# Homework Help: Conduction power loss

1. Sep 26, 2016

### jaus tail

1. The problem statement, all variables and given/known data

2. Relevant equations
power loss = I2R
power loss = V2/R

3. The attempt at a solution
voltage across diode is 0.7V, current is 100A. resistance = 0.01 ohm.
Power loss is I2R - 100 * 100 * 0.01
= 100 W
Power loss across IGBT = V2/R. V across IGBT = V across Diode = 0.7.
R for IGBT = 0.02 ohm
So loss is 0.7 * 0.7 / 0.02 = 24.5W
Total loss is 100 + 24.5 is 124.5 W

But they've done something like:
No current through IGBT
Conduction loss = VtIav + Irms2Ron
= 0.7 * 100 + 1002 * 0.01 = 170W.

Why did they do V * T + I2R? Shouldnt they do either one? and i'm not sure which one.

2. Sep 26, 2016

### Staff: Mentor

Look at the VI curve for the diode. As the current increases the potential drop across the diode junction will increase beyond 0.7 V. So the effective resistance of the diode rises and the power loss increases, too.

See if you can't write an expression for the junction potential drop versus current.

3. Sep 26, 2016

### jaus tail

Oh yeah... (v2-v1)/(i1-i1) = .01

So (v2 - 0.7 / (100 - 0) = .01
This gives V2 as 1.7 V.
So power loss is V * I = 1.7 * 100 is 170 W.
Well we get same answer. I guess the way the book has gone about solution is confusing.

4. Sep 26, 2016

### Staff: Mentor

Good.

They chose to treat the diode as two components in series. The first being the fixed potential drop 0.7V and the second a resistance. They summed the power associated with each.