Regulated power supply 0. Simple power supply. About the wires included in the kit
This power supply is based on the LM317 chip and does not require any special knowledge for assembly, and after correct installation from serviceable parts, does not require adjustment. Despite its apparent simplicity, this unit is a reliable power source for digital devices and has built-in protection against overheating and overcurrent. The microcircuit inside itself has over twenty transistors and is a high-tech device, although from the outside it looks like an ordinary transistor.
The power supply of the circuit is designed for voltages up to 40 volts alternating current, and the output can be obtained from 1.2 to 30 volts of constant, stabilized voltage. Adjustment from minimum to maximum with a potentiometer occurs very smoothly, without jumps or dips. Output current up to 1.5 amperes. If the current consumption is not planned to exceed 250 milliamps, then a radiator is not needed. When consuming a larger load, place the microcircuit on a heat-conducting paste to a radiator with a total dissipation area of 350 - 400 or more square millimeters. The selection of a power transformer must be calculated based on the fact that the voltage at the input to the power supply should be 10 - 15% greater than what you plan to receive at the output. It is better to take the power of the supply transformer with a good margin, in order to avoid excessive overheating, and be sure to install a fuse at its input, selected according to the power, to protect against possible troubles.
To us, to make this desired device, you will need the following details:
- Chip LM317 or LM317T.
- Almost any rectifier assembly or four separate diodes with a current of at least 1 ampere each.
- Capacitor C1 from 1000 μF and higher with a voltage of 50 volts, it serves to smooth out voltage surges in the supply network and the larger its capacitance, the more stable the output voltage will be.
- C2 and C4 – 0.047 uF. There is a number 104 on the capacitor cap.
- C3 – 1 µF or more with a voltage of 50 volts. This capacitor can also be used with a larger capacity to increase the stability of the output voltage.
- D5 and D6 - diodes, for example 1N4007, or any others with a current of 1 ampere or more.
- R1 – potentiometer for 10 Kom. Any type, but always a good one, otherwise the output voltage will “jump”.
- R2 – 220 Ohm, power 0.25 – 0.5 watts.
Assembling an adjustable stabilized power supply
I assembled it on a regular breadboard without any etching. I like this method because of its simplicity. Thanks to it, the circuit can be assembled in a matter of minutes.
Checking the power supply
By rotating the variable resistor you can set the desired output voltage, which is very convenient.Good day, forum users and site guests. Radio circuits! Wanting to put together a decent, but not too expensive and cool power supply, so that it has everything and it doesn’t cost anything. In the end, I chose the best, in my opinion, circuit with current and voltage regulation, which consists of only five transistors, not counting a couple of dozen resistors and capacitors. Nevertheless, it works reliably and is highly repeatable. This scheme has already been reviewed on the site, but with the help of colleagues we managed to improve it somewhat.
I assembled this circuit in its original form and encountered one unpleasant problem. When adjusting the current, I can’t set it to 0.1 A - at least 1.5 A at R6 0.22 Ohm. When I increased the resistance of R6 to 1.2 Ohms, the current during a short circuit turned out to be at least 0.5 A. But now R6 began to heat up quickly and strongly. Then I used a small modification and got a much wider current regulation. Approximately 16 mA to maximum. You can also make it from 120 mA if you transfer the end of the resistor R8 to the T4 base. The bottom line is that before the resistor voltage drops, a drop is added B-E transition and this additional voltage allows you to open T5 earlier, and as a result, limit the current earlier.
Based on this proposal, I conducted successful tests and eventually received a simple laboratory power supply. I'm posting a photo of mine laboratory block power supply with three outputs, where:
- 1-output 0-22v
- 2-output 0-22v
- 3-output +/- 16V
Also, in addition to the output voltage regulation board, the device was supplemented with a power filter board with a fuse block. What happened in the end - see below.
In fact, this is the simplest adjustable block food in the world!
Having spent less than an hour for its assembly you will receive a stabilized regulated power supply with output voltage 0...12 V and maximum load current 1 A to power your structures.
