Buffer Question

[seten]

Is there any practical difference between the first two op amp buffer schematics other than one extra part? Also, I know the second one has an input impedance of 500k because youre taking the two resistors forming the voltage divider in parallel but I cant figure out why R2 affects the input impedance at all.

http://www.muzique.com/lab/buffers.htm

[FiveseveN]

The difference only depends on how you obtain that "Vr": if it's only a resistive divider then there's no practical difference. But it's usually bypassed with a large cap which shunts noise and ripple that your bias resistor(s) would otherwise inject into the signal. And if that's not enough you could buffer the bias voltage itself.

I cant figure out why R2 affects the input impedance at all.

As far as the AC signal is concerned, any DC looks like "ground". So in terms of input impedance, R1 and R2 are in parallel, giving us 500 K.

[ElectricDruid]

The only practical difference is that you'd use (1) when you've already got a Vr point in your circuit, and you'd use (2) when you haven't. Aside from that, there's nothing to choose between them.

[antonis]

IMHO, (with all respect to Jack..), first op-am buffer uses an extra 1M "isolator" resistor for keeping initial voltage divider resistor values at low level (10k each) where second buffer connects directly Vr to non-inverting input + signal path, saving one resistor and one cap but risking Vr ripple & noise issues..

[seten]

IMHO, (with all respect to Jack..), first op-am buffer uses an extra 1M "isolator" resistor for keeping initial voltage divider resistor values at low level (10k each) where second buffer connects directly Vr to non-inverting input + signal path, saving one resistor and one cap but risking Vr ripple & noise issues..

I think this was the root of my question. So how does that 1M resistor allow you to keep the initial voltage dividers low (10k each)? And theres no voltage drop across it because theres no current flowing, its 4.5v on either side of the 1M resistor?

[FiveseveN]

theres no current flowing, its 4.5v on either side of the 1M resistor?

There's hardly any current flowing because the op amp input is very high impedance.

how does that 1M resistor allow you to keep the initial voltage dividers low (10k each)?

Think of the reference voltage "generator" as a separate, 'black box', a 4.5 V battery if you will.
You want to set the op amp's input to this potential (see 'biasing'), so you connect it with a resistor to this 4.5 V. The value of this resistor then sets your input impedance: if it were a straight short (0 Ω), all your signal would be shunted to 4.5 V. If it's less than infinite, some of your signal gets shunted. So we pick a value where that 'some' part is negligible. The value can be high (i.e. 1 MΩ) because again the op amp's input impedance is very high.
Alternatively you can say the op amps need to draw very little current to set their bias (see 'input bias current'), less than what can flow through that 1 MΩ resistor. The simple BJT buffers need more current, so the biasing resistor(s) can't be as high.

Then comes the separate issue of what exactly is in that '4.5 V black box', which I touched on in my initial reply.

[antonis]

theres no voltage drop across it because theres no current flowing, its 4.5v on either side of the 1M resistor?

Absolutelly true & correct..
(for "ideal" op-amp..)

For real world op-amps, the above is almost true for Fet inputs and fairly close to truth for Bipolar inputs..
(of course, you can always count for bias resistor voltage drop by off-setting a bit Vref, but this is a perfectionist point of view - and prerfectionists don't stop on this particular level only..)

[Gus]

seten

have you searched "buffer" or "buffer design" at the forum? you should find some good threads

[seten]

Finally came back to this at a place where I'm ready to understand. Yall we're super helpful - one thing is bothering me though on what FiveseveN said. How come in the second schematic the input doesn't get shunted to 4.5V? Isn't that voltage divider essentially a "black box" of 4.5v that we short directly to the input?

[GibsonGM]

Some of the input DOES get shunted to ground...an ultra-tiny almost immeasurable voltage, thru that 1M resistor.

Tube stages often use the same 1M to ground, with the same un-noticeable result.   The signal is riding above one MEGOHM...you're not gonna hear it.

This is how the 1 MEG resistor isolates those 10k resistors in the bias network mentioned above - without it, you will both short to AC gnd via the filter cap in there, and if the cap wasn't there, you'd shunt thru the 10k resistors, which ARE low enough to have a noticeable effect...

[seten]

Hmmm I dont think that answers my question - Fiveseven had said that in the first schematic if we had connected vr directly to the input then our input signal would be entirely shunted to 4.5v - in the second schem if I just redraw it so the voltage divider is separate and thats our "Vr" (minus the filter cap), well then its clear now that we're connecting the input directly to a black box providing 4.5v - just like in the first schem - so why wont the input signal be shunted to 4.5v? I'm guessing its because the resistors used in the second case are 1M and in the first case they were 10k? But in both cases theyre directly connected to a node being held constant at 4.5v.

I noticed you explained why its not shunted to ground (which I understand - 1M is big enough but 10k wouldnt have been) but my question was about shunting to 4.5v - is "shunting to 4.5v" not a thing? If so what did fiveseven mean by that?

