Diode Anaylsis Problems

Hello. I'm starting to do some diode analysis, but my answer came up wrong.
The problem asked to find V.Here's the diagram




Originally I thought all the diodes were conducting and the voltage at V was 8v but the answer was 2v. In other words, D1 is not conducting, and D2 is. Originally I used superposition for each diode and see if the current flowing across the diode was greater than zero.

Can someone please help me clarify some way of realizing when diodes are conducting and not?

D2 anode is at the highest potential with respect to the rest of the circuit so D2 at least must conduct.

With D2 conducting its cathode will be at +2V-Vf. Where Vf is the notional diode forward drop.

D1 cathode will also be at +2-Vf. But D1 anode is at +1V so the anode-cathode bias across D1 is +1-(2-Vf)=Vf-1. Since Vf will be less than 1V diode D1 will be reverse biased with its anode at a lower potential than the cathode.

If the diodes are considered ideal with Vf=0 then D2 cathode [=V] will be at 2V.

To be valid, superposition requires a linear system. Your diode circuit is non-linear, so analysis using superposition will not work.

What happens with the two diodes is that conduction through one of them can suppress the conduction which would otherwise occur through the other. If you try to consider the two inputs in isolation, this interaction is lost, and you get the false impression that both diodes will conduct.

t_n_k said:

D1 cathode will also be at +2-Vf. But D1 anode is at +1V so the anode-cathode bias across D1 is +1-(2-Vf)=Vf-1. Since Vf will be less than 1V diode D1 will be reverse biased with its anode at a lower potential than the cathode.

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Okay. If +1-(2-vf) is the voltage across the diode then why did you set it to vf-1 (which I assume is also the voltage drop across d1). Basically, I would like to know why you're setting it to itself?

crazyengineer said:

Okay. If +1-(2-vf) is the voltage across the diode then why did you set it to vf-1 (which I assume is also the voltage drop across d1). Basically, I would like to know why you're setting it to itself?

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Well I guess because if you re-arrange the terms then +1-(2-vf) = vf-1.

Is that what you are asking?

If you built this circuit with silicon diodes with a vf = 0.7 volt then I would expect the voltage V to be about 2-0.7=1.3V. With 1.3 V on D1 cathode and +1 V on D1 anode then D1 is reverse biased (@ -0.3V) - by virtue of the D1 anode being at a lower potential than the D1 cathode.

Does that help?

t_n_k said:

Well I guess because if you re-arrange the terms then +1-(2-vf) = vf-1.

Is that what you are asking?

If you built this circuit with silicon diodes with a vf = 0.7 volt then I would expect the voltage V to be about 2-0.7=1.3V. With 1.3 V on D1 cathode and +1 V on D1 anode then D1 is reverse biased (@ -0.3V) - by virtue of the D1 anode being at a lower potential than the D1 cathode.

Does that help?

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I think so. So one way to tell if a diode conducts or not is to see if the voltage across the anode is greater than the voltage across the cathode or greater than the .7 voltage drop?

crazyengineer said:

I think so. So one way to tell if a diode conducts or not is to see if the voltage across the anode is greater than the voltage across the cathode or greater than the .7 voltage drop?

Click to expand...

Your terminology needs some work but that's the idea.
Perhaps I would state that the voltage drop (or potential difference) from Anode to Cathode must equal or exceed the nominal on-state forward conduction voltage of the diode for the diode to conduct.

The test to tell if a diode conducts is as t_n_k said to check if it's forward biased. The methodology to actually make the initial selection to confirm is rather discrete:

For a circuit with n diodes, start with all of them non-conducting and check if any of them is actually forward biased. If the results don't agree with your hypothesis, move on and assume that the first diode conducts. Recalculate the voltages and check again.
Repeat for all of the 2^n diode configurations until you find the unique one that can be confirmed.
In this case you have 2^2=4 possible combinations.