Understanding this AC coupled circuit

Question

I'm having a little trouble understanding this circuit and could use some help.

So far I understand that this is an AC coupled circuit, and the capacitor is a blocking capacitor. I also know that this is a high pass filter, but if I was to find the cutoff frequency, what resistance would I use? Would I simply use the 2k lower resistor or combine both resistors for a 4k resistor here?

Also what does this circuit do to an input signal? Is everything offset by +10V here, or does the capacitor block this so any signal from V(t) have an average voltage centered around 0?

Answer 1

DC circuit:

\$V_{out}\$ is just the output of a voltage divider, so you have 5V out. The \$1 \mu F\$ capacitor behaves as an open circuit, so you can ignore V(t) because it's unconnected.

AC circuit:

The 10V DC voltage source is shorted and the 1uF capacitor behaves like a "wire" (short-circuit or low impedance). So you have \$V(t)\$ in parallel with a 1k resistor: \$V_{out} = V(t)\$.

Thus, when considering DC+AC:

$$ V_{out} = 5V + V(t) $$

EDIT:

For the expression above to be an approximation good enough, you must check that the frequency of \$V(t)\$ is well above the cut-off frequency of the high pass section formed by the capacitor and the parallel of the 2k resistors, thus:

$$ f \gg \frac {1} {2 \pi \cdot 1000 \cdot 10^{-6}} \approx 159.15 Hz $$

(thanks go to Vladimir for pointing this to me in the comments)

Answer 2

For DC you only have ordinary voltage divider \$Vo = 10V*\frac{2k}{2k+2k} = 5V\$. And now if we connect our capacitor into the circuit the capacitor will charged to 5V after $$T >5RC = 1\mu F * 1k\Omega = 5ms $$

Why \$1k\Omega \:?\$ What equivalent resistance is seen at the terminals of the capacitor?

Interesting things will start to happen if we connect AC signal source. But you must remember one thing that the current in the capacitor is proportional the rate of voltage change across it (proportional to how quickly the voltage across the capacitor is changing). The faster the voltage change (frequency of an AC signal is high) the large the current flow through the capacitor.

So if AC signal frequency is high enough (\$F_{sig} > \frac{1}{2 \pi RC} = 160Hz\$) the capacitor will act like a short in the circuit. The capacitor is sufficiently large value (for the frequency of AC signal) so that it does not have time to charge or discharge. That means that voltage across capacitor will be some average \$ 5V \$. And the situation will look like this if AC signal is a sine wave (Fsing > 160Hz and 1Vpeak).

Now the input AC voltage will change the voltage at the voltage divider output from 4V to 6V in "rhythm" of input AC voltage. Because voltage across capacitor is fixed (5V), because it does not have time to charge or discharge, we can consider that (at least short term) it acts as if it was a battery (whose voltage we fixed (5V) by choosing values of voltage divider).

Answer 3

Yes you combine the resistors but you combine them as parallel components hence, as far as the filter is concerned the value of resistance is 1 kohm. The capacitor blocks the dc voltages from getting onto the input. The average value of Vout is 5 volts.

It may not seem obvious as to why you parallel the resistors but, if you imagine the 10 volt dc source as a short circuit to ac signals it should become clear.