# Ideal diode circuit analysis

I am having trouble figuring out how to analyze this circuit and check which ideal diode is in forward bias or not. Why is my analysis wrong?

If your ideal diode is conducting it will have the same voltage on both sides. If the diode is not conducting it must be reverse biased. In no case would one voltage add to the other. You also cannot have a positive bias greater than 0V.

So if D1 is conducting V = 1V. However that means that D2 would be forward biased with unlimited current flowing through it, so we know that's not true. So let's see if D2 is forward biased and D1 is reverse biased. That works, and we can see that current is flowing from +3 to -3 through the resistor so it will actually be biased on (if the bottom source was +4V rather than -3V then both would be reverse biased).

So now you know the voltages and can easily calculate the current.

Really with this kind of circuit you have to guess what the situation is then check if it is real or not. This one can be done in your head, with some practice, but more complex ones with many resistors would require some calculations.

I am having trouble figuring out how to analyze this circuit and check which ideal diode is in forward bias or not.

D1 is reverse biased because its anode has only 1 volt on it compared to the other diode that has 3 volts on it. D1 can be removed from the circuit.

• Why does that happen? Do I take out the minimum voltage? What is the value of V? May 5, 2017 at 18:03
• The value of "V" for ideal diodes is 3 volts i.e. the diode that has the greater input voltage always "wins" in this type of circuit. May 5, 2017 at 18:04

You said ideal diodes so I'll express this with no diode drop voltages...

The diode from 3 V will put 3 V at the top of resistor R.

This will reverse-bias the 1 V diode, as it has Va = 1 V and Vk = 3 V.

So V is 3 V.

You can now work out the resistor current easily enough from Ohm's Law.