This set was created based on a wonderful article on a well-known cat website. The article (see below...) describes the simplest stabilized power supply that you can imagine. And it’s not just described - the second part of this article describes all the calculations that need to be performed when designing such a power supply.
The developers just added an LED to the circuit D2 and ballast resistor Rd for LED. The LED will indicate that voltage is being supplied to the power supply.
And yes, a small radiator for the transistor is added to the kit VT2 and fasteners for it so that you can test your power supply immediately after assembly.
Characteristics:
Input voltage: 12...15 V;
Output voltage: 0...12 (±1) V;
Maximum load current: 1 A;
Difficulty: 1 point;
Assembly time: About 1 hour;
Dimensions printed circuit board: 81 x 31 x 2 mm;
Packing: OEM;
OEM packaging dimensions: ~255 x 123 x 35 mm;
Device dimensions: ~81 x 31 x 35 mm;
total weight set: ~200 g.
Contents of delivery:
Printed circuit board;
Set of radio components;
A coil of mounting wire for a variable resistor (~0.5 m);
Radiator for microcircuit;
Radiator fasteners (~M3x20 screw; M3 nut; M3 washer);
BONUS! Roll of tubular solder POS-61 (~0.5 m);
Component pinout diagram;
Resistor color marking scheme;
Assembly and operation instructions.
Notes:
This power supply requires a step-down transformer with a voltage on the secondary winding of 12...15 V and a current of at least 1 A.
Connect the transformer to the power supply via terminal block X1.
Connect the transformer to the network.
LED D2 should light up, indicating the arrival of DC voltage to the power supply.
Using variable resistor R2, set the required output voltage.
Connect the load - everything works!
Click on the picture to enlarge
(navigate through the pictures using the arrows on the keyboard)
PART 1
power unit
Yes, yes, I already understood that you are impatient - you have already read the theory, read what it is electricity, what resistance is, I found out who Comrade Om is and much more. And now you want to reasonably ask: “So what? What’s the point of all this? Where can all this be applied?” Or perhaps you haven’t read any of this, because it’s terribly boring, but you still want to get your hands on something electronic. I hasten to please you - now we will do just that: we will apply all this properly and solder the first real structure, which will be very useful to you in the future.
We will make a power supply to power various electronic devices that we will assemble in the future. After all, if we first assemble, for example, a radio receiver, it still won’t work until we give it power. So, to paraphrase the well-known proverb - “the power supply is the head of everything” (c) by Author of the article.
So let's get started. First of all, let's set the initial parameters - the voltage that our power supply will produce and the maximum current that it will be able to supply to the load. That is, how powerful a load can be connected to it - can we connect only one radio receiver to it or can we connect ten? Don’t ask me why turn on ten radios at the same time - I don’t know, I just said it as an example.
First, let's think about the output voltage. Let's assume that we have two radios, one of which operates on 9 Volts, and the second on 12 Volts. We won’t make two different power supplies for these devices. Hence the conclusion - you need to make the output voltage adjustable so that it can be adjusted to different meanings and power a wide variety of devices.
Our power supply will have an output voltage adjustment range from 1.5 to 14 Volts - quite enough for the first time. Well, we will take the load current equal to 1 Ampere.
It couldn't be simpler, could it? So, what parts do we need to solder this circuit?
First of all, we need a transformer with a voltage on the secondary winding of 13...16 Volts and a load current of at least 1 Ampere. It is designated in the diagram as T1.
We will also need a diode bridge VD1 - KTs405B or any other with a maximum current of 1 Ampere.
Let's move on - C1 - electrolytic capacitor, which we will use to filter and smooth the voltage rectified by the diode bridge; its parameters are indicated in the diagram.
D1 is a zener diode - it manages voltage stabilization - after all, we don’t want the voltage at the output of the power supply to fluctuate along with mains voltage. We will take a Zener diode D814D or any other with a stabilization voltage of 14 volts.
We also need a constant resistor R1 and a variable resistor R2, with which we will regulate the output voltage.
