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]
