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Topic: ATTiny Power (Read 2690 times) previous topic - next topic

retrolefty


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


LarryD

I have used the diode idea several times with success.
I added a 100uF capacitor from the cathode to 0 volts in these applications.
The way you have it in your schematic isn't the same as how you have it wired up!

dc42

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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fungus


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...?

No, I don't answer questions sent in private messages (but I do accept thank-you notes...)

dhenry

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.

SirNickity

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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