And also two transistors - KT315 with any letter in the name and KT817 also with any letter.
For convenience, I put all the necessary elements into a plate that you can print out and, together with this piece of paper, go to the store to purchase (or find these components or their analogues).
| Designation on the diagram | Denomination | Note |
| T1 | Any with a secondary winding voltage of 12...13 Volts and a current of 1 Ampere | |
| VD1 | KTs405B | Diode bridge. Maximum rectified current not less than 1 Ampere |
| C1 | 2000 uF x 25 Volts | Electrolytic capacitor |
| R1 | 470 Ohm | |
| R2 | 10 kOhm | Variable resistor |
| R3 | 1 kOhm | Fixed resistor, dissipation power 0.125...0.25 W |
| D1 | D814D | Zener diode. Stabilization voltage 14 V |
| VT1 | KT315 | |
| VT2 | KT817 | Transistor. With any letter index |
All this can be soldered either on the board or by surface mounting - fortunately there are very few elements in the circuit, but it is recommended (to debug the circuit) to assemble it on solderless breadboard
.
Transistor VT2 must be installed on the radiator. The optimal radiator area can be selected experimentally, but it must be at least 50 square meters. cm.
When installed correctly, the circuit requires no adjustment at all and starts working immediately.
We connect a tester or Voltmeter to the output of the power supply and set resistor R2 to the voltage we need.
That's basically all. Any questions?
Well, for example: “Why is resistor R1 100 Ohms?” or, “why two transistors - is it really impossible to get by with one?” No?
Well, whatever you want, but if they do appear, read next part This article explains how this power supply was calculated and how to calculate your own.
PART 2
Power supply "It couldn't be simpler"
Yeah, did you come in yet? What, curiosity tormented you? But I'm very happy. No, really.
Make yourself comfortable, now together we will make some simple calculations that are needed to assemble the power supply that we have already done in the first part of the article.
Although it must be said that these calculations can be useful in more complex schemes.
So, our power supply consists of two main components:
A rectifier consisting of a transformer, rectifying diodes and a capacitor;
Stabilizer, consisting of everything else.
Like real Indians, let's start from the end and calculate the stabilizer first.
Stabilizer
The stabilizer circuit is shown in the figure:
This is the so-called parametric stabilizer. It consists of two parts:
The stabilizer itself on a zener diode D with a ballast resistor R b ;
Emitter follower on transistor VT.
The stabilizer ensures that the voltage remains what we need, and the emitter follower allows you to connect a powerful load to the stabilizer.
It plays the role of an amplifier or, if you like, a booster.
The two main parameters of our power supply are the output voltage and the maximum load current.
Let's call them: Uout(this is tension) and Imax(this is current).
For the power supply, which we discussed in the last part, Uout = 14 Volts, and Imax = 1 Ampere.
First, we need to determine what voltage Uin we must apply to the stabilizer in order to obtain the required Uout at the output.
This voltage is determined by the formula: Uin = Uout + 3
Where did the number 3 come from? This is the voltage drop across the collector-emitter junction of the VT transistor. Thus, for our stabilizer to operate, we must supply at least 17 volts to its input.
Let's determine what kind of transistor VT we need. To do this, we need to determine how much power it will dissipate.
We consider: Pmax=1.3(Uin-Uout)Imax
One point needs to be taken into account here. For the calculation, we took the maximum output voltage of the power supply. However, in this calculation, on the contrary, we must take the minimum voltage that the power supply produces. And in our case it is 1.5 Volts. If this is not done, the transistor may be covered with a copper basin, since the maximum power will be calculated incorrectly.
Take a look yourself:
If we take Uout = 14 Volts, we get P max=1.3*(17-14)*1=3.9 W.
And if we take Uout = 1.5 Volts, then P max=1.3*(17-1.5)*1=20.15 W
That is, if this had not been taken into account, it would have turned out that the calculated power was FIVE times less than the real one. Of course, the transistor would not like this very much.
