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## Electrical engineering

### 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? • 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"?
• 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
• if the + side of a power supply is hooked to a circuit and the - side is simply grounded, how can we expect it to work? I mean how is the current going to flow, and in what path?
and is grounded the same as "just not connected to anything"? • Without those extra resistors will the gain of the non-inverting op amp be 1 ? If yes, can you explain how it is 1 ?
(1 vote) • I know v- have the same frequency that v+ has because of the linear nature of our circuits, but I wonder if we can assume that the phase is the same in the two both of them. In a mixer circuit we would have an v(out) deformed in relation with v+... Isn't it?

Thank you very much for this site and this course!
(1 vote) • As we start studying opamps, we talk about how they work at very low frequency. Real opamps always have a certain amount of time delay for signals passing through them. When we limit the inputs and outputs to change slowly compared to the delay through the opamp, we can think about how the circuit works when all the signals are in phase.

Later on, as you push the frequency higher, the output signal starts to lag behind the input signals, resulting in a phase difference between output and input. In the extreme (but not rare) case, the phase delay can be so large the feedback changes from negative feedback to positive feedback and the circuit "rings" with its own oscillations (this is almost always a bad thing, unless you want to build an oscillator).

The mixer circuit you mentioned is a particular kind of circuit function that multiplies two sine waves together. It's operation is usually not based on phase delay, but rather on other design features.