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.
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
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.
In the above sircuit, the common mode component isThe applied voltages have both a differential component and a common mode component.
Apply the differential component Vd=1 like this?If you want to determine the differential resistance, you must apply voltages that have only a differential component such as a floating voltage.
You mentioned "applying a floating voltage", and crutschow mentioned "a floating load". Could you please explain a little bit what they mean?Only the differential input resistance measurement of the right circuit for a floating load is useful.
Thank you, Ratchit.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.
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?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.
I see, Thanks a lot, 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.
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............................ 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.
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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.
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.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.
Ratch, since you always seem to enjoy getting in the last word, you can reply to this and I'll leave you with that.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.
Ratch since you asked two questions I will reply.Pretty poor definition of "virtual", don't you think? Do they want money for that publication? ...............................
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