Unequivalent Input Impedance in Differential OpAmp Circuit

[Rick C]

I have an existing design that uses a single op amp stage with differential inputs and AC coupling caps on the inputs. The circuit acts as a 5:1 divider so the two input resistors are 51 kohms and the feedback resistor as well as the ground resistor in the non-inverting input leg are 10 kohms. The input caps are sized to maintain a good low frequency response.

Many years ago I recall going through the equations for such an amplifier design to understand why the voltage divider was required to get equal gain on the two input legs. I guess I never really looked at the unbalanced input impedances. In the circuit above the difference in impedance is not large on the two legs. One is 51 kohms and the other is 61 kohms. The input filter calculations were done with 51 kohms, so the imbalance never showed up.

Now a customer is asking for these resistors to be changed to give a voltage gain of 1 rather than 1/5. This will give an imbalance in input impedance of 2:1. Still, this should not be a problem as long as the minimum frequency response is met. I just never realized a differential op amp circuit had this issue.

This circuit depends on a low pass filter to perform anti-aliasing. But that is elsewhere in the design and is not impacted by this issue. However, if there were a cap across the feedback resistor for this low pass filter, the filtering would also be unbalanced. Am I correct in thinking to maintain a balanced frequency response for a low pass filter in a differential op amp circuit would require a similar capacitor in the ground leg of the non-inverting input to have equal cutoff frequencies on the two inputs?

I can't find much info on this on the web, so I guessing this is messy enough that it is not done very often. Unfortunately there just isn't enough space available to make a more complex circuit. It worked pretty well with the larger input resistors. With smaller input resistors the compromises start to be a bigger problem.

--

Rick C.

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[whit3rd]

On Friday, March 6, 2020 at 11:39:27 PM UTC-8, Rick C wrote:

I have an existing design that uses a single op amp stage with differential inputs...
Now a customer is asking for these resistors to be changed to give a voltage gain of 1 rather than 1/5. This will give an imbalance in input impedance of 2:1.

The usual problem in imbalancing input impedance is (in bipolar op amps) a coupling of
temperature-of-chip to the offset voltage; in one memorable instance, I had to
add series resistance in order to get an offset trim to stop drifting.

A secondary problem is transient response, because of input capacitance loading.

In either case, a series input resistance (directly into the high-impedance input) can restore
balance, at least at low frequencies.

 

[Rick C]

On Saturday, March 7, 2020 at 4:41:20 PM UTC-5, whit3rd wrote:

On Friday, March 6, 2020 at 11:39:27 PM UTC-8, Rick C wrote:
I have an existing design that uses a single op amp stage with differential inputs...
Now a customer is asking for these resistors to be changed to give a voltage gain of 1 rather than 1/5. This will give an imbalance in input impedance of 2:1.

The usual problem in imbalancing input impedance is (in bipolar op amps) a coupling of
temperature-of-chip to the offset voltage; in one memorable instance, I had to
add series resistance in order to get an offset trim to stop drifting.

A secondary problem is transient response, because of input capacitance loading.

In either case, a series input resistance (directly into the high-impedance input) can restore
balance, at least at low frequencies.

Thanks for the info. I'm not sure I understand the series resistance you are talking about. I think this is already provided for in this design.

The differential input op amp circuit has equivalent voltage divider circuits on both the inverting and non-inverting inputs. The difference is the non-inverting input divider is referenced to ground and the inverting input divider is referenced to the output. But dividers use the same value resistances Ra and Rb. I believe this presents equal impedances to the op amp inputs so bias currents result in equal offsets other than the input offset current.

The impedance I am referring to is the impedance seen by the circuit driving the op amp circuit. In this case the driving circuit is the AC coupling cap. As best I can tell the input coupling cap on the non-inverting input see the sum of Ra and Rb as the input resistance. The input coupling cap on the inverting input sees just the input resistor Rb as the input resistance. When Rb is much larger than Ra (reduction of voltage between input and output of this stage) the two input resistances are roughly equal.

When the gain of the op amp circuit is close to unity or larger Ra is larger than Rb and the two inputs have very different input resistances. I suppose this could be compensated for by using equal ratios, but unequal values in the two halves of the circuit. Unfortunately I don't think this can be used in this design because the customer wants to add external series resistors to change the gain setting.

There is a later low pass filter in the circuit to provide anti-aliasing. Until now I've allowed the input coupling cap operate as a second pole for this anti-aliasing filter. I may have to settle for a single pole.

Actually, this has given me an idea. The audio circuits are 600 ohm impedance, so rather than add a resistor to the input to couple into the op amp gain computation, there is enough drive to use a 5:1 divider with a 600 ohm input impedance which will have about a 100 ohm output impedance. Then the divider can be added or not without impacting the operation of the op amp and the input coupling caps.

Thanks for stimulating some new thoughts.

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[Rick C]

I mocked up a simulation and I don't get the result I expected. The op amp uses four 20 kohm resistors so the two dividers are 1:1. I expected this to give 40 kohm input impedance for the non-inverting input impedance and 20 kohm input impedance for the inverting input. Combine that with 1 uF input coupling caps and I expected to see uneven frequency response on the resulting filters of the two inputs. I didn't. The attenuation at 20 Hz was very even between the two inputs.

I also tried this with 40 kohm resistors in the inverting divider. I expected the two input impedances to then be equal resulting in matched frequency responses, but they were not matched.

Can anyone explain what I am thinking wrong about the circuit? I am attaching the LTspice code below.

I'm not unhappy. I just can't find a web page that adequately analyzes this so I can understand what is happening. When I look at this as two separate single ended inputs I expect the impedance to be twice as high on the non-inverting input. When I look at it as differential inputs with a differential signal the non-inverting input seems to have three times the impedance as the inverting input. When I look at the common mode signal the two inputs seem to have the same impedance.

I don't get it...

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--

Rick C.

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[John S]

On 3/8/2020 11:34 PM, Rick C wrote:

I mocked up a simulation and I don't get the result I expected. The op amp uses four 20 kohm resistors so the two dividers are 1:1. I expected this to give 40 kohm input impedance for the non-inverting input impedance and 20 kohm input impedance for the inverting input. Combine that with 1 uF input coupling caps and I expected to see uneven frequency response on the resulting filters of the two inputs. I didn't. The attenuation at 20 Hz was very even between the two inputs.

