Feedback topology: Is output voltage or current being sampled in this circuit?

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In this op-amp feedback circuit, is the output quantity being sampled voltage or current, and what feedback topology is it?

The circuit samples output voltage, not output current; it is an inverting voltage amplifier with negative feedback. Converting the current source and its shunt resistor to the Thevenin equivalent gives a basic inverting op-amp circuit with Vs = IsRs, and the feedback voltage is Vo = IoRf [#21667566] Because the op-amp input current is approximately zero, Is ≈ Io, so the transfer function is a voltage gain of about Rf/Rs [#21667566] The current through Rf is not the load current, because Rf is part of the feedback network and the output is still a voltage referenced to ground; without a load there is no meaningful output current path [#21667568] [#21667570] In short, the topology is negative voltage feedback, commonly used as a voltage amplifier or voltage follower depending on the feedback network [#21667575]

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#1 21667565 12 Apr 2013 10:55

AMAN GUPTA AMAN GUPTA

Anonymous

Post #1
21667565 12 Apr 2013 10:55

What is the feedback topology in the given circuit ??
At the input i can see that input and feedback currents are mixing, but what quantity is being sampled at the output here.
Is it voltage because i am getting If = -Vo / Rf
or is it current because i am also getting Io = -If......... ??
pls help i am confused
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#2 21667566 12 Apr 2013 11:15

George Duval George Duval

Anonymous

Post #2
21667566 12 Apr 2013 11:15

Would it help you to you convert the current source and shunt resistance to it's thevenin equivalent.?

You get a basic inverting op-amp circuit with.

Vs=IsRs
Vo = IoRf
Voltage gain then: Av= - (IoRf)/(IsRs)

Since no current between the inverting and non inverting inputs of the opamp ( due to high impedance), you can almost say Is~=Io. It creates a virtual ground.
.
So the gain is the ratio of Rf/ Rs as a transfer function of the output voltage over the input.

The output will be a voltage
#3 21667567 12 Apr 2013 11:41

AMAN GUPTA AMAN GUPTA

Anonymous

Post #3
21667567 12 Apr 2013 11:41

If i consider current as the output then i get Is = Vs / Rs and Is = Io , so Io / Vs = Rs
so why can't i consider current as the output ??
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#4 21667568 12 Apr 2013 11:44

George Duval George Duval

Anonymous

Post #4
21667568 12 Apr 2013 11:44

Ohm's law ... you can't have current without a voltage. and a resistance What is R_load? Right now, you have none... so ... where is the current going to on the output with respect to ground?
#5 21667569 12 Apr 2013 11:47

AMAN GUPTA AMAN GUPTA

Anonymous

Post #5
21667569 12 Apr 2013 11:47

we can take load as Rf .... can't we ?
#6 21667570 12 Apr 2013 11:54

George Duval George Duval

Anonymous

Post #6
21667570 12 Apr 2013 11:54

No... Rf is not the load. Even though the Non-inverting input acts a "virtual GND" .. there is no current to ground at that Junction..

In your diagram the path to ground is through the Nortan Eq Current source and it's shunt.

You have an open Circuit.... what is the application? Or is this a homework assignment you are trying to understand better?
#7 21667571 12 Apr 2013 11:57

AMAN GUPTA AMAN GUPTA

Anonymous

Post #7
21667571 12 Apr 2013 11:57

I was trying to understand feedback topologies , so i found this diagram and i was self questioning when i got this doubt
#8 21667572 12 Apr 2013 12:00

AMAN GUPTA AMAN GUPTA

Anonymous

Post #8
21667572 12 Apr 2013 12:00

why is it necessary that one of the terminals of the load should be at ground ?? we can have floating load also
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#9 21667573 12 Apr 2013 12:16

George Duval George Duval

Anonymous

Post #9
21667573 12 Apr 2013 12:16

Floating out is referenced to what?

