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I came across a problem which I cannot calculate by theory.

How does this circuit with diodes and resistors work?

diode in series with resistor

In the above diahram, I can calculate the current simply by subtracting the voltage drop of the diode from the source voltage which gives 11.4V (the green value in picture.) 11.4 divided by 2k (the total of the resistors) gives 5.7 mA that is drawn and we can work out the voltage drop over each resistor.

The next circuit confuses me.

diode in series with resistor in parallel with resistor

Can someone please explain how we get the current through every component and the voltage drops? The other thing a normal voltage divider of 1k and 1k, 2k in total on a 12V supply will draw 6mA but in the diode circuit R3 only draws 4.2 mA.

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  • \$\begingroup\$ Does this answer your question? diode and resistor in parallel with resistor \$\endgroup\$ Nov 3 '20 at 11:47
  • \$\begingroup\$ No not really but wil do research about the shocky equation. Thanks. \$\endgroup\$
    – Gideon
    Nov 3 '20 at 11:51
  • \$\begingroup\$ Turn that circuit upside down then rethink what you said. Else use Millman's theorem. \$\endgroup\$
    – Andy aka
    Nov 3 '20 at 12:05
  • \$\begingroup\$ Apparently, this problem wants you to ignore Shockley, and to assume that D1 always drops 0.6 V. \$\endgroup\$
    – CL.
    Nov 3 '20 at 12:13
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    \$\begingroup\$ Use Kirchhoff's laws, and algebra. \$\endgroup\$
    – CL.
    Nov 3 '20 at 12:16
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The voltage drop over R1+D1 and R3 is the same: V1 + 0.6 = V3
The currents through R1 and R3 add up in R2: I1 + I3 = I2
The total voltage drop is 12 V: V3 + V2 = 12
And Ohm's law: 1 kΩ × I1 = V1, 1 kΩ × I2 = V2, 1 kΩ × I3 = V3

These six equtions are enough to compute the results,

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  • \$\begingroup\$ Using Kirchhoff's law is so much simpler and makes sense. Thank you \$\endgroup\$
    – Gideon
    Nov 3 '20 at 15:43

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