# Analyzing a circuit with diodes

In this circuit: Assuming ideal diodes

We have 8 combinations of assumptions. I have proved that 5 of them are wrong. The three remaining assumptions are:

1. All diodes are on.
2. D1 is on, D2 is off, and D3 is on.
3. D1 is on, D2 is on, and D3 is off.

1. When all diodes are on, the circuit will look like this: I = (16 - 12) / 4.7k = 0.85 mA.

Vo = 16v.

1. When D1 is on, D2 is on, and D3 is off, the circuit will look like this: I = (16 - 12) / 4.7k = 0.85 mA.

Vo = 16v

3 is identical to 2.

So, is it possible that we have three valid assumptions in this circuit? and if not, what is the wrong with this solution?

• An ideal diode in parallel with an ideal diode = an ideal diode. An ideal diode in series with an ideal diode = an ideal diode. – Andy aka Oct 18 '14 at 13:45
• So the three cases are in fact one case ? – ammar Oct 18 '14 at 14:14
• @ Andy aka but what to do with @Asger Bjørn Jørgensen answer? – ammar Oct 18 '14 at 14:31
• Basicly what we say yields the same result. @Andy aka says that the ideal diodes in series or parallel will all be in the same state, so your options 2. and 3. will be invalid because they must be in the same state. I provide you with a general answer which is applicable to all circuits containing ideal diodes. – asgerbj Oct 18 '14 at 15:08

Your options 2. and 3. are not valid, because the voltage across the OFF diode is not negative.

For your ideal diode to be OFF it must have a negative voltage and zero current, while it should have a positive current and zero voltage when it is ON. Always check for these 2 conditions.

You are stating that your ideal diode has zero current and zero voltage, which is none of its 2 available states when you are analyzing your circuit.

(Please note that this is being theoretically strict (as we should be when analyzing a circuit of ideal components). Practically your diode can have V = 0, I = 0, but it is not to much use when you are analyzing your circuit)

All the diodes will be on. Your conditions 2 and 3 are false. How would 2 conduct and 3 not conduct, and vice versa? The diodes are in parallel.

For this problem best way to approach is to short all the diodes and calculate current in all the paths. If you get a result with -ve current, then you can infer that, the diode in that particular path is reverse biased.

But in this case you get positive current in all the paths while calculating under short circuited conditions. So all the diodes must be, as assumed, forward biased.

And, yes the output voltage will be 16V.

All diodes are "on." Assuming that these are silicon diodes then the voltage drop across any one diode follows an exponential function. As a rule of thumb you can estimate this voltage as about 0.7 volts for a silicon pn diode when the current varies from about 0.1 milliampere to 10's of milliamperes. This is what they mean by "forward drop." It is about 0.3 volts for a silicon Schottky diode or a germanium pn diode. Since there can be 16-12=4 Volts across the three diodes then the diodes will be in forward bias and conduct. Now, whatever current is in D1 will split and be equally shared by D2 and D3. So the voltage across D2 and D3 will be slightly less than that in D1. But roughly, D1 will have 0.7 volts of drop and D2 and D3 will have 0.7 volts of drop. The net drop is 1.4 volts. The output terminal must be 16-1.4 volts lower in electric potential, so the output will be about 14.6 volts. Since we now have an estimate for this voltage we can find the current in the resistor. It is (14.6-12)/4700=0.55 milliamperes. Our earlier assumption about diode current is valid and the above estimates will be accurate. Lastly, no diode can be truly off as long as the lower voltage source is less than 16 volts. Some folks will argue that "on" means "appreciable current." Then "on" versus "off" is a matter of assertion. But there is definitely current if there is a positive potential difference as shown.

• I said in bold: assuming ideal diodes – ammar Oct 18 '14 at 13:58
• @ammarx: If you're assuming ideal diodes, then - since they're all forward biased - all of your assumptions about different combinations of ON or OFF are wrong except the one about that they're all ON. End of story. – EM Fields Oct 18 '14 at 18:13

We appear to be talking here about an ideal diode, which is a mathematical abstraction. The question is: what state is such a diode in at zero volts? I'd like to assume that it is undefined. Because of the voltage sources there is no question that D1 is on. The question then becomes which of D2 and D3 turns on first? Assume D2 turns on first and has zero volts drop across it. Zero volts across D3 keeps it in an indeterminate state in which you don't know if it is on or off. If you were to place a resistor in either leg to measure current in that leg, you upset the current in that leg to assure the diode in that leg stays off. If you place an identical resistor in both legs you would still not be able to predict which diode turns on first, nor would you be able to tell after the fact because the voltage at the junction of the two sense resistors will rise enough to instantaneously turn on the other diode. (Heisenberg's Uncertainty Principle?) The bottom line is that you know for sure that at least one of them is on giving you zero volts and that's all you care about. Or if you do care, you could say that all ideal diodes are identical, therefore they have identical turn-on time and both will turn on simultaneously, resulting in zero volts and putting them both in an indeterminate state, which conflicts with the fact that at least one has to be on. So if I change my mind that zero volts is an on state rather than being indeterminate as I assumed at first, then the only state that can exist is for all three diodes to be on and the other two states are invalid. And how many angels can dance on the point of an ideal pin?

• Assuming "ideal" silicon diodes, all three will be turned on (allowing for doping variations, D2 or D3 will be slightly more "turned on" than the other) and the output voltage will be about 16-0.7-0.7 = 14.6V. Because of their V-I characteristics they have more influence on the output than R1. – blarz Jan 3 '16 at 19:27