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If the power source is connected to another power source and input voltage is higher than output voltage a bypass diode usually used to protect the regulator:

schematic

simulate this circuit – Schematic created using CircuitLab

In all power supply schematics that I've seen so far, none of the designers used a series diode:

schematic

simulate this circuit

While the voltage drop can be easily compensate for, heat might be the main reason.

The 12SQ045 schottky diode is capable of passing 12 A, thermal resistance is 3° C/W. at 3 A forward voltage is only 0.4 V;

3 x 0.4 = 1.2 W -- (1.2 x 3) + 25° = 28.6° C

Well that's not so much heat (if I did the math right...)!

The efficiency is important but not for a 3 A linear power supply which with or without a diode always dissipating some power in the main regulator.

Why designer don't use a series diode to protect the regulator and circuitry around it? it's easy and cheap to implant.

Update:

The first example in this question can't protect the components before the transistor unless we add another diode or some extra circuitry, while in the second example a single diode can protect the whole circuit against reverse voltage.

In two answers provided to this question they mentioned the capacitor charge may damage the transistor which is not going to happen because no current can pass through the diode (maybe very little), and the capacitors can be discharged quickly if it's in 10-100 uF region with a 10mA current sinking.

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    \$\begingroup\$ "While the voltage drop can be easily compensate for ..." You don't do this just by increasing the regulator output voltage; you do it by monitoring the actual output voltage. This requires feedback from after the diode and you haven't shown that in your circuit snippet. That will affect the complexity of your protection circuit as now you have to protect against backfeed via the feedback. \$\endgroup\$
    – Transistor
    Commented Apr 19, 2020 at 8:56
  • \$\begingroup\$ @Transistor Yes, when I said it can be compensate for I meant using a voltage feedback op amp which a 100K ohm resistor can protect the inverting input of the op amp against reasonably high voltages. is there any other disadvantages that I'm missing? \$\endgroup\$ Commented Apr 19, 2020 at 9:07
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    \$\begingroup\$ OK, I see where you confused me; your question was too short. You only mentioned the normal situation, without mentioning the abnormal situation Your edit helped, but you should re-edit your question to include the problem in those first 2 sentences, otherwise it feels you are making us go around in order to understand you. Again, drawing an example schematic around your central parts will help us understand you better. \$\endgroup\$ Commented Apr 19, 2020 at 9:14

2 Answers 2

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Your first drawing is an example of protecting the regulator (in this case the transistor) from reverse voltage and current when the power on the left is removed (turned off) or shorted to ground, and there is a charge left in the capacitors of the load circuit on the right (the emitter end of the transistor). And maybe in the case as you described (2 power sources connected together) but in the opposite direction (an input is considered to be on the left and an output on the right).

There is no reason to add a diode in series as you did in your second example because:
1. If you place a diode as in your first example, there is no need for this one in series because the first one already does the job;
2. The first example will protect both the emitter and the base of the transistor, but the second example will leave the base unprotected if there is a charged capacitor on it and the input power is removed or shorted out;
3. The first example will neither use nor waste any power during normal operation, as it will only (maybe) conduct in the brief moment when it is necessary to protect the transistor or regulator. The second example will waste power the whole time the circuit is in operation while providing less protection, and even in a linear regulator you try as much as possible to reduce the amount of power wasted, so that's why you don't see it used.

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  • \$\begingroup\$ In first example the regulator is safe but there will be high current at the circuitry before the regulator, the current should go some where! but in the second example no current will pass through. not even the capacitors charge. the charge in capacitors can be discharged fairly quickly, in most designs capacitors are not bigger than 47uF and in a good design shouldn't be more than 10uF, so a bleeder resistor should be able to drain the capacitors. \$\endgroup\$ Commented Apr 19, 2020 at 8:02
  • \$\begingroup\$ I still fail to see exactly what you mean because it doesn't sound like conventional direction to me. If you add just enough components to the left and right, it would be a lot more clear. \$\endgroup\$ Commented Apr 19, 2020 at 8:23
  • \$\begingroup\$ See the update please. \$\endgroup\$ Commented Apr 19, 2020 at 8:39
  • \$\begingroup\$ OK, it's a little more clear, but you should edit your question as I suggested in my comment to your question. Additionally, some of the problems and solutions depend on the voltages and the specific circuitry around the transistor/regulator, and your question is too broad without giving enough to get the whole picture. \$\endgroup\$ Commented Apr 19, 2020 at 9:24
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Maybe I don't understand what you're asking, but I don't think that diode does what you think it does...

There will be capacitance at VOUT of the regulator. Probably 100uF or more. Then there will also often be a bunch of other capacitors at other places, all in parallel.

So when the input power is removed, all this capacitance will hold a voltage at VOUT until the load finally dissipates it. If the load is a CPU in "sleep" or low-power mode, that capacitance may not drain very fast. It may be so slow, you end up with a 'brown out' or a poor reset. It's best to drain the capacitance fast as possible.

Many voltage regulators will not pass current from "VOUT" to "VIN", even if the voltage at VOUT is higher than VIN. So to give the capacitance a path to drain fast, that reverse polarity diode is added. Of course, for this to work there has to be some resistance from VIN to ground. That's easy to implement if the power switch is a 1P2T.

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  • \$\begingroup\$ But the capacitor charge can be discharged with a bleeder resistor (or current sink) to avoid all those problem that you described. \$\endgroup\$ Commented Apr 19, 2020 at 6:50
  • \$\begingroup\$ Yes it could. But it's gotta be fast enough to discharge them before the user can restore power. That could be pretty quick. A bleed resistor sized to do that would waste enormous amount of power. \$\endgroup\$
    – Kyle B
    Commented Apr 19, 2020 at 23:22

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