I'm sorry for posting a textbook question, but it does not provide any explanation and I am pulling hair trying to make sense of this.

I tried converting each of the diodes to a 0.7V DC source with positive terminals up. Then I thought I found the equivalent resistance of the 3 diodes by

Rd = Vd/Id = (0.7 * 3 V)/(1 mA) = 2100 ohms

Then I did a voltage divider

Vo = Vin * Rd / (Rd + R)

Rearranged for R

R = Vin * Rd / Vo - Rd

Plugged in the values

R = (10 - 0.7 * 3) * 2100 / 2.4 - 2100
R = 4812.5

Which isn't even remotely close to the supplied answer of 139.

I've been at this for over an hour and I'm too stubborn to ignore it but obviously not learned enough to figure it out yet. Can you help me see what I'm missing?

Question and circuit diagram

EDIT: Full Solution

$$ I = I_s e^{\frac{v}{v_t}} $$

$$ I_s = \frac{I}{e^{\frac{v}{v_t}}} = \frac{0.001}{e^{\frac{0.7}{0.025}}} \approx 6.914 \cdot 10^{-16} $$

$$ I = I_s e^{\frac{v}{v_t}} = (I_s) e^{\frac{2.4 / 3}{0.025}} = 0.0546\, A $$

$$ R = \frac{v}{i} = \frac{10 - 2.4}{0.0546} = 139 $$


1 Answer 1


I'm going to assume they want you to use the diode V-I equation which involves exponentials. They tell you a starting point of the diode at 0.7V at 1mA so you can find a set point on the diode curve. Then you'll have to solve the simultaneous equations (10-2.4)/R=Idiode and 2.4=3*Vdiode.

https://en.wikipedia.org/wiki/Diode_modelling#Shockley_diode_model Vt of that equation is ~25 mV. Use the approximate equation since Vd>>nVt.

Really you'll end up with 3 equations to solve simultaneously. The first two I gave above. The last is the diode equation once you've found Is using 0.7V at 1mA.

  • \$\begingroup\$ If I use that equation with I = 1mA and Vd = 0.7 then all there is left to solve for is Is, which is irrelevant is it not? Or am I missing something? I was under the impression that this circuit needed to be solved using the constant-voltage-source model. \$\endgroup\$
    – TW80000
    Commented Feb 4, 2016 at 4:31
  • \$\begingroup\$ You use 1mA and 0.7 to find Is. Then you use the diode equation again with the variables V and I intact. Also assume n=1 as an approximation. \$\endgroup\$
    – horta
    Commented Feb 4, 2016 at 4:34
  • \$\begingroup\$ @TW80000 If it was to be solved by a constant voltage drop model, then Vo=3*0.7V=2.1V which is clearly wrong. The "0.7V drop at 1mA" indicates that it is not a constant V drop problem, and needs to be solved using the exponential model for a diode. \$\endgroup\$
    – uint128_t
    Commented Feb 4, 2016 at 4:34
  • \$\begingroup\$ Thanks a ton horta, right after I added my comment I realized that's what you meant! I went through and it all makes sense now. Thanks so much for your help! \$\endgroup\$
    – TW80000
    Commented Feb 4, 2016 at 4:38
  • \$\begingroup\$ @TW80000 Np, and nj for working through it and posting your solution it to help others. \$\endgroup\$
    – horta
    Commented Feb 4, 2016 at 5:33

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