Finding the input resistance of an op amp?

The first thing to understand is that to a signal, all power rails are ground. In addition, the inverting input of the opamp is a virtual ground (if you don't understand that, then google it) so the input resistance (to an ac signal) is

RinA + (RinB || R3) where || means 'in parallel with'

Cheers

PeteS

Continuing from that, the cutoff frequency (-3dB) is

Cheers

PeteS

You need to consider the AC (signal) and DC (power / bias) conditions separately.

Power supply rails are ground as far as AC is concerned - so the 27k resistors are connected in parallel between the input and AC ground.

In the other case, the -ve op-amp input is held at Vcc/2 by the action of feedback. This is known as a "virtual earth" and is another AC ground. So the input impedance is RinA + (RinB || R3)

Thanks!

I am familiar with simple op amp analysis. The concept of the power rails being viewed as "ground" by AC signals is new to me. I thought that author of the site got the 27k ohm resistors to be in parallel from a Thevenin analysis, but at the same time, I get lost sometimes with Thevenin.

A power rail (voltage) is designed to have a constant voltage. So for any Delta I (change in current), there should be zero change in voltage, so we get:

R = V/I = 0/x =0. In addition, power rails are decoupled to ground through capacitors, which have (simplistically) zero impedance at AC.

Cheers

PeteS

The 50uF smoothing capacitor across the power supply is an AC short circuit. Good decoupling makes the power supply an AC ground.

Ahhh.. ... That is right. Sorry, I haven't quite racked up on experience years to be this insightful.

Thanks!

That's the purpose of s.e.b :)

Cheers

PeteS

Think of all fixed voltages (ground, Vcc and Vcc/2) as zero ohm points as far as signal current is concerned. An inverting opamp also uses feedback to hold its - input at the same voltage as is applied to its + input, so that node can be considered to be a zero ohm to ground node as far as signals are concerned.

As to the best values for all those resistors, it depends on what attenuation or gain you want, and whether or not the output can absorb all the input current through the feedback resistor, or if you want some of the input current to pass through R3 to Vcc/2. If you are not concerned with keeping the input impedance low input impedance to some low value with R3, R3 is not needed. So the best answer depends on all the design constraints. lots of combinations will work under some conditions.

Each of them loads AC signals with a path to some fixed voltage. From an AC analysis point of view, all fixed voltages are ground.

Hello again,

In reference to this circuit:

How did the input impedance come to equal R1 in parallel to R2? Just by inspection, I was assuming that the input impedance would be the input resistance of the op amp since the incoming signal is just "seeing" the

• input.

Thanks!

The input impedance as shown is the input impedance of the amp, which will be very high.

The notes say 'Input impedance = R1 || R2 for minimum error due to input bias current'

It really means the source impedance of the signal (as seen 'looking out' of the non-inverting input) should be R1 || R2, perhaps by means of a resistor, divider or the output impedance of the driver. This would make the source impedance as seen by both inputs equal.

Input bias current offsets (which is what you would have here) translate to input errors, which show up at the output multiplied by the gain.

For this reason, it's common to see voltage followers with a resistor in the feedback path, although most texts show it as a direct connection.

Cheers

PeteS

The circuit is incomplete. You need an additional resistor from the + input to ground, to provide a path for the opamp bias current (very small) and to set the bias voltage for the + input. The note is suggesting that the value of that resistor should be the parallel combination of R1 and R2, if you want minimum DC error from the opamp bias currents, since this value would produce equal and canceling voltage drops across the input resistor networks on both the + and - inputs, if you assume that each input has a similar bias current.

If you follow this advice on selecting the + input bias resistor, then its resistance constitutes essentially the entire input impedance (and would be the value specified for that resistor.

If DC accuracy is not real important to you (for instance, if the output if the opamp is capacitor coupled to its load) then the value of the bias resistor on the + input is not nearly so tricky. Then you can select the input bias resistor to produce whatever input impedance you want.

Is this all clear?

I suppose so. From PeteS and your post, I think I have a good idea. I actually wanted to use this circuit to test another circuit that I found. It's a basic mic pre-amp. I wanted to boost the output of it. So I was trying to figure out from the input resistance as to what value I should design the input bypass capacitor to roll off at 16 Hz.

Now, the additional resistor at the + input can also be connected in series, right? If that's the case, then would the series resistor introduce more noise voltage or would it be the same as a shunt resistor at the + input?

... But then again, I did read on another data sheet that adding a series resistor at the input would limit the input current to the op amp so it will be able to tolerate input voltages greater than the VCC supply. Is this true for most bipolar opamp?

Thanks!

MRW wrote: (snip)

That would be the input coupling capacitor.

No. It connects as a shunt to ground. It is effectively in series with the capacitor, as far as signal current is concerned, but it is in parallel with the + input.

Lets not worry about noise till we get straight what you need to hook up.

That is a different function, entirely. and a separate resistor, also.

You want an RC product (for the input coupling capacitor and this bias resistor) of 1/(2*pi*16) to have a 3 db roll off at 16 Hz. For instance, if the coupling capacitor was .1 uF, the resistor to ground that would start to roll the input signal voltage off by 3 db at 16 Hz would be 100k. This is because 100k*.1*10^-6F = 0.01 seconds which also equals 1/(2*pi*16)

MRW: Can you access alt.binaries.schematics.electronic?

If so, I might throw a circuit diagram together to show you what John is suggesting and PDF it, then post it.

Cheers

PeteS

Hi PeteS!

It took me sometime to figure out, but I think I have access to it. I was able to see someone's bench setup in one of the posts.

Thanks!

I'll do it tomorrow; I have had sufficient wine to make use of a capture tool 'not the right thing' right now :)

Cheers

PeteS

Enjoy your drink! :-)

Posted to a.b.s.e.

Cheers

PeteS