.l.l.l.l.l.

 

Your diode conducts whenever the cathode is .7 volts less positive than the anode. That should make 1b & c a snap.

 

how would i go about solving for vout then if i dont know the current through the diode, and assumming the diode has no resistance?

 

Once it's conducting, it has effectively no resistance. Treat it as a short with a .7 volt drop.

 

so in that case, if the right most node has a v of 8 volts, then vout is simply 7.3 volts?

 

beenthere, I think we've "been had". This is the same problem electo101 asked about last Friday. And this doesn't look like a "practice" test to me. At the top of the page it says it's due tomorrow. The first paragraph says "You may use any non-human and non-internet references to help you solve the problems" and "Naturally, you are on your honor in these respects; I am certain none of you will let me down in that department".

 

beenthere, I think we've "been had". This is the same problem electo101 asked about last Friday. And this doesn't look like a "practice" test to me. At the top of the page it says it's due tomorrow. The first paragraph says "You may use any non-human and non-internet references to help you solve the problems" and "Naturally, you are on your honor in these respects; I am certain none of you will let me down in that department".

man that's sad. you know, talking electronics might cheer us up. for instance i've got a couple of questions regarding the foregoing dialogue:

*) "Your diode conducts whenever the cathode is .7 volts less positive than the anode."

ok so clearly the diode conducts when anode-cathode is exactly 0.7 but what about these cases?:

1) anode - cathode < 0.7
2) anode - cathode > 0.7

*) "Once it's conducting, it has effectively no resistance. Treat it as a short with a .7 volt drop."

as i understand it resistance is (volt drop)/current. we're told the diode is conducting so presumably current must be nonzero. we're also told the volt drop is 0.7 volts which is likewise nonzero. how does this square with the device being "a short" with "no resistance"?

peace
stm

 

You're not taking this same test, are you?

Go read this thread:
http://forum.allaboutcircuits.com/showthread.php?t=7922

Does that help?

I'm relatively new here, and I've been helping a number of people. I thought I was helping with homework, not exams. Will one of the moderators tell me, is this a big problem, namely people representing exams as homework?

I think what I would prefer to do in such a case, is invent a new circuit, not on the exam, but similar. Then discuss that circuit, with the goal of teaching principles, not solving an exam problem.

 

man that's sad. you know, talking electronics might cheer us up. for instance i've got a couple of questions regarding the foregoing dialogue:

*) "Your diode conducts whenever the cathode is .7 volts less positive than the anode."

ok so clearly the diode conducts when anode-cathode is exactly 0.7 but what about these cases?:

1) anode - cathode < 0.7
2) anode - cathode > 0.7

*) "Once it's conducting, it has effectively no resistance. Treat it as a short with a .7 volt drop."

as i understand it resistance is (volt drop)/current. we're told the diode is conducting so presumably current must be nonzero. we're also told the volt drop is 0.7 volts which is likewise nonzero. how does this square with the device being "a short" with "no resistance"?

peace
stm

We don't know the resistance of the forward biased diode. But if you include a 0.7 volt drop across the diode in your calculations, your implicit assumption is that it has some finite resistance. The relationship between diode voltage and diode current is not linear. Therefore, the diode's resistance is not directly proportional to the current through it.

John

 

You're not taking this same test, are you?

Go read this thread:
http://forum.allaboutcircuits.com/showthread.php?t=7922

Does that help?

I'm relatively new here, and I've been helping a number of people. I thought I was helping with homework, not exams. Will one of the moderators tell me, is this a big problem, namely people representing exams as homework?

I think what I would prefer to do in such a case, is invent a new circuit, not on the exam, but similar. Then discuss that circuit, with the goal of teaching principles, not solving an exam problem.

That doesn't really tell me much, other than the answer.

 

Electrician,

That is a judgement call on behalf of those responding.

Yes, we could be responding to test questions or homework questions. One never knows.

We could be offering design advise to those you normally wouldn't offer design advise.

It's always a tough call, after all, this is the internet.

 

We don't know the resistance of the forward biased diode. But if you include a 0.7 volt drop across the diode in your calculations, your implicit assumption is that it has some finite resistance. The relationship between diode voltage and diode current is not linear. Therefore, the diode's resistance is not directly proportional to the current through it.

John

If you know the characteristics of the diode you can use the Ebers-Moll Equation to plot the I-V characteristic. The non-linear behaviour arises out of the exponential function implicit in Ebers-Moll. For a given characteristic you can calculate the conductance for a given voltage by taking the derivative of the Ebers-Moll Equation - the resistance is the reciprocal.

Beyond 0.7V for silicon, and 0.3V for Germanium \(\partial\)I/\(\partial\)V \(\rightarrow\) \(\infty\), therefore the resistance tends to 0.

Dave

 

Electrician,

That is a judgement call on behalf of those responding.

Yes, we could be responding to test questions or homework questions. One never knows.

My practice is to never give the short answers but to give information which helps understand what the answer should be. I do not care whether it is homework or an exam, I do not want to do anybody's homework for them. I am willing to help them with their study and understanding but I am not directly going to do their homework (well, I can discuss a fair price ) as that serves no purpose.

 

I do not want to do anybody's homework for them. I am willing to help them with their study and understanding but I am not directly going to do their homework (well, I can discuss a fair price ) as that serves no purpose

GS3,

At that philosphy is congruent with the posted Rules.

 

Electrician,

That is a judgement call on behalf of those responding.

Yes, we could be responding to test questions or homework questions. One never knows.

We could be offering design advise to those you normally wouldn't offer design advise.

It's always a tough call, after all, this is the internet.

In this case, I think we knew. Apparently you didn't see the OP's original question and attachment, and he edited his original post to delete his question and attachment. As it happens, I downloaded his attachment, and here it is. It's an exam, due today.

As for the actual question he wanted answered, it was dealt with a couple of days ago, in this thread: http://forum.allaboutcircuits.com/showthread.php?t=7922

The question said to take the diode voltage as .6 volts and assume it has zero resistance.