Voltage Shifting Question

The 1uF capacitor provides what is known as "AC coupling" at the output. The AC component of the signal is faithfully coupled but the DC level is shifted. This is standard practice in (for example) multi-stage amplifiers to isolate the biasing networks between stages. The resistor to ground establishes the average output voltage at 0V (ground). A DC voltage of 8V develops across the capacitor providing the level shift from -12/-4 up to -4/+4. The size of the capacitor must be large enough that it's reactance is small enough at the operating frequency for the phase shift to cause minimal degradation to the wave shape.

A basic high pass RC network always behaves this way. The AC wavform stays the same but is level shifted such that the average voltage is equal to the voltage that the resistor is conected to, in this case 0v. If you were to return the resistor to -8v the wave would go back to -12v to -4v.

Considering the other answers were unrelated to this particular question, I would say additional responses were warranted.

Not a bad idea, although your response would have been more helpful if you had mentioned specific books to look for. I own a couple different electronics books and I would be happy to look at others. It's entirely possible that I simply overlooked something in one of my books which would have answered my question, but I didn't see the answer so it seemed entirely appropriate to pose the question to a 'basics' newsgroup. Thankfully there were a couple other posters who actually bothered to answer the question.

Capacitors are, basically, two plates, that are put close enough so that any charge imbalance can affect the other plate, causing a similar but opposite charge imbalance.

By changing the amount of charge, you can change the voltage across the plates. Capacitance is defined by this:

C = Q/V

That means that if there is Q more colombs of charge on one side of a capacitor than the other, then with a capacitor of size C farads, you get a voltage difference of V volts.

However, there is a wrinkle. In order to change the voltage across the capacitor, you need a source of charge. With your setup, when you change the voltage quickly, the resistor to ground on the other side restricts the flow of charge onto the capacitor. Thus, the other side of the capacitor can't keep adding and subtracting charge as quickly as the opamp side. That means that the voltage across the capacitor cannot change quickly, which means that the voltage on the other side tracks the voltage on the opamp side.

The side opposite of the opamp will eventually equalize with ground if the opamp side is left unchanged, but no, you have to keep changing it don't you?

What this all means is that a capacitor can be used to track changes in voltage, even though there is no DC connection between the two sides.

So, the question is, what is the absolute voltage on the other side of the capacitor from the opamp? Well, the answer is that it is anyplace you want it to be. Only the changes will be reflected.

Thus, with your connection through the resistor to ground, the voltage will start off at ground, and move with the capacitor. So, if the opamp starts at 0V, and is discharged to -12V, the other side of the capacitor will start at 0V, and go down somewhere near -12V. Then, when the opamp goes up to -4V from there, the other side will track it, but the whole time, it's slowly equalizing voltage with ground, moving charge across the resistor. After a few cycles, the average value will be at ground, and it will be oscillating between 4 and -4 volts.