Diode Analysis

Thread Starter

crazyengineer

Joined Dec 29, 2010
156
I need help trying to solve this diode problem

This is a large current device whose reverse leakage is reasonably independent of voltage. If Voltage at the resistor is 1V at 20 degrees Celsius, find the voltage of the resistor at 40 degrees Celsius.



(excuse the bad picture)

At first, I tried to solve for Is. But in order to do that, I needed to find the current flowing through the diode

i=V/R=>1V/(1*10^6)=1*10^-6 A,

but I'm totally lost from there. I tried using Is=i*e^(-V/vt) but I realized that the diode should be in reverse bias mode (i=-Is).

The answer came out to be 4V
 

jegues

Joined Sep 13, 2010
733
I need help trying to solve this diode problem

This is a large current device whose reverse leakage is reasonably independent of voltage. If Voltage at the resistor is 1V at 20 degrees Celsius, find the voltage of the resistor at 40 degrees Celsius.



(excuse the bad picture)

At first, I tried to solve for Is. But in order to do that, I needed to find the current flowing through the diode

i=V/R=>1V/(1*10^6)=1*10^-6 A,

but I'm totally lost from there. I tried using Is=i*e^(-V/vt) but I realized that the diode should be in reverse bias mode (i=-Is).

The answer came out to be 4V
The diode is going to be operating in the reverse bias-region.

Real diodes will exhibit a reverse current although small, is still much larger than Is.

A large part of the reverse current is due to leakage effects.

The rule of thumb for the temperature dependence of the reverse current is that it doubles for every 10°C rise in temperature.

I'm assuming this question was taken from the Sedra/Smith text, as all this information is directly above the identical question you've posted.

Is it clear how the answer is 4V now?
 

Thread Starter

crazyengineer

Joined Dec 29, 2010
156
The diode is going to be operating in the reverse bias-region.

Real diodes will exhibit a reverse current although small, is still much larger than Is.

A large part of the reverse current is due to leakage effects.

The rule of thumb for the temperature dependence of the reverse current is that it doubles for every 10°C rise in temperature.

I'm assuming this question was taken from the Sedra/Smith text, as all this information is directly above the identical question you've posted.

Is it clear how the answer is 4V now?
Yes it was taken from the book, and now I see the relationship. Thank you!
 
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