I also tried this with 40 kohm resistors in the inverting divider. I expected the two input impedances to then be equal resulting in matched frequency responses, but they were not matched.

Can anyone explain what I am thinking wrong about the circuit? I am attaching the LTspice code below.

I'm not unhappy. I just can't find a web page that adequately analyzes this so I can understand what is happening. When I look at this as two separate single ended inputs I expect the impedance to be twice as high on the non-inverting input. When I look at it as differential inputs with a differential signal the non-inverting input seems to have three times the impedance as the inverting input. When I look at the common mode signal the two inputs seem to have the same impedance.

I don't get it...

(snip LTSpice code)

At what point in your circuit do you want input impedance? Perhaps at
INA and INB?

 

[John S]

On 3/8/2020 11:34 PM, Rick C wrote:

I mocked up a simulation and I don't get the result I expected. The op amp uses four 20 kohm resistors so the two dividers are 1:1. I expected this to give 40 kohm input impedance for the non-inverting input impedance and 20 kohm input impedance for the inverting input. Combine that with 1 uF input coupling caps and I expected to see uneven frequency response on the resulting filters of the two inputs. I didn't. The attenuation at 20 Hz was very even between the two inputs.

I also tried this with 40 kohm resistors in the inverting divider. I expected the two input impedances to then be equal resulting in matched frequency responses, but they were not matched.

Can anyone explain what I am thinking wrong about the circuit? I am attaching the LTspice code below.

I'm not unhappy. I just can't find a web page that adequately analyzes this so I can understand what is happening. When I look at this as two separate single ended inputs I expect the impedance to be twice as high on the non-inverting input. When I look at it as differential inputs with a differential signal the non-inverting input seems to have three times the impedance as the inverting input. When I look at the common mode signal the two inputs seem to have the same impedance.

I don't get it...

See my earlier post, please. And at what frequency? Do you want the
impedance as R,theta or R+jX?

 

[Rick C]

On Monday, March 9, 2020 at 12:32:04 PM UTC-4, Rick C wrote:

On Monday, March 9, 2020 at 9:05:56 AM UTC-4, John S wrote:
On 3/8/2020 11:34 PM, Rick C wrote:
I mocked up a simulation and I don't get the result I expected. The op amp uses four 20 kohm resistors so the two dividers are 1:1. I expected this to give 40 kohm input impedance for the non-inverting input impedance and 20 kohm input impedance for the inverting input. Combine that with 1 uF input coupling caps and I expected to see uneven frequency response on the resulting filters of the two inputs. I didn't. The attenuation at 20 Hz was very even between the two inputs.

I also tried this with 40 kohm resistors in the inverting divider. I expected the two input impedances to then be equal resulting in matched frequency responses, but they were not matched.

Can anyone explain what I am thinking wrong about the circuit? I am attaching the LTspice code below.

I'm not unhappy. I just can't find a web page that adequately analyzes this so I can understand what is happening. When I look at this as two separate single ended inputs I expect the impedance to be twice as high on the non-inverting input. When I look at it as differential inputs with a differential signal the non-inverting input seems to have three times the impedance as the inverting input. When I look at the common mode signal the two inputs seem to have the same impedance.

I don't get it...

(snip LTSpice code)

At what point in your circuit do you want input impedance? Perhaps at
INA and INB?

Actually, the point crossed by the dotted line between Cable and Input Amp. The decoupling caps are on my board.

The circuitry in the Cable section can be ignored or cut out other than R18 which should be 600 ohms I expect. This will be supplied by the customer once we decide how best to do the two variations.

My main concern now is that the decoupling caps C100 and C101 have an equal impact on the frequency response of the circuit. While these do not need to be what sets the frequency response at the low end, unless I want to up their value significantly they will be what sets the low end frequency response in the input circuit. I'm also concerned about upsetting the common mode rejection due to an imbalance for low frequencies like power line noise.

Let me change what I wrote. I'm looking to analyze the input impedance at points INA and INB without the caps in the circuit. That will allow me to understand the impact of adding the capacitors C100 and C101.

--

Rick C.

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[Rick C]

On Monday, March 9, 2020 at 10:54:35 AM UTC-4, John S wrote:

On 3/8/2020 11:34 PM, Rick C wrote:
I mocked up a simulation and I don't get the result I expected. The op amp uses four 20 kohm resistors so the two dividers are 1:1. I expected this to give 40 kohm input impedance for the non-inverting input impedance and 20 kohm input impedance for the inverting input. Combine that with 1 uF input coupling caps and I expected to see uneven frequency response on the resulting filters of the two inputs. I didn't. The attenuation at 20 Hz was very even between the two inputs.

I also tried this with 40 kohm resistors in the inverting divider. I expected the two input impedances to then be equal resulting in matched frequency responses, but they were not matched.

Can anyone explain what I am thinking wrong about the circuit? I am attaching the LTspice code below.

I'm not unhappy. I just can't find a web page that adequately analyzes this so I can understand what is happening. When I look at this as two separate single ended inputs I expect the impedance to be twice as high on the non-inverting input. When I look at it as differential inputs with a differential signal the non-inverting input seems to have three times the impedance as the inverting input. When I look at the common mode signal the two inputs seem to have the same impedance.

I don't get it...

See my earlier post, please. And at what frequency? Do you want the
impedance as R,theta or R+jX?

Sorry, I guess I'm looking for more of an explanation that will let me understand what makes it balanced. The simulations seems to show it works when I expect it to not work. I'm willing to bet the real circuit works ok too..

In the LTspice group someone indicated the impedances are indeed not equal, but the unequal gains of the two inputs result in the filter having the same result on each leg. I can't say I understand that mostly because with the ratio of resistors being equal in the inverting and non-inverting legs, the resulting gains of the two inputs are equal. So I'm just not clear on what is going on with the impedance.

--

Rick C.

