Technical Question: buffering

[burningman]

Can anyone confirm my following assumptions?
Using Ohm's law: The signal source sees a high impedance at the input of the buffer. The 'higher' impedance at the input, increases the voltage. The increase in voltage, in tandem with the low impedance output of the op amp, increases the current flowing through the open-loop feedback network to the output. The increased current permits more drive capability.

[MullisMan]

Yes that's pretty much it.  The voltage between the input terminals of an "ideal" opamp is always zero, thus the huge input impedances we see.  If we look at ohm's law with the water hose analogy, then yes, the voltage does try to force more and more current through the opamp and we get the huge closed loop gains from the feedback loop.  If only we lived in an ideal world.

[burningman]

sorry I meant closed-loop gain, voltage-follower - not open-loop

[R.G.]

Can anyone confirm my following assumptions?
Using Ohm's law: The signal source sees a high impedance at the input of the buffer. The 'higher' impedance at the input, increases the voltage.

Not exactly. The signal source sees a high to very high impedance (and this is the critical part!) and is not loaded down by the buffer input.. This lets the signal source, in the parlance of a US Army recuiting poster, be all it can be. It's like the signal source is not connected to anything but an open circuit.

In fact, that's the best definition of a perfect buffer - it drives other stuff, but makes the signal source think it's not connected to anything except several inches of air or other insulator.

The buffer does NOT increase the voltage at the signal source over what it would otherwise be, except that it does not *decrease* it. That's an important distinction. The buffer does not make the signal source MORE. It does not make it less. There is a difference.

The increase in voltage, in tandem with the low impedance output of the op amp, increases the current flowing through the open-loop feedback network to the output. The increased current permits more drive capability.

Again, not exactly.

The NON-LOADING of the signal source does not decrease the signal source voltage. The buffer - if it is an opamp, it does not have to be an opamp - has a low output impedance and can drive more output current. However, this has nothing much to do with the current flowing in the feedback network. The current in the feedback network can be nothing more than the output voltage divided by the input impedance of the inverting input, which can be huge, and therefore the current in the feedback network can be so nearly zero that it's hard to measure.

Imagine: an opamp buffer made from JFET input opamps, like the TL072. The input resistance at the inputs of the TL072 is (from the datasheet) is ten to the *twelfth*power ohms. That's a million megohms. If you connect the output to the inverting input with a wire - a perfectly valid buffer arrangement - the current in the feedback network is the output voltage divided by the input resistance. There is no other place for the current to go but into the inverting input other than off into the air. So the current that "flows" in the feedback network is (for a one-volt output, for example) one volt divided by ten to the twelfth power, or a millionth of a micro amp.

(to the folks who read the datasheets and correct these little explanations: yes, I know that this is oversimplified and that bias currents and high frequency effects change this; let the guy understand the basics before he has to chase the details~)

So no, the current in the feedback network does not necessarily have anything to do with buffer operation. If you're using an opamp, and If you have a gain resistance/network connected to the inverting input as well as the feedback network, yes, some current may flow through the feedback network, but in a properly designed voltage-mode buffer, this current is inconsequential compared to the output current capability of the opamp.

[burningman]

Thank you very much for clarifying this R.G.
It seems like a subject that is often glossed over. It's nice to have it broken down like that.