Electronics and Communication Engineering - Analog Electronics - Discussion

Option B is correct.

First, consider both the diodes forward bias.
Then with the help of nodal analysis find the voltage at junction 1k,1k and 10k.

{((V-8)/1k) +((V-4)/1k)+(V/10k)} = 0.
On solving the above equation we get, V = 5.71v.

Given that V1=8v and V2=4v which are positive terminal voltage for D1 and D2 respectively
for both diodes, we have a negative terminal voltage V=5.71v.

So D1 will conduct because Vp>Vn (condition for forwarding bias) (Vp=8v, Vn=5.71v)
D2 will not conduct because Vp<Vn (condition for reverse bias) (Vp=4v, Vn=5.71v).

First take: I(1) + I(2) = I.
where I(1) = (V-8)/1k ohm.
similarly I(2) = (V-4)/1k ohm.
And I = V/10k ohm.

Next solve for 'V' ( USING KCL AT NODE V).
You will get V = 40/7 which is 5.7 something.

Hence, d1 will conduct and d2 will be reverse bias and won't conduct.

After Diode D1 conducts the potential creat at the junction between 1k and 10k is 6.36V (assume both diodes are Silicon diode) so at the diode, D2 anode potential is lesser than the cathode potential so D2 will not conduct.

So, Answer B is the right answer.

The diode which experiences the maximum potential differences between the two nodes, that diode will conduct. As in this case between v1 and ground Diode D1 experiences the potential difference of 8v but D2 experiences only 4v. So the diode D1 conducts.

Assume both diodes are in forward bias, then after doing node analysis, the node at the intersection of 3 resistors, has the potential of 5 V (assuming cut off voltage 0.7V), which contradicts, so only one diode is in forward bias.

Simple apply kcl at the node.

V-8+v-4=v/10,
2v-12=v/10,
20v-120=v,
19v=120,
V=120/19,
V=6.31v.

You will get 6.31v at that node. So diode D1 is FB and D2 is Rb.
The answer should be B.

@Nikkie

You will get V = 40/7 which is 5.7.

Hence, d1 will conduct and d2 will be reverse bias and won't conduct.
But from above statement how can tell that d1 will conduct?

@KCL.

i total = i1 + i2.
(V-8-0.7)/1k + (V-4-0.7)/1k = V/10k.
V = 7.05 so only D1 will conduct since it exceeds the threshold 7.05 and D2 is reverse biased.

B is correct answer. First you suppose that both are forward biased and then find out the voltage at node. You will see that diode D2 will reverse biased.

i total = i1 + i2.

(V-8+0.7)/1k + (V-4+0.7)/1k = V/10k.
V = 5.05 so only D1 will conduct since it exceeds the threshold 5.05 and D2 is reverse biased.