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## Electrical engineering

### Course: Electrical engineering>Unit 3

Lesson 1: Operational amplifier

# Inverting op-amp

We analyze the inverting op-amp configuration, doing all the algebra from first principles. Created by Willy McAllister.

## Want to join the conversation?

• Why is Vout crossed out if Vout = A(v+ - v-)? Doesn't the definition of Vout make it along the same order of magnitude as A? • what is the difference between positive and negative feedback? • In positive feedback circuits, part of the output signal is added to the input signal. In negative feedback circuits, part of the output is subtracted from the input signal. Most feedback circuits use negative feedback, because the circuit gain stays under control. Positive feedback is less common. When someone with a microphone stands too close to the loudspeaker, you might hear a loud squeal caused by positive feedback. The amplified sound from the loudspeaker goes back into the microphone and gets amplified again, even louder.
• Why is VR1 = Vin - V- and why is VR2 = V- - Vout ? • Looking at around in the video:
Let's review the definitions of the voltages at each node. The circuit has 4 nodes: Upper left, upper middle, upper right output node, and ground node. Now identify/name the voltages at each node The upper left node is at a voltage of v_in with respect to ground. The upper middle node is at some (for now) unknown voltage we will give the name v- (suggested by its connection to the v- input of the opamp). The upper right node we will call v_out because it's connected to the output of the opamp. We don't know v_out yet either, but we will by the end of the video. The voltage of the ground node is 0 V by definion, and we label it with the triangle symbol. All these voltage names are called node voltages.

If we look at the two resistors we can give names to the voltages across each resistor: v_R1 and v_R2. These are called element voltages, and they will be useful later on in the analysis. An element voltage is the difference in the voltages of the two terminals of a resistor. We've already given node voltage names to all the points in the circuit, so let's use those names to figure out the element voltages on R1 and R2 in terms of node voltages:

For R1: v_R1 = v_in - v-
For R2: v_R2 = v- - v_out

I picked a particular orientation of the + and - signs when I defined the element voltages on the schematic. That gives rise to the particular order of the subtraction. If you flip the signs on the element voltage definition, it would flip the subtractions around the other way.
• could you please explain the formula for current at 3.20?
why does it involve v_in - v- ? • By Ohm's law the current flowing through R1 is equal to the voltage drop across R1 divided by the resistance of R1. The voltage drop is computed as the difference of electric potential between the terminals of the component. The left terminal (at a higher potential) is at Vin volts, while the right terminal (lower potential) is at V_ volts. Thus, the voltage drop equals Vin - V_.
• When simplifying AVin = -R1/R2vo(1 + A) - vo, why didn't we simplify (1+A) by A directly? • At the simplification steps are to first drop the trailing vo term, and next divide through by A. We notice that (1+A)/A is very close to unity, so it can be left out of the equation. I think you are asking if it would be ok to do the divide through by A as the first step. Yes it is. Then you simplify by noticing that vo/A is a very tiny number and can be left out. I thought my order was a little bit simpler, but your suggestion is fine, too.
• If I'm seeing this correct a non-inverting op-amp only effects current on the Vout terminal but an inverting op-amp effects current on both Vin and Vout.
I suppose with a robust Vin available current this wouldn't matter but with a robust Vin why bother with an op-amp at all? Without robust Vin current this Vin parasitic current could be suppressed to an insignificant level with very high value resistors but then you increase sensitivity to both external interference and the stray micro-current of a real op-amp.
(1 vote) • I like your term "robust". You are correct about the difference between the inverting and non-inverting opamp configurations. The non-inverting version has very high input impedance, equal to that of the opamp chip itself.

The inverting version has the input connected to R1 and then to the inverting input of the opamp. We know that when this circuit is working properly the difference in voltage between the two opamp input pins (+/-) is very small. So in the circuit shown here the right side of R1 is "virtually" connected to ground.

That means v_in is required to drive R1 connected to ground, making the input impedance = R1. (Interestingly, R2 has no impact on the input impedance.)

The question of "robustness" is a matter of scale. Suppose R1 = 10kOhm and suppose the load on the output of the opamp is 100 Ohm. The opamp presents much less load on v_in than if you asked v_in to drive the ultimate 100 Ohm load. It is a matter of degree.

If you have a really feeble v_in you probably don't choose an inverting opamp as the first stage. You would prefer to place a non-inverting opamp in that position for the reason you stated.
• at how did he take Vo out of the fraction by multiplying in R1?
(1 vote) • What is a multistage amplifier? Please explain.
Thank you
(1 vote) • is A is a symbol for gain? So can I say that A = -R2/R1?
(1 vote) • So the both the inverting op amp and the non inverting op amp are on the inverting terminal, but the difference is that the inverting op amp is connected to a power source?
(1 vote) 