Well, now we go into the directory and choose a transistor for ourselves.
In addition to the power just received, it must be taken into account that the maximum voltage between the emitter and the collector must be greater than Uin, and the maximum collector current must be greater than Imax.
I chose KT817 - a pretty decent transistor...
First, let's determine the maximum base current of a freshly selected transistor (what did you think? in our cruel world consume everything - even the transistor bases).
I b max=I max/h21 E min
h21 E min- this is the minimum current transfer coefficient of the transistor and it is taken from the reference book. If the limits of this parameter are indicated there - something like 30...40, then the smallest one is taken. Well, in my reference book there is only one number written - 25, we will count with it, but what else is left?
I b max=1/25=0.04 A (or 40 mA), which is not small.
Well, let's now look for a zener diode.
You need to look for it using two parameters - stabilization voltage and stabilization current.
The stabilization voltage should be equal to the maximum output voltage of the power supply, that is, 14 Volts, and the current should be at least 40 mA, that is, what we calculated.
Let's go back to the directory...
In terms of voltage, a zener diode is terrible for us D814D, besides, I had it at hand. But the stabilization current... 5 mA is not suitable for us. What are we going to do? We will reduce the base current of the output transistor.
And to do this, we’ll add another transistor to the circuit. Let's look at the drawing. We added transistor VT2 to the circuit.
This operation allows us to reduce the load on the zener diode by h21E times. h21E, of course, the transistor that we just added to the circuit. Without thinking too much, I took the KT315 from the pile of pieces of hardware.
Its minimum h21E is 30, that is, we can reduce the current to 40/30=1.33 mA, which suits us quite well.
Now let's calculate the resistance and power of the ballast resistor R b :
R b=(Uin-Ust)/(I b max+I st min),
Where:
Ust - stabilization voltage of the zener diode,
Ist min - zener diode stabilization current.
R b= (17-14)/((1.33+5)/1000) = 470 Ohm.
Now let's determine the power of this resistor:
P rb= (U input-U st)*2/R b ,
That is:
P rb= (17-14)2/470=0.02 W.
That's all. Thus, from the initial data - output voltage and current, we obtained all the elements of the circuit and the input voltage that should be supplied to the stabilizer.
However, let's not relax - the rectifier is still waiting for us. I think so, I think so (pun intended, however).
Rectifier
So, let's look at the rectifier circuit:
Well, everything is simpler here and almost on your fingers.
Considering that we know what voltage we need to supply to the stabilizer - 17 volts, let's calculate the voltage on the secondary winding of the transformer. To do this, let's go, as in the beginning - from the tail. So after the filter capacitor we should have a voltage of 17 volts.
Considering that the filter capacitor increases the rectified voltage by 1.41 times, we find that after the rectifier bridge we should have 17/1.41=12 Volts.
Now let’s take into account that on the rectifier bridge we lose about 1.5-2 Volts, therefore, the voltage on the secondary winding should be 12+2=14 Volts. It may well happen that such a transformer will not be found, no big deal - in this case you can use a transformer with a voltage on the secondary winding of 13 to 16 Volts.
C f= 3200*I n/(U n*K n ,
Where:
Iн - maximum load current;
Un - load voltage;
Kn - pulsation coefficient.
In our case:
In = 1 Ampere;
Un=17 Volts;
Kn=0.01.
C f = 3200*1/17*0,01=18823.
However, since there is also a voltage stabilizer behind the rectifier, we can reduce the calculated capacity by 5...10 times. That is, 2000 uF will be quite enough.
All that remains is to choose rectifier diodes or a diode bridge.
To do this, we need to know two main parameters - the maximum current flowing through one diode and the maximum reverse voltage, also through one diode.
The required maximum reverse voltage is calculated as follows:
U arr max= 2U n, that is, U arr max=2*17=34 Volts.
And the maximum current for one diode must be greater than or equal to the load current of the power supply. Well, for diode assemblies, reference books indicate the total maximum current that can flow through this assembly.
