Electronic – Finding necessary voltage for diode conduction

circuit analysisdiodes

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In the following, I need to find \$V_a(v_{in})\$ such that \$D_3\$ is on. Assume a forward voltage drop of \$0.7V\$. From past computations, it turned out to be around \$10.75V\$ (as you can check from the sim, whilst putting \$10V\$ would turn it off); however, I can't find the issue in my computation.

Since $$i_1=i_2+i_3,\quad \frac{v_i-0.7-v_a}{R_1}=\frac{v_a}{R_2}+\frac{v_a-0.7+6}{R_3}$$
In order for the diode to turn on, I then set \$v_a=6.7\$ and solve for \$v_{in}\$; however, I find a bigger value (\$14.75V\$) than it should be. Where's that flaw?

Best Answer

If you assume that D3 is not quite on the verge of conduction then there is zero current passing through it and, for this ideal diode (that can have a 0.7 volt drop across it before conduction) then the current from Va is all flowing through R2 and this current is 6.7 volts / 10 kohm = 0.67 mA. This current also flows through R1 and therefore the voltage on the left node of R1 is 6.7 volts + 3.35 volts = 10.05 volts. And this means that the voltage on the left hand node of D2 must be (assuming the same ideal diode) is 10.75 volts.

Maybe the flaw is in your equation where you wrote "Va - 0.7 + 6" when it should be "Va - (0.7 + 6)"?

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