the input resistance of a single opamp difference amplifier

[Heidi]

Dear friends,

Here's the difference amplifier I'm looking at, assuming the op amp to be ideal:

Its input resistance is defined as the resistance seen by Vi, as shown below, that is Ri=R1+R1.

For the right circuit below, knowing the input resistance as 2kΩ, I can tell that before the op-amp output voltage saturates, the ratio of the input voltage and the input current is equal to 2KΩ. But when I look back to its original circuit on the left, what can it tell us knowing the circuit's input resistance is 2kΩ? If the circuit's input voltage is V1-V2=3-1=2 Volts, what is defined as its input current? What is the purpose of defining the input resistance like that anyway? Thank you!

 

[ericgibbs]

hi Heidi,
The two circuits you have posted are quite different.
Left is a standard OPA and the right is a OPA connected a Difference amp.

Not sure what your question is.?

E

 

[Heidi]

Hi ericgibbs,

Thank you for your soon response.

It was said in my textbook that the input resistance of the left circuit can be calculated via the right circuit, by applying a input voltage which is equal to the voltage difference (V1-V2), divided by the input current in the right circuit. My question is "what can such an input resistance definition for the left circuit tell us?"

When I think of an input resistance for a given circuit, I think of applying a voltage across the input terminals of the circuit, measuring/calculating the resulting current that enters one of its input terminals and leaves the other, then divide the voltage by the current and I get the input resistance. I can tell the input voltage is propotional to the input current. In the left circuit, the input voltage might be (V1-V2), but what is its input current? The current entering R1 and the current entering R3 are different.

 

[crutschow]

Since the input current direction is into R3 and out of R1 for the given input voltages in the left circuit, a resistance calculation will give a positive resistance value for the bottom input and a negative resistance value for the top input.
Since the current and apparent resistance of the top input is modified by the input voltage of the bottom input, I don't think a resistance calculation of the top input separately from the bottom for that circuit is meaningful.

Only the differential input resistance measurement of the right circuit for a floating load is useful.

The circuit on the right has a floating input so the voltage referenced to ground has to float to a value where the two input currents are always opposite and equal.
For the given source voltage and resistor values that would be 4V.

 

[steveB]

Heidi,

Crutschow is directing you to the right way to think of this . I think there is an article on differential amplifiers here. You have to break input signals into differential mode and common mode signals. Each type of signal behaves differently. There are various books and online articles about this.

There is a general mathematical theorem which says that any input signals, such as that shown in your first schematic, can be represented as a sum of a unique differential mode signal and common mode signal. This is a very powerful tool to understand and simplify some systems.

 

[Ratchit]

Dear friends,

Here's the difference amplifier I'm looking at, assuming the op amp to be ideal:

View attachment 90630

Its input resistance is defined as the resistance seen by Vi, as shown below, that is Ri=R1+R1.
View attachment 90628

For the right circuit below, knowing the input resistance as 2kΩ, I can tell that before the op-amp output voltage saturates, the ratio of the input voltage and the input current is equal to 2KΩ.

I can tell you that the op-amp output voltage does not saturate. I can also tell you that the output voltage is 6 volts, and the voltages at the plus and minus terminals of the op-amp are both 3 volts. The output current of the op-amp is 1 ma. You can follow the sum of voltages from the output voltage, through all 8k of resistance plus the 2 volts input voltage using 1 ma and get zero volts according to Kirchoff's law. The diagram that says there is a virtual short between the input pins of the op-amp is wrong. No significant current exists between those input pins. Those pins have an extremely high resistance between them, not a virtual short.

But when I look back to its original circuit on the left, what can it tell us knowing the circuit's input resistance is 2kΩ? If the circuit's input voltage is V1-V2=3-1=2 Volts, what is defined as its input current? What is the purpose of defining the input resistance like that anyway? Thank you!
View attachment 90632

Different circuit, different analysis.

Ratch

 

[steveB]

Heidi, This may not be the best reference for you, but I've attached my old electronics notes on this subject. It may provide a basis for you to track down better references.

You will see that this information is a transistor version of a differential amplifier, which is not the same as your op-amp circuit. However, you should look at the approach used. Notice how the common mode and differential mode analysis is done. I'll look for a reference that deals better with the opamp circuit, but you may find a good one before I do.

 

[Heidi]

Hello experts,

Thank you so much for all your warm help. I'll need some time to digest the information you gave me. After that, if I still have something I can't understand, I'll get back to you. Thanks a lot!

 

[The Electrician]

When I think of an input resistance for a given circuit, I think of applying a voltage across the input terminals of the circuit, measuring/calculating the resulting current that enters one of its input terminals and leaves the other, then divide the voltage by the current and I get the input resistance. I can tell the input voltage is propotional to the input current. In the left circuit, the input voltage might be (V1-V2), but what is its input current? The current entering R1 and the current entering R3 are different.

You can indeed determine the input resistance by applying a voltage and calculating (or measuring) the current. BUT, you can't apply more than one voltage at a time. In the left hand circuit, if you want to know the resistance at the left end of R1, you can apply a voltage there and determine the current, but you can't also have a voltage applied to the left end of R3; you must ground the left end of R3. Similarly, if you want to know the resistance at the left end of R3, ground the left end of R1 before applying a test voltage to the left end of R3.