+- Get 1,000 miles of free Supercharging
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[Rick C]

On Monday, March 9, 2020 at 9:05:56 AM UTC-4, John S wrote:

On 3/8/2020 11:34 PM, Rick C wrote:
I mocked up a simulation and I don't get the result I expected. The op amp uses four 20 kohm resistors so the two dividers are 1:1. I expected this to give 40 kohm input impedance for the non-inverting input impedance and 20 kohm input impedance for the inverting input. Combine that with 1 uF input coupling caps and I expected to see uneven frequency response on the resulting filters of the two inputs. I didn't. The attenuation at 20 Hz was very even between the two inputs.

I also tried this with 40 kohm resistors in the inverting divider. I expected the two input impedances to then be equal resulting in matched frequency responses, but they were not matched.

Can anyone explain what I am thinking wrong about the circuit? I am attaching the LTspice code below.

I'm not unhappy. I just can't find a web page that adequately analyzes this so I can understand what is happening. When I look at this as two separate single ended inputs I expect the impedance to be twice as high on the non-inverting input. When I look at it as differential inputs with a differential signal the non-inverting input seems to have three times the impedance as the inverting input. When I look at the common mode signal the two inputs seem to have the same impedance.

I don't get it...

(snip LTSpice code)

At what point in your circuit do you want input impedance? Perhaps at
INA and INB?

Actually, the point crossed by the dotted line between Cable and Input Amp. The decoupling caps are on my board.

The circuitry in the Cable section can be ignored or cut out other than R18 which should be 600 ohms I expect. This will be supplied by the customer once we decide how best to do the two variations.

My main concern now is that the decoupling caps C100 and C101 have an equal impact on the frequency response of the circuit. While these do not need to be what sets the frequency response at the low end, unless I want to up their value significantly they will be what sets the low end frequency response in the input circuit. I'm also concerned about upsetting the common mode rejection due to an imbalance for low frequencies like power line noise..

--

Rick C.

-+ Get 1,000 miles of free Supercharging
-+ Tesla referral code - https://ts.la/richard11209

 

[Clive Arthur]

On 09/03/2020 16:37, Rick C wrote:

<snip>

In the LTspice group someone indicated the impedances are indeed not equal, but the unequal gains of the two inputs result in the filter having the same result on each leg. I can't say I understand that mostly because with the ratio of resistors being equal in the inverting and non-inverting legs, the resulting gains of the two inputs are equal. So I'm just not clear on what is going on with the impedance.

Just looking without any Cs, four equal resistors, dual supplies...

Consider the case where the circuit's +ve input is at 2V. That means
the non-inverting op-amp terminal is at 1V.

Now put 1V on the circuit's -ve input. No current will flow though the
input R because there's no voltage across it, so the -ve input impedance
V/I is infinite. Try with some other voltages.

The -ve input impedance varies with signal level. The -ve input
impedance equals the +ve input impedance (2R) only when both inputs are
at the same voltage. If the inputs are equal voltage but opposite
polarity, I think the -ve input impedance is 1/3 the +ve, or 2R/3.
--
Cheers
Clive

 

[Phil Hobbs]

On 2020-03-09 00:34, Rick C wrote:

I mocked up a simulation and I don't get the result I expected. The op amp uses four 20 kohm resistors so the two dividers are 1:1. I expected this to give 40 kohm input impedance for the non-inverting input impedance and 20 kohm input impedance for the inverting input. Combine that with 1 uF input coupling caps and I expected to see uneven frequency response on the resulting filters of the two inputs. I didn't. The attenuation at 20 Hz was very even between the two inputs.

I also tried this with 40 kohm resistors in the inverting divider. I expected the two input impedances to then be equal resulting in matched frequency responses, but they were not matched.

Can anyone explain what I am thinking wrong about the circuit? I am attaching the LTspice code below.

I'm not unhappy. I just can't find a web page that adequately analyzes this so I can understand what is happening. When I look at this as two separate single ended inputs I expect the impedance to be twice as high on the non-inverting input. When I look at it as differential inputs with a differential signal the non-inverting input seems to have three times the impedance as the inverting input. When I look at the common mode signal the two inputs seem to have the same impedance.

I don't get it...

The circuit has different impedances for differential and common-mode
signals.


CM: The output stays still, and so does ground, so the input impedances
are both Rin+Rf.

Differential: The output moves the same amount as the differential
signal, and both inputs move half that amount.

Going pos -> pos + 1, neg -> neg -1 (2V differential) makes the noninv
input increase by 0.5V, so its differential input impedance is 2*Rin.
Feedback forces the inverting input to follow, so the voltage across its
input resistor increases by 1.5V, so its differential resistance is 2/3
* Rin.

Moving just one input is half differential mode and half common mode, so
you have to figure the contributions separately. In other words, the
neg input current depends on both voltages.

Cheers

Phil Hobbs


--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC / Hobbs ElectroOptics
Optics, Electro-optics, Photonics, Analog Electronics
Briarcliff Manor NY 10510

http://electrooptical.net
http://hobbs-eo.com

 

[Rick C]

On Monday, March 9, 2020 at 1:25:31 PM UTC-4, Clive Arthur wrote:

On 09/03/2020 16:37, Rick C wrote:

snip

In the LTspice group someone indicated the impedances are indeed not equal, but the unequal gains of the two inputs result in the filter having the same result on each leg. I can't say I understand that mostly because with the ratio of resistors being equal in the inverting and non-inverting legs, the resulting gains of the two inputs are equal. So I'm just not clear on what is going on with the impedance.


Just looking without any Cs, four equal resistors, dual supplies...

Consider the case where the circuit's +ve input is at 2V. That means
the non-inverting op-amp terminal is at 1V.

Now put 1V on the circuit's -ve input. No current will flow though the
input R because there's no voltage across it, so the -ve input impedance
V/I is infinite. Try with some other voltages.

The -ve input impedance varies with signal level. The -ve input
impedance equals the +ve input impedance (2R) only when both inputs are
at the same voltage. If the inputs are equal voltage but opposite
polarity, I think the -ve input impedance is 1/3 the +ve, or 2R/3.
--
Cheers
Clive

Let me talk about the difference between DC voltages and AC voltages. I care about the input impedance because of how it will interact with the impedance of the input capacitors. At DC the input capacitors have infinite impedance and the amp circuit will self bias to the reference voltage applied to the termination on the non-inverting input (LCOM which we can think of as ground). That's the DC analysis.

