Two parallel diodes in a circuit

[Jahnavi]

OK

 

[cnh1995]

Yes, at the instant the switch gets closed the initial current is zero, and 12 V appears across the diodes, but the diodes and all the circuit do not behave as ideal ones. The p-n junctions of the diodes behave as capacitors and all wires behave as resistors and inductors. There will be some transient current in the circuit (till the equilibrium charge distribution builds up at the p-n junctions) which ends in the stationary voltage drop across the diodes and stationary current in the circuit in a very short time.

Whie I agree with the rest of your explanation, I don't see how we can say 12V will appear across the diodes initially. When the equilibrium charge distribution builds up at the pn junction and surface charge distribution builds up on Ohmic components (resistors, wires), there is no way we can compute voltage across any component. It's all chaos for that very very short time. Even if we assume the diode junctions to be capacitances, I don't think we can say initially all the 12V appear across the junctions. I believe this process is very different, extremely short-lived and hence, can't be described by the steady state quantities like V and I.

 

[cnh1995]

@Jahnavi, you might enjoy this old thread.
https://www.physicsforums.com/posts/5309445/

 

[ehild]

Whie I agree with the rest of your explanation, I don't see how we can say 12V will appear across the diodes initially. When the equilibrium charge distribution builds up at the pn junction and surface charge distribution builds up on Ohmic components (resistors, wires), there is no way we can compute voltage across any component. It's all chaos for that very very short time. Even if we assume the diode junctions to be capacitances, I don't think we can say initially all the 12V appear across the junctions. I believe this process is very different, extremely short-lived and hence, can't be described by the steady state quantities like V and I.

Yes, you are right, we can not use the steady-state quantities in the circuit, but the 12 V emf is applied at the instant of switching up. Even the diode have inductance and resistance because of the leads and the bulk of the semiconductor, so not all the voltage drop appears across the pn-junction or even across the diode because of the inductance of all the wires and the resistor.

 

[CWatters]

A lot comes down to which model of a diode you are using. The problem statement provides very little information so you have to assume a very crude model, almost an ideal diode but with a defined forward voltage.

 

[NTL2009]

A lot comes down to which model of a diode you are using. The problem statement provides very little information so you have to assume a very crude model, almost an ideal diode but with a defined forward voltage.

Agreed. And the earlier discussion of a 'perfect 12V battery' applied to a 'perfect 0.3V Vf diode' isn't a useful real-world model. It's the same as a 'perfect battery' connected to a 'perfect conductor', or the theoretical infinite force applied to an infinite mass. It is undefined. One or the other has to 'give'.

So either we assume enough internal resistance in the battery so that the diode will clamp the battery terminal's voltage to 0.3V w/o burning up, or we assume the battery can maintain 12 V, and the diode burns out, or somehow the diode can survive with 12V forward voltage (I don't think any real diode can - if you extrapolate the I/V curve I think you will soon reach mega-amps!)