Op Amp input impedance

[tom_pay]

Hi Guys,

I have been looking around on the internet for a while on how to calculate the input impedance of a simple op amp circuit. From the sticky op amp page and PDF I have found a formula but I have no idea of how to use it. Would someone please be able to explain it to me?

Thanks so much, in advance.

Tom

PS I have attached a picture of the circuit/formula.

 

[Nigel Goodwin]

You're missing the vital resistor on the input, this provides bias and essentially sets the impedance.

 

[audioguru]

Many modern opamps have FET inputs that have an extremely high impedance because they need no bias current, just tiny leakage currents.

The sketch of the opamp that is missing an important input resistor is from a Texas Instruments tutorial about opamps.

 

[crutschow]

As noted, you need a resistor from the (+) input to Vcc/2 for proper bias. (It's Vcc/2 since the op amp is operating with only one power supply voltage).

And as stated on the schematic, the value for this resistor should be equal to (R1 in parallel with R2) for minimum output offset due to the input bias current.

 

[Ratchit]

tom_pay,

I have been looking around on the internet for a while on how to calculate the input impedance of a simple op amp circuit.

Which op circuit? The noninverting circuit show in the attachment? If so, the impedance is extremely high in accordance with the leakage currents.

From the sticky op amp page and PDF I have found a formula but I have no idea of how to use it.

And no wonder. That is what you use to compensate input bias current offset, not calculate input impedance.

Would someone please be able to explain it to me?

Would you explain exactly what you are trying to find out?

Ratch

 

[Winterstone]

Would someone please be able to explain it to me?
.

Hi Tom,

at first - I agree to all comments related to the missing bias circuitry that is necessary in case of single supply operation.
The opamp input impedance Zin(+) at the non-inv. terminal - degraded by the impedance Zbias of the bias network in parallel - will give the final input impedance Zin of the whole circuit.
More than that, I assume that your question mainly concerns the input impedance Zin(+) of the opamp.
In this regard you should know that the impedance as given in the data sheet (opamp without any feedback) is further enlarged if negative feedback is applied. This is one of the benefits connected with negative voltage feedback. The input impedance is increased by the factor k=1+LG (LG=loop gain). In practice, the resulting value Zin(+) will be very large and, thus, can be neglected against Zbias (as long as no capacitive influences come into the play).

 

[tom_pay]

Hi, Thanks for your help, however I still have a few questions,

As noted, you need a resistor from the (+) input to Vcc/2 for proper bias. (It's Vcc/2 since the op amp is operating with only one power supply voltage).

And as stated on the schematic, the value for this resistor should be equal to (R1 in parallel with R2) for minimum output offset due to the input bias current.

I understand that this resistor needs to be added, but what does it do?

Also does a high input impedance increase the noise in this simple circuit, with the added bias resistor?

Thanks so much for your help,

Tom

 

[audioguru]

When the opamp has a dual-polarity supply, its (+) input connects to 0VDC (half the total supply voltage) so that its output idles at 0VDC andf can swing equally up and down producing AC (audio).

When the opamp has only a positive supply then if its (+) input is at 0VDC then its output can swing only positive producing a rectified output which is horrible for audio.
But if the (+) input is set at half the supply voltage with a voltage divider then its output can swing up and down perfectly producing AC (audio). Then an input coupling capacitor, output coupling capacitor and maybe a capacitor to 0V in its negative feedback are needed.

The resistor or resistors that bias the (+) input at half the total supply voltage do not produce much noise because they are swamped by the much lower impedance of the source signal.

 

[unclejed613]

in the two circuits below, R3 and R6 are the input termination resistors. note that they are 9k (10k||100k). making them 9k balances the input bias currents and minimizes offset, and the input impedance is 9k. R4 and R5 are the voltage divider required to get Vcc/2 for the single supply op amp. C1 is the input coupling cap, and C2 is the output coupling cap (required for single supply operation). C3 makes the Vcc/2 point an AC ground.

 

[tom_pay]

Ahh, I see.

By having this resistor (R3/R6) will there be a greater load on the source? I ask this as the input source is a tuned coil, receiving a very weak signal.