Now apply a 1 Vpp signal to the non-inverting input. The impedance seen will be the series resistance of R42 and R43. Now instead, apply a 1 Vpp signal to the inverting input. The inverting op amp input will not move due to the operation of the negative feedback resulting in the input impedance being R41 alone. When I simulate this with an AC analysis on one input at a time I get the same results. Or maybe not. I have three voltage sources wired in and apply AC 1 to one at a time. Do the other two voltage sources act as shorts for AC analysis?

Apply opposite polarity 1 Vpp signals (differential) to the two circuit inputs. The op amp non-inverting input will see 0.5 Vpp and again the impedance is R42 + R43. The output will be 2 Vpp with an opposite polarity of the inverting input, so the current in R39 and R41 will be 3 times that of the current through R42 and R43 resulting in 3 times lower input impedance. Simulating this by applying AC 1 to the + and - inputs at the same time gives a 6 dB higher signal out, but the same filter response.

Apply same polarity 1 Vpp signals to both inputs (common mode). The op amp non-inverting input is 0.5 Vpp as is the op amp inverting input. The output is 0 volts and both inputs have the same current so the same impedance. In this case the impedance "seen" by the input caps will be the same in both legs, so the filters will have the same characteristics and the common mode rejection will be preserved. The simulation shows -56 dB attenuation through the pass band with roll off in the higher and lower frequencies... as I would expect. But I wonder why -56 dB? Why not -66 dB or -106 dB? Perhaps this is an artifact of the Zener diode reference having a non-zero impedance?

I just wish I could analyze the differential mode input impedances to be the same. I suppose the combined effects of the two sides of the signal path result in the same result on the two sides somehow.

I hate stumbling around like this.

--

Rick C.

--- Get 1,000 miles of free Supercharging
--- Tesla referral code - https://ts.la/richard11209

 

[John S]

On 3/9/2020 11:39 AM, Rick C wrote:

On Monday, March 9, 2020 at 12:32:04 PM UTC-4, Rick C wrote:
On Monday, March 9, 2020 at 9:05:56 AM UTC-4, John S wrote:
On 3/8/2020 11:34 PM, Rick C wrote:
I mocked up a simulation and I don't get the result I expected. The op amp uses four 20 kohm resistors so the two dividers are 1:1. I expected this to give 40 kohm input impedance for the non-inverting input impedance and 20 kohm input impedance for the inverting input. Combine that with 1 uF input coupling caps and I expected to see uneven frequency response on the resulting filters of the two inputs. I didn't. The attenuation at 20 Hz was very even between the two inputs.

I also tried this with 40 kohm resistors in the inverting divider. I expected the two input impedances to then be equal resulting in matched frequency responses, but they were not matched.

Can anyone explain what I am thinking wrong about the circuit? I am attaching the LTspice code below.

I'm not unhappy. I just can't find a web page that adequately analyzes this so I can understand what is happening. When I look at this as two separate single ended inputs I expect the impedance to be twice as high on the non-inverting input. When I look at it as differential inputs with a differential signal the non-inverting input seems to have three times the impedance as the inverting input. When I look at the common mode signal the two inputs seem to have the same impedance.

I don't get it...

(snip LTSpice code)

At what point in your circuit do you want input impedance? Perhaps at
INA and INB?

Actually, the point crossed by the dotted line between Cable and Input Amp. The decoupling caps are on my board.

The circuitry in the Cable section can be ignored or cut out other than R18 which should be 600 ohms I expect. This will be supplied by the customer once we decide how best to do the two variations.

My main concern now is that the decoupling caps C100 and C101 have an equal impact on the frequency response of the circuit. While these do not need to be what sets the frequency response at the low end, unless I want to up their value significantly they will be what sets the low end frequency response in the input circuit. I'm also concerned about upsetting the common mode rejection due to an imbalance for low frequencies like power line noise.

Let me change what I wrote. I'm looking to analyze the input impedance at points INA and INB without the caps in the circuit. That will allow me to understand the impact of adding the capacitors C100 and C101.

Very well. Using you circuit but disconnecting the capacitors and
inserting an AC source (0 ohm source impedance) to supply the inputs in
parallel shows that the resistance of the two inputs at R41 and R42 are
equal and ~40k. These are each with respect to ground.

Remember that source impedance and differential operation will make
things much different from this. If you want my analysis of other parts
of your circuit and under other conditions, please let me know.

 

[Rick C]

On Monday, March 9, 2020 at 1:52:39 PM UTC-4, Phil Hobbs wrote:

On 2020-03-09 00:34, Rick C wrote:
I mocked up a simulation and I don't get the result I expected. The op amp uses four 20 kohm resistors so the two dividers are 1:1. I expected this to give 40 kohm input impedance for the non-inverting input impedance and 20 kohm input impedance for the inverting input. Combine that with 1 uF input coupling caps and I expected to see uneven frequency response on the resulting filters of the two inputs. I didn't. The attenuation at 20 Hz was very even between the two inputs.

I also tried this with 40 kohm resistors in the inverting divider. I expected the two input impedances to then be equal resulting in matched frequency responses, but they were not matched.

Can anyone explain what I am thinking wrong about the circuit? I am attaching the LTspice code below.

I'm not unhappy. I just can't find a web page that adequately analyzes this so I can understand what is happening. When I look at this as two separate single ended inputs I expect the impedance to be twice as high on the non-inverting input. When I look at it as differential inputs with a differential signal the non-inverting input seems to have three times the impedance as the inverting input. When I look at the common mode signal the two inputs seem to have the same impedance.

I don't get it...


The circuit has different impedances for differential and common-mode
signals.


CM: The output stays still, and so does ground, so the input impedances
are both Rin+Rf.

Differential: The output moves the same amount as the differential
signal, and both inputs move half that amount.