Sure if you have no load, then it's all potential. What is potential?
#10 21667574 12 Apr 2013 12:30

AMAN GUPTA AMAN GUPTA

Anonymous

Post #10
21667574 12 Apr 2013 12:30

i am still unclear why is it necessary that one of the terminals of the load should be at ground ?
#11 21667575 12 Apr 2013 12:48

Mark Harrington Mark Harrington

Anonymous

Post #11
21667575 12 Apr 2013 12:48

If you apply output from the op amps output in your drawing back to the inverting input Its negative feedback

So the toplogy is an inverting voltage amplifier or voltage follower depending on RF (Your feedback network)

As george says voltage gain is Av= – (IoRf)/(IsRs) Its normal use is as voltage amplifier , follower

Try this book for further reference Audio Ic Op amp applications ISBN 0-672-22452-6 by Walter G Jung its an excellent primer into op amp design and shows worked examples plus calculations

I highly recommend this book to engineers whishing to study op amp basics plus design
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#12 21667576 13 Apr 2013 02:37

Floy Viola Floy Viola

Anonymous

Post #12
21667576 13 Apr 2013 02:37

maybe this could further help you understand.
#13 21667577 13 Apr 2013 03:23

AMAN GUPTA AMAN GUPTA

Anonymous

Post #13
21667577 13 Apr 2013 03:23

ok
wat i want to clarify is why can't we use the current flowing through Rf as the input current to a load ......... in other words why Rf can't be a load ?? if yes then in that case Io is our input current and the feedback will be shunt series..............
Most probably i am wrong , but i want to clear this
#14 21667578 13 Apr 2013 04:09

AMAN GUPTA AMAN GUPTA

Anonymous

Post #14
21667578 13 Apr 2013 04:09

tell me what kind of feedback is there in the new figure ?
#15 21667579 13 Apr 2013 08:53

Rohit Dubla Rohit Dubla

Anonymous

Post #15
21667579 13 Apr 2013 08:53

Hi Aman,

Referring to the single transistor circuit, it has only local emitter degeneration in that the current through the emitter resistor (i.e., the collector or emitter current) causes a voltage drop across that resistor which opposes the voltage at the base.

Now, I would not call that "feedback" simply because there nothing being "fed back" from the o/p to the input via a feedback network.

Regards,
Rohit
#16 21667580 13 Apr 2013 16:38

Mark Harrington Mark Harrington

Anonymous

Post #16
21667580 13 Apr 2013 16:38

There is slight feed back with respect to RE Rohit

This you could loosely classify as negative feedback The value of Rc for example depends on whether the transistor is required to deliver voltage or current output

If the output were to be derived from the collector Rc being the load resistor then the inclusion of a electrolytic capacitor in parallel with RE would improve the overall ac gain and would be selectively chosen for example, in an audio amplifier stage, for suitable roll off
#17 21667581 17 Apr 2013 03:45

Rohit Dubla Rohit Dubla

Anonymous

Post #17
21667581 17 Apr 2013 03:45

Hello Mark, yes there is local feedback - one that I termed "degeneration". The point here is that the presence of an external Re helps linearise the transistor's voltage gain. Without this, the voltage gain will be a function of the transistor's internal emitter resistance (re or hie). Having Re (much larger than re) swamps out re thereby linearising the gain - however now the closed loop gain might be too low to be useful, and needs bypassing with a capacitor to extract as much gain as possible. When bypassed by a capacitor, we are, in effect defeating this local feedback and Re only serves to stabilise the DC operating point of the device. Also, local feedback has no effect upon the stage's input or output impedance and therefore is not true negative feedback. True negative feedback will stabilise the gain as well as increase input impedance and reduce output impedance.
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Topic summary

✨ The discussion centers on identifying the feedback topology in an op-amp circuit where input and feedback currents mix. Analysis using Thevenin equivalents shows the circuit behaves as a basic inverting amplifier with voltage gain Av = -Rf/Rs, indicating the output quantity sampled is voltage, not current. The virtual ground at the inverting input ensures input current (Is) approximately equals output current (Io), but the output is a voltage across the feedback resistor Rf, which is not considered a load. The necessity of a grounded reference for the load is debated, clarifying that without a proper load to ground, current flow at the output is undefined. Local feedback in transistor circuits, such as emitter degeneration via Re, is distinguished from true negative feedback, as it stabilizes gain but does not feed output back to input through a feedback network. Recommended literature includes "Audio IC Op Amp Applications" by Walter G. Jung for further understanding of op-amp feedback and design.
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FAQ

TL;DR: In this inverting op-amp, a fraction of the output voltage is fed back; gain magnitude equals Rf/Rs, and “The output will be a voltage.” [Elektroda, George Duval, post #21667566]

Why it matters: It clarifies whether you design and measure this stage as a voltage amplifier or a current-mode block.

For: students and engineers asking how to classify feedback in a current-driven, inverting op-amp and what the output actually represents.

