2
\$\begingroup\$

I am working on part (c) from the following problem:!enter image description here

Vbias = 3.3V which should have been given.

Trying to solve for Vout, I first tried a voltage divider ignoring the diode. $$3.3V * \frac{1k\Omega}{1k\Omega+2.7k\Omega} = .892V$$ which would mean that the voltage drop across the diode is .892V (?). Which would mean Vout = .892V.

This doesn't seem right to me. In the previous problem (b), I said Vout = .6V which is just Vd and it seems strange to me that adding a resistor in parellel would increase the voltage at Vout node especially if it was smaller, say 100ohm.

\$\endgroup\$
  • 4
    \$\begingroup\$ You should first transform the topology of (b) to (a) by replacing the battery and the two resistors with their Thevenin-equivalent. \$\endgroup\$ – Wouter van Ooijen Oct 15 '13 at 21:00
2
\$\begingroup\$

Forward biased normal PN junction diode would act as a voltage regulator.so the voltage drop across the diode may be even 0.7V. when we decrease the load resistance(connected across the diode) the current drawn by the diode decrease and the voltage drop also reduces(when a lighter load is connected across the diode it would behave like a good regulator even when there are changes in the supply voltage this is known as LINE REGULATION;but when there is a change in the load it would not hold good LOAD REGULATION).

In other words the dynamic forward resistance of the forward biased diode would increase, if there is a reduction in the current flowing through the diode(the current flowing through the diode would reduce, when the load resistance is decreased) enter image description here

\$\endgroup\$
1
\$\begingroup\$

The comment by Wouter gives the best approach. The equivalent circuit is:

schematic

simulate this circuit – Schematic created using CircuitLab

If this isn't clear, all that has been done is to swap the position of the diode and R2 and then replace the circuit to the left of the diode with the Thevenin equivalent circuit.

Now you have a circuit just like in part (b) to be solved in the same way.

\$\endgroup\$
0
\$\begingroup\$

They have given you all the necessary numbers so that you can model the diode in closed form. Using KCL (Kickoff's Current Laws) and KVL (.... Voltage laws) you should be able to come up with a set of equations that when solved give you the answer. Since you started with a resistor divider (which is a good start) you know that the answer is going to be intermediate between the 0.6 and 0.892 because at 0.892 that diode is going to be sucking a lot of the current!

Part a) gets you to calculate \$ I_s \$ and ideality factor \$\eta\$ is given as 1. You can use the simplified version of \$I_D = I_se^\dfrac{V_D}{V_{t}}\$.

\$\endgroup\$
  • \$\begingroup\$ I calculated Is to be .95fA (femto). I calculated the current to be 75A going through the diode .95fA * e^(.892/.026) . Which seems unreasonably high. \$\endgroup\$ – Nick Oct 15 '13 at 20:36
  • \$\begingroup\$ Like my house might catch on fire if I set this up. \$\endgroup\$ – Nick Oct 15 '13 at 20:41
0
\$\begingroup\$

As others have said consider how the voltage divider behaves in the absence of the diode or when the diode is not forward biased. the equation for the circuit is below:

vout = (r2/(r1+r2))*(Vin - Id*R1)

From this you can see that when there is little current through the diode the equation is reduced to a simple divider.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.