Electronics and Communication Engineering - Analog Electronics - Discussion

Thank you @Aatif.

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).

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.

(8-v)/1 +(4-v)/1=v/10,
v=5.71, where v is nodal voltage.
so D1 forward biased and D2 is reverse biased.
So, D1 is conducting.

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.

@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.

Thanks @Mukkiii Nautiyal.

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.

Thanks @Nikkie.

@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?