This power supply, based on the LM317 chip, does not require any special knowledge for assembly, and after proper 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 make this necessary device, we will need the following parts:
- 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.In fact, this is the simplest adjustable power supply in the world!
After spending less than an hour assembling it, you will receive a stabilized, regulated power supply with an 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;
PCB dimensions: 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 of the 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 operating 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 that DC voltage is supplied 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 a lot of theory, read what electric current is, what resistance is, 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 values 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 is an electrolytic capacitor with which we will filter and smooth out 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 the 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 does not require any 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 the next part of this article, which talks about 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, we must, on the contrary, 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, everyone consumes - even the bases of transistors).
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 - stabilization current of the zener diode.
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 how 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 tricky 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 would like to draw special attention to the fact that although the LM317 is a complete functional analogue of the 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.
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 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 a 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 if the power supply is used as a 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
The master whose device was described in the first part, having set out to make a power supply with regulation, did not complicate things for himself and simply used boards that were lying idle. The second option involves the use of an even more common material - an adjustment has been added to the usual block, perhaps this is a very promising solution in terms of simplicity, given that the necessary characteristics will not be lost and even the most experienced radio amateur can implement the idea with his own hands. As a bonus, there are two more options for very simple schemes with all the detailed explanations for beginners. So, there are 4 ways for you to choose from.
We'll tell you how to make an adjustable power supply from an unnecessary computer board. The master took the computer board and cut out the block that powers the RAM.
This is what he looks like.
Let's decide which parts need to be taken and which ones not, in order to cut off what is needed so that the board has all the components of the power supply. Typically, a pulse unit for supplying current to a computer consists of a microcircuit, a PWM controller, key transistors, an output inductor and an output capacitor, and an input capacitor. For some reason, the board also has an input choke. He left him too. Key transistors - maybe two, three. There is a seat for 3 transistors, but it is not used in the circuit.
The PWM controller chip itself may look like this. Here she is under a magnifying glass.
It may look like a square with small pins on all sides. This is a typical PWM controller on a laptop board.
This is what a switching power supply looks like on a video card.
The power supply for the processor looks exactly the same. We see a PWM controller and several processor power channels. 3 transistors in this case. Choke and capacitor. This is one channel.
Three transistors, a choke, a capacitor - the second channel. Channel 3. And two more channels for other purposes.
You know what a PWM controller looks like, look at its markings under a magnifying glass, search for a datasheet on the Internet, download the pdf file and look at the diagram so as not to confuse anything.
In the diagram we see a PWM controller, but the pins are marked and numbered along the edges.
Transistors are designated. This is the throttle. This is an output capacitor and an input capacitor. The input voltage ranges from 1.5 to 19 volts, but the supply voltage to the PWM controller should be from 5 volts to 12 volts. That is, it may turn out that a separate power source is required to power the PWM controller. All the wiring, resistors and capacitors, don’t be alarmed. You don't need to know this. Everything is on the board; you do not assemble a PWM controller, but use a ready-made one. You only need to know 2 resistors - they set the output voltage.
Resistor divider. Its whole point is to reduce the signal from the output to about 1 volt and apply feedback to the input of the PWM controller. In short, by changing the value of the resistors, we can regulate the output voltage. In the case shown, instead of a feedback resistor, the master installed a 10 kilo-ohm tuning resistor. This was sufficient to regulate the output voltage from 1 volt to approximately 12 volts. Unfortunately, this is not possible on all PWM controllers. For example, on PWM controllers of processors and video cards, in order to be able to adjust the voltage, the possibility of overclocking, the output voltage is supplied by software via a multi-channel bus. The only way to change the output voltage of such a PWM controller is by using jumpers.
So, knowing what a PWM controller looks like and the elements that are needed, we can already cut out the power supply. But this must be done carefully, since there are tracks around the PWM controller that may be needed. For example, you can see that the track goes from the base of the transistor to the PWM controller. It was difficult to save it; I had to carefully cut out the board.
