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### Course: Electrical engineering>Unit 3

Lesson 1: Operational amplifier

# Non-inverting op-amp

An op-amp can be used to amplify a voltage. The gain is determined by the ratio of two resistors. Created by Willy McAllister.

## Want to join the conversation?

• It the current is zero, what is the point of the circuit?
also isnt it impossible to have 0 current when there is a voltage.
And what does a -negative voltage even mean?
• Good questions. For your first question, the current is not literally zero, but it is so close to zero that the concepts and equations work well while assuming zero current.

Your point about the impossibility of zero current when there is a voltage is addressed by considering the power outlets in your house. The outlets have voltage, but no current will flow until there is a complete circuit, e.g. plugging in an electric razor or hair dryer.

• At , how can you say V+ is Vin ? Wasn't Vin=V+-V-?
• Hello Kunal,

Assuming the gain is infinite then feedback will make V+ = V-.

This is a good assumption as long as the signal has a low frequency e.g. audio. Things start to fall apart when the signals get higher.

Let us know when you reach these more interesting levels with op-amp.

Quick question for you - have you had the chance to build and experiment with an op-amp?

Regards,

APD
• Why cant the inverting input (v-) simply be connected to v_out?
Why do we need tgose extra resistors?
• Hi Atharva,

Those extra resistors are used to set the gain of the op-amp stage. Without them the non-inverting amp has a gain of 1. While this is useful in certain applications there are many where you want more gain.

For example, if you had a microphone you may want a gain of 100 or even more.

Regards,

APD
• There were two gains referred to, the A and the R2/(R2+R2). Could you please tell me more about them? I understand A is large and a property of the particular op-amp whereas R1 and R2 are decided by us and that which we actually use? So the resisters decide the 'gain' in the voltage? Then whats the working of A?If so could please elaborate on the difference between their workings? Is it that A does not matter as long as it's very large?
Thank you for the video, Khan Academy has never let me down and I am immensely indebted to you. All of you provide a immeasurable service to the word!
• This is a good question. You have it all right. Big A is the gain of the opamp itself. The gain of the whole circuit (opamp plus resistors) is the resistor ratio expression. The resistor ratio answer is valid as long as Big A is huge. A's job is to be huge enough to make it disappear from the gain expression. The value of A does not matter as long as it is very large. It is large enough when we decide that A and (A+1) are so close together we can cancel them in a fraction.

You can explore this more by using specific values for A. Go back to the equation just before I decided to ignore the "1" right at in the video. Plug in real A values and get a "feeling" for how appropriate the cancellation is. Try A = 5 and A = 100 and A = 10,000. How much of a mistake do you make when you ignore the "1"?
• In the previous video you said Vin=A(Vplus-Vminus). In this video you say Vin = Vplus. I suppose this would be true if Vminus was ZERO, but here, Minus is NOT zero. Why did you label Vplus as Vin to begin with? You keep using Vin and Vplus interchangeably. Please explain
• The previous video "What is an operational amplifier" gave the general equation for what an opamp does,

V_out = A(v+ - v-)

In this video the circuit shown has a signal named v_in coming from the input voltage source. It is connected to the v+ input of the opamp symbol. So the signal name v_in is interchangeable with the opamp pin v+.
• At , Why can't the voltage divider equation be Vo = V in*(R2)/R1+R2 ? Why does R2 go on the numerator and not R1?
• At , it is stated that v+ = v_in. But isn't v_in = v+ - v-? How are these two statements consistent with each other?
• In this video, v_in is the name of the voltage created by the voltage source on the left side of the circuit. It is connected directly to the v+ input of the opamp. In other opamp videos, the same variable name might be assigned a different meaning. like v_in = v+ - v- as you suggest. There are no standard rules or conventions for which signal gets to be called v_in. It all depends on the person who draws the schematic.
• @, (R1 + R2) / R2 is not 2 if R1 is not close to being equal to R2; right?
• Hi Julio,

Correct, the op-amp circuit's gain is determined by the resistors as :

Vout = [(R1 + R2) / R2] x Vin

There is nothing special about the gain of 2. It simply occurs when R1 = R2:

(A + B)/B = 2 when A = B

Regards,

APD
• Is the non-inverting amplifier ALWAYS designed to have two resistors stacked on one another while connected to Vout? And how do you know the size of the resistors to use for this?
• Vanessa - Good question. The non-inverting opamp configuration pretty much always looks like the one in this video, with a feedback resistor connected between the output pin and the - input, and another resistor to ground. (Another way to do non-inverting is called an "instrumentation amplifier". That's a pretty fancy circuit.)

Designing resistor values is a balance. Small resistors mean the opamp output will have to drive a lot of current all the time (lots of power wasted). Really big resistors (1M and above) mean the current in the resistor stack is nice and low, but you have to be wary of the tiny (but never 0) current that actually flows into the negative input pin. Even a tiny input current like 1 uA produces a significant voltage drop across a 1 MOhm resistor. That gives you errors in the gain you expect. So a good balanced approach is to select resistors in the range of 10K to 100K ohms. You can find the input current for any opamp on its data sheet.