Also is there any reason to put a small capacitor across R2/R8? As I have seen it in many other designs.

Thanks heaps,

Tom

 

[unclejed613]

is this audio or RF? if it's RF, an op amp isn't going to be much use. back up the truck a bit and tell us what you're trying to do....

 

[tom_pay]

No, no its audio, the coil is picking up a 1khz signal and amplifying it to some earphones.

 

[Nigel Goodwin]

Ahh, I see.

By having this resistor (R3/R6) will there be a greater load on the source? I ask this as the input source is a tuned coil, receiving a very weak signal.

That essential resistor sets the input impedance (as I mentioned at the beginning of this thread), you choose it accordingly.

Also is there any reason to put a small capacitor across R2/R8? As I have seen it in many other designs.

It's to provide HF roll off.

 

[tom_pay]

Thanks Nigel

That essential resistor sets the input impedance (as I mentioned at the beginning of this thread), you choose it accordingly.



It's to provide HF roll off.

Is it possible to calculate the roll off point?

Also what happens if the resistor is omitted? Will the op-amp oscillate or create noise? (or smoke!!)

 

[Nigel Goodwin]

Thanks Nigel



Is it possible to calculate the roll off point?

Yes, simply work out the reactance of the capacitor at the frequency in question and apply that value in parallel with the feedback resistor.

There's a basic opamp 'sticky' at the top of this forum.

Also what happens if the resistor is omitted? Will the op-amp oscillate or create noise? (or smoke!!)

It probably won't work at all - as I've said, it's a VITAL component.

 

[tom_pay]

Ok, Thanks Nigel, and everyone else.

I had used the sticky, but the TI PDF was, in places, way over my head!! But, I shall get building!!!

Thanks So much

Tom

 

[tom_pay]

Oops, still more questions, sorry

I understand that there must be a third resistor, to minimize input bias current, one side must connect to the input pin, but does the other end have to go to ground, can it be connected in series with the input? Like in the picture.

Sorry for all the questions, but I am learning!!!

Tom

 

[ericgibbs]

Oops, still more questions, sorry

I understand that there must be a third resistor, to minimize input bias current, one side must connect to the input pin, but does the other end have to go to ground, can it be connected in series with the input? Like in the picture.

Sorry for all the questions, but I am learning!!!

Tom

hi Tom,
Attached pdf, keep it near your pillow for a little bedtime reading, a useful reference document.

 

[Nigel Goodwin]

Oops, still more questions, sorry

I understand that there must be a third resistor, to minimize input bias current, one side must connect to the input pin, but does the other end have to go to ground, can it be connected in series with the input? Like in the picture.

No, it's doing nothing there - like I said, look at the basic opamp sticky at the top of this forum - it shows the basic minimum you need.

 

[Winterstone]

Oops, still more questions, sorry
I understand that there must be a third resistor, to minimize input bias current, one side must connect to the input pin, but does the other end have to go to ground, can it be connected in series with the input? Like in the picture.
Sorry for all the questions, but I am learning!!!
Tom

Hi Tom, I am afraid, you are mixing something (bias voltage and bias current).
Let me try to explain the situation with words (example: non-inverting operation):
* For single supply with Vcc you need an operating point (output dc voltage) approx. at Vcc/2. Only then the output can swing up and down.
* This can be accomplished by biasing the non-inv. input with Vcc/2 (using a simple voltage divider or any other method) .
* This dc voltage must be transferred to the output. Therefore, the dc gain must be unity. This is accomplished with 100% negative feedback (to be realized for dc only !!).
* For larger ac gain values (above a certain cut-off frequency) there is a resistor R2 from the output to the inv. terminal and a series combination R1-C to ground.
* Then, the dc gain is unity (because of C) and the ac gain is (1+R2/R1)
* Instead of using this capacitor you can use the scheme as shown in post#9 (right side). In this case, the ac gain is (slightly) determined by the divider resistors also.
* Because the pos. input terminal is biased with a dc voltage the ac input signal must be applied using an additional coupling capacitor.
____________
I hope these explanations can help you to design and to understand the circuit.