Where is the current coming from?

Thread Starter

MikeJacobs

Joined Dec 7, 2019
201
Ok so, say you have a summing amplifier that is inverted. So the output is a negative voltage.
Nothing to extraordinary. The non-inverting terminal is grounded
The inverting terminal is the collection point obviously for the voltages to be summed.

the voltage at the inverting terminal is ~0 and the output voltage is say -5v

The feedback resistor is is 5k

So the feedback current is just 1 mA

So here is my question. on a normal op amp circuit that has some positive output voltage i would tell you that the current consumed by the op amp was just the supply current plus the feedback current. Normally obviously the output would have to supply the feedback current.

In this case, the current is supplied the other way. Its coming from the voltage devices it is trying to sum. So i would of thought that the supply current is just the supply current since the feedback current is not coming from the op amp source.
However i am not seeing that in the sim. The op amp supply current (into the supply pin) is still much higher then just what the supply current should be
Why is this?
 

Ian0

Joined Aug 7, 2020
1,933
The positive supply current will just be the specified op-amp supply current.
The negative supply current will be the specified op-amp supply current, plus 5mA (5V/1k).
(Almost) no current goes into the inverting input pin. The input current is the same current that flows in the feedback resistor.
 

dl324

Joined Mar 30, 2015
12,825
I don't know what you're seeing in your simulation, but here's an example with an inverting opamp:
clipimage.jpg
From the zero differential input theorem, the current in Rin is Es/Rin. Since, in an ideal opamp, no current flows into the inverting input, KCL says that the current in Rf is Iin. The direction of that current (out of the node between the resistors) means that the output voltage will be negative.

This is how the formula for an inverting amplifier was derived. Since the current flowing in Rin and Rf is the same, gain is -Rf/Rin.
 

crutschow

Joined Mar 14, 2008
27,028
In this case, the current is supplied the other way. Its coming from the voltage devices it is trying to sum. So i would of thought that the supply current is just the supply current since the feedback current is not coming from the op amp source.
Why do you think that?
The feedback current has to be supplied by the op amp output which then flows to the input.
The output current always equals the input current.
Where else would it come from?
 

Thread Starter

MikeJacobs

Joined Dec 7, 2019
201
Why do you think that?
The feedback current has to be supplied by the op amp output which then flows to the input.
The output current always equals the input current.
Where else would it come from?
Because i can very clearly see that the current being supplied by the voltages being summed is equal to the feedback current.
 

dl324

Joined Mar 30, 2015
12,825
A schematic would be beneficial:
clipimage.jpg
If you assume I2 < I1, the opamp would have to sink current.

If you assume that I2 > I1, the opamp would have to source current.

If you assume I2 = I1, the opamp wouldn't be supplying any current load to the load.

So which condition are you talking about?
 

DarthVolta

Joined Jan 27, 2015
521
In a basic 2-port model of an op-amp, with a few MOhm between V+ and V-, a uA might flow from V- to the grounded V+

So if the external circuit might naturally carry a few mA, the uA's get ignored by comparison. Which is why an op-amp, with a super high gain, can get tamed, to the controllable gain, of the external circuitry, a voltage divider, with V- being driven to GND. And the BJT/MOSFET/JFET internals, just use a feedback loop, to end up with the result being that, V- can be driven to match V+ as best as possible.

And with the way the op-amp is built, such that V- drives to V+ to GND

And ignoring the current that enters or leaves the actual op-amp at node V-.
Vin=R1*Iin(entering node V-)=R2*(-Iin)(leaving node V-)=-Vout
Or
-Vin+R1*Iin=0-R2*Iin=-Vout
that could be wrong, bed time for me
 
Last edited:

MisterBill2

Joined Jan 23, 2018
8,505
Ok so, say you have a summing amplifier that is inverted. So the output is a negative voltage.
Nothing to extraordinary. The non-inverting terminal is grounded
The inverting terminal is the collection point obviously for the voltages to be summed.

the voltage at the inverting terminal is ~0 and the output voltage is say -5v

The feedback resistor is is 5k

So the feedback current is just 1 mA

So here is my question. on a normal op amp circuit that has some positive output voltage i would tell you that the current consumed by the op amp was just the supply current plus the feedback current. Normally obviously the output would have to supply the feedback current.

In this case, the current is supplied the other way. Its coming from the voltage devices it is trying to sum. So i would of thought that the supply current is just the supply current since the feedback current is not coming from the op amp source.
However i am not seeing that in the sim. The op amp supply current (into the supply pin) is still much higher then just what the supply current should be
Why is this?
For starters, every opamp IC has a quiescent current that is not zero. So there is at least part of your answer. Next, most simulator software does not use a complete model of the active device being simulated. That means that the results are seldom totally correct. Bob Pease had a big complain about simulators in that regard, and he was incredibly knowlegable about seniconductors and electronics in general.
 

MisterBill2

Joined Jan 23, 2018
8,505
If you examine the circuit of most op-amps as it is presented in some of the data books, or at least used to be, there is a lot of circuit between the input and the output and all of that circuit consumes power that is not from the input or output connections. Thus the supply current will always be greater than them. That is the nature of complex active devices.
 
Top