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#1
01-21-2007, 01:05 AM
 unlisted Junior Member Join Date: Jan 2007 Posts: 11
Diode problem

Hey guys,

I am looking for a point in the right direction for this problem.

The problem asks for the value of R for which the voltage across the resistor is 40mV. I assume I need to workout the current passing through the resistor. Since the resistor is in series with D1, which is in parallel with D2 the current should be divided. Once I have the current I can use Ohm's law to find R.

So I think what I need to find is the current passing through D1 and R.

Thanks
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#2
01-21-2007, 02:17 AM
 Ron H E-book Developer Join Date: Apr 2005 Location: Idaho, USA (GMT-7) Posts: 7,050

I know how to solve this, and it involves the diode equation. Since it's homework, I won't give you the answer, but I can give you some clues.

But first - have you been exposed to the diode equation? If so, tell us what it is.
#3
01-21-2007, 03:07 AM
 unlisted Junior Member Join Date: Jan 2007 Posts: 11

Thanks for the help.

I have been exposed to the diode equation i = Is[exp(v/nVT - 1)]
At room temperature VT = .025V
n is 1, or 2, but we are not given a value for n.
#4
01-21-2007, 03:17 AM
 Ron H E-book Developer Join Date: Apr 2005 Location: Idaho, USA (GMT-7) Posts: 7,050

Quote:
 Originally Posted by unlisted Thanks for the help. I have been exposed to the diode equation i = Is[exp(v/nVT - 1)] At room temperature VT = .025V n is 1, or 2, but we are not given a value for n.
In order to solve this, you first have to solve for n. It is not 1 or 2, but it is somewhere in between. You can solve for n from the information given.
After you solve for n, you can then solve for the current through the two diodes, and then, of course, you can solve for R.
If you get stuck, post your efforts here and I'll give you some more clues.
Hint: the "-1" in the equation is insignificant at the current levels you are working with. You may already be aware of this.
#5
01-21-2007, 04:02 AM
 unlisted Junior Member Join Date: Jan 2007 Posts: 11

Ok, I solved for n and Is, and got an equation for i.

The V given in the problem statement is the drop across R, right? So
I can't just plug that into my diode equation. Do I need to first solve
for the drops across the diodes?
#6
01-21-2007, 04:30 AM
 Ron H E-book Developer Join Date: Apr 2005 Location: Idaho, USA (GMT-7) Posts: 7,050

Quote:
 Originally Posted by unlisted Ok, I solved for n and Is, and got an equation for i. The V given in the problem statement is the drop across R, right? So I can't just plug that into my diode equation. Do I need to first solve for the drops across the diodes?

I1~Is*e^(v1/nVt)
I2~Is*e^(v2/nVt)
therefore,
(1) I1/I2=e^((v1-v2)/nVt)
Note that Is has fallen out of the equation.
From this relationship, and the information about the diodes given in the problem, you can find n.
Once you find n, remember that you know that in the actual circuit, v1-v2=40mV. From this fact, and using eq. (1), you can find the currents through the two diodes.

Another hint: Naperian logarithims are required to solve for n, and again for I1 and I2.

Last edited by Ron H; 01-21-2007 at 04:32 AM.
#7
01-21-2007, 04:48 AM
 unlisted Junior Member Join Date: Jan 2007 Posts: 11

I used the natural log to find n. Does v1-v2=40mV come from Kirchoff's voltage law?
#8
01-21-2007, 03:18 PM
 Ron H E-book Developer Join Date: Apr 2005 Location: Idaho, USA (GMT-7) Posts: 7,050

Quote:
 Originally Posted by unlisted I used the natural log to find n. Does v1-v2=40mV come from Kirchoff's voltage law?
It comes from the fact that the anodes are tied together, the cathode of D2 is at zero volts, and the cathode of D1 is at 40mV. Therefore (VD2-VD1)=40mV. I guess that's Kirchhoff's voltage law. When you've been an engineer for 40 years, the names of the laws (in my case, anyway) become irrelevant. Who else had a voltage law?

Did you understand the ratio equation, and why Is falls out of the equation? I suppose you could find Is, but you don't need it.

Last edited by Ron H; 01-21-2007 at 03:20 PM.
#9
01-21-2007, 06:08 PM
 unlisted Junior Member Join Date: Jan 2007 Posts: 11

I understand the ratio equation, that is basically how I solved for n. I just don't get the v1-v2 = 40mV. To me, the two diodes look like they are hooked up in parallel. In which case they should share the same voltage drop, right?

Ok, assuming v1-v2=40 mV I am still unclear on how to find the currents through D1 and D2. I'll look back through the post and see if I can work it out.

Wow, 40 years? Ever think about teaching this stuff?
#10
01-21-2007, 07:08 PM
 Ron H E-book Developer Join Date: Apr 2005 Location: Idaho, USA (GMT-7) Posts: 7,050

Quote:
 Originally Posted by unlisted I understand the ratio equation, that is basically how I solved for n. I just don't get the v1-v2 = 40mV. To me, the two diodes look like they are hooked up in parallel. In which case they should share the same voltage drop, right? Ok, assuming v1-v2=40 mV I am still unclear on how to find the currents through D1 and D2. I'll look back through the post and see if I can work it out. Wow, 40 years? Ever think about teaching this stuff?
I work full time as an engineer. I don't have the time or the inclination to teach (except on forums like this).

How can you think the diodes are in parallel when R has 40mV across it? D1 obviously (to me, anyway) has 40mV less across it than D2.
If you accept that, do you understand where this equation comes from?

ln(I2/I1)= (v2-v1)/nVt

You can solve for (I2/I1).
Your other equation is (I2+I1)=10ma (from the schematic).

You now have two simultaneous equations. Solve for I2 and I1.

 Tags diode, problem

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