Going pos -> pos + 1, neg -> neg -1 (2V differential) makes the noninv
input increase by 0.5V, so its differential input impedance is 2*Rin.
Feedback forces the inverting input to follow, so the voltage across its
input resistor increases by 1.5V, so its differential resistance is 2/3
* Rin.

It would seem these distinct input impedances in differential mode would result in different corner frequencies with the input coupling caps. I can't seem to m measure that, but then perhaps my circuit doesn't work the way I expect in AC mode. Wiring the two voltage sources between the input and ground clearly isolates them and driving one at a time still gives exactly the same corner frequency. It seems to be around Req of 36 kohms. 4.54 Hz measured vs 4 Hz with the more expected 40 kohms.

--

Rick C.

--+ Get 1,000 miles of free Supercharging
--+ Tesla referral code - https://ts.la/richard11209

 

[Rick C]

On Monday, March 9, 2020 at 2:34:34 PM UTC-4, John S wrote:

On 3/9/2020 11:39 AM, Rick C wrote:
On Monday, March 9, 2020 at 12:32:04 PM UTC-4, Rick C wrote:
On Monday, March 9, 2020 at 9:05:56 AM UTC-4, John S wrote:
On 3/8/2020 11:34 PM, Rick C wrote:
I mocked up a simulation and I don't get the result I expected. The op amp uses four 20 kohm resistors so the two dividers are 1:1. I expected this to give 40 kohm input impedance for the non-inverting input impedance and 20 kohm input impedance for the inverting input. Combine that with 1 uF input coupling caps and I expected to see uneven frequency response on the resulting filters of the two inputs. I didn't. The attenuation at 20 Hz was very even between the two inputs.

I also tried this with 40 kohm resistors in the inverting divider. I expected the two input impedances to then be equal resulting in matched frequency responses, but they were not matched.

Can anyone explain what I am thinking wrong about the circuit? I am attaching the LTspice code below.

I'm not unhappy. I just can't find a web page that adequately analyzes this so I can understand what is happening. When I look at this as two separate single ended inputs I expect the impedance to be twice as high on the non-inverting input. When I look at it as differential inputs with a differential signal the non-inverting input seems to have three times the impedance as the inverting input. When I look at the common mode signal the two inputs seem to have the same impedance.

I don't get it...

(snip LTSpice code)

At what point in your circuit do you want input impedance? Perhaps at
INA and INB?

Actually, the point crossed by the dotted line between Cable and Input Amp. The decoupling caps are on my board.

The circuitry in the Cable section can be ignored or cut out other than R18 which should be 600 ohms I expect. This will be supplied by the customer once we decide how best to do the two variations.

My main concern now is that the decoupling caps C100 and C101 have an equal impact on the frequency response of the circuit. While these do not need to be what sets the frequency response at the low end, unless I want to up their value significantly they will be what sets the low end frequency response in the input circuit. I'm also concerned about upsetting the common mode rejection due to an imbalance for low frequencies like power line noise.

Let me change what I wrote. I'm looking to analyze the input impedance at points INA and INB without the caps in the circuit. That will allow me to understand the impact of adding the capacitors C100 and C101.


Very well. Using you circuit but disconnecting the capacitors and
inserting an AC source (0 ohm source impedance) to supply the inputs in
parallel shows that the resistance of the two inputs at R41 and R42 are
equal and ~40k. These are each with respect to ground.

Remember that source impedance and differential operation will make
things much different from this. If you want my analysis of other parts
of your circuit and under other conditions, please let me know.

I'm looking for understanding. The simulation tells me what it is, so I know that. But I don't understand why the inverting input has the same input impedance as the non-inverting input. How did you come to your conclusion?

The source impedance is low compared to these values. It is nominally a 600 ohm circuit so assume a 600 ohm source impedance and a 600 ohm termination.

Thanks for your comments.

--

Rick C.

-+- Get 1,000 miles of free Supercharging
-+- Tesla referral code - https://ts.la/richard11209

 

[John S]

On 3/9/2020 1:46 PM, Rick C wrote:

On Monday, March 9, 2020 at 2:34:34 PM UTC-4, John S wrote:
On 3/9/2020 11:39 AM, Rick C wrote:
On Monday, March 9, 2020 at 12:32:04 PM UTC-4, Rick C wrote:
On Monday, March 9, 2020 at 9:05:56 AM UTC-4, John S wrote:
On 3/8/2020 11:34 PM, Rick C wrote:
I mocked up a simulation and I don't get the result I expected. The op amp uses four 20 kohm resistors so the two dividers are 1:1. I expected this to give 40 kohm input impedance for the non-inverting input impedance and 20 kohm input impedance for the inverting input. Combine that with 1 uF input coupling caps and I expected to see uneven frequency response on the resulting filters of the two inputs. I didn't. The attenuation at 20 Hz was very even between the two inputs.

I also tried this with 40 kohm resistors in the inverting divider. I expected the two input impedances to then be equal resulting in matched frequency responses, but they were not matched.

Can anyone explain what I am thinking wrong about the circuit? I am attaching the LTspice code below.

I'm not unhappy. I just can't find a web page that adequately analyzes this so I can understand what is happening. When I look at this as two separate single ended inputs I expect the impedance to be twice as high on the non-inverting input. When I look at it as differential inputs with a differential signal the non-inverting input seems to have three times the impedance as the inverting input. When I look at the common mode signal the two inputs seem to have the same impedance.

I don't get it...

(snip LTSpice code)

At what point in your circuit do you want input impedance? Perhaps at
INA and INB?

Actually, the point crossed by the dotted line between Cable and Input Amp. The decoupling caps are on my board.

The circuitry in the Cable section can be ignored or cut out other than R18 which should be 600 ohms I expect. This will be supplied by the customer once we decide how best to do the two variations.

My main concern now is that the decoupling caps C100 and C101 have an equal impact on the frequency response of the circuit. While these do not need to be what sets the frequency response at the low end, unless I want to up their value significantly they will be what sets the low end frequency response in the input circuit. I'm also concerned about upsetting the common mode rejection due to an imbalance for low frequencies like power line noise.