Well, that seems to be all about rectifiers and parametric stabilizers.
Ahead we have a stabilizer for the laziest - on an integrated circuit and a stabilizer for the most hardworking - a compensation stabilizer.
PART 3
power unit
In this part, as promised, we will talk about another type of stabilizers - compensatory. As the name suggests (the name is obvious, no?), their operating principle is based on compensation of something by something, somehow, somewhere. What and with what we find out now.
To begin with, let's look at the circuit of the simplest compensation stabilizer. Its circuit is more complex than a regular parametric one, but just a little:
The circuit consists of the following nodes:
- Reference voltage source (VS) on R 2, D 1, which itself is a parametric stabilizer.
- Voltage divider R3-R5.
- Direct current amplifier (DCA) on transistor VT1.
- Regulating element on transistor VT2.
This whole zoo works as follows. The ION produces a reference voltage equal to the voltage at the output of the stabilizer to the emitter VT1. The voltage from the divider is supplied to the base of VT1. As a result, this poor guy has to decide what to do with the voltage on the collector - either leave everything as it is, or increase it, or decrease it. And in order not to fool too much, he does this - if the voltage at the base is less than the reference (which is at the emitter), he increases the voltage at the collector, thus opening the transistor VT2 more strongly and increasing the voltage at the output, but if the voltage at the base is greater than the reference, then the reverse process occurs.
As a result of all this fuss, the output voltage remains unchanged, that is, stabilized, which is what is required. Moreover, compared to parametric stabilizers, the stabilization coefficient of compensatory stabilizers is much higher. The efficiency is also higher.
Resistor R4 is needed to adjust the output voltage of the stabilizer within small limits.
Well, now let's move on to the sweet stuff - to stabilizers on microcircuits. I call them stabilizers for the lazy, because soldering such a stabilizer takes about two minutes, if not less. In order not to drag too much, let’s go straight to the diagram, although the diagram is...
So, here is a diagram that is disgustingly simple. There are only three elements in it, and only one is required - the DA1 chip. By the way, integral stabilizers are compensatory in nature. Well, sir, what do we need? There is only one thing - to know the voltage that we want to get from the stabilizer. Next we go to the table and choose a microcircuit to our liking.
The voltage at the input of the microcircuit must be at least 3 Volts higher than the output, but should not exceed 30 Volts. Well, that's all.
I'm sorry, what? Do you need not 15 Volts, but 14? How capricious you are. Anyway. As an incentive prize (though I don’t know why yet), I’ll tell you about one more scheme.
Of course, in addition to stabilizers with a fixed voltage, there are integrated stabilizers specially designed for adjustable voltage. So, pay attention to the diagram!
We meet - KREN12A (B is also possible) - an adjustable voltage stabilizer of 1.3...30 Volts and a maximum current of 1.5 A.
By the way, it also has a bourgeois analogue - LM317 (in the diagram the pin numbering for it is given in brackets). Input voltage no more than 37 Volts.
If you really want to, there is something to calculate in this scheme. In any case, if you don’t have a 240 Ohm resistor, you can plug in another one, while recalculating resistor R2.
There is a clever formula for this:
The formula includes:
U reference = 1.25 V - internal reference voltage of the microcircuit between the 2nd and 8th pins, see diagram;
I support - control current flowing through resistor R2.
Generally speaking, the formula can be simplified due to the fact that this same control current is very, very small - about 0.0055A, that is, it has practically no effect on the result:
Well, now let's count.
First, let's take the MINIMUM value of the output voltage that you want to get.
So, R1=240 Ohm, Uout=1.3 V, Uref=1.25 V. Then:
R2=240(1.3-1.25)/1.25 = 9.6 Ohm
Afterwards, we take the MAXIMUM voltage that our stabilizer should produce:
R1=240 Ohm, Uout=30 V, Uref=1.25 V
R2=240(30-1.25)/1.25=5500 Ohm, which is 5.5 kOhm.
