Op amp basics

[Mr D]

Hi folks,

Taking my first look at op amps, many questions spring to mind.

But let's start with the basics.

Below is an GIF for your inspection.

I'm trying to get my head around the signal flow in this simplest of all op amp circuits.

Ok, so in the real world, the op amp has power leads coming in and out of it, which are not shown on the diagram.

In this circuit, because the difference in voltage across the signal pins is great, the op amp is setting it's output voltage as high as possible, because that's what it's designed to do.

What i don't understand is: the circuit actually shows current flowing from the output down to ground.

But how do I / we know where that ground is connected to with regard to the invisible power source that is supplying the op amps power?

That ground must be connected somewhere to the invisible power source, otherwise no current would flow, right?

But maybe it's connected to the op amp power source's positive terminal, in which case no current would flow??!!

[malagas_on_fire]

well that looks like a voltage to current conversion topology, but i might be wrong since i've rebooting electronics  aka, learned electronics but went to the firmware path .. so going back to basics has well

How about some first steps into opamps principles ?

https://www.analog.com/media/en/training-seminars/design-handbooks/Basic-Linear-Design/Chapter1.pdf

Well current flows to the easiest path to ground Not the high impedance inputs , not to the power source , unless apply a load such a LED

[xavier60]

That is not a practical op-amp application. There is supposed to be another resistor between the Inverting input and ground, not a wire.
The inputs should be close to the same voltage, the op-amps output controlling the voltage on the Inverting input via the feedback path.

[Nerull]

Simple opamp models often have 'assumed' power supplies. I believe it can be configured in everycircuit and defaults to a +/- 15V split supply, which means your positive rail is connected to +15V relative to ground and your negative rail is connected to -15V relative to ground.

If ground (0V) was connected to the opamps positive rail, than the opamp could only output negative voltages. That yours is outputing 10V tells you it is not.

[MarkF]

Here are some basics

[David Hess]

That ground is assumed to be connected somewhere between the positive and negative supplies such that the operational amplifier's maximum input and output voltage ranges are not compromised.  For circuit analysis, "ground" is just a reference point for all measurements.

[hamster_nz]

There is no way that the op-amp's output can move the '-' input as it is tied directly to ground, so there is no feedback.

Because of this the circuit will perform much the same as if there was no feedback resistor:

 - if the '+' input is higher the (fixed) '-' input then it will swing to full positive output.

 - if the '+' input is higher the (fixed) '-' input then it will swing to full negative output.

(but of course additional power will be dissipated in the resistor and the op-amp itself, due to current flowing from the op-amp through the resistor to ground).

In this case, (when the + input is positive with reference to the - one) the current would be flowing from the power supply, through op-amp's positive supply. out through the output of the Op amp, through the resistor and to ground (which is also the power supply's 0V/Gnd connection).

[pianovt]

Study this. If you get stuck and can't find the current path, ask.

[Zero999]

The op-amp is open loop as there's no way its output can make its input voltages equal. It's simply acting as a comparator. The simulator model has an output current limit of 10mA, which is pretty typical of an op-amp, which is why there is 10mA through the 1k resistor and 1V across it.

[Mr D]

Thanks all!

Study this. If you get stuck and can't find the current path, ask.

So that image you posted is just the full version of my initial circuit, showing the unseen connections?

And the power supply of the op amp consists of two power supplies in series, with ground set between them, so you get min and max V values centered around 0V?

[Zero999]

If you built his circuit in real life, it would probably not give exactly the same results. The current limit of an op-amp is not a a tightly controlled parameter. It's just there to protect the output from short circuits which can be handy for interfacing with external circuitry or using another circuit to pull the output to zero, thus disabling it.

For example, look at the short circuit current specification for the NE5532, a common general purpose op-amp. It's typically 38mA, but could be as low as 10mA, or as high as 60mA. In fact if you tried building this circuit with the NE5532 and there was nothing to limit the current from the 1V power supply (a series resistor for example) the op-amp would be destroyed. The NE5532 has diode connected transistors connected back-to-back between its inverting and non-inverting inputs, which would burn out, probably damaging the rest of the op-amp IC, if they were connected directly to a low impedance 1V source.

Refer to the internal schematic on page 7 of the datasheet paying attention to the +IN and -IN inputs. There are two transistors, each with their base and collector junctions connected. They simply act as diodes which limit the voltage between the two inputs to under 0.6V. This is done to protect the next pair of transistors, a differential pair, from having reverse voltages applied between their base and emitter junctions. The diodes don't conduct in normal operation, as the two inputs should theoretically be at the same voltage.
https://www.ti.com/lit/ds/symlink/ne5532a.pdf

This is just one of many ways real life components can differ from simplified simulator models.