Determine Diode Bias


Joined Mar 24, 2008
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Are these still ideal diodes?


Last edited:


Joined Feb 17, 2009
To determine whether D1 or D2 conduct in this circuit D:
First you must remember that diode conducts current in only one direction.
From anode to cathode. The anode mus by connect to the higher potential then cathode. And if diode conduct then is a voltage across the diode equals 0.6V.
So first we remove D1 and then calculate the voltage on Vo.
Vo = I2*5K +0.6V
I2=I1 = (5V - 0.6V)/ 15KΩ = 293uA ----> Vo = 2V.
So there is a chance to D1 to conduct. If we connect D1 to the circuit we now know that D1 is conduct.
So Vo = Vd1 = 0.6V and this voltage is to small to supply D2 + 5KΩ.
So finally D1 is conduct and D2 is cut-off.
I1 = (5V - 0.6V)/10K = 440uA.

Proceeding the same method for circuit E, we find Vo for circuit when D1 is disconnect.

Vo = 1.867V
So there is a condition to both diode to conduct.
So if we connect D1 then :
Vo = 0.6V
I1 = (5V-0.6V)/5K = 880uA
and I2 = (5.6V - 0.6V)/10K = 5V / 10K = 500uA


Joined Mar 24, 2008
If D1 conducts then the point it connects to is ground. Call this point OUT. If OUT is negative then D1 would be the reverse biased.

The first order of business is to establish whether OUT is negative. Since we are talking a voltage divider do the math, and see if OUT would be negative.

BTW, did you notice these were AND gates?

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