# Diodes, potentials

## Homework Statement

Find the current I and potential V for a simple diode circuit from the picture using
a) ideal diode model
b) CVD diode model (Vd=0,6 V)
[PLAIN]http://pokit.etf.ba/get/36406f902c2c24bd8a3c4ffa8fe97b78.jpg [Broken]

U=RI
I1+I2=I

## The Attempt at a Solution

This diode can be on or off right? I do the calculations for the ON method, and i get that I=I1+I2=-0,2 mA, now my mates said that they get that right result is that the diode is OFF. I do not understand this, this isn't a real assignment, more like a theoretical question.

Can you help me?

Thanks

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berkeman
Mentor

## Homework Statement

Find the current I and potential V for a simple diode circuit from the picture using
a) ideal diode model
b) CVD diode model (Vd=0,6 V)
[PLAIN]http://pokit.etf.ba/get/36406f902c2c24bd8a3c4ffa8fe97b78.jpg [Broken]

U=RI
I1+I2=I

## The Attempt at a Solution

This diode can be on or off right? I do the calculations for the ON method, and i get that I=I1+I2=-0,2 mA, now my mates said that they get that right result is that the diode is OFF. I do not understand this, this isn't a real assignment, more like a theoretical question.

Can you help me?

Thanks
One way to approach problems like this is to first substitute a short for the diode and solve for the voltages, and then substitute an open for the diode and solve for the voltages. In those two cases, what is the output voltage? What does that tell you about which state the diode will be in?

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I know the procedure, with replacing ideal diodes with a wire etc. But when I give it an open switch case, I cannot calculate anything, at least not by my knowledge... Because I have that unknown V potential

berkeman
Mentor
I know the procedure, with replacing ideal diodes with a wire etc. But when I give it an open switch case, I cannot calculate anything, at least not by my knowledge... Because I have that unknown V potential
I believe that V is only the output voltage. In problems like this, at least as shown above, V is not an additional unknown input voltage. So when the diode is open, what is the output voltage V?

I believe that V is only the output voltage. In problems like this, at least as shown above, V is not an additional unknown input voltage. So when the diode is open, what is the output voltage V?
That's the problem, i only know how to calculate when you replace the potentials as input voltage... How do you do that then? Is there a method or something like that?

berkeman
Mentor
That's the problem, i only know how to calculate when you replace the potentials as input voltage... How do you do that then? Is there a method or something like that?
Picture yourself hooking a voltmeter to that point V. When the diode is open circuit, all you have is 7V connected to your voltmeter through a resistor. Assume that your voltmeter is ideal, so its input resistance is infinite. What does the voltmeter read?

Picture yourself hooking a voltmeter to that point V. When the diode is open circuit, all you have is 7V connected to your voltmeter through a resistor. Assume that your voltmeter is ideal, so its input resistance is infinite. What does the voltmeter read?
Ooooh i get it, so there is no current running, and the potential just transfers to that point? And the voltmeter would read 7 V also?

berkeman
Mentor
Ooooh i get it, so there is no current running, and the potential just transfers to that point? And the voltmeter would read 7 V also?
Correct. So if the diode is reverse biased, it stays reverse biased (off).

Is there any reason for it to get forward biased in that circuit? Instead of substituting a short for the diode, what happens if you substitute a 20k resistor?

Correct. So if the diode is reverse biased, it stays reverse biased (off).

Is there any reason for it to get forward biased in that circuit? Instead of substituting a short for the diode, what happens if you substitute a 20k resistor?
I came to the conclusion that in the case that we discussed before, u get -4 V drop on the diode, which we want! Reverse polarized has negative voltage. So there is no need to check the other case...

berkeman
Mentor
I came to the conclusion that in the case that we discussed before, u get -4 V drop on the diode, which we want! Reverse polarized has negative voltage. So there is no need to check the other case...
Sounds good to me.

Thank you very much!