Question about determing if a diode is on or not.

Psf1575

Joined Mar 2, 2018
16
Hey Everyone, im scared that the way im going about determining if a diode is on or off might be wrong. In the circuit I posted, I know the diode is forward biased so it will conduct if it has atleast 0.7 V across from it. The way I set out to try and find the answer was to assume the diode is off, so its an open circuit there, then use thevenin to find the voltage on it, and if its atleast 0.7 then its on. After i determine the thevenin path, i know that the V thevenin will be 2 Volts, can I stop there and say its on, or do i have to go all the way into the series thevenin circuit with the thevenin resistance to do this. Thank you everyone.

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JoeJester

Joined Apr 26, 2005
4,390
You have to go all the way. Draw your diagrams and show your work.

Psf1575

Joined Mar 2, 2018
16
You have to go all the way. Draw your diagrams and show your work.
Ok, heres what I think is the thevenin equivalent circuit with the element of interest back in it. Now, do I use this circuit to see if my diode is on? Meaning, assume the diode is 0.7V, and see if current is flowing in the all series circuit? My reasoning is that if theres a voltage across the thevenin resistance, then certainly the diode must be on, or else the series circuit would have no current. Am I wrong?

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freak101

Joined Aug 7, 2017
37
Hi,
In problems involving resistors and diodes, books follow a different approach. They assume that a current is flowing from P side to N side of a diode. After performing calculations, if it occurs that that assumed current is actually negative, then diode is off (remember diode conducts in only one direction).
What you are saying will also work as long as you keep an eye on the direction of current in the branch containing diode.

WBahn

Joined Mar 31, 2012
30,067
Ok, heres what I think is the thevenin equivalent circuit with the element of interest back in it. Now, do I use this circuit to see if my diode is on? Meaning, assume the diode is 0.7V, and see if current is flowing in the all series circuit? My reasoning is that if theres a voltage across the thevenin resistance, then certainly the diode must be on, or else the series circuit would have no current. Am I wrong?
You have one error that didn't bite you.

In your work you say that Voc = V34 + V24

This is not correct. It is also ambiguous because while you indicate where Voc is measured, you don't define the polarity. Thus, if Voc turns out to be positive, you don't know whether this is forward biasing or reverse biasing the diode.

By convention, the diode voltage is the anode relative to the cathode. So explicitly define your Voc accordingly.

This would mean that

Voc = V23.

This, in turn, is

Voc = V23 = V24 + V43 = V24 - V34.

NOW you can say that V34 = 0 and so

Voc = V24

That you happen to get the same result is pure coincidence and not worthy of much credit. You need to be more careful.

Once you have Voc you can determine if the diode is forward biased or not. If Voc is greater than Vd (~= 0.7 V usually) then the diode is conducting. If it is less, then it is not. That's if you are using the basic diode-as-switch-with-fixed-forward-voltage-drop model, which is adequate for the vast majority of analysis work.

There no need to continue further unless you want to know the actual current in the diode.

An alternative is to assume the diode is on and replace it with a voltage source outputting Vd and then analyze the circuit to see if the current through the diode (defined from anode to cathode) is positive. If it is, the diode is on, otherwise it is off.

You generally do not need to do both, although in theory you could probably design a circuit (but I don't think if would be linear) in which if you assume the diode is off it will tell you it is on but if you assume it is on then it will tell you it is off. That would be an oscillator. Or you could probably design a circuit in which if you assume it is off it will tall you it is off, but if you assume it is on then it will tell you it is on. That would be a bistable latch.

Finally, I notice that you appear to be tracking your units throughout your work (though that's really only based on one equation). If so, great job on that -- it's so rare to see.

Psf1575

Joined Mar 2, 2018
16
You have one error that didn't bite you.

In your work you say that Voc = V34 + V24

This is not correct. It is also ambiguous because while you indicate where Voc is measured, you don't define the polarity. Thus, if Voc turns out to be positive, you don't know whether this is forward biasing or reverse biasing the diode.

By convention, the diode voltage is the anode relative to the cathode. So explicitly define your Voc accordingly.

