biasing question about non-inverting opamps

[Derringer]

ok ... several questions here


First, here's what (I think) I know.
assume a voltage divider creating 1/2 V+ called Vb which serves as a virtual ground
You have R1 which connects the "+" input to Vb and sets the input Z (unless your input signal is riding on Vb already)
You have R2 which connects the op-amp output to the "-" input
R2 "sets" the amount of negative feedback to attenuate the input signal in order to achieve the desired output gain.
then you have R3 which connects the "-" input to Vb.

Gain is R2/R3+1


Questions:

1. What is the purpose of R3 and how/why does it factor into the gain equation? How low can R3 go?
   

2. (and maybe this one will be answered with the response to question 1)
    Why do I often see R3 connected in series to a large cap in series to ground?

3. How does that cap to ground (and its size) affect gain characteristics?



Thanks!

Bill

[R.G.]

assume a voltage divider creating 1/2 V+ called Vb which serves as a virtual ground
You have R1 which connects the "+" input to Vb and sets the input Z (unless your input signal is riding on Vb already)
You have R2 which connects the op-amp output to the "-" input
R2 "sets" the amount of negative feedback to attenuate the input signal in order to achieve the desired output gain.
then you have R3 which connects the "-" input to Vb.
Gain is R2/R3+1

That's not how I'd say that, exactly. It's more like R2 sets the amount of voltage the output has to swing to get the output to pull the input current off the inverting input to keep it in balance. R3 sets the amount of current which goes into the inverting input for a given input voltage.

1. What is the purpose of R3 and how/why does it factor into the gain equation?

R3 sets the amount of current which goes into the inverting input for a given input voltage. The inverting input is always driven to the same voltage as the + input as long as the amplifier power supply and gain can do this. The inverting input is viewed as a "virtual ground" as the signal voltage there is the output signal divided by approximately the open loop gain in most cases.

R3 factors into the gain equation by forcing the input voltage on the outboard end of R3 to be a current into/out of the inverting input pin. We'll see this in your next question:

How low can R3 go?

It can be zero. The input to the inverting input can be a current supplied by a current source, like perhaps a photodiode. In this case, the current is what it is. In that case, the output voltage is forced by the amplifier's gain to be whatever is necessary to pull that same amount of current OFF the inverting input through R2. Another way of looking at the same thing is to say that R3 can be zero, plus whatever output impedance the signal source has.
   

2. (and maybe this one will be answered with the response to question 1)
    Why do I often see R3 connected in series to a large cap in series to ground?

It makes the circuit have a DC gain of unity. A capacitor has an infinite DC resistance, in theory. So at DC, the capacitor is an open circuit and the gain reduces to just being unity - it's a voltage follower, following the + input. The amplifier output deposits enough charge/voltage on that cap to bring its DC level to equal the + input.

3. How does that cap to ground (and its size) affect gain characteristics?

A capacitor is a frequency-variable impedance. R3 and the cap in series (let's call it C3) form a frequency-variable impedance of value R3 + j*2*pi*f*C3 where f is the frequency. Another way of looking at it is that above the frequency where the capacitor impedance X = 2*pi*f*C equals R3, the capacitor looks more and more like a short  circuit. So at frequencies above f = 1/(2*pi*R3*C3), the AC gain of the circuit becomes just 1+(R2/R3). At frequencies below that, it becomes just unity, as noted in the equation in 2 above.

So the capacitor size in conjunction with the value of R3 sets the frequency where the voltage gain drops from G =1+(R2/R3) towards G = 1 by making the R2/R3 term go to zero by R3 - which should really be (R3 + j*2*pi*f*C3) - tend toward infinity as C3's impedance comes up.

That is - it sets the frequency where the AC gain rolls off from its mid-band gain toward that DC unity gain follower.

[Derringer]

Thank you man.

[amptramp]

It may confuse the issue to say that R3 can be zero.  This would make the stage gain infinite.  A phtodiode as a current source has a very high impedance and the virtual ground would tend to linearize its performance. (I used to design amplifiers for military IR surveillance and targeting systems used on ships.)  If the voltage across a photodiode is held at zero, the current is directly proportional to the light input.  This is the classic transimpedance amplifier and there is no need for a separate resistor corresponding to R3, but the resistance of a currrent source is infinite.

The way I visualize this in my own mind is that no current flows into or out of the amplifier input, so R2 and R3 form a see-saw where the voltage across each resistor is proportional to the resistance itself, because the same current is flowing through both of them.  It follows similar equations to leverage.

R2 can be zero, which gives you a gain of 1, the classic unity gain buffer.

[Gurner]

   

2. (and maybe this one will be answered with the response to question 1)
   Why do I often see R3 connected in series to a large cap in series to ground?

I had a 'bada bing' moment when I was trying to get to grips with all this some time ago....so I'll lob it in here fwiw....

Your opamp output pin is normally sitting at a DC levelof  about half the supply - so let's say 4.5V.

Ok, so we now need to connect t R2 & R3  .,....specifically the 'other' end of R3 potential divider chain to a low impedance point. Ground is about as low impedance as you can get, but it's sitting at 0V, so if you use it then you need a cap to decouple the DC.

*But* if you happened to have 4.5V handy that's sufficently low impedance (eg take two equal value resistors into a +ve pin of an opamp with 100% negative feedback ....the output from that opamp arrangement is sufficiently low impedance), you could readily connect the other end of R3 to that point and then you wouldn't need the capacitor.

Also, you wouldn't need the capacitor if you used a dual supply circuit, because then the DC level of your opamp would be 0V ...therefore the lower end of R3 could readily connect to ground without a decoupling capacitor.