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#1




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 
#2




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




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




Quote:
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




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




Quote:
I1~Is*e^(v1/nVt) I2~Is*e^(v2/nVt) therefore, (1) I1/I2=e^((v1v2)/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, v1v2=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; 01212007 at 04:32 AM. 
#7




I used the natural log to find n. Does v1v2=40mV come from Kirchoff's voltage law?

#8




Quote:
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; 01212007 at 03:20 PM. 
#9




I understand the ratio equation, that is basically how I solved for n. I just don't get the v1v2 = 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 v1v2=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




Quote:
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)= (v2v1)/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. 
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