Using the tester in dial mode and focusing on the diagram, I soldered the wires. Also using the tester, I found pin 6 of the PWM controller and the feedback resistors rang from it. The resistor was located in the rfb, it was removed and instead of it, a 10 kilo-ohm tuning resistor was soldered from the output to regulate the output voltage; I also found out by calling that the power supply of the PWM controller is directly connected to the input power line. This means that you cannot supply more than 12 volts to the input, so as not to burn out the PWM controller.
Let's see what the power supply looks like in operation
I soldered the input voltage plug, voltage indicator and output wires. We connect an external 12 volt power supply. The indicator lights up. It was already set to 9.2 volts. Let's try to adjust the power supply with a screwdriver.
It's time to check out what the power supply is capable of. I took a wooden block and a homemade wirewound resistor made from nichrome wire. Its resistance is low and, together with the tester probes, is 1.7 Ohms. We turn the multimeter into ammeter mode and connect it in series with the resistor. See what happens - the resistor heats up to red, the output voltage remains virtually unchanged, and the current is about 4 amperes.
The master had already made similar power supplies before. One is cut out with your own hands from a laptop board.
This is the so-called standby voltage. Two sources of 3.3 volts and 5 volts. I made a case for it on a 3D printer. You can also look at the article where I made a similar adjustable power supply, also cut from a laptop board (https://electro-repair.livejournal.com/3645.html). This is also a PWM power controller for RAM.
How to make a regulating power supply from a regular printer
We will talk about the power supply for a Canon inkjet printer. Many people have them idle. This is essentially a separate device, held in the printer by a latch.
Its characteristics: 24 volts, 0.7 amperes.
I needed a power supply for a homemade drill. It's just right in terms of power. But there is one caveat - if you connect it like this, the output will only get 7 volts. Triple output, connector and we get only 7 volts. How to get 24 volts?
How to get 24 volts without disassembling the unit?
Well, the simplest one is to close the plus with the middle output and we get 24 volts.
Let's try to do it. We connect the power supply to the 220 network. We take the device and try to measure it. Let's connect and see 7 volts at the output.
Its central connector is not used. If we take it and connect it to two at the same time, the voltage is 24 volts. This is the easiest way to ensure that this power supply produces 24 volts without disassembling it.
A homemade regulator is needed so that the voltage can be adjusted within certain limits. From 10 volts to maximum. It's easy to do. What is needed for this? First, open the power supply itself. It is usually glued. How to open it without damaging the case. There is no need to pick or pry anything. We take a piece of wood that is heavier or have a rubber mallet. Place it on a hard surface and tap along the seam. The glue is coming off. Then they tapped thoroughly on all sides. Miraculously, the glue comes off and everything opens up. Inside we see the power supply.
We'll get the payment. Such power supplies can be easily converted to the desired voltage and can also be made adjustable. On the reverse side, if we turn it over, there is an adjustable zener diode tl431. On the other hand, we will see the middle contact goes to the base of transistor q51.
If we apply voltage, then this transistor opens and 2.5 volts appears at the resistive divider, which is needed for the zener diode to operate. And 24 volts appears at the output. This is the simplest option. Another way to start it is to throw away transistor q51 and put a jumper instead of resistor r 57 and that’s it. When we turn it on, the output is always 24 volts continuously.
How to make an adjustment?
You can change the voltage, make it 12 volts. But in particular, the master does not need this. You need to make it adjustable. How to do it? We throw away this transistor and replace the 57 by 38 kilo-ohm resistor with an adjustable one. There is an old Soviet one with 3.3 kilo-ohms. You can put from 4.7 to 10, which is what it is. Only the minimum voltage to which it can lower it depends on this resistor. 3.3 is very low and not necessary. The engines are planned to be supplied at 24 volts. And just from 10 volts to 24 is normal. If you need a different voltage, you can use a high-resistance tuning resistor.
Let's get started, let's solder. Take a soldering iron and hair dryer. I removed the transistor and resistor.