Thus, in order for the voltage at the output of the stabilizer to change from minimum to maximum, we need the resistance of resistor R2 to change from 9.6 Ohms to 5.5 kOhms.
We select the one closest to this value - I found it to be 4.8 kOhm.
These are the pies. By the way, before I forget, the microcircuits must be placed on a radiator, otherwise they will die, and quite quickly. Really sad.
Externally, the microcircuit in the KT28-2 package looks like this:
I want to convert Special attention the fact that although LM317 is a complete functional analogue of KREN12A, the pin layout of these microcircuits DOES NOT MATCH, if KREN12 is made in the above-mentioned housing.
Pin layout of the LM317 chip. The terminals of KREN12 are also located if it is made in the TO-200 housing:
That's it now.
We assemble an adjustable power supply 0...30V / 5A.
Have you decided to assemble a power supply, but don’t know which circuit to choose? But indeed, on the Internet you can find many schematic diagrams of these devices. Well, in this article we will look at a power supply circuit implemented on a domestic element base; these components from which the circuit is assembled are quite widespread and not at all in short supply, and this is a big advantage of this option. The second advantage of this circuit is that the output voltage of the power supply is adjustable over a wide range, ranging from 0 to 30 Volts, while the output current can reach 5 Amps. And one more important point, this circuit has overload protection and short circuit under load. Schematic diagram shown in the figure below:
Let's look at what nodes the circuit consists of:
A step-down transformer. Its power should be about 150 watts. For example, you can rewind the secondary windings of the TS-160 transformer, or use similar iron. When remaking the TS-160, the primary winding remains unchanged. The second winding is designed for a voltage of 28...30 Volts, and a current of at least 5...6 Amperes. The third winding should produce 5...6 Volts with a current of at least 1 Ampere.
Rectifier assembly. It consists of a diode bridge VD1...VD4, and a smoothing capacitance C1. The printed circuit board provides for the use of an imported diode assembly RS603 (RS602) for a current of 10 Amps, but you can also assemble a bridge from individual domestic diodes, for example, D242, although the dimensions of the device will naturally increase.
The KTs407 diode bridge and two integrated stabilizers 7805 and 7905 form the power supply unit for the control and protection unit. Instead of KTs407 you can put KTs402 or KTs405.
The protection is assembled on the KU101E thyristor, the VD9 LED indicates its status, and in case of overload and short circuit it lights up. Resistor R4 is installed as a current sensor; in the circuit it is designed for a current of 3 Amps; for 5 Amps it must be recalculated.
The regulating element is a powerful silicon transistor VT1 (KT827A). It must be installed on a radiator with a cooling area of at least 1500 square meters. see. If difficulties arise in purchasing the KT827A, then instead you can install a pair of transistors connected according to the following diagram:
Resistor R7 regulates the minimum voltage of the power supply output. The potentiometer R13 handle is located on the front panel of the power supply and is an output voltage regulator. Rotate R14 to adjust the upper limit of the output voltage. R7 and R14 are multi-turn type SP5.
The pictures below show a version of the power supply circuit board:
The printed circuit board has dimensions of 110x75 mm.
Setting up the power supply:
The entire setup of the power supply comes down to setting the necessary limits for adjusting the output voltage, as well as the current value at which the protection will operate. As mentioned above, the protection current depends on the value of resistor R4.
To determine the output voltage regulation range, perform the following steps:
Set potentiometers R7 and R13 to the middle position.
Measuring Uout with a voltmeter. Using resistor R14, set the value to 15 Volts.
Turn resistor R13 to minimum, and use R7 to set the output to zero volts.
Now R13 to maximum, and using R14 set the output to 30 Volts. If necessary, instead of R14 (by measuring its readings), you can solder a constant resistance.
At this point, the setup is complete, if everything is assembled without mistakes and errors, the power supply will work “like a clock.” This is where we end the article, good luck with your repetition.