This would mean that

Voc = V23.

This, in turn, is

Voc = V23 = V24 + V43 = V24 - V34.

NOW you can say that V34 = 0 and so

Voc = V24

That you happen to get the same result is pure coincidence and not worthy of much credit. You need to be more careful.

Once you have Voc you can determine if the diode is forward biased or not. If Voc is greater than Vd (~= 0.7 V usually) then the diode is conducting. If it is less, then it is not. That's if you are using the basic diode-as-switch-with-fixed-forward-voltage-drop model, which is adequate for the vast majority of analysis work.

There no need to continue further unless you want to know the actual current in the diode.

An alternative is to assume the diode is on and replace it with a voltage source outputting Vd and then analyze the circuit to see if the current through the diode (defined from anode to cathode) is positive. If it is, the diode is on, otherwise it is off.

You generally do not need to do both, although in theory you could probably design a circuit (but I don't think if would be linear) in which if you assume the diode is off it will tell you it is on but if you assume it is on then it will tell you it is off. That would be an oscillator. Or you could probably design a circuit in which if you assume it is off it will tall you it is off, but if you assume it is on then it will tell you it is on. That would be a bistable latch.

Finally, I notice that you appear to be tracking your units throughout your work (though that's really only based on one equation). If so, great job on that -- it's so rare to see.
Hey, thanks for the reply, it was very helpful. I know i should have labeled the polarity of my VOC, i just didnt know how to do it properly. Attached is my attempt at labeling it. I put plus on the anode side and negative on the cathode, then i just followed my path and labeled those other polarities. Im not sure how to add those voltages accordingly after that. Technically, dont we know the diode is forward biased because of the DC voltage source polarity, so we only have to check if it has atleast .7V forward voltage across of it?

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WBahn

Joined Mar 31, 2012
30,067
Hey, thanks for the reply, it was very helpful. I know i should have labeled the polarity of my VOC, i just didnt know how to do it properly. Attached is my attempt at labeling it. I put plus on the anode side and negative on the cathode, then i just followed my path and labeled those other polarities. Im not sure how to add those voltages accordingly after that. Technically, dont we know the diode is forward biased because of the DC voltage source polarity, so we only have to check if it has atleast .7V forward voltage across of it?
Once you've got the circuit reduced to what you show here, then as you say it is obvious that it is forward biased and the only question is whether it is forward biased enough.

Why did you choose the polarity for the right hand resistor that you did?

You should also label the voltages. It's very inconvenient when people have to describe things like, "the voltage across the right hand resistor". It's a lot easier if they can just say, "v5" because you put "v5" on the diagram between the polarity marks associated with it.

Psf1575

Joined Mar 2, 2018
16
Once you've got the circuit reduced to what you show here, then as you say it is obvious that it is forward biased and the only question is whether it is forward biased enough.

Why did you choose the polarity for the right hand resistor that you did?

You should also label the voltages. It's very inconvenient when people have to describe things like, "the voltage across the right hand resistor". It's a lot easier if they can just say, "v5" because you put "v5" on the diagram between the polarity marks associated with it.
Im not sure why i picked the polarity for that resistor, i only know forsure that the polarity across the diode is like I put it because thats where the anode and cathode were. Then after that I just followed the positive polarity all the way back to the VOC, since i know i have to label the shortest path the electrons take back to the VOC.

MrAl

Joined Jun 17, 2014
11,489
Hey Everyone, im scared that the way im going about determining if a diode is on or off might be wrong. In the circuit I posted, I know the diode is forward biased so it will conduct if it has atleast 0.7 V across from it. The way I set out to try and find the answer was to assume the diode is off, so its an open circuit there, then use thevenin to find the voltage on it, and if its atleast 0.7 then its on. After i determine the thevenin path, i know that the V thevenin will be 2 Volts, can I stop there and say its on, or do i have to go all the way into the series thevenin circuit with the thevenin resistance to do this. Thank you everyone.
Hello,

Here is a cleaned up diagram so others can read your circuit a little easier. It helps to make a clear circuit and of course rotated to the right angle for proper display.