. If the current via R 1 D 1 is too small - the stabilion voltage will be significantly higher than the stabilization voltage, respectively, the output voltage will be significantly higher than the desired and LED can burn. So, it is necessary that at u p MIN current through R 1 D 1 was greater than or equal to the I ST MIN (the minimum stabilization current learn from the Datashet to Stabilod).

R 1 max \u003d (U n MIN -U ST) / I ST MIN

Secondly, with the maximum power supply voltage, the current via Stabytron should not be higher than I st Max (our stabilion should not burn). I.e

R 1 min \u003d (U n Max -U ST) / I ST MAX

And finally, thirdly, the voltage on the real stabilion is not exactly equal to U Art, - it, depending on the current, changes from u st min to the U Art Max. Accordingly, the drop on the R 3 resistor also varies from u st min to U st Max. It should also be taken into account, since the larger ΔU st is the greater the current control error, depending on the supply voltage.

Well, okay, with little currents figured out, but what should I do if we need a current through the LED not 20, and 500 mA, what exceeds the possibility of the operator? Here, too, everything is enough simple - the output can be able to be able to use the usual bipolar or field transistor, all calculations remain unchanged. The only obvious condition is the transistor to withstand the required current and maximum supply voltage.

Well, perhaps, everything. Good luck! And in no case, do not throw out old raids - we still have a lot of cool things ahead.