# Power Electronics - Diode

[Power electronics] The measured values of a diode at a temperature of 25 Degrees Celsius are: Vd=1V at Id=50A, Vd=1.5V at Id=600A.

Determine (i) the emission coefficient n, and (ii) the leakage current.

==============================================

Answers:

a. (a) n=6, (b) Is=0.2A

b. (i) n=5, (ii) Is=1A

c. (i) n=8, (ii) Is=0.347A

d. (i) n=8, (ii) Is=1A

The reverse recovery time of a diode is trr=5 usec, and the rate of fall of the diode current is di/dt = 80A/usec. If the softness factor is SF = 0.5, determine (i) the storage charge Qrr, and the peak reverse current Irr:

================================================

a. (i) Qrr=900 uC, (ii) Irr = 200A

b. (i) Qrr=500 uC, (ii) Irr = 500A

c. (i) Qrr=200 uC, (ii) Irr = 100A

d. (i) Qrr=1000 UC, (ii) Irr = 400A

OTA: please take this only if you can show step-by-step problem solving. Thank you.

Â© BrainMass Inc. brainmass.com March 4, 2021, 6:00 pm ad1c9bdddfhttps://brainmass.com/engineering/power-engineering/power-electronics-diode-23438

#### Solution Preview

[Power electronics] The measured values of a diode at a temperature of 25 Degrees Celsius are: Vd=1V at Id=50A, Vd=1.5V at Id=600A.

Determine (i) the emission coefficient n, and (ii) the leakage current.

==============================================

Answers:

a. (a) n=6, (b) Is=0.2A

b. (i) n=5, (ii) Is=1A

c. (i) n=8, (ii) Is=0.347A

d. (i) n=8, (ii) Is=1A

Id = Is*e^[Vd/(nVt)] where Vt = 27 mV at 25 deg. C

So we get,

50 = Is*e^[1000/27n] for Vd =1V @ Id = 50A ...

#### Solution Summary

The equations are given step-by-step.