It's because you have two voltages applied at the same time that the currents in R1 and R3 are different. The applied voltages have both a differential component and a common mode component. If you want to determine the differential resistance, you must apply voltages that have only a differential component such as a floating voltage.

The right hand circuit has only one voltage applied, differentially, so you would be ok there.

 

[Heidi]

Thank you, The Electrician.

The applied voltages have both a differential component and a common mode component.

In the above sircuit, the common mode component is
Vcm=(V1+V2)/2 = 1.6
and the differential component is
Vd=(V1-V2)=1.

And
V1=Vcm + Vd/2 = 2.1
V2=Vcm - Vd/2 = 1.1

Correct? --------- (1)

If you want to determine the differential resistance, you must apply voltages that have only a differential component such as a floating voltage.

Apply the differential component Vd=1 like this?

So the input resistance seen by Vd is (Vd/500u)=2Ω.

And by definition, this 2Ω resistance is the input resistance of the first circuit. Correct? ------- (2)

Only the differential input resistance measurement of the right circuit for a floating load is useful.

You mentioned "applying a floating voltage", and crutschow mentioned "a floating load". Could you please explain a little bit what they mean?

If both answers to my questions (1) and (2) are yes, this is the primary question I want to ask :
Knowing the input resistance of the first circuit as being 2Ω, which is calculated by using the second circuit, then what?
Please let me try to clarify my question a little bit.

The above diagram is showing a signal voltage source Vs with an internal resistance Rs applying across the input terminals (p,n) of a circuit which has an input resistance Rin. If we want most of the signal Vs is applying at the circuit's input terminals, then we should design the circuit to have an input resistance Rin that is much larger than Rs.

So returnning back to the first circuit (in which R1=R3 and R2=R4), what plays the role of a signal source and what plays the internal resistance and how do we define its input current? Can the first circuit be represented by a conceptual diagram like the third one?

 

[Heidi]

I can tell you that the op-amp output voltage does not saturate. I can also tell you that the output voltage is 6 volts, and the voltages at the plus and minus terminals of the op-amp are both 3 volts. The output current of the op-amp is 1 ma. You can follow the sum of voltages from the output voltage, through all 8k of resistance plus the 2 volts input voltage using 1 ma and get zero volts according to Kirchoff's law. The diagram that says there is a virtual short between the input pins of the op-amp is wrong. No significant current exists between those input pins. Those pins have an extremely high resistance between them, not a virtual short.

Thank you, Ratchit.

The results you gave me seems to be the responds for the right circuit in post #1, I have no problem obtaining the same results. For the left circuit in post #1, how would you define the input resistance for it? What would come to your mind if you know a circuit's input resistance/impedance?

Since the input current direction is into R3 and out of R1 for the given input voltages in the left circuit, a resistance calculation will give a positive resistance value for the bottom input and a negative resistance value for the top input.
Since the current and apparent resistance of the top input is modified by the input voltage of the bottom input, I don't think a resistance calculation of the top input separately from the bottom for that circuit is meaningful.
Only the differential input resistance measurement of the right circuit for a floating load is useful.
The circuit on the right has a floating input so the voltage referenced to ground has to float to a value where the two input currents are always opposite and equal.
For the given source voltage and resistor values that would be 4V.

Sorry, crutschow. I can't understand almost all of your explanations. Maybe you would like to tell me what a "floating load", a "floating input" means first?

 

[The Electrician]

In post #10, you made an arithmetic error. Where you have "So the (differential) input resistance seen by Vd is (Vd/500u)=2Ω.", the result is 2000Ω, not 2Ω.

Three different input resistances can be defined for your left hand circuit in post #1.

If you ground the left end of R3 and apply a test voltage at the left end of R1, you will be determining the "driving point" resistance at that node.
If you ground the left end of R1 and apply a test voltage at the left end of R3, you will be determining the "driving point" resistance at that node.

If you connect a floating voltage source between the left end of R1 and the left end of R3, you will be determining the differential input resistance of your circuit. This is what you show for the right hand circuit.

If you connect the left end of R1 to the left end of R3 and apply a test voltage to both the left end of R1 and the left end of R3 together, you will be determining the common mode input resistance of your circuit.

So when you say, as you did in the title of your thread, "the input resistance of a single opamp difference amplifier", you have not fully specified what you're looking for. Just saying "input resistance" is ambiguous. You need to say "common mode input resistance", or "differential input resistance", or something fully specified.

 

[Heidi]

In post #10, you made an arithmetic error. Where you have "So the (differential) input resistance seen by Vd is (Vd/500u)=2Ω.", the result is 2000Ω, not 2Ω.

Three different input resistances can be defined for your left hand circuit in post #1.

If you ground the left end of R3 and apply a test voltage at the left end of R1, you will be determining the "driving point" resistance at that node.
If you ground the left end of R1 and apply a test voltage at the left end of R3, you will be determining the "driving point" resistance at that node.