Let me change what I wrote. I'm looking to analyze the input impedance at points INA and INB without the caps in the circuit. That will allow me to understand the impact of adding the capacitors C100 and C101.


Very well. Using you circuit but disconnecting the capacitors and
inserting an AC source (0 ohm source impedance) to supply the inputs in
parallel shows that the resistance of the two inputs at R41 and R42 are
equal and ~40k. These are each with respect to ground.

Remember that source impedance and differential operation will make
things much different from this. If you want my analysis of other parts
of your circuit and under other conditions, please let me know.

I'm looking for understanding. The simulation tells me what it is, so I know that. But I don't understand why the inverting input has the same input impedance as the non-inverting input. How did you come to your conclusion?

LTSpice came to that conclusion under the conditions I listed above.
Caps disconnected. Left ends of R41 and R43 tied together and connected
to a 1V AC source with 0 ohm internal impedance. I measured the current
through each resistor and divided the source voltage by each current. If
you make the vertical axis linearly scaled, it will read directly in ohms.

Under these conditions both opamp inputs will be equal. Therefore the OA
output is zero. If the OA output is zero, that means that each input is
the same impedance and equal to 40k.

This is immediately apparent by inspection and no spice simulation is
needed.

> The source impedance is low compared to these values. It is nominally a 600 ohm circuit so assume a 600 ohm source impedance and a 600 ohm termination.

I set my source impedance to 600 ohms. The input impedance remained the
same. Where do you want to put the 600 ohm termination?

Thanks for your comments.

My pleasure. I just hope I understand your quest.

 

[Rick C]

On Tuesday, March 10, 2020 at 8:07:26 AM UTC-4, John S wrote:

On 3/9/2020 1:46 PM, Rick C wrote:
On Monday, March 9, 2020 at 2:34:34 PM UTC-4, John S wrote:
On 3/9/2020 11:39 AM, Rick C wrote:
On Monday, March 9, 2020 at 12:32:04 PM UTC-4, Rick C wrote:
On Monday, March 9, 2020 at 9:05:56 AM UTC-4, John S wrote:
On 3/8/2020 11:34 PM, Rick C wrote:
I mocked up a simulation and I don't get the result I expected. The op amp uses four 20 kohm resistors so the two dividers are 1:1. I expected this to give 40 kohm input impedance for the non-inverting input impedance and 20 kohm input impedance for the inverting input. Combine that with 1 uF input coupling caps and I expected to see uneven frequency response on the resulting filters of the two inputs. I didn't. The attenuation at 20 Hz was very even between the two inputs.

I also tried this with 40 kohm resistors in the inverting divider. I expected the two input impedances to then be equal resulting in matched frequency responses, but they were not matched.

Can anyone explain what I am thinking wrong about the circuit? I am attaching the LTspice code below.

I'm not unhappy. I just can't find a web page that adequately analyzes this so I can understand what is happening. When I look at this as two separate single ended inputs I expect the impedance to be twice as high on the non-inverting input. When I look at it as differential inputs with a differential signal the non-inverting input seems to have three times the impedance as the inverting input. When I look at the common mode signal the two inputs seem to have the same impedance.

I don't get it...

(snip LTSpice code)

At what point in your circuit do you want input impedance? Perhaps at
INA and INB?

Actually, the point crossed by the dotted line between Cable and Input Amp. The decoupling caps are on my board.

The circuitry in the Cable section can be ignored or cut out other than R18 which should be 600 ohms I expect. This will be supplied by the customer once we decide how best to do the two variations.

My main concern now is that the decoupling caps C100 and C101 have an equal impact on the frequency response of the circuit. While these do not need to be what sets the frequency response at the low end, unless I want to up their value significantly they will be what sets the low end frequency response in the input circuit. I'm also concerned about upsetting the common mode rejection due to an imbalance for low frequencies like power line noise.

Let me change what I wrote. I'm looking to analyze the input impedance at points INA and INB without the caps in the circuit. That will allow me to understand the impact of adding the capacitors C100 and C101.


Very well. Using you circuit but disconnecting the capacitors and
inserting an AC source (0 ohm source impedance) to supply the inputs in
parallel shows that the resistance of the two inputs at R41 and R42 are
equal and ~40k. These are each with respect to ground.

Remember that source impedance and differential operation will make
things much different from this. If you want my analysis of other parts
of your circuit and under other conditions, please let me know.

I'm looking for understanding. The simulation tells me what it is, so I know that. But I don't understand why the inverting input has the same input impedance as the non-inverting input. How did you come to your conclusion?

LTSpice came to that conclusion under the conditions I listed above.
Caps disconnected. Left ends of R41 and R43 tied together and connected
to a 1V AC source with 0 ohm internal impedance. I measured the current
through each resistor and divided the source voltage by each current. If
you make the vertical axis linearly scaled, it will read directly in ohms..

Under these conditions both opamp inputs will be equal. Therefore the OA
output is zero. If the OA output is zero, that means that each input is
the same impedance and equal to 40k.

This is immediately apparent by inspection and no spice simulation is
needed.

What you just described is common mode which is the case where my paper analysis says the two input impedances are equal with high rejection.

The source impedance is low compared to these values. It is nominally a 600 ohm circuit so assume a 600 ohm source impedance and a 600 ohm termination.

I set my source impedance to 600 ohms. The input impedance remained the
same. Where do you want to put the 600 ohm termination?

It should be between the two input terminals of course.

Really, the input signal should be between the two inputs as well since it is a differential signal. I also ran simulations with the signal on each input separately.

Thanks for your comments.


My pleasure. I just hope I understand your quest.

I'm more interested in understanding the theory of the differential operation. My simulations have shown no situation where the input impedances are not equal and in fact the value appears to be around 40 kohms on each input leg. I just can't explain that for a differential input signal.

--

Rick C.