Quick Facts

Topology in context: inverting op-amp with negative feedback; effective voltage gain ≈ −Rf/Rs. [Elektroda, George Duval, post #21667566]
Virtual-ground node makes Is ≈ Io at the summing junction (high input impedance assumption). [Elektroda, George Duval, post #21667566]
Without an external load path, output current is undefined (open circuit). [Elektroda, George Duval, post #21667570]
Rf is the feedback element, not the load; load current must return to a defined reference. [Elektroda, George Duval, post #21667570]
Local emitter degeneration stabilizes a BJT stage’s gain but isn’t “true” global negative feedback. [Elektroda, Rohit Dubla, post #21667581]

Is the output quantity in this circuit a voltage or a current?

Voltage. The feedback network returns a proportion of the output voltage to the inverting input, creating a voltage-defined output with gain set by Rf and Rs. “The output will be a voltage.” [Elektroda, George Duval, post #21667566]

What feedback topology label fits this inverting op-amp?

It operates as a voltage amplifier with negative feedback. You analyze it by returning output voltage through Rf to the summing node and using Av ≈ −Rf/Rs after source conversion. [Elektroda, George Duval, post #21667566]

Why do people convert the source to a Thevenin/Norton equivalent first?

Source conversion exposes a standard inverting op-amp form. With Vs = Is·Rs, the stage looks like a classic voltage amplifier where Av ≈ −(Io·Rf)/(Is·Rs) and Is ≈ Io at the virtual ground. [Elektroda, George Duval, post #21667566]

Can I treat Rf as the load and call it a current output?

No. Rf is part of the feedback network, not the load. Without a defined return path for load current, the output current isn’t well-defined, even with a virtual ground at the summing node. [Elektroda, George Duval, post #21667570]

Do I need a load referenced to ground for current to flow?

You need a defined reference path. With no load path to the reference, the circuit’s output current has nowhere to go, giving an open-circuit condition at the output node. [Elektroda, George Duval, post #21667570]

What does “virtual ground” mean here?

The op-amp drives its output so the inverting input sits near the reference potential. That makes input and feedback currents sum to nearly zero at the node, hence Is ≈ Io. [Elektroda, George Duval, post #21667566]

What gain should I expect numerically?

Magnitude equals Rf/Rs. Example: if Rf = 10 kΩ and Rs = 1 kΩ, |Av| = 10. The sign is negative due to inversion at the summing node. [Elektroda, George Duval, post #21667566]

Edge case: what fails if the load is floating?

With a floating load and no return, output current is undefined and only a potential difference exists. This is effectively an open circuit and won’t deliver power. [Elektroda, George Duval, post #21667570]

Is there any current flowing into the op-amp inputs?

Ideally no. High input impedance means negligible input currents, so the source and feedback currents balance at the inverting node, enabling the virtual-ground analysis. [Elektroda, George Duval, post #21667566]

How do I quickly classify the feedback in three steps?

Convert the source to its Thevenin equivalent (Vs = Is·Rs).
Identify Rf from output to inverting input.
Write Av; with Is ≈ Io, get Av ≈ −Rf/Rs (voltage output). [Elektroda, George Duval, post #21667566]

Why isn’t the single-transistor example considered global feedback?

That circuit shows local emitter degeneration. The emitter resistor drops voltage proportional to current, linearizing gain, but it doesn’t feed output back through a network to the input node. [Elektroda, Rohit Dubla, post #21667579]

Does emitter degeneration change input or output impedance like global feedback?

No. Local degeneration mainly stabilizes gain and bias. “True negative feedback” between input and output also raises input impedance and lowers output impedance, which degeneration alone does not. [Elektroda, Rohit Dubla, post #21667581]

Is the op-amp always a voltage follower in this setup?

Only if Rf equals Rs in the Thevenin form and is arranged for unity gain magnitude. Otherwise, it’s an inverting voltage amplifier with |Av| = Rf/Rs. [Elektroda, Mark Harrington, post #21667575]

How do Ohm’s law and the load path affect my interpretation?

Ohm’s law links voltage, current, and resistance. Without a defined load and reference, you cannot claim a usable output current, only a potential difference. [Elektroda, George Duval, post #21667568]

What practical takeaway should I remember when “current is my input”?

Even with a Norton input, the stage’s controlled quantity is output voltage. Treat it as a voltage amplifier set by Rf and Rs, not as a current source. [Elektroda, George Duval, post #21667566]