R3 10k (4k7 – 22k) reostat R6 0.22R 5W (0.15-0.47R) R8 100R (47R – 330R) | C1 1000 x35v (2200 x50v) C2 1000 x35v (2200 x50v) C5 100n ceramic (0.01-0.47) | T1 KT816 (BD140) T2 BC548 (BC547) T3 KT815 (BD139) T4 KT819(KT805,2N3055) T5 KT815 (BD139) VD1-4 KD202 (50v 3-5A) VD5 BZX27 (KS527) VD6 AL307B, K (RED LED) |
Adjustablestabilizedpower supply – 0-24V, 1 – 3A
with current limitation.
The power supply unit (PSU) is designed to obtain an adjustable, stabilized output voltage from 0 to 24v at a current of about 1-3A, in other words, so that you don’t buy batteries, but use it to experiment with your own designs.
The power supply provides so-called protection, i.e. maximum current limitation.
What is it for? In order for this power supply to serve faithfully, without fear of short circuits and not require repairs, so to speak, “fireproof and indestructible”
A zener diode current stabilizer is assembled on T1, that is, it is possible to install almost any zener diode with a stabilization voltage less than the input voltage by 5 volts
This means that when installing a VD5 zener diode, let’s say BZX5.6 or KS156 at the output of the stabilizer, we get an adjustable voltage from 0 to approximately 4 volts, respectively - if the zener diode is 27 volts, then the maximum output voltage will be within 24-25 volts.
The transformer should be selected something like this - the alternating voltage of the secondary winding should be about 3-5 volts greater than what you expect to receive at the output of the stabilizer, which in turn depends on the installed zener diode,
The current of the secondary winding of the transformer must at a minimum be no less than the current that needs to be obtained at the output of the stabilizer.
Selection of capacitors by capacity C1 and C2 - approximately 1000-2000 µF per 1A, C4 - 220 µF per 1A
It is somewhat more complicated with voltage capacitances - the operating voltage is roughly calculated using this method - the alternating voltage of the secondary winding of the transformer is divided by 3 and multiplied by 4
(~ Uin:3×4)
That is, let’s say that the output voltage of your transformer is about 30 volts - divide 30 by 3 and multiply by 4 - we get 40 - which means the operating voltage of the capacitors should be more than 40 volts.
The level of current limitation at the output of the stabilizer depends on R6 at a minimum and R8 (at a maximum until shutdown)
When installing a jumper instead of R8 between the base of VT5 and the emitter of VT4 with a resistance of R6 equal to 0.39 ohms, the limiting current will be approximately 3A,
How do we understand “limitation”? It’s very simple - the output current, even in short circuit mode, will not exceed 3 A, due to the fact that the output voltage will be automatically reduced to almost zero,
Is it possible to charge car battery? Easily. It is enough to set the voltage regulator, I apologize - with potentiometer R3 the voltage is 14.5 volts at idle (that is, with the battery disconnected) and then connect the battery to the output of the unit, and your battery will be charged with a stable current to the level of 14.5 V, Current as it charges will decrease and when it reaches 14.5 volts (14.5 V is the voltage of a fully charged battery) it will be zero.
How to adjust the limiting current. Set the idle voltage at the output of the stabilizer to about 5-7 volts. Then connect a resistance of approximately 1 ohm with a power of 5-10 watts to the output of the stabilizer and an ammeter in series with it. Use trimmer resistor R8 to set the required current. Correctly set limiting current can be checked by turning the output voltage adjustment potentiometer all the way to the maximum. In this case, the current controlled by the ammeter should remain at the same level.
Now about the details. Rectifier bridge - it is advisable to select diodes with a current reserve of at least one and a half times. The indicated KD202 diodes can operate without radiators for quite a long time at a current of 1 ampere, but if you expect that this is not enough for you, then by installing radiators you can provide 3-5 amperes, that's just what you need Look in the directory which of them and with which letter can carry up to 3 and which up to 5 amperes. If you want more, look at the reference book and choose more powerful diodes, say 10 amperes.