We soldered the variable resistor and will try to turn it on. We applied 220 volts, we see 7 volts on our device and begin to rotate the variable resistor. The voltage has risen to 24 volts and we rotate it smoothly and smoothly, it drops - 17-15-14, that is, it decreases to 7 volts. In particular, it is installed on 3.3 rooms. And our rework turned out to be quite successful. That is, for purposes from 7 to 24 volts, voltage regulation is quite acceptable.
This option worked out. I installed a variable resistor. The handle turns out to be an adjustable power supply - quite convenient.
Video of the channel “Technician”.
Such power supplies are easy to find in China. I came across an interesting store that sells used power supplies from various printers, laptops and netbooks. They disassemble and sell the boards themselves, fully functional for different voltages and currents. The biggest plus is that they disassemble branded equipment and all power supplies are high quality, with good parts, all have filters.
The photos are of different power supplies, they cost pennies, practically a freebie.
Simple block with adjustment
A simple version of a homemade device for powering devices with regulation. The scheme is popular, it is widespread on the Internet and has shown its effectiveness. But there are also limitations, which are shown in the video along with all the instructions for making a regulated power supply.
Homemade regulated unit on one transistor
What is the simplest regulated power supply you can make yourself? This can be done on the lm317 chip. It almost represents a power supply itself. It can be used to make both a voltage- and flow-regulated power supply. This video tutorial shows a device with voltage regulation. The master found a simple scheme. Input voltage maximum 40 volts. Output from 1.2 to 37 volts. Maximum output current 1.5 amperes. 
Without a heat sink, without a radiator, the maximum power can be only 1 watt. And with a radiator 10 watts. List of radio components. 
Let's start assembling 
Let's connect an electronic load to the output of the device. Let's see how well it holds current. We set it to minimum. 7.7 volts, 30 milliamps.
Everything is regulated. Let's set it to 3 volts and add current. We will only set higher restrictions on the power supply. We move the toggle switch to the upper position. Now it's 0.5 ampere. The microcircuit began to warm up. There is nothing to do without a heat sink. I found some kind of plate, not for long, but enough. Let's try again. There is a drawdown. But the block works. Voltage adjustment is in progress. We can insert a test into this scheme.
Radioblogful video. Soldering video blog.
I took this diagram from the Internet many years ago. The reason I decided to post it is that there are errors in the original that I corrected. Therefore, you can safely take the circuit and make this power supply. It has been working for me for four years now.
This power supply is built on a common radio element base and does not contain scarce parts. A special feature of the unit is that the regulated DA4 microcircuit does not require bipolar power. On the DA1 chip, a smooth adjustment of the output current has been introduced in the range of 0 ... 3A (according to the diagram). This limit can be expanded to 5A by recalculating resistor R4. In the author's version, resistor R7 is replaced with a tuning one, because Smooth current adjustment was not required. The current limitation with the set ratings of the parts occurs at a current of 3.2A and the output voltage drops to 0. The current limitation is selected by resistor R7. During current limitation, the HL1 LED turns on, signaling a short circuit in the power supply load or the selected current value by resistor R7 being exceeded. If resistor R7 selects a response threshold of 1.5A, then if this threshold is exceeded, a low voltage (-1.4V) will appear at the output of the microcircuit and 127mV will be established at the base of transistor VT2. The voltage at the output of the power supply becomes equal to » 1 µV, which is normal for most amateur radio applications, and the voltage indication unit will read 00.0 volts. The HL1 LED will light up. During normal operation of the overcurrent unit based on the DA1 chip, the voltage will be 5.5V and the HL1 diode will not light up.
The characteristics of the power supply are as follows:
The output voltage is adjustable from 0 to 30 V.
Output current 4A.