-++ Get 1,000 miles of free Supercharging
-++ Tesla referral code - https://ts.la/richard11209

 

[John S]

On 3/10/2020 7:54 AM, Rick C wrote:

On Tuesday, March 10, 2020 at 8:07:26 AM UTC-4, John S wrote:
On 3/9/2020 1:46 PM, Rick C wrote:
On Monday, March 9, 2020 at 2:34:34 PM UTC-4, John S wrote:
On 3/9/2020 11:39 AM, Rick C wrote:
On Monday, March 9, 2020 at 12:32:04 PM UTC-4, Rick C wrote:
On Monday, March 9, 2020 at 9:05:56 AM UTC-4, John S wrote:
On 3/8/2020 11:34 PM, Rick C wrote:
I mocked up a simulation and I don't get the result I expected. The op amp uses four 20 kohm resistors so the two dividers are 1:1. I expected this to give 40 kohm input impedance for the non-inverting input impedance and 20 kohm input impedance for the inverting input. Combine that with 1 uF input coupling caps and I expected to see uneven frequency response on the resulting filters of the two inputs. I didn't. The attenuation at 20 Hz was very even between the two inputs.

I also tried this with 40 kohm resistors in the inverting divider. I expected the two input impedances to then be equal resulting in matched frequency responses, but they were not matched.

Can anyone explain what I am thinking wrong about the circuit? I am attaching the LTspice code below.

I'm not unhappy. I just can't find a web page that adequately analyzes this so I can understand what is happening. When I look at this as two separate single ended inputs I expect the impedance to be twice as high on the non-inverting input. When I look at it as differential inputs with a differential signal the non-inverting input seems to have three times the impedance as the inverting input. When I look at the common mode signal the two inputs seem to have the same impedance.

I don't get it...

(snip LTSpice code)

At what point in your circuit do you want input impedance? Perhaps at
INA and INB?

Actually, the point crossed by the dotted line between Cable and Input Amp. The decoupling caps are on my board.

The circuitry in the Cable section can be ignored or cut out other than R18 which should be 600 ohms I expect. This will be supplied by the customer once we decide how best to do the two variations.

My main concern now is that the decoupling caps C100 and C101 have an equal impact on the frequency response of the circuit. While these do not need to be what sets the frequency response at the low end, unless I want to up their value significantly they will be what sets the low end frequency response in the input circuit. I'm also concerned about upsetting the common mode rejection due to an imbalance for low frequencies like power line noise.

Let me change what I wrote. I'm looking to analyze the input impedance at points INA and INB without the caps in the circuit. That will allow me to understand the impact of adding the capacitors C100 and C101.


Very well. Using you circuit but disconnecting the capacitors and
inserting an AC source (0 ohm source impedance) to supply the inputs in
parallel shows that the resistance of the two inputs at R41 and R42 are
equal and ~40k. These are each with respect to ground.

Remember that source impedance and differential operation will make
things much different from this. If you want my analysis of other parts
of your circuit and under other conditions, please let me know.

I'm looking for understanding. The simulation tells me what it is, so I know that. But I don't understand why the inverting input has the same input impedance as the non-inverting input. How did you come to your conclusion?

LTSpice came to that conclusion under the conditions I listed above.
Caps disconnected. Left ends of R41 and R43 tied together and connected
to a 1V AC source with 0 ohm internal impedance. I measured the current
through each resistor and divided the source voltage by each current. If
you make the vertical axis linearly scaled, it will read directly in ohms.

Under these conditions both opamp inputs will be equal. Therefore the OA
output is zero. If the OA output is zero, that means that each input is
the same impedance and equal to 40k.

This is immediately apparent by inspection and no spice simulation is
needed.

What you just described is common mode which is the case where my paper analysis says the two input impedances are equal with high rejection.

Yes, I agree. I thought that was what you wanted.

The source impedance is low compared to these values. It is nominally a 600 ohm circuit so assume a 600 ohm source impedance and a 600 ohm termination.

I set my source impedance to 600 ohms. The input impedance remained the
same. Where do you want to put the 600 ohm termination?

It should be between the two input terminals of course.

Really, the input signal should be between the two inputs as well since it is a differential signal. I also ran simulations with the signal on each input separately.


Thanks for your comments.


My pleasure. I just hope I understand your quest.

I'm more interested in understanding the theory of the differential operation. My simulations have shown no situation where the input impedances are not equal and in fact the value appears to be around 40 kohms on each input leg. I just can't explain that for a differential input signal.

Ok, I can't help with the theory. Sorry.

 

[Rick C]

On Tuesday, March 10, 2020 at 10:12:50 AM UTC-4, John S wrote:

On 3/10/2020 7:54 AM, Rick C wrote:
On Tuesday, March 10, 2020 at 8:07:26 AM UTC-4, John S wrote:
On 3/9/2020 1:46 PM, Rick C wrote:
On Monday, March 9, 2020 at 2:34:34 PM UTC-4, John S wrote:
On 3/9/2020 11:39 AM, Rick C wrote:
On Monday, March 9, 2020 at 12:32:04 PM UTC-4, Rick C wrote:
On Monday, March 9, 2020 at 9:05:56 AM UTC-4, John S wrote:
On 3/8/2020 11:34 PM, Rick C wrote:
I mocked up a simulation and I don't get the result I expected. The op amp uses four 20 kohm resistors so the two dividers are 1:1. I expected this to give 40 kohm input impedance for the non-inverting input impedance and 20 kohm input impedance for the inverting input. Combine that with 1 uF input coupling caps and I expected to see uneven frequency response on the resulting filters of the two inputs. I didn't. The attenuation at 20 Hz was very even between the two inputs.

I also tried this with 40 kohm resistors in the inverting divider. I expected the two input impedances to then be equal resulting in matched frequency responses, but they were not matched.

Can anyone explain what I am thinking wrong about the circuit? I am attaching the LTspice code below.

I'm not unhappy. I just can't find a web page that adequately analyzes this so I can understand what is happening. When I look at this as two separate single ended inputs I expect the impedance to be twice as high on the non-inverting input. When I look at it as differential inputs with a differential signal the non-inverting input seems to have three times the impedance as the inverting input. When I look at the common mode signal the two inputs seem to have the same impedance.

I don't get it...

(snip LTSpice code)

At what point in your circuit do you want input impedance? Perhaps at
INA and INB?

Actually, the point crossed by the dotted line between Cable and Input Amp. The decoupling caps are on my board.

