Input impedance of a non-inverting op amp?

[glennb]

I have some doubts in my mind about the input impedance of a common non-inverting op amp circuit which appears in the data sheet for the LM4780 chip amp, and my research has not revealed a definitive answer.

Here's a couple of simple questions for which I am seeking the correct answer, and why:

1. What is the AC input impedance of the circuit below, for frequencies above the f3 point determined in part by Cin ?

2. What value should RB actually be to balance the DC input offset currents, so that the output DC offset is minimised?

Please reply if you know the correct answer. I would prefer not to waste bandwidth with any off-the-top-of-my-head guesses.

[moamps]

Hi,
your question is about non-inv. opamp but your picture shows inv.amp.........wasting bandwidth?

Answers are here http://en.wikipedia.org/wiki/Operational_amplifier

Regards

[Ouroboros]

DC offset is caused by other things as well as by the PD due to input currents!

But to answer your question, 14K for Rb is about right (20k in parallel with 48k). But frankly, unless your input bias currents are high, I wouldn't have though it will make much difference.

But why is this an inverting amp when you call it non-inverting?

[Ouroboros]

Oh, and your input impedance will be about 980 Ohms in the audio range.

[peranders]

Glenn, first I think you should take a close look at the datasheet of LM4780 and also AN-1192 and also application notes for the LM4780.

The 47 k is totally unnecessary. If you want to do it really right the 953 ohms resistors should be 20 kohms if you want to minimize the offset voltage. This resistor had also a purpose of protection during power down and the recommend value was 1 kohms.

The input impedance = Rin => 1 kohms in your case => the input cap is way too small. 10 uF at least should it be. f = 1 /(2*pi*R*C)

EDIT: I have checked the datasheet and I see that you have taken the exact circuit as suggested but something is wrong, according to me. Am I wrong?

[glennb]

Whoops, I actually meant an inverting amp circuit. My apologies.

The wikipedia says Zin = Rin for the inverting amp, as does every other reference I could find. In which case, Cin = 1 uF is indeed way too small for this circuit as audio amplifier:

f3 = 1 / ( 2 . pi . C . Rin || Ri ) = 163 Hz (no bass!)

(where Rin || Ri = 979 ohms)

Also, RB should really be Rf || ( Ri + Rin) = 14118 ohms, rather than the Rf || Ri = 953 ohms given in the circuit.

Does anyone think that NS stuffed up these two aspects of this particular app circuit for the LM4780?

[Ouroboros]

953 Ohms would be correct if you didn't have the input capacitor or course.

[peranders]

Originally posted by glennb
Does anyone think that NS stuffed up these two aspects of this particular app circuit for the LM4780?

I think NS has done a little too much copy and pasting I'm afraid.

[glennb]

Originally posted by Ouroboros
953 Ohms would be correct if you didn't have the input capacitor or course.

Yes, because then Ri would be DC coupled to a low source resistance to ground, which is the general assumption made for calcuating RB = Ri || Rf in a classic inverting opamp circuit. I think NS failed to realise that adding Cin actually changes the DC considerations around the opamp inverting input.

[glennb]

Originally posted by peranders
...The 47 k is totally unnecessary....

Yes, without the 47K there is still a DC charge path for Cin, which is through Ri and Rf.

The resistive load on the input which determines the f3 then becomes Ri alone, ie.

f3 = 1 / ( 2 . pi . Ri . C )

It then follows that RB = Rf, as there is now no DC path through Ri.

My beef with the inverting circuit is that to have a reasonable input impedance (say >20K) and a resonable gain (say 20) requires Ri = 20K and Rf = 400K. In my mind, 400K is too large for a feedback resistor. The overall noise would be higher and the DC offset input currents would make the DC output offset voltage much higher.

Originally posted by peranders
..I have checked the datasheet and I see that you have taken the exact circuit as suggested but something is wrong, according to me. Am I wrong?

I cut/paste the application circuit from the LM4780 PDF, wiped out the power supply connections, and moved the Rf label from below the resistor to above.