» Electronics »Linear voltage regulator with regulation on TL431 and NPN transistors

Linear voltage regulator adjustable on TL431 and NPN transistors

Hello!
Recently, I became interested in the assembly of linear voltage stabilizer circuits. Such schemes do not require rare details, and the selection of components and tuning also does not cause any special difficulties. This time I decided to assemble a linear voltage stabilizer circuit on the "regulated zener diode" (microcircuit) TL431. TL431 acts as a reference voltage source, and the power role is played by a powerful NPN transistor in the TO -220 package.

With an input voltage of 19V, the circuit can serve as a source of stabilized voltage in the range from 2.7 to 16 V at a current of up to 4A. The stabilizer is designed as a module assembled on a breadboard. As follows:



Video:


The stabilizer requires a DC power supply. It makes sense to use such a stabilizer with a classic linear power supply, consisting of an iron transformer, a diode bridge and a large capacitor. The voltage in the network can vary depending on the load and as a result, the voltage at the transformer output will change. This circuit will provide a stable output voltage with a varying input. It must be understood that the stabilizer of the lowering type, as well as in the circuit itself, drops 1-3 V, so the maximum output voltage will always be less than the input.

In principle, switching power supplies can be used as a power supply for this stabilizer, for example, from a 19 V laptop. But in this case, the role of stabilization will be minimal, because factory switching power supplies and so on output stabilized voltage.

Scheme:


Selection of components
The maximum current that the TL431 chip can pass through itself, according to the documentation, is 100 mA. In my case, I limited the current with a margin to about 80 mA using the resistor R1. It is necessary to calculate the resistor according to the formulas.

First you need to determine the resistance of the resistor. At a maximum input voltage of 19 V, according to Ohm's law, the resistance is calculated as follows:
R = U / I = 19V / 0.08A = 240 Ohm

It is necessary to calculate the power of the resistor R1:
P = I ^ 2 * R = 0.08 A * 0.08 A * 240 Ohms = 1.5 Watts

I used a Soviet 2-watt resistor

Resistors R2 and R3 form a voltage divider that “programs” TL431, and the resistor R3 is variable, which allows you to change the reference voltage, which is then repeated in a cascade of transistors. I used R2 - 1K ohm, R3 - 10K ohm. The power of resistor R2 depends on the output voltage. For example, with an output voltage of 19V:
P = U ^ 2 / R = 19 * 19/1000 = 0.361 watts

I used a 1 watt resistor.

Resistor R4 is used to limit the current based on the transistor VT2. It is better to select the rating experimentally, controlling the output voltage. If the resistance is too large, this will significantly limit the output voltage of the circuit. In my case, it is 100 Ohms, any power is suitable.

As the main power transistor (VT1), it is better to use transistors in the TO - 220 or more powerful case (TO247, TO-3). I used transistor E13009, purchased on Ali Express. Transistor for voltage up to 400V and current up to 12A. For such a circuit, a high-voltage transistor is not the most optimal solution, but it will work fine. The transistor is most likely fake and 12 A will not stand, but 5-6A is quite. In our circuit, the current is up to 4A, therefore, suitable for this circuit. In this scheme, the transistor must be able to dissipate power up to 30-35 watts.

The power dissipation is calculated as the difference between the input and output voltage multiplied by the collector current:
P = (U output -U input) * I collector
For example, the input voltage is 19 V, we set the output voltage to 12 V, and the collector current is 3 A
P = (19V-12V) * 3A = 21 watts - this is a completely normal situation for our transistor.

And if we continue to reduce the output voltage to 6V, the picture will be different:
P = (19V-6V) * 3A = 39 watts, which is not very good for a transistor in a TO-220 package (you also need to take into account that when the transistor is closed, the current will also decrease: by 6V the current will be about 2-2.5A, and not 3). In this case, it is better to either use another transistor in a more massive case, or reduce the difference between the input and output voltage (for example, if the power supply is transformer, by switching the windings).

Also, the transistor must be rated for a current of 5A or more. It is better to take a transistor with a static current transfer coefficient of 20. The Chinese transistor fully meets these requirements. Before sealing in the circuit, I checked it (current and power dissipation) on a special stand.

Because TL431 can produce a current of not more than 100 mA, and to power the base of the transistor requires more current, you will need another transistor, which will amplify the current from the output of the TL431 chip, repeating the reference voltage. For this, we need a transistor VT2.
Transistor VT2 must be able to supply sufficient current to the base of transistor VT1.

It is possible to roughly determine the required current through the static current transfer coefficient (h21e or hFE or β) of the transistor VT1. If we want to have a current of 4 A at the output, and the static current transfer coefficient VT1 is 20, then:
I base = I collector / β = 4 A / 20 = 0.2 A.