The circuitry in the Cable section can be ignored or cut out other than R18 which should be 600 ohms I expect. This will be supplied by the customer once we decide how best to do the two variations.

My main concern now is that the decoupling caps C100 and C101 have an equal impact on the frequency response of the circuit. While these do not need to be what sets the frequency response at the low end, unless I want to up their value significantly they will be what sets the low end frequency response in the input circuit. I'm also concerned about upsetting the common mode rejection due to an imbalance for low frequencies like power line noise.

Let me change what I wrote. I'm looking to analyze the input impedance at points INA and INB without the caps in the circuit. That will allow me to understand the impact of adding the capacitors C100 and C101.


Very well. Using you circuit but disconnecting the capacitors and
inserting an AC source (0 ohm source impedance) to supply the inputs in
parallel shows that the resistance of the two inputs at R41 and R42 are
equal and ~40k. These are each with respect to ground.

Remember that source impedance and differential operation will make
things much different from this. If you want my analysis of other parts
of your circuit and under other conditions, please let me know.

I'm looking for understanding. The simulation tells me what it is, so I know that. But I don't understand why the inverting input has the same input impedance as the non-inverting input. How did you come to your conclusion?

LTSpice came to that conclusion under the conditions I listed above.
Caps disconnected. Left ends of R41 and R43 tied together and connected
to a 1V AC source with 0 ohm internal impedance. I measured the current
through each resistor and divided the source voltage by each current. If
you make the vertical axis linearly scaled, it will read directly in ohms.

Under these conditions both opamp inputs will be equal. Therefore the OA
output is zero. If the OA output is zero, that means that each input is
the same impedance and equal to 40k.

This is immediately apparent by inspection and no spice simulation is
needed.

What you just described is common mode which is the case where my paper analysis says the two input impedances are equal with high rejection.

Yes, I agree. I thought that was what you wanted.

Yes, my concern is the functionality for common mode signals. I have already verified the functionality in the simulation. I am trying to understand what is going on. I suspect the problem lies in the questions I am asking rather than the answers. I just tried using a differential input - not ground referenced - to analyze the voltage and current at the inputs. The inverting input shows nearly zero volts of AC signal and the other shows the full half volt. That makes it hard to calculate the impedance. lol I guess I'll have to provide ground referenced inputs.

When I do that I get 40 kohms for the non-inverting input impedance and 13.3 kohms for the inverting input impedance. It would seem that should not provide equal filter response of the input coupling caps and the two input impedances, but I can't measure any difference.

The source impedance is low compared to these values. It is nominally a 600 ohm circuit so assume a 600 ohm source impedance and a 600 ohm termination.

I set my source impedance to 600 ohms. The input impedance remained the
same. Where do you want to put the 600 ohm termination?

It should be between the two input terminals of course.

Really, the input signal should be between the two inputs as well since it is a differential signal. I also ran simulations with the signal on each input separately.


Thanks for your comments.


My pleasure. I just hope I understand your quest.

I'm more interested in understanding the theory of the differential operation. My simulations have shown no situation where the input impedances are not equal and in fact the value appears to be around 40 kohms on each input leg. I just can't explain that for a differential input signal.


Ok, I can't help with the theory. Sorry.

Yeah, thanks.

--

Rick C.

+-- Get 1,000 miles of free Supercharging
+-- Tesla referral code - https://ts.la/richard11209

 

[Clive Arthur]

On 10/03/2020 12:54, Rick C wrote:
<snip>

I'm more interested in understanding the theory of the differential operation. My simulations have shown no situation where the input impedances are not equal and in fact the value appears to be around 40 kohms on each input leg. I just can't explain that for a differential input signal.

Differential - say +ve input at +1V and -ve at -1V. +ve input impedance
is 40kR as is obvious. Op-amp non-inverting pin is +0.5V as must be
inverting pin. So -ve input is -1V through 20kR to +0.5V so -1.5V
across 20kR which is -75uA and -1V supplying -75uA = 13.33kR, see
circuit below, run it and measure V and I at the two voltage sources...


Version 4
SHEET 1 880 680
WIRE 80 16 32 16
WIRE 224 16 160 16
WIRE 352 16 304 16
WIRE 400 16 352 16
WIRE 528 16 480 16
WIRE -176 64 -176 32
WIRE 416 160 416 128
WIRE -176 176 -176 144
WIRE -176 176 -224 176
WIRE 352 176 352 16
WIRE 384 176 352 176
WIRE -224 192 -224 176
WIRE 528 192 528 16
WIRE 528 192 448 192
WIRE -176 208 -176 176
WIRE 32 208 32 16
WIRE 80 208 32 208
WIRE 224 208 160 208
WIRE 352 208 304 208
WIRE 384 208 352 208
WIRE 416 256 416 224
WIRE 32 272 32 208
WIRE 352 272 352 208
WIRE -176 320 -176 288
WIRE 352 384 352 352
FLAG 352 384 0
FLAG -224 192 0
FLAG 416 128 +S
FLAG 416 256 -S
FLAG -176 320 -S
FLAG -176 32 +S
FLAG 32 272 0
SYMBOL Opamps\\AD822 416 128 R0
SYMATTR InstName U1
SYMBOL voltage -176 48 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value 10
SYMBOL voltage -176 192 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V2
SYMATTR Value 10
SYMBOL res 496 0 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R1
SYMATTR Value 20k
SYMBOL res 320 0 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R2
SYMATTR Value 20k
SYMBOL res 320 192 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R3
SYMATTR Value 20k
SYMBOL res 368 368 R180
WINDOW 0 36 76 Left 2
WINDOW 3 36 40 Left 2
SYMATTR InstName R4
SYMATTR Value 20k
SYMBOL voltage 176 16 R90
WINDOW 0 -32 56 VBottom 2
WINDOW 3 32 56 VTop 2
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V3
SYMATTR Value -1
SYMBOL voltage 176 208 R90
WINDOW 0 -32 56 VBottom 2
WINDOW 3 32 56 VTop 2
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V4
SYMATTR Value +1
TEXT -248 408 Left 2 !.tran 1m

--
Cheers
Clive