# Finding the input resistance of an op amp?

Discussion in 'Electronic Basics' started by MRW, Nov 29, 2006.

1. ### MRWGuest

Sorry for all the questions. I'm just curious.

In reading, Texas Instrument's Single Supply Op Amp Circuit Collection
guide, I found this attenuator circuit:
http://img360.imageshack.us/img360/8594/opampattenuatorpj3.jpg .

It says that the input resistance should be split into two: RinA and
RinB. So I figure that the input resistance would be RinA + RinB.

In general, however, how do I go about my analyses to find the input
resistance?

I get confused easily when looking at the circuits. An example is this
page: http://www.geocities.com/ferocious_1999/md/micpreamp2.html ... It
says that the two 27k ohm resistors are in parallel. How?

The reason why I want to know about the input resistance is because
according to http://tangentsoft.net/audio/input-cap.html, the input
capacitor value will be dependent on the series resistor and the
specified cut off frequency.

Thanks!

MRW, Nov 29, 2006

2. ### PeteSGuest

MRW wrote:
> Sorry for all the questions. I'm just curious.
>
> In reading, Texas Instrument's Single Supply Op Amp Circuit Collection
> guide, I found this attenuator circuit:
> http://img360.imageshack.us/img360/8594/opampattenuatorpj3.jpg .
>
> It says that the input resistance should be split into two: RinA and
> RinB. So I figure that the input resistance would be RinA + RinB.
>
> In general, however, how do I go about my analyses to find the input
> resistance?
>
> I get confused easily when looking at the circuits. An example is this
> page: http://www.geocities.com/ferocious_1999/md/micpreamp2.html ... It
> says that the two 27k ohm resistors are in parallel. How?
>
> The reason why I want to know about the input resistance is because
> according to http://tangentsoft.net/audio/input-cap.html, the input
> capacitor value will be dependent on the series resistor and the
> specified cut off frequency.
>
> Thanks!
>

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

PeteS, Nov 29, 2006

3. ### PeteSGuest

PeteS wrote:
> MRW wrote:
>> Sorry for all the questions. I'm just curious.
>>
>> In reading, Texas Instrument's Single Supply Op Amp Circuit Collection
>> guide, I found this attenuator circuit:
>> http://img360.imageshack.us/img360/8594/opampattenuatorpj3.jpg .
>>
>> It says that the input resistance should be split into two: RinA and
>> RinB. So I figure that the input resistance would be RinA + RinB.
>>
>> In general, however, how do I go about my analyses to find the input
>> resistance?
>>
>> I get confused easily when looking at the circuits. An example is this
>> page: http://www.geocities.com/ferocious_1999/md/micpreamp2.html ... It
>> says that the two 27k ohm resistors are in parallel. How?
>>
>> The reason why I want to know about the input resistance is because
>> according to http://tangentsoft.net/audio/input-cap.html, the input
>> capacitor value will be dependent on the series resistor and the
>> specified cut off frequency.
>>
>> Thanks!
>>

>
> 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

1/2 x pi x Rin(eff)x C where Rin(eff) is the effective input resistance
above.

Cheers

PeteS

PeteS, Nov 29, 2006
4. ### Andrew HolmeGuest

"MRW" <> wrote in message
news:...
> Sorry for all the questions. I'm just curious.
>
> In reading, Texas Instrument's Single Supply Op Amp Circuit Collection
> guide, I found this attenuator circuit:
> http://img360.imageshack.us/img360/8594/opampattenuatorpj3.jpg .
>
> It says that the input resistance should be split into two: RinA and
> RinB. So I figure that the input resistance would be RinA + RinB.
>
> In general, however, how do I go about my analyses to find the input
> resistance?
>
> I get confused easily when looking at the circuits. An example is this
> page: http://www.geocities.com/ferocious_1999/md/micpreamp2.html ... It
> says that the two 27k ohm resistors are in parallel. How?
>
> The reason why I want to know about the input resistance is because
> according to http://tangentsoft.net/audio/input-cap.html, the input
> capacitor value will be dependent on the series resistor and the
> specified cut off frequency.
>
> Thanks!
>

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)

Andrew Holme, Nov 29, 2006
5. ### MRWGuest

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.

MRW, Nov 29, 2006
6. ### PeteSGuest

MRW wrote:
> 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

PeteS, Nov 29, 2006
7. ### Andrew HolmeGuest

"MRW" <> wrote in message
news:...
> 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.
>

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

Andrew Holme, Nov 29, 2006
8. ### MRWGuest

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

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

Thanks!

MRW, Nov 29, 2006
9. ### PeteSGuest

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

>
>
> Ahhh.. <light bulb on>... 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

PeteS, Nov 29, 2006
10. ### John PopelishGuest

MRW wrote:
> Sorry for all the questions. I'm just curious.
>
> In reading, Texas Instrument's Single Supply Op Amp Circuit Collection
> guide, I found this attenuator circuit:
> http://img360.imageshack.us/img360/8594/opampattenuatorpj3.jpg .
>
> It says that the input resistance should be split into two: RinA and
> RinB. So I figure that the input resistance would be RinA + RinB.
>
> In general, however, how do I go about my analyses to find the input
> resistance?

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.

> I get confused easily when looking at the circuits. An example is this
> page: http://www.geocities.com/ferocious_1999/md/micpreamp2.html ... It
> says that the two 27k ohm resistors are in parallel. How?

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.

> The reason why I want to know about the input resistance is because
> according to http://tangentsoft.net/audio/input-cap.html, the input
> capacitor value will be dependent on the series resistor and the
> specified cut off frequency.
>
> Thanks!
>

John Popelish, Nov 29, 2006
11. ### MRWGuest

Hello again,

In reference to this circuit:
http://img119.imageshack.us/img119/6721/noninvertingsinglesupplyh9.jpg

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!

MRW, Nov 30, 2006
12. ### PeteSGuest

MRW wrote:
> Hello again,
>
> In reference to this circuit:
> http://img119.imageshack.us/img119/6721/noninvertingsinglesupplyh9.jpg
>
> 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

PeteS, Nov 30, 2006
13. ### John PopelishGuest

MRW wrote:
> Hello again,
>
> In reference to this circuit:
> http://img119.imageshack.us/img119/6721/noninvertingsinglesupplyh9.jpg
>
> 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.

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.

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?

John Popelish, Dec 1, 2006
14. ### MRWGuest

John Popelish wrote:
> 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, Dec 1, 2006
15. ### John PopelishGuest

MRW wrote:
(snip)

> ... 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.

That would be the input coupling capacitor.

> Now, the additional resistor at the + input can also be connected in
> series, right?

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.

> 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?

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

> ... 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?

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)

John Popelish, Dec 1, 2006
16. ### PeteSGuest

John Popelish wrote:
> MRW wrote:
> (snip)
>
>> ... 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.

>
> That would be the input coupling capacitor.
>
>> Now, the additional resistor at the + input can also be connected in
>> series, right?

>
> 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.
>
>> 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?

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

>
> 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

PeteS, Dec 1, 2006
17. ### MRWGuest

PeteS wrote:
> 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.

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!

MRW, Dec 1, 2006
18. ### PeteSGuest

MRW wrote:
> PeteS wrote:
>> 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.

>
>
> 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

PeteS, Dec 1, 2006
19. ### MRWGuest

PeteS wrote:
> 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

MRW, Dec 1, 2006
20. ### PeteSGuest

MRW wrote:
> PeteS wrote:
>> 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

>
>
>

Posted to a.b.s.e.

Cheers

PeteS

PeteS, Dec 2, 2006