If you connect a floating voltage source between the left end of R1 and the left end of R3, you will be determining the differential input resistance of your circuit. This is what you show for the right hand circuit.

If you connect the left end of R1 to the left end of R3 and apply a test voltage to both the left end of R1 and the left end of R3 together, you will be determining the common mode input resistance of your circuit.

So when you say, as you did in the title of your thread, "the input resistance of a single opamp difference amplifier", you have not fully specified what you're looking for. Just saying "input resistance" is ambiguous. You need to say "common mode input resistance", or "differential input resistance", or something fully specified.

I see, Thanks a lot, The Electrician.

Your definitions are simple enough for me to understand, although Steve (and other people) was also trying to show me those definitions in his transistor version notes. I think I haven't had fully considered my questions before I address them.

WTW, I guess I have grasped the idea about what a "floating" voltage or load is. Thanks!

 

[crutschow]

........................... The diagram that says there is a virtual short between the input pins of the op-amp is wrong. No significant current exists between those input pins. Those pins have an extremely high resistance between them, not a virtual short.
.................

You are referring to an actual short, not a virtual short. A virtual short is similar to the virtual ground that the minus input is said to have in an inverting op amp circuit with the plus terminal connected to ground. Certainly there is no current flow between the virtual ground and ground (or between the op amp input terminals) but, in a closed-loop circuit operating in the linear region, the two voltages stay close, making the minus input voltage virtually at ground.
By extension, a virtual short exists between the two op amp terminals in a differential circuit since feedback forces the minus input voltage to stay close to the plus input voltage.

 

[Ratchit]

You are referring to an actual short, not a virtual short. A virtual short is similar to the virtual ground that the minus input is said to have in an inverting op amp circuit with the plus terminal connected to ground. Certainly there is no current flow between the virtual ground and ground (or between the op amp input terminals) but, in a closed-loop circuit operating in the linear region, the two voltages stay close, making the minus input voltage virtually at ground.
By extension, a virtual short exists between the two op amp terminals in a differential circuit since feedback forces the minus input voltage to stay close to the plus input voltage.

If it cannot do what a short does by passing current, then it should not be called a short of any kind. It should instead be called something like a "voltage equivalency".

Ratch

 

[crutschow]

If it cannot do what a short does by passing current, then it should not be called a short of any kind. It should instead be called something like a "voltage equivalency".

The word "virtual" would seem to make the meaning clear. If you want to use such an awkward phrase as "voltage equivalency", fine, but virtual short works for me. I see no difference between that and "virtual ground" in its usage.

 

[Ratchit]

The word "virtual" would seem to make the meaning clear. If you want to use such an awkward phrase as "voltage equivalency", fine, but virtual short works for me. I see no difference between that and "virtual ground" in its usage.

I don't see how "virtual" implies equal voltage. "Virtual ground" implies the same voltage as ground reference, but it is not obvious until explained. "Voltage equivalency" implies the same voltage between any two points and covers all cases. Why is voltage equivalency any more awkward that virtual ground or short? Does not VE contain two correct English language words?

Ratch

 

[crutschow]

I don't see how "virtual" implies equal voltage. "Virtual ground" implies the same voltage as ground reference, but it is not obvious until explained. "Voltage equivalency" implies the same voltage between any two points and covers all cases. Why is voltage equivalency any more awkward that virtual ground or short? Does not VE contain two correct English language words?

Ratch

Ratch, since you always seem to enjoy getting in the last word, you can reply to this and I'll leave you with that.

Among other things, virtual means "very close to being something without actually being it" (https://www.merriam-webster.com/dictionary/virtual) which would seem to apply to either of the phrases, "virtual ground" or "virtual short".
So a virtual short would imply equal voltage since that's the end result of a short even if it doesn't actually "short" the two points.

In any case, don't expect to see "VE" in common engineering use anytime soon. Just because it uses two "correct English language words" doesn't make it a usefull phrase.

 

[Ratchit]

Ratch, since you always seem to enjoy getting in the last word, you can reply to this and I'll leave you with that.

Among other things, virtual means "very close to being something without actually being it" (https://www.merriam-webster.com/dictionary/virtual) which would seem to apply to either of the phrases, "virtual ground" or "virtual short".
So a virtual short would imply equal voltage since that's the end result of a short even if it doesn't actually "short" the two points.

In any case, don't expect to see "VE" in common engineering use anytime soon.

Pretty poor definition of "virtual", don't you think? Do they want money for that publication? Virtual in our case really means "appear" or "apparent". Because of equal voltages, a conduction path appears to exist between a pin and ground, even though there is much resistance between them. So much for the dictionary's "very close" description. Or stand 6 feet in front of a plain flat mirror. You will see an image of yourself at what appears to be 12 feet away from you and 6 feet behind the mirror. Your image is not really there, it just appears like it is. That is why opticians call it a "virtual" image.

I don't expect to see "current exists" instead of "current flow", either. Even though current flow is semantically incorrect.

Ratch

 

[crutschow]

Pretty poor definition of "virtual", don't you think? Do they want money for that publication? ...............................

Ratch since you asked two questions I will reply.
No.
No.