# Diode in parallel with two DC sources

Determine the voltage across the diode in the following figure, using the complete diode model with r'd = 10 Ohms and r'R = 100 Ohms. Since the lower point is grounded, voltage at upper node is -10 V. Hence diode is forward biased. Then in complete diode model we can consider the diode as a closed switch with a DC source and resister inside.

Is this explanation is right and if so how may I proceed from here?

• I suggest you combine the two voltage sources and two resistors into a single voltage source and a single resistor (Thévenin equivalent) and proceed from there. Jul 15, 2014 at 4:22

As Spehro said, I would prefer combining the 2 sources and resistors into a single one. This can be done by noting that the two sources + resistances are in parallel branches.

The formula for equivalent Voltage in parallel branches is

V_eq = ( V1/R1 + V2/R2 + .....Vn/Rn) / ( 1/R1 + 1/R2 + .......1/Rn)


Be careful about the polarities of the batteries though. In your case the two batteries will have opposite polarities and so V2 will have negative sign.

The equivalent resistance is then given by

1/R_eq = (1/R1 + 1/R2 + .....1/Rn)


Using the above formulas, your circuit would be a single voltage + resistance branch and a diode.

Alternatively ,if I am correct, in your question r'd denotes the forward resistance of the diode and r'R denotes the reverse bias resistance. Then you can also use Superposition theorem taking one source at a time. In that case, when considering the 10V source, you would replace the diode with reverse bias resistance ( since it is reverse biased) and when considering the 20V source, you would replace the diode with a forward bias resistance ( since it is forward biased).

I don't know what the complete diode model you refer to is, but for an ideal diode:

Consider the circuit without the diode. What is the voltage between the resistors? 15V falls across each resistor, so it will be 10-15 = -5 volts. So when we add the diode back into the circuit it will be forward biased (or "on") and the cathode will move to just below zero volts: about -0.6V or -0.7V in practise.

Of course you probably still need to do it mathematically using your diode model, for an exact answer.

I wanted to comment to Spehro, (But I lack in reputation.)

As a first pass guess, assume Vdiode ~ zero.
Then there is 1 mA into the diode node from the 10 V supply and 2mA drawn from the 20V supply.
So there's something like 1mA going through the diode. In practice that would be enough. For your class you'll have to work out the "exact" answer. (for whatever diode model you are using.)