The static current transfer coefficient will vary depending on the collector current, so this value is indicative. Measurement in practice showed that it is necessary to supply about 170 mA to the base of transistor VT1 so that the collector current is 4A. The transistors in the TO-92 package start to warm up noticeably at currents above 0.1 A, so in this circuit I used the KT815A transistor in the TO-126 package. The transistor is designed for current up to 1.5A, the static current transfer coefficient is about 75. A small heatsink for this transistor will be appropriate.
Capacitor C3 is needed to stabilize the voltage on the basis of the transistor VT1, the nominal value is 100 μF, the voltage is 25V.

Filters from capacitors are installed at the output and input: C1 and C4 (electrolytic at 25V, 1000 μF) and C2, C5 (ceramic 2-10 μF).
The diode D1 serves to protect the transistor VT1 from reverse current. Diode D2 is needed to protect against a transistor when supplying collector motors. When the power is turned off, the engines spin for a while and in the braking mode work as generators. The current generated in this way goes in the opposite direction and can damage the transistor.The diode in this case closes the motor to itself and the current does not reach the transistor. Resistor R5 plays the role of a small load for stabilization in idle mode, a nominal value of 10k Ohm, any power.

Assembly
The circuit is assembled as a module on a breadboard. I used a radiator from a switching power supply.

With a radiator of this size, you should not load the circuit as much as possible. With a current of more than 1 A, it is necessary to replace the radiator with a more massive one, blowing with a fan will also not hurt.

It is important to remember that the greater the difference between the input and output voltage and the greater the current, the more heat is generated and the more cooling is needed.
It took about an hour to solder. In principle, it would be a good form to make a board using the LUT method, but since I only need a board in one copy, I did not want to waste time designing the board.

The result is such a module:

After assembly, I checked the characteristics:


The circuit has virtually no protection (meaning that there is no short circuit protection, reverse polarity protection, soft start, current limitation, etc.), so you need to use it very carefully. For the same reason, it is not recommended to use such schemes in "laboratory" power supplies. For this purpose, ready-made microcircuits in the TO-220 package are suitable for currents up to 5A, for example, KR142EN22A. Or at least for this circuit, you need to make an additional module for protection against short circuit.

The circuit can be called classical, like most linear stabilizer circuits. Modern pulse circuits have many advantages, for example: higher efficiency, much less heating, smaller dimensions and weight. At the same time, linear circuits are easier to master for beginner hams, and if the efficiency and dimensions are not particularly important, they are quite suitable for supplying devices with stabilized voltage.

And of course, nothing beats the feeling when I powered some device from a home-made power source, and linear circuits for beginner hams are more accessible, whatever one may say.
9.1
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10 comments
Author
As promised, I checked the stabilization of the output voltage when the load current changes.
Stabilization works fine. Test conditions: input voltage -16V, output - 12V. The load current changed from about 1.5A to 3A (one lamp is constantly on, connecting / disconnecting an additional lamp to it). The voltage of 12V holds stably.
Quote: EandV
I can of course be mistaken, but in my opinion, if field workers in the linear (ajar) mode are used, they will also warm up.
According to the science of physics, too. )) What is the difference, which transistor is bipolar, field, IGBT, BSIT? If a current passes through it and the voltage drops, then the power allocated to it does not depend on the type of transistor. )))
Author
Thanks!
Yes, linear circuits on bipolar transistors are essentially boilers. You can’t argue with that.
Personally, I used fieldwork only for PWM control circuits. I can of course be mistaken, but in my opinion, if field workers in the linear (ajar) mode are used, they will also warm up. They do not heat only when fully open.
Guest Alexander
Hats off a lot of work done. But there is one thing, but bipolar transistors without (water) cooling are difficult to cool. For this reason, it has long gone over to the field. Try not to regret with the same parameters a large radiator is not needed
Author
It's nice to talk to people who give comments on the merits. On a number of issues I have to admit that you are right.
But still I want to clarify a couple of points.
This is less than 1%, but usually when we talk about stabilizers, we mean the load on the output.

I agree with you; I needed to formulate it more precisely. I think the general meaning is clear: the voltage in the network walks, albeit not very much, it will also walk at the output of the transformer.
This is your data. In the video, when the voltage decreases, the current really decreases (Ohm's law), but at a specific, active load. Reduce the load resistance - the current will increase.

