feedback topology

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

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

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 ??

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?

we can take load as Rf .... can't we ?

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?

I was trying to understand feedback topologies , so i found this diagram and i was self questioning when i got this doubt

why is it necessary that one of the terminals of the load should be at ground ?? we can have floating load also

Floating out is referenced to what?

Sure if you have no load, then it's all potential. What is potential?

i am still unclear why is it necessary that one of the terminals of the load should be at ground ?

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

maybe this could further help you understand.

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

tell me what kind of feedback is there in the new figure ?

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

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

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.