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schematic

simulate this circuit – Schematic created using CircuitLab

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?

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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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  • \$\begingroup\$ It definitely is. The thing that made me exclude that was because then \$i_3=0\$ for va = 6.7 so "it's off". But it actually means that for any voltage higher than that, the diode is on. \$\endgroup\$ – edmz Oct 26 '18 at 17:41
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Assuming D3 is forward biased with a constant forward voltage of 0.7 V, the current through resistor R3 is,

$$i_3 = \frac{v_a-(0.7+6)}{R_3} $$

Note the brackets.

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  • \$\begingroup\$ The thing that made me exclude that was because then \$i_3=0\$ for va = 6.7 so "it's off". But it actually means that for any voltage higher than that, the diode is on. \$\endgroup\$ – edmz Oct 26 '18 at 17:40
  • \$\begingroup\$ @edmz You are quite right. There are some quirks to approximating VD as a constant 0.7 V, but it also makes for simple hand analysis. You could replace VD as \$\eta V_T ln( I_3/ I_s)\$ but the solution becomes very difficult. \$\endgroup\$ – sstobbe Oct 26 '18 at 17:49
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If \$V_a = 6.7V\$ and \$D_3\$ diode threshold voltage is \$0.7V\$ the \$I_3\$ current is \$0A\$. Therefore we can find \$V_{IN}\$ using voltage divider equation.

$$V_{IN} = V_a \cdot (1+ \frac{R_1}{R_2}) + 0.7V = 6.7V \cdot 1.5 + 0.7V = 10.75V$$

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  • \$\begingroup\$ You surely meant \$6.\color{red}{7}V\cdot 1.5\$. Still, thanks for your approach. \$\endgroup\$ – edmz Oct 26 '18 at 18:01
  • \$\begingroup\$ Yep. You are right. \$\endgroup\$ – G36 Oct 26 '18 at 18:13
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Assume that diode has the usual logarithmic behavior

Vdiode Idiode

0.600 1mA (Assumed)

0.542 0.1mA

0.484 0.01mA

0.658 10mA

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