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Left Coast, CA (USA)
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Brattain Member
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Measurement changes behavior
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I can't seem to find any concrete reference specifying the typical diode loss as a resistive loss at all.  I was under the impression it's a carrier loss and is not represented by Ohm's Law, and therefore does not have a heat byproduct (other than natural resistive losses in any real-world conductive material).  Pedantic as it may be, anyone know for sure?

Not ohms law as applied to resistance, more like a constant voltage drop so power dissipation is voltage drop X forward current = real power dissipated as waste heat wattage. So not like a resistor as voltage drop doesn't increase with increasing current flow, but still real power loss as in Watts = voltage drop X current.

Lefty

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The last thing you did is where you should start looking.
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I have used the diode idea several times with success.
I added a 100uF capacitor from the cathode to 0 volts in these applications.
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The way you have it in your schematic isn't the same as how you have it wired up! That goes for me too.

United Kingdom
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Hofstadter's Law: It always takes longer than you expect, even when you take into account Hofstadter's Law.
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I wouldn't call most of the power loss in a forward-biased diode resistive, but it certainly generates heat (or, in the case of an LED, heat + light).
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Valencia, Spain
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I can't seem to find any concrete reference specifying the typical diode loss as a resistive loss at all.  I was under the impression it's a carrier loss and is not represented by Ohm's Law, and therefore does not have a heat byproduct...

It has to go somewhere (the laws of physics don't allow energy to vanish)

In the absence of electromagnetic radiation, where do you think it goes...?

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Quote
I can't seem to find any concrete reference specifying the typical diode loss as a resistive loss at all.

If * Vfwd.

Always true, no exception, even in vectorized forms.
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Anchorage, AK
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In the absence of electromagnetic radiation, where do you think it goes...?

Not lost, just not used?  I don't know exactly.  I guess I rationalized the idea somewhat like this:  Take two halves of a transformer and separate them.  The induced current in the secondary coil will be lower than if they're more closely coupled.  Maybe even that can be described as a resistive loss, since there's resistance from the air gap between coils.

Ugh.  This is where being self-taught fails compared with having a formal education.
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