I'm curious to why in the above analysis and circuit that the input state is an emitter-follower instead of directly going into the non-inverting input of the op amp?
It mentions that the input impedance is large enough to prevent loading of guitar pickups but surely the op amp has a higher input impedence?
Is there any particular reason they did what they did here? Only thing I can see is because of the bypassing but surely using a SPDT switch would work just as fine?
Thanks, Jon
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J
John Popelish
Some of the opamp circuits have fairly low impedance networks, like the tone control stage, though I don't think that one connects directly to the input.
It may just be that the circuit evolved from earlier experiments, and the emitter follower was grandfathered in, without really being needed.
You have to consider the network connected to the non inverting input, as well as the input impedance of the op amp.
M
Michael Black
You don't know what it was like in the old days. Not very good specs on op-amps. You could get something better, but they cost more and were thus less in circulation. You could get "high impedance op-amps", but they were generally merely better compared to what was common.
Hence you'd see all kinds of schemes to get better input impedance. FETs and even bipolar transistors, one to each of the inputs, and only when you really needed it.
Then CMOS and bifet opamps appeared, and worrying about input impedance pretty much went away.
Michael
R
Rich Grise
Where does "dry side of bypass" go? That might need the buffering.
Good Luck! Rich
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Jon Slaughter
yeah, I forgot that. But in my case the op amps have FET inputs so surely this means that the inputs have huge impedences?
I know in a bipolar amp that the configuration and biasing can effect the impedences but in an op amp I thought that the inputs were made to have high impedences no matter what? (no matter the configuration and close loop gain)
Thanks, Jon
J
John Popelish
The opamp inputs can usually be assumed to be high impedance nodes (that is one of the definitions of an opamp). But the effect of the connected network certainly cannot. For instance, in the inverting amplifier configuration, even though the opamp inverting input does not load the circuit, the output connected back to it, does. That feedback connection forces the input node to look like a short circuit to whatever voltage is applied to the non inverting input, a condition referred to as a virtual ground.
But all the configurations on the web page you linked to are non inverting, and these generally have high impedance inputs, except for the impedance of the bias resistors connected to the input. I think these could certainly be designed to eliminate either the input follower or output follower, or to replace them with opamp versions of those followers. I think there should be at least one follower in the circuit for he total bypass case, where the downstream cable capacitance is driven by the follower, to prevent the high frequencies from being rolled off by the loading effect of that capacitance on the high source impedance.
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