# Having a really hard time with Diodes and transistors...

#### domandrzejczuk

Joined Nov 19, 2007
9
.l.l.l.l.l.

#### beenthere

Joined Apr 20, 2004
15,819
Your diode conducts whenever the cathode is .7 volts less positive than the anode. That should make 1b & c a snap.

#### domandrzejczuk

Joined Nov 19, 2007
9
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?

#### beenthere

Joined Apr 20, 2004
15,819
Once it's conducting, it has effectively no resistance. Treat it as a short with a .7 volt drop.

#### domandrzejczuk

Joined Nov 19, 2007
9
so in that case, if the right most node has a v of 8 volts, then vout is simply 7.3 volts?

#### The Electrician

Joined Oct 9, 2007
2,793
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".

Joined May 1, 2007
19
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

#### The Electrician

Joined Oct 9, 2007
2,793
You're not taking this same test, are you?

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.

#### chesart1

Joined Jan 23, 2006
269
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

#### domandrzejczuk

Joined Nov 19, 2007
9
You're not taking this same test, are you?

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.

#### JoeJester

Joined Apr 26, 2005
4,369
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.

#### Dave

Joined Nov 17, 2003
6,970
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

#### GS3

Joined Sep 21, 2007
408
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.

#### JoeJester

Joined Apr 26, 2005
4,369
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.

#### The Electrician

Joined Oct 9, 2007
2,793
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.