There are two linear regulators in the bag of most people dealing with hobby electronics: LM317. One opamp (LM318) and plenty of resistance, capacitor and so on. Let us ask this; Let us design a power supply with short circuit protection, without spending a lot of money. In particular, by paralleling LM317s, we should increase our current draw ratio.

The first question we will ask before starting the project can be why we use 2 LM317. If you explore the Internet, you can see a lot of power supply projects. However, in most projects, the regulators are unconsciously connected in parallel, while the high current draws are first and the other regulators burn after the weaker regulator. Even if there is no short circuit protection, even after a certain time, the 50Hz transformer that we use at the entrance is damaged. In order to prevent this, the most intelligent work to be done is to share equal current to the regulators, regardless of the output load.

Why do the regulators not share currents evenly when connected in parallel?

Today, many advanced linear regulators can be observed in parallel. But due to both thermal and production differences, different currents will be drawn from the regulators.

Another problem is that most linear regulators are short-circuited. Linear regulators actually work like current-controlled resistance. In other words, if our input is 12V and our output is 5V, then the load value is 1A, while the resistance value is 7ümüz, 2A, while the resistance value is 3.5 12. Of course, linear regulators are not a resistance, but they can be shown in this way when modeling. This explanation also shows why the linear values of linear regulators increase linearly as the currents increase. Therefore, we will add 2.5A short circuit protection in our circuit.

**Design Phase**

We start with the power electronics industry, which will be recognized by everyone. Specifically when you go to the electronics shop for beginners, please indicate to the vendor 50Hz, 220V / 24V and 24 × 2.5V = 60W transformer. It is not enough to tell the transformer to say only the voltage ratios, like most first-time persons. Hence, hundreds of questions on the internet are n why the transformer is burned O.

*There is a big question here, especially in the heads of new friends; There is a transformer 90kg draws 100W,a transformer have 250gr 200W , so how does this happen? For those who do not want to think about this part of the work, I would like to say that it is because of the frequency difference and leave other details to you.*

After acquiring the transformer, the 24V AC output must be converted to DC. Therefore, bridge diodes should be used at the entrance. Since the 2.5A will be drawn at the exit, a value of 3A and above of the bridge diode should be selected. Large diameters are placed on the bridge diode outputs in order to convert the voltage to DC after the sine signal has been rectified. Output capacities are also determined according to the desired voltage ripple value. The formula is as follows.

**I = C * dV / dt**

Here dV = 2V, output current I = 2.5A, f = 50Hz full wave direction dt = 10ms for C = 12.5mF output. Let’s assume that we have installed 4 units of capacitor value of 3300uF.

*Polar capacitors are deformed for a long time and slowly lose their values. Therefore, the large selection of the input capacity block will be your advantage.*

Our next aim is to allocate the flows to LM317 regulators equally. For this I will adjust the output voltage level with an LM317 and then check the current of the other parallel LM317 by comparing the currents that will pass through the LM317s.

The general circuit of the LM317 is as above and the output voltage is set to the desired value by adjusting the R1, R2 resistors. The formula given in the datasheet of LM317 is as follows.

**Vout = 1.25 (1 + R2 / R1) + Iadj**

If Iadj is ignored above, R1 = 120 çıkış, R2 = 1032 ih can be selected for 12V output. Of course, 1032arak in the market can not be found, but this value can be obtained by inserting 1k + 33 fakat.

Now let’s make the following circuit by using current measurement and opamp to share the currents evenly.

Briefly explain the circuit; The opamp, which wants to equalize the voltages on the shunt resistors, will generate voltage accordingly and adjust the output current due to the voltage of the upper LM317 and thus ensure equal current flow from the two regulators. I mean, the brain of that department is kind of an opamp. It should be noted that the opamps try to keep the voltages at the + and – inputs equal to each other. Thus, the LM317s take equal loads and the regulators are less heated than the regulator with equal load on them.

The last part is output short circuit protection. Although there are many methods on the internet, the easiest thing to do is always to cut the tension of the main circuit. For this, let us assume that the voltage at the regulator outputs is fixed at 12V. Then, by attaching a shunt resistor to the outlet, we try to fix the tension on an opamp.

I compared the current reference input I received from the middle of the two resistors I divided by 12V, with the voltage falling above 0.1 ım shunt resistance. I have limited the current to the desired level by means of the mosfet compared with the observed voltages. As a result of the 12V voltage and resistance sections, I got 0.25V at the middle end. When the shunt has passed 2.5A over the resistance, the tension on the shunt resistance is 0.25V. Over 2.5A, the opamp starts to gradually switch off the output voltage and when the output is short-circuited, the opamp mosfeti fully protects the regulators by opening. The current limit can be adjusted by replacing it with a 470 ayar or 10erek multi-turn pot. The reference voltage divider resistors 470 and 10 minimum are selected to provide the minimum load current when the output is open. This current is given as datasheette 10mA.

I would like to say that the shunt resistors must be wattage resistance and that the mosfetins and regulators must be connected to the cooler. In addition, if a large voltage is observed at the output, I recommend that they output a large capacity to the output. Since the capacities are empty at first, excess current draws can be seen in the first energization of the circuit. Therefore, necessary precautions can be taken at the entrance. It can also be fitted with a protective glass fuse.

You can see the complete operation of the circuit below. If you click on the image, the diagram will grow.

You can ask all your questions in the comments section.

I wish everyone success in their work.