voltage - Electronics I problem

Suppose that Vin= -∞, that forces D1 to open(OFF) and D2 to close (ON).

Given E is 2 , then Vout=-2. Then I applied KVL to find the equation for Vd1 ( Off). Now I know that I have to find the current through Vd2 since it is ON. I applied KCL on the top right node.

$$I_{D1}=I_{R3}+I_{R2}$$ Now I know that I need to find the current going through R2 and R3 but I am not sure if Vin is involved in in R3 and R2 or not , if so, why? Please give me a hint.

Answer

You left out some details, so I'm making some assumptions.

1. The diodes are ideal diodes.


2. The problem is the standard one -- determine whether each diode is on or off, then find the current through the on diodes and the voltage across the off diodes.

If $V_{in} = -\infty$, the answer won't make much sense. You're correct that this condition causes $D_1$ to turn off and $D_2$ to turn on, and that this makes $V_{out} = -2\ V$. Now you're trying to solve a KCL equation (which had a typo):

$$I_{D2} = I_{R3} + I_{R2}$$

$I_{R2}$ is pretty straightforward:

$$I_{R2} = \frac{V_{out}}{R_2} = \frac{-2\ V}{R_2}$$

But $I_{R3}$ is not:

$$I_{R3} = \frac{V_{out} - V_{in}}{R_3} = \frac{-2\ V - \infty}{R_3} = -\infty$$

That's what happens when you put an infinite voltage in your circuit. $V_{in}$ tries to pull an infinite amount of current from ground through $R_3$. That current is almost entirely supplied by $E$, which is an ideal voltage source. The rest of the current comes through $R_2$. How much is determined entirely by $V_{out}$, which is capped at $-2\ V$ by $E$ and $D_2$.