# Heat from electronic components

Discussion in 'Electronic Basics' started by Rookie, Nov 26, 2003.

1. ### RookieGuest

Can someone help me understand heat generation in electronic
components (IC's, caps, inductors etc)

I just can't seem to get a handle on the concept. Here is what I know:

power = V * I (this should work for anything?) - does this mean that
you can calculate the heat loss from any IC just by knowing its V and
I requirements? Or do we need to know the nitty gritty details of what
components are on the IC?

The only easy one for me is a simple resistor: P = V * I or P = I^2 *
R etc - this IS the heat dissipation for this component

I guess the question is if a circuit (IC or whatever) is being fed P =
V * I power, then does all this energy get dissipated as heat, by
definition? Electrical power in = heat out?

thanks much
confused...

2. ### RDGuest

Unless the circuit produces another form of energy i.e. light (LED) then you
can apply P=VI just as a resistor as almost all power will converted to
heat.

Regards RD

3. ### John PopelishGuest

That is all there is to it except for a couple details. The I and V
have to be DC to be measured as average values and then multiplied.
If AC, they have to be multiplied as instantaneous values and that
product averaged. This accounts for the energy storage of inductors
and returns to the driving source twice per cycle.

There is also an exception for things that convert electric energy to
some other form besides heat (like motors and lamps). If they were
100% efficient (which none are) then I*V would just measure their
energy output in some other form. As it is, I*V measures the power
entering them, and you have to have some other information to tell
what fraction of that power is not turning into heat, inside them.

4. ### Ben SchaefferGuest

I was taught that anything that draws alot of current typically heats up,
regardless of voltage drops. Something like a fuse is the best example,
vaporizing at a certain point when too much current is drawn.

5. ### Fernan BolandoGuest

This is more or less how it works, But It depends on the circuit. The
idea is Power in = Power out. Due to losses in the circuit some of those
Power get's converted to heat, So it becomes Power out = Power useful +
heat out.

What you should realize is that all conducters has a very small
resistance which causes losses in power in the form of heat.

,Fernan