# Does the power drop on a diode become entirely heat?

Let's consider this simple circuit with:

1. A DC voltage source (for instance of 24V)

2. 3 Diodes in series (with total $$0.65\times3 \approx 2 V$$ of voltage drop)

3. A resistor of $$10\Omega$$

This device gives 22V of voltage to the resistor starting from a 24 voltage source.

Now,

I) The resistor receives 22V, so the current on it is equal to 2.2A. The power it absorbs is equal to $$2.2A \times 22V = 48.4W$$

II) The DC voltage source provides 24V, and the current on it is always 2.2A. So the power it provides is equal to $$2.2A \times 24V = 52.8W$$

Obviously the remaining power (52.8 - 48.4 = 4.4W) goes to the 3 diodes.

So, my question is: does this power entirely become heat (so 4.4W of heat) because of the diode parasitic resistances? Or some part of it is stored in some ways inside the diodes?

It's heat.

Unless your diodes are LED, where 10-30% of the energy goes out as light. The rest is heat, as usual.

Diodes don't store energy. At least, not intentionally and not in the meaning usually accepted in electronics.

The diode may have some parasitic capacitance, inductance or even a parasitic Peltier effect that can, in some cases, return some energy back to the circuit. The capacitance may even not be considered parasitic, as in varicap diodes, but they are a corner case.

Elements that are expected to store (and return) energy:

• capacitors
• inductors
• batteries

At DC it's basically entirely heat (unless you grossly overload the diode, in which case there might be acoustic and visible light emitted). A tiny, tiny bit is stored charge.

Yes, the recombination energy is released as phonons which are essentially heat.

• Do you mean "phonons"? Dec 11, 2020 at 18:51
• Yes, what did I say? Photons?
– vir
Dec 11, 2020 at 18:53
• It can be both photons and phonons. Dec 11, 2020 at 18:54