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I have a question regarding how diode clamps work in a ideal circuit.

enter image description here

  • The voltage source with infinite output impedance will force Vout to 6V
  • The diode will be forward biased with a vf of 0.6V. Thus clamping Vout to 5.6V

Is my issue that I am taking a non ideal model of the diode with a Vf of 0.6V while taking a fully ideal model of a voltage source with no output impedance to separate the two nets ?

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  • \$\begingroup\$ @DKNguyen I see. If my diode was fully ideal with a Vf =0V, then I assume I have another problem with my VDD voltage source in parallel with Vin, \$\endgroup\$ Apr 7, 2021 at 20:03
  • \$\begingroup\$ You end up with unstoppable force vs immovable object in that case, whereas in your original case it is trying to stop unstoppable force with very movable object. \$\endgroup\$
    – DKNguyen
    Apr 7, 2021 at 20:03

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Yes that is the problem. You need series resistance somewhere. It is not optional, ideal or not, and you often add extra series resistance to limit the current through the diodes so they don't fry the diodes or the driver, and to limit disruption from current being injected into the power rail.

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Assuming you mean "zero output impedance" and assuming Vdd is actually 5.0V.

You have the equivalent of immovable object meets irresistible force. Your models are not accurate enough to describe what will happen.

In reality there is no such thing as zero output impedance for either the 6.0V or 5.0V sources. In reality the diode will not have a 0.6V voltage drop at high current.

Let's simulate what would happen assuming the voltage sources were ideal, and assuming a real diode 1N4004 or 1N4148:

schematic

simulate this circuit – Schematic created using CircuitLab

As you can see, the voltage across a real diode simply increases to meet the "immovable" ideal voltage sources. In both cases the diode current exceeds the maximum DC current rating so the diodes will fail, but for a short time they will do what is shown, fairly closely.

Adding some series resistance will decrease diode current and allow the clamping to occur without excessive current, however note that the voltage sources also may not be ideal. In particular the 5V source may not be capable of sinking current.

enter image description here

Here you can see that the 10V input is not clamped to 5.6V but rather the 5V output increases to 9V+. The Circuitlab LM7805 model is not accurate enough to demonstrate this behavior, by the way, so I used LTspice.

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  • \$\begingroup\$ In practice, the current of the input source will be sinked by the load (the supplied device)... and it will be added to the current of the power supply... that, in its turn, will slightly decrease its current. Of course, there is a need of a limiting resistor in series. \$\endgroup\$ Apr 8, 2021 at 5:39
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    \$\begingroup\$ @Circuitfantasist You are making the assumption that whatever load there is on the regulator always exceeds that of the current through the limiting resistor (plus minimum load on the regulator, if applicable). That assumption may well not be valid. For example, if your MCU is in sleep mode, it maybe it draws 10uA. So the MCU is only destroyed under special conditions. Not very nice or easy to troubleshoot. \$\endgroup\$ Apr 8, 2021 at 5:46
  • \$\begingroup\$ I agree that it is so... but it was still good to discuss. And what to do in such a case? The simplest looks a resistor in parallel to the device (to the supply output as some initial load)? \$\endgroup\$ Apr 8, 2021 at 6:02
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    \$\begingroup\$ @Circuitfantasist That's certainly one possibility. Maybe just a resistor and power-on LED. Another is to clamp the power supply with an active device or zener. Yet another (my preferred way) is to clamp the input directly with an active device or zener so that the current flows to ground. \$\endgroup\$ Apr 8, 2021 at 6:04

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