Transistors - VT1 and VT4 should be installed on radiators. VT1 will heat up slightly, so a small radiator is needed, but VT4 will heat up quite well in current limiting mode. Therefore, you need to choose an impressive radiator, you can also adapt a fan from the computer power supply to it - believe me, it won’t hurt.
For those who are especially inquisitive, why does the transistor get hot? Current flows through it and the greater the current, the more the transistor heats up. Let's do the math - 30 volts at the input, across the capacitors. At the output of the stabilizer, let’s say 13 volts. As a result, 17 volts remain between the collector and emitter.
From 30 volts we minus 13 volts, we get 17 volts (who wants to see mathematics here, but one of the laws of grandfather Kirgoff, about the sum of voltage drops, somehow comes to mind)
Well, the same Kirgoff said something about the current in the circuit, like what kind of current flows in the load, the same current flows through the VT4 transistor. Let's say about 3 amperes flow, the resistor in the load heats up, the transistor also heats up, So this is the heat with which we heat the air and can be called power that is dissipated... But let's try to express it mathematically, that is
school physics course
Where R is the power in watts, U is the voltage across the transistor in volts, and J- the current that flows through our load and through the ammeter and, naturally, through the transistor.
So 17 volts multiplied by 3 amperes we get 51 watts dissipated by the transistor,
Well, let’s say we connect a resistance of 1 ohm. According to Ohm's law, at a current of 3A, the voltage drop across the resistor will be 3 volts and the dissipated power of 3 watts will begin to heat the resistance. Then the voltage drop across the transistor is: 30 volts minus 3 volts = 27 volts, and the power dissipated by the transistor is 27v×3A = 81 watts... Now let’s look in the reference book, in the transistors section. If we have a pass-through transistor, ie VT4, say KT819 in a plastic case, then according to the reference book it turns out that it will not withstand the dissipation power (Pk*max) it has 60 watts, but in a metal case (KT819GM, analogue 2N3055) - 100 watts - this one will do, but a radiator is required.
I hope it’s more or less clear about transistors, let’s move on to fuses. In general, a fuse is the last resort, reacting to gross mistakes made by you and preventing it “at the cost of your life.” Let’s assume that for some reason a short circuit occurs in the primary winding of the transformer, or in the secondary. Maybe it’s because it’s overheated, maybe the insulation is leaky, or maybe it’s just an incorrect connection of the windings, but there are no fuses. The transformer smokes, the insulation melts, the power cable, trying to perform the valiant function of a fuse, burns, and God forbid if you have plugs with nails instead of fuses on the distribution panel instead of a machine.
One fuse for a current of approximately 1A greater than the limiting current of the power supply (i.e. 4-5A) should be placed between the diode bridge and the transformer, and the second between the transformer and the 220 volt network for approximately 0.5-1 ampere.
Transformer. Perhaps the most expensive thing in the design Roughly speaking, the more massive the transformer, the more powerful it is. The thicker the secondary winding wire, the more current the transformer can deliver. It all comes down to one thing - the power of the transformer. So how to choose a transformer? Again a school physics course, electrical engineering section.... Again 30 volts, 3 amperes and ultimately a power of 90 watts. This is the minimum, which should be understood as follows - this transformer can briefly provide an output voltage of 30 volts at a current of 3 amperes. Therefore, it is advisable to add a current reserve of at least 10 percent, and better yet 30-50 percent. So 30 volts at a current of 4-5 amperes at the output of the transformer and your power supply will be able to supply a current of 3 amperes to the load for hours, if not days.
Well, for those who want to get the maximum current from this power supply, let’s say about 10 amperes.
First - a transformer that matches your needs
Second - 15 ampere diode bridge and for radiators
Third, replace the pass-through transistor with two or three connected in parallel with resistances in the emitters of 0.1 ohms (radiator and forced airflow)
Fourth, it is desirable, of course, to increase the capacity, but in the event that the power supply will be used as Charger– this is not critical.
Fifth, reinforce the conductive paths along the path of large currents by soldering additional conductors and, accordingly, do not forget about the “thicker” connecting wires
Connection diagram for parallel transistors instead of one