[GibsonGM]

If I understand you correctly, yes - there is a 1M resistor between the source and the input in EACH case.    AC does not care if you are talking about "+" or " - ".  Each is a potential pathway to 'shunt' the signal.   It cannot 'shunt' either way with the two 1M resistors in the divider, of course. 

I noticed you explained why its not shunted to ground (which I understand - 1M is big enough but 10k wouldnt have been) but my question was about shunting to 4.5v - is "shunting to 4.5v" not a thing? If so what did fiveseven mean by that?

In both cases there is a high value resistor between the 4.5V and the signal.   In the 'Vr' case, a 1M isolates the signal from the 4.5v.  The 4.5 volts at "Vr" is created with the two 10k divider, just a slightly more elegant (and quiet) way of doing this.     In the voltage divider case, both 1M resistors do the same, as they directly provide the 4.5V AND isolation, but the 4.5v bias there is not at 'clean' (and that may not matter)...hope that helps a bit.

[seten]

Okay, so if in the Vr case the voltage divider used two 1M resistors then it would be okay to connect it directly to the input?

So its not the Vr point we're worried about shunting to if we connect it direcrtly to the input, its the 9v and ground points - and the fact that there are only 10k resistors between them and the vr point?

[FiveseveN]

So its not the Vr point we're worried about shunting to if we connect it direcrtly to the input, its the 9v and ground points

AC is the wiggly stuff that we need to buffer, DC is a flat line, yes? Any DC voltage between ground and 9 V and beyond is "ground" or "silence" for the AC, no wiggles. When you want to mix them, what comes out depends on the source impedance: put a 9V battery in parallel with a dynamic mic and you'll hardly notice anything other than 9V. Stick that battery in a mains socket and (after the smoke clears) you'll never know a battery was "shorting" the mains AC.

The first version has a low impedance Vr source (because of the capacitor, not just the smaller resistor values), which we "isolate" through R1, i.e. we make the impedance high enough that it doesn't shunt/dominate/drown out the signal.
In the second version the source already has a high impedance. Handy but has its drawbacks.

[antonis]

its clear now that we're connecting the input directly to a black box providing 4.5v

There are quite a few versions of that "black box"..
1. Simple resistor voltage divider..
2. The above with smoothing/reservoir cap on their junction..
3. A simple Zerner diode..
4. All the above followed by buffer op-amp..

2, 3 & 4 are "no-no" for for direct signal coupling due to their low impedance..

1 might (or might not) be suitable for direct signal coupling.. It only depends on both divider resistors values 'cause they are effectively set in parallel..
(lower resistor goes directly to GND while upper one goes to AC GND either via supply filter capacitor or via voltage source low internal resistance..)

P.S.
The later can also be considered as Thevenin equivalent where Vcc with divider resistors are replaced with Vref in series with resistors parallel combination.. 

[GibsonGM]

Okay, so if in the Vr case the voltage divider used two 1M resistors then it would be okay to connect it directly to the input?

So its not the Vr point we're worried about shunting to if we connect it direcrtly to the input, its the 9v and ground points - and the fact that there are only 10k resistors between them and the vr point?

All of the 2 replies above, and for clarity - remember, at the bias network there is a cap.  Fiveseven mentions it but I'll go further, as I did before:   the 10u cap is a low impedance path directly to ground for AC.   You may not be considering it because it's only shown in the first image on the page.  This is the 'black box'.   A capacitor is effectively an open for DC and blocks it, but AC sure loves it (at varying levels of reactance; worry about that after you understand this   )



If you connect a path from your signal to this cap (where the 4.5V is located!), it will go thru that cap and 'shunt to ground'.  A 1 Meg series resistor prevents this from happening, and as Antonis called it, it is an 'isolation resistor'.     

[antonis]

as Antonis called it, it is an 'isolation resistor'.

Tried by best to be politically correct..

cheers Mike..

[GibsonGM]

as Antonis called it, it is an 'isolation resistor'.

Tried by best to be politically correct..

cheers Mike..

I'm not sure of the 'proper' name for that resistor, either, but 'isolation resistor' sounds very appropriate.  I do not think many will take offense; of course, one never knows...

[seten]

Ahhhh that was all super helpful thanks so much yall. Let me see if I get what fiveseven is saying about combining AC and DC signals - if you put a 9v in parallel with a dynamic mic you wont notice the mic bc the impedance is so much higher than the low internal impedance of the battery (and not because 9v is so much higher than the mic output in mV? thats what i wouldve thought.) but then if you put the battery across the mains, it might as well have been a short because the battery impedance, while low, is still much higher than the mains impedance (again, not because the 9v DC is so much lower than the 120v AC)

Also would like to share this thread I found while double checking that mains impedance was indeed very low. Glad I'm here and not there.

https://www.electronicspoint.com/forums/threads/ac-mains-input-impedance.110409/

[GibsonGM]

Neat, Seten, and glad you understand things a little better now.  Mains? I thought we were talking about buffers, ha ha?!   

I have never come across anything where I needed to know that, although it IS interesting.