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Diode

In this question we can easily find the solution,but here(in this particular question) we can make out the most forward bias voltage and solve it, but in some circuits where we cannot identify which diode is connected to most forward bias voltage,that time which conditions to assume first when there is more than 1 diode.

I believe one assumption is more than enough to conclude about the diodes,isn't it??

For example in the above case,i assume first case as both diode on but at the same time two different voltages in parallel are not possible which is nonsense,so i took the condition as both diode off,dere it make a conclusion that both diode should be on,but we should be capable of proving that one diode is on and one diode is off,which am unable to do??

So how to go for assumption when we have no idea which diode is more forward bias??

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short span time answer (Assuming VF=0.7V):

a conducting D1 would give 1V-0.7V = 0.3V

a conducting D2 would give 3V-0.7V = 2.3V

2.3V override 0.3V, so D2 is the winner and will turn on (conducting), D1 is off (because its driving voltage is not as strong as D2's driving voltage).

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You simply have to consider all possible cases, and choose the one that doesn't lead to a logical contradiction:

Case 1: Both diodes off. Then point "V" is at -3 V, and the voltage across D1 is 4 V and the voltage across D2 is 6 V, contradicting the assumption that the diodes are off. So we reject this case.

Case 2: D1 is on, D2 is off. Then the voltage at "V" is about 0.3 V, and there is 2.7 V across D2, again contradicting the assumption that D2 is off. So we reject this case.

Case 3: D1 and D2 are both on. Then the voltage at "V" is both 0.3 and 2.3 V at the same time, again a contradiction, so we reject this case.

That leaves the correct solution: D2 is on and D1 is off, which indeed produces no contradiction and leads to the solution of the circuit.

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  • \$\begingroup\$ --can you please tell me how to deal these questions in a short span of time as am preparing for an exam where we get just two minutes to solve the ckt ,so i believe taking all four cases will take time. I would appreciate your advice on it \$\endgroup\$ – Jay Maclean Aug 21 '15 at 5:04

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