Of course, you’re right, it’s useless to argue, but in practice, people who have at least minimal experience in electronics will probably not use linear stabilizers with a large voltage drop and at maximum currents (I at least hope).
I tried to convey this with the following text, perhaps not very well:
"P = (19V-6V) * 3A = 39 watts, which is not very good for a transistor in a TO-220 package (you also need to consider that when the transistor is closed, the current will also decrease: by 6V the current will be about 2-2.5A, but not 3). In this case, it is better either to use another transistor in a more massive case, or to reduce the difference between the input and output voltage (for example, if the power supply is transformer, by switching the windings). "
The capacitor cannot stabilize the voltage.

Again, the wording is not the same ... In general, with a capacitor is better.

Thanks again for the constructive comments. In our time, this does not often happen, they usually just say that the circuit is unusable, and the author is sent to read some classic textbook on electronics.
mains voltage dipped a couple of volts
This is less than 1%, but usually when we talk about stabilizers, we mean the load on the output.
There will be no 4A when limited to 2.7V. Watch the video for the article
With an input voltage of 19V, the circuit can serve as a source of stabilized voltage in the range from 2,7 up to 16 V at current up to 4A.
This is your data. In the video, when the voltage decreases, the current really decreases (Ohm's law), but at a specific, active load. Reduce the load resistance - the current will increase.
Without C3, the output voltage after the transistor walks a little +/- 50 mA
Capacitor C3 is needed to stabilize the voltage based on transistor VT1
The capacitor cannot stabilize the voltage. scratch
Author
I have not tried such a scheme, to be honest. An interesting variant with a field worker.
Initially, in the circuit described in the article, I wanted to use exactly bipolar, or rather Darlington's transistor (I had a couple of extra KT827A lying around). Then I thought that not everyone has KT827A, but they are not cheap. Therefore, he made an option with a composite transistor from cheap parts.
Author
Greetings!
Well, you directly analyzed the whole article, I am already pleased that someone read everything so carefully. No wonder it means I wrote so much beech.
I will try to answer essentially:
A cardboard trance is not suitable?

Cardboard does not fit unambiguously, feritic - pulsed, too. I once thought that feritic - high-frequency and 50 Hertz iron - is the same thing, now I know that it’s not, and I hasten to mention it. Suddenly I’m not the only one in the whole world.
What load?

People came in the evening after work, turned on TVs, electric stoves, computers, etc., the voltage in the network dipped a couple of volts (more relevant for small cities and rural areas, although small fluctuations are not uncommon in Moscow)
And when the load current changes?

This is the right question, I’ll check.
And the voltage drop on the TL431 does not count?

He did not take into account, because in my case, it is not significant, the transistor VT2 amplifies with a margin. The task of selecting a resistor for TL is to limit the current so that it does not burn, if it is a little less, it is not fatal
Pmax = (19-2.7) * 4 = 65.2 W

Dear, this is a linear circuit, the limitation of the output voltage is obtained due to the limitation of current by the transistor. There will be no 4A when limited to 2.7V. Watch the video for the article, where the tests clearly show: at 2.7V, the current is 1.7A, i.e. (19-2,7) * 1,7 = 27,7 W, so we live!
???

Without C3, the output voltage after the transistor walks a little +/- 50 mA. In such schemes, I saw how for this purpose they also put a resistor compensating the collector reverse current of about 4.7k between the base and the emitter of the transistor.
Regarding the diode - a typo (well, excuse me), "to protect the transistor" - I mean, we are talking about this:
If you are too lazy to read, here is a quote:
“Since the motor is an inductive load, we must be careful. If current flows through the winding and we suddenly stop this flow, then a large voltage will temporarily appear on the terminals of the winding. This voltage can damage the transistor (in the above diagram), causing a breakdown of the junction collector base. In addition, this can cause significant interference. To prevent this, it is necessary to connect a diode in parallel with the inductive load. "
But I absolutely agree with this!

After the rest of the comments, this one is like a balm for wounds)
This option, I think, is more acceptable.
consisting of iron transformer
A cardboard trance is not suitable?
Voltage in the network may vary depending on the load
What load?
This circuit will provide a stable output voltage with a varying input
And when the load current changes?
R = U / I = 19V / 0.08A = 240 Ohm
And the voltage drop on the TL431 does not count?
voltages ranging from 2.7 to 16 V at current up to 4A.
the transistor must be able to dissipate power up to 30-35 watts
Pmax = (19-2.7) * 4 = 65.2 W!
Capacitor C3 is needed to stabilize voltage based on transistor VT1
Diode D2 is needed for transistor protection when powered by collector motors
???
For this purpose, ready-made microcircuits in the TO-220 package are suitable for currents up to 5A, for example, KR142EN22A
But I absolutely agree with this! smile

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