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DESIGN PROJECT!

Many hobbyists in the field of electronics are familiar with two common linear regulators: LM317. These are typically accompanied by an op-amp (LM318), along with various resistors and capacitors. Now, let’s consider this question: how can we design an affordable power supply with short circuit protection? One effective approach is to parallel LM317 regulators to enhance the current output.

Before delving into the project, the initial query arises: why use 2 LM317 regulators? Numerous power supply projects can be found online, but a common issue persists. Often, regulators are unwittingly connected in parallel, leading to unequal current distribution. This discrepancy results in the stronger regulator drawing higher currents, causing the weaker regulator to burn out. Furthermore, even without short circuit protection, prolonged imbalances can damage the 50Hz transformer used at the input. To mitigate this risk, the most prudent approach is to ensure equal current distribution among regulators, regardless of the output load.

So, why do regulators fail to share currents evenly when connected in parallel? Despite the availability of advanced linear regulators designed for parallel operation, variations in thermal and production characteristics cause different currents to be drawn from each regulator.

Another challenge stems from the fact that most linear regulators behave like current-controlled resistors. For instance, if our input is 12V and the output is 5V with a load of 1A, the resistance value is 7Ω. As the current increases to 2A, the resistance value reduces to 3.5Ω. While linear regulators aren’t actual resistors, this analogy helps illustrate their behavior. It also explains why the linear values of these regulators increase proportionally with current. Therefore, incorporating 2.5A short circuit protection into our circuit becomes essential.

Design Phase

Let’s begin with a fundamental aspect of the power electronics industry, a topic that should be familiar to everyone, especially beginners. When you venture into an electronics store, especially if you’re new to the field, make sure to specify certain details to the vendor. Rather than merely stating the voltage ratios, as many beginners tend to do, be precise. For instance, if you’re looking for a transformer, mention that you need one with specifications like 50Hz, 220V primary voltage, 24V secondary voltage, and a power output of 60W (calculated as 24 × 2.5V).

This seemingly straightforward task often raises numerous questions online, with many wondering why transformers burn out. A significant query arises, particularly among newcomers: how can a transformer weighing 90kg handle 100W, while another weighing just 250g manages 200W? For those uninterested in delving deep into the technicalities, I’ll simplify it by attributing these differences to frequency variations and leave the intricacies for you to explore.

Once you have the transformer, the next step involves converting the 24V AC output into DC. To achieve this, bridge diodes must be employed at the input. Given that a current draw of 2.5A is expected at the output, select a bridge diode with a rating of 3A or higher. After the sine signal has been rectified, large diodes are used to convert the voltage to DC. The selection of output capacities depends on the desired voltage ripple value and can be calculated using the following formula.

I = C * dV / dt

LM317

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.

LM317

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.

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