"true gain", or, what the resistance of potentiometers for output do.

[Thecomedian]

I've played around with enough circuits now, that a pattern has emerged, whereby current is converted into voltage by the resistance of the output potentiometer.

If you have an absurdly high resistance for the output pot, you will have a very large voltage gain, in appearance, but this is at a cost of current.

Fig 1. Example circuit. 1K potentiometer simulated for the output. Voltage gain of 60/20, or, 3.



Fig 2. Current of ~=60uA, for both input and output.



Fig 3. 100K pot sim. voltage gain of 170/20.



Fig 4. Input current @ 60uA, Output @ ~=1.7uA




The conclusion is that you are technically getting a high voltage gain in the latter scenario, but you're losing current to do so. Considering the voltage and current together would be the "power gain" of the circuit. How does this loss of current affect the next stages? Is the "true gain" based on simply getting the voltage as high as possible, or does loss of current also play a role, as the signal is passed to the next amplifier circuit?

[tubegeek]

It depends on the current requirements of the next input, and its input impedance.

If the next device has a LOW input impedance, it will draw current that a HIGH output impedance cannot supply.

For this reason, the convention is to use LOW output impedance and HIGH input impedance. If the ratio of input impedance to output impedance is 10:1 or higher, the connection is said to be "bridged." Then more than 90% of the output voltage developed will be present at the input of the next device. (This is equivalent to saying that less than 1 dB of loss will occur due to the interconnection.)

If for some reason (like, you have a gain pot on the output which doesn't have a low-output-impedance buffer on it, as your sims are ) you will indeed lose signal in the handoff, because the current required to develop a voltage across the input is not available. You have now seen EXACTLY why outputs often have buffers, or "drivers," on them, and why pots rarely show up at the end of a circuit.

PS: a different aspect of the exact same concept here: http://www.diystompboxes.com/smfforum/index.php?topic=103634.0

[Thecomedian]

hum. Would a current mirror with the 60uA input be able to apply a 60uA current after output cap while the pot provides the 170mV? Most DIY pedal schems (and even things like the AMZ treble/bass boosters) tend to only have a pot at output.

is always putting a buffer on the end of a circuit before output "the best choice", or is it YMMV, depending on what you're intending to drive, etc?

[GibsonGM]

That was great, comedian!  Nice and different way of looking at the output issue.
Perhaps a unity-gain buffer after the vol. pot would be the answer?  That should retain the 1:1 relationship between E and I at the output?   IOW, we will have as low an output Z as we can get. 

[tubegeek]

Sadly, I do not understand current mirrors (yet?) so I don't know.

I'm more familiar with the kinds of things GibsonGM suggested, but that's not to say your idea is not as good. I just dunno.

Reading with interest... I'll try and look in Small Signal Audio Design (D. Self) or Horowitz & Hill for info on current mirrors and stop back in. But today will be kind of busy, so it might not be today.

[R.G.]

I've played around with enough circuits now, that a pattern has emerged, whereby current is converted into voltage by the resistance of the output potentiometer.
...
If you have an absurdly high resistance for the output pot, you will have a very large voltage gain, in appearance, but this is at a cost of current.
...
The conclusion is that you are technically getting a high voltage gain in the latter scenario, but you're losing current to do so. Considering the voltage and current together would be the "power gain" of the circuit. How does this loss of current affect the next stages? Is the "true gain" based on simply getting the voltage as high as possible, or does loss of current also play a role, as the signal is passed to the next amplifier circuit?

You're ready to snatch the pebble, grasshopper. Or at least to snatch one pebble.

You observation is mostly correct. In bipolar transistors, or any device where the output is best thought of as a current source, the size of the output *voltage* depends on the load resistance the output sees.

The collector of a bipolar transistor acts very much like a controlled current source. That is, it sucks in a current from whatever is connected to it (given that the DC conditions are set up so that this is possible); if that source is simply a voltage source, it just takes in current and passes it to the emitter, as in an emitter follower. If that source is a voltage with a resistor in series, it still sucks in the same current, but the resistor converts that current flow to a voltage.

So for a bipolar transistor, simply changing the collector resistance (carefully, including tinkering with the DC conditions to not upset the bias conditions) can change gain.

The pebble here is that electrically, the output potentiometer is in parallel with the collector resistor. So the transistor's collector really sees a load of the DC-connected collector resistor that sets its bias conditions, but also the AC-only connected output potentiometer. These resistances combine in parallel, and that is the effective resistance that determines AC gain as far as the collector resistor takes part in it.

It is worth noting that this does not stop with the output potentiometer. Whatever is connected to the output pot loads the collector, too. This may be a very large resistance, if it's the input of a tube amp with a 1M grid resistor, or it can be the input of a fuzz face clone, with a very small resistance. When the pot is turned all the way up, the input of the next stage appears in parallel with the output pot and the DC collector resistor.

If you're considering AC voltage gain, you must *always* think about the effect of whatever stage is connected to the output. It can affect the gain.

As a sub-pebble, look up "Norton equivalent circuits". That is the most useful model of what's going on to me.