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I am doing a question related to diodes from Sedra Smith Micro electronics and here is the question snip: enter image description here

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In the second part of the question where it says that if 1mV is being drawn by a load attached to output terminal, so for this I do the following:::

enter image description here

And for the value of 'I' I use the result that I get in the first part of the question which is 3.81mV. But the answer I am getting is not correct. The correct value of the output voltage in the second part is -22mV. My answer is not the same. Can someone guide me as to how to do this question thoroughly. Thanks in advance!

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  • \$\begingroup\$ You seem to be mixing up volts and amps. \$\endgroup\$
    – Hearth
    May 24, 2017 at 2:25
  • \$\begingroup\$ I meant mA in the diagram. Changed it now. \$\endgroup\$
    – Moeen Ahmed
    May 24, 2017 at 2:32
  • \$\begingroup\$ How are you trying to do it, and what is your answer? I get -23mV-- and please fix the "3.81mV" in your question. \$\endgroup\$
    – Spehro Pefhany
    May 24, 2017 at 2:51
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    \$\begingroup\$ Watch out. The answer to part 1 is extremely sensitive to the exact value of Vt you use. You used Vt = 25 mV. However, if you use Vt = 25.25 mV, then the answer to part 1 is only 2.93 mA. Make sure you are using the same values in part 2 as part 1. \$\endgroup\$
    – Evan
    May 24, 2017 at 3:07
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    \$\begingroup\$ Assuming the emission coefficient is 1, then I get \$V_T\approx 24.897023\:\textrm{mV}\$ in order to get exactly \$-22\:\textrm{mV}\$ as the difference. To the OP: Are you going to show some of your work? Where did that \$3.81\:\textrm{mV}\$ come from, for example? \$\endgroup\$
    – jonk
    May 24, 2017 at 4:40

1 Answer 1

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For the first part , Since Vo is 2 V so potential drop across each diode is 2/3 V. Hence for calculating I we can use Shockley equation. I = Is (exp^(2e/3KT)- 1). Because same I flows across each diode.(diodes connected in series).

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