While this is simpler it brings up some things I left out on purpose.
On July 25, 2015 at 18:31, alihashemi said...
This can be made much simpler. IDEAL amplifiers swing rail to rail. This means they can take a small signal and take it all the way down to their lowest voltage or all the way up to their highest voltage (these are the +/- voltage supplies that are connected to the amplifier). Now, that's the ideal case. Transistors can never actually output EXACTLY that highest or lowest voltage due to leakage current.
I think it's not leakage current, but minimum voltage drop across the transistor when it's on full. With NPN and PNP transistors, that's in the range of 0.6 volts, germanium 0.3 volts. But we don't use germanium for output devices because they can't pass as much current as the other technologies. I do't know what it is for other kinds of devices.
For example, we can just say this: If we have an amplifier that is connected to a +/-15V supply, it'll be able to output up to +/-14.7V. When your amplified signal tries to go past the 14.7V limit, clipping occurs. Generally, that extra voltage drop is due to something called threshold voltage.
I must go look up threshold voltage, though that description doesn't include any "extra voltage drop...." wait a minute: that's probably the standard voltage drop across a junction.
There's another issue that makes this kind of amp look like it outputs less than rail to rail. If it could swing completely from rail to rail, that's its instantaneous voltage. If the signal is a sine wave, its RMS voltage is 0.707 times its peak voltage. So in this particular perfect world, an amp that outputs 10 volts peak to peak, with a plus and minus 10 volt power supply (20 volts rail to rail), outputs 7.07 volts RMS. I threw in the 20 volt number because it's true that rail to rail is twice the peak voltage of the RMS voltage, and I didn't want someone coming back to say "but wait, there's 20 volts there! How does it become 7.07 at full volume?"
As Ernie said, these amplifiers have fixed gain, so the max value you can input will always be in terms of voltage. Also, most Audio Amplifiers have FET inputs meaning they have infinite input impedance and no current flows in (ideally no current, there is always leakage), therefore this value will never be a maximum in terms of current.
I haven't investigated actual input circuits for years, but I doubt that most inputs are FET inputs. Back in plain old transistor days, some line inputs went directly to an input buffer, some went through switching, volume and tone without any amplification at all! In other words, things varied.
Another high input impedance device is the op amp. The inverting input is a short to ground and the noninverting input is (ideally) an open circuit! Both of those facts dramatically change the details of how the line level signal is handled.
Ernie is right, however, on loading the amplifier with a low impedance. Ideal amplifiers have 0 output impedance. Since you're always loading an amplifier in parallel, whatever load will be in parallel with 0 ohms which can then be approximated as ~0 load. This value is actually closer to ~10 Ohms for good amplifiers and ~100 Ohms for the crappy ones. If you load an amplifier with something that is less than its output impedance, you'll generally get distortion due to higher current draw.
The output impedance of an amp is an impedance in series with the output, not in parallel. An amp with zero output impedance will not decrease in output voltage due to low impedance loads. Those are the kind of amps that blow up or weld when their active outputs are shorted with a signal coming out of them.
Output impedance limits the amount of current that can be delivered, I agree. But that means it can't be in parallel with the load. An output impedance of zero ohms per that concept is a short circuit across the output. Look up buildout resistance for an example of adding resistance in series with an output in order to get a better result.
The conversation past here gets very complicated. Audio amplifiers (talking down at the IC level here) have multiple stages, each designed for a specific purpose. The Class A,B,C, AB, etc. amplifiers we all know of represent the output stage, etc.
Yes.
crosen, maybe you can ask another question if all this has been so much as to obscure the thing you wanted to ask about!
