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Equivalent Circuit?

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crutschow

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Well it does confuse the issue (for me) :)
I admit the "virtual ground" at the op-amp input is a mystery to me.
I know it is related to the negative feedback etc. and (for me) a lot of math but what I have trouble with is even if we are feeding a bjt common emitter stage, if the cap is large enough,
whatever is on the input side, impedance wize, it's the output side of the cap that will determine if the circuits will behave differently.
Do you follow me? Am I wrong?
Not sure by what you mean by impedance wise.

But to try to clarify, a virtual ground point means that the apparent impedance at that point to ground is essentially zero ohms. Thus you really need to look at the input and output currents to determine the circuit behavior. For example, if you have a 1kΩ input resistor going to a virtual ground input on the op amp, then a +1V input will give +1ma through the resistor to the virtual ground. The op amp absorbs this current by the negative feedback from the output to the virtual ground. Thus, in this example, if you had a 10k ohm resistor from the output to the virtual ground, then the op amp output will be at -10V (inverting gain of 10) to sink the 1ma of current from the virtual ground and keep the virtual ground at zero volts.

Of course, if you add capacitors or other impedances into the circuit, then you have to account for their effect at the frequency of interest.

Does that make any more sense?
 
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MrAl

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Hi again,


To help understand the virtual ground thing, it might help to know
that there is a virtual "short circuit" between the non inverting terminal (+)
and the inverting terminal (-).

The reason this happens is not because there is really a short there,
but because the op amp is made such that it automatically tries its hardest
to keep the voltage between the (+) and (-) terminals exactly the same
ie 0v between the two (when it is connected in a circuit correctly)
and any two terminals that have 0v across them appear almost like they
are shorted together. This is one of the main features about an op amp like this.

The op amp does this by sensing the difference in voltage between (+) and (-)
inputs and when it sees a change, it forces its output to ramp in the direction
that will remedy the situation (ie make 0v between the inputs again).
It's this way that the voltage difference stays close to 0v, but because of
internal limitations (the op amp isnt perfect) there is always some small
voltage across the two but it can many times be ignored.

In your circuit since one of the terminals of the op amp is connected to ground, the other
terminal is sometimes said to be "at virtual ground" and this is just a way to
describe what really takes place inside and outside of the op amp.

All this seems to make things harder to analyze, but really it makes things a lot
easier because we can ground the (-) terminal in our minds and then we only
have to think about the currents flowing or the voltages from the other nodes
to ground instead of a more complex node that the (-) terminal would be otherwise.
Once you get used to thinking like this you will be happy to see how much easier
it makes everything about op amps.
 
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MrAl

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Hi again,


If we start to do the actual network analysis we find that we end
up with a voltage source:

VN*RNb/(RNa+RNb)

where

VN is any input voltage and
RNa is the 'top' resistor of the pot and
RNb is the 'bottom' resistor of the pot

and an impedance:

RNa||RNb+10k+1/sCN

for each input channel in the three capacitor circuit, and
for the single capacitor network we end up wtih the same
only the impedance is

RNa||RNb+10k+(1/sCN)||RNN

where

RNN is a bunch of resistors from the other two inputs lumped,

but then since we also have the requirement that C is large enough
to swamp the effects of frequency that means we can take the
limit as C --> +infinity, so guess what? We end up with the same
network:

RNa||RNb+10k

For both circuits!
(and of course the same equivalent input voltage).

That makes both circuits the same, given C is large.


But hey, you know what? Why dont we just let C approach
infinity to start with, without doing anything else, then look
at the two networks with these large C's in place...

With all C's approaching infinity, they become short circuits.
With all C's shorted out, both networks are:

EXACTLY THE SAME :)

The best way to see any differences when the C's are not large
(that is frequency affects operation) is to do a network analysis
of both circuits and then compare.
The simplest case for the three cap ciruit is to let all pots be set at
maximum, so that we end up with three networks of R+1/sC
in parallel, which is electrically equivalent to three caps in parallel
in series with three resistors in parallel (when they are all the same value).
This tells us that the total C from the three cap circuit is three times the
value of that of the one cap circuit, which means the frequency response
will be different.
 
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