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I am asked to find the current through the second diode, as shown in the image below.

I have gone through my procedure to find this current, and can not figure out why my answer is wrong.

What I did is as follows:

I assumed that both diodes were conducting. I then checked this assumption by seeing if the current through diode 1 and diode 2 is greater than zero.

current through diode 1 = (12 - 0.7 - (-0.7) - 4) / 4000 = 2mA (which is greater than 0)

current through the second resistor = (4 - 0.7)/ (1.3 x 10^3) = 2.538 x 10^-3 A

And by KCL, current through diode 2 = 5.38 x 10^-4, which is greater than zero.

I appreciate your help and time,

thank you very much.

enter image description here

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  • \$\begingroup\$ You may wish to look at the conditions in which the diodes would conduct. Consider the voltage of the center point to ground when D1 has a volt drop of 0.7V. Which way is D2 biased? \$\endgroup\$ – K H Apr 7 at 6:03
  • \$\begingroup\$ I gave this a vote, because it has question with a diagram and shows effort towards a solution. Good job. \$\endgroup\$ – Solar Mike Apr 7 at 6:33
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You have to solve the circuit for every possible combination of diode states. Both on, both off, D1 off and D2 on, D1 on and D2 off. After solving each case using KVL or KCL, you check to see if your assumption was correct. If it was, you're done. If it's not then you solve for the next possible scenario.

The reason you are not getting the right answer is your two loop equations are formulated and solved completely independently of each other.

All loop equations must be solved simultaneously since both 12V and 4V supplies could possibly be pushing current through R2, OR one supply might be pushing so much current through R2 that the voltage drop across R2 is high enough that the other diode is never able to be forward biased and conduct.

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  • \$\begingroup\$ Hi Toor. I found the current through each diode to be greater than zero, so is it correct to assume that both diodes are conducting? \$\endgroup\$ – Michel Apr 7 at 6:06
  • \$\begingroup\$ No. Because you making the assumption that the diodes are conducting will inevitably result in a solution where the current through each diode is greater than zero. it's circular reasoning. What you actually have to check is if the voltages across the diode forward bias the diode (by using your calculated currents to trace the voltage drops to both sides of the diode and taking the difference. \$\endgroup\$ – DKNguyen Apr 7 at 6:08
  • \$\begingroup\$ Your passive sign convention is also wrong. As are your loop equations (as I already stated where you are formulating and solving the two loop equations completely independently of each other). \$\endgroup\$ – DKNguyen Apr 7 at 6:21
  • \$\begingroup\$ I am afraid you are wrong. Assuming a diode is conducting and then check the current to be greater than zero is by no means circular reasoning. If you try assuming D2 ON on the circuit with the R2=6k you'll soon find a negative ID2. The same applies when assuming the diode is OFF and then check voltage to be less than zero (or threshold upon model used). In short @Michel is right \$\endgroup\$ – carloc Apr 7 at 7:43
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While solving this problem it will pay to remain aware that you are using a simplified diode model in which the diode is either fully conducting (in which case the diode Vf = 0.7V) or fully "off" (when diode Vf <= 0V and in which case Id = 0).

Starting with the assumption that D1 and D2 are both conducting, the key to solving this problem is to consider that when D2 is conducting then the voltage across R2 is fixed:

VR2 = 4V - 0.7V = 3.3V

That then gives:

IR2 = (VR2 / R2)

and:

ID1 = (12V - 0.7V - VR2)/R1

and finally:

ID2 = IR2 - ID1

Solving this for R2=1.3k should give you a sensible answer.

I would suggest then also solving it for the original R2=6k which should reveal that the assumptions don't hold then modify the analysis appropriately.

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Changing R2 to 1.30 kOhm results without D2 as;

\$ V_{R2}= 11.3V*\dfrac{1.3k}{1.3k+4k} = 2.77 V\$

for a Norton Equiv. with;

\$ R1=4k, ~~~ R2=1.3k, ~~~ R1//R2 = 981 Ω, ~~~ Id1=11.3V/(4k+1.3k)=2.13 mA \$

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schematic

simulate this circuit – Schematic created using CircuitLab

\$I_{{D_2}}= 538~ \mu A\$

We compute the Norton model above with the D2 diode switch open then check the voltage is less than the right side after the diode drop to 3.3V and then (imagine) close the switch to solve for Id2.

In this solution, I treated the diode as a 0.7V battery and switch if forward biased then closed.

In reality near 1~2mA a Silicon diode is about Vf=0.6V for future reference a 1N4448 small signal diode is Vf=0.7V @ 23mA, Vf=1V@100mA pulsed (100mW) at 25'C

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  • \$\begingroup\$ But R2 is 1.3k ohms not 6k \$\endgroup\$ – G36 Apr 7 at 6:13
  • \$\begingroup\$ I couldn't read the fine print oops \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 7 at 6:14

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