The operation of the DA4 microcircuit has no special features and it operates in single-supply mode. 9V is supplied to leg 7, leg 4 is connected to a common bus. Unlike most microcircuits of the 140UD series... it is very difficult to achieve a zero level at the output of the power supply when switched on in this way. Experimentally, the choice was made on the KR140UD17A microcircuit. With this circuit design, it was possible to obtain a voltage of 156 μV at the output of the power supply, which will be displayed on the indicator as 00.0V.
Capacitor C5 prevents excitation of the power supply.
With serviceable parts and error-free installation, the power supply starts working immediately. Resistor R12 sets the upper level of the output voltage, within 30.03V. The VD5 zener diode is used to stabilize the voltage across the regulating resistor R16 and, if the power supply operates without failures, the zener diode can be dispensed with. If resistor R7 is used as a tuning resistor, then it sets the operating threshold when the maximum current is exceeded.
Transistor VT1 is installed on the radiator. The radiator area is calculated by the formula: S = 10In*(Uin. - Uout.), where S is the surface area of the radiator (cm 2); In - maximum current consumed by the load; Uin. - input voltage (V); Uout. - output voltage (V).
The power supply circuit is shown in Fig. 1, the printed circuit board is shown in Figures 2 and 3.
What is highlighted in red are the errors that I corrected. If you don't do this, the scheme won't work.
Resistors R7 and R12 are multi-turn SP5-2. Instead of the RS602 diode assembly, you can use the RS407, RS603 diode assembly, depending on the current consumption, or 242 diodes with any letter index, but they must be placed separately from the printed circuit board. The input voltage on capacitor C1 can vary within 35... 40V without changing the ratings of the parts. Transformer T1 must be designed for a power of at least 100 W, the current of winding II is not less than 5 A at a voltage of 35 ... 40 V. The current of winding III is not less than 1 A. Winding III MUST (otherwise the circuit will not work, this is one of the mistakes) be with a tap from the middle, which is connected to the common bus of the power supply. The printed circuit board is provided with a contact pad for this purpose. The size of the power supply circuit board is 110 x 75 mm. The KT825 transistor is composite and costs a lot, so it can be replaced with transistors, as shown in Figure 4.
Transistors can be with letter indices B - G, connected according to a Darlington circuit.
Resistor R4 is a piece of nichrome wire with a diameter of 1 mm and a length of about 7 cm (selected experimentally). Microcircuits DA2, DA3 and DA5 can be replaced with domestic analogues K142EN8A, KR1168EN5 and K142EN5A. If the digital display panel is not used, then instead of the DA2 chip you can use KR1157EN902, and exclude the DA5 chip. Resistor R16 is variable with group A dependence. In the author's version, a variable resistor PPB-3A with a nominal value of 2.2K - 5% is used.
If you do not place great demands on the protection unit, and it will only be required to protect the power supply from overcurrent and short circuit, then such a unit can be used according to the diagram in Fig. 6, and the printed circuit board can be slightly reworked.
The protection unit is assembled on transistors VT1 and VT2 of different structures, resistors R1 - R3 and capacitor C1. Short circuit current 16mA. Resistor R1 regulates the response threshold of the protective block. During normal operation of the unit, the voltage at the emitter of transistor VT2 is about 7 V and does not affect the operation of the power supply. When the protection is triggered, the voltage at the emitter of transistor VT2 drops to 1.2 V and is supplied through the diode VD4 to the base of transistor VT2 of the power supply. The voltage at the output of the power supply drops to 0 V and the HL1 LED signals that the protection has tripped. During normal operation of the power supply and the protection unit, the LED lights up, and when the protection is triggered, it goes out. When using the protection unit in Fig. 6, the DA3 microcircuit and capacitors C3, C5 can be excluded from the circuit.
The digital panel serves to visually monitor the voltage and current of the power supply. It can be used separately from the power supply with other designs, performing the above tasks.
I took the voltmeter and ammeter from here.
Here are a few photos of my power supply, which show that I also attached a fan for cooling, the power for which I took from the third winding of the transformer, having previously wound it with this calculation.
(click on images to enlarge)
Alexander, thank you for the work you have done!
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