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.
- The diodes are ideal diodes.
- 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\$.
No comments:
Post a Comment