How op amp can keep V-=V+??? - Page 1
Forum section: electronics / general

author: kgavionics
author_rank: Regular Contributor
timestamp: November 12, 2012, 12:22:39 am
content: hii'm wondering how op amp can make the inverting input equal to non inverting input? is this due to the negative feedback or the internal structure of the op amp it self?i know thatvout=Aol( v+ - V-)and we know that Aol is infinite so vout/Aol =v+ - V- =0 so v+ =V-i'm wondering technically how this can be made by the op amp to always keep this difference is zero?thank you in advance for clearifying this for me

author: Alana
author_rank: Frequent Contributor
timestamp: November 12, 2012, 12:34:00 am
content: Its the feedback, without it opamp becomes easly saturated [works as comparator].

author: tom66
author_rank: Super Contributor
timestamp: November 12, 2012, 12:38:03 am
content: The different isn't zero. In an ideal op-amp with a finite gain, the difference is equal to Vout / Open Loop Gain. In reality, the inputs have a small voltage offset which also influences the differential voltage.

author: Mechatrommer
author_rank: Super Contributor
timestamp: November 12, 2012, 02:03:13 am
content: Quote from: kgavionics on November 12, 2012, 12:22:39 amthis is the question that i choose to ignore for so long. in order to know the answer you need to learn how the internal structure works. current mirror thing and differential input bjt thing.Quote from: kgavionics on November 12, 2012, 12:22:39 amyes it is due to the internal structure of the opamp it self.Quote from: kgavionics on November 12, 2012, 12:22:39 amno Aol is not infinite in real world. and the whole formula starts to collapse at the edge of the opamp's BW limit.

author: ejeffrey
author_rank: Super Contributor
timestamp: November 12, 2012, 02:51:44 am
content: Quote from: Mechatrommer on November 12, 2012, 02:03:13 amNo.QuoteNo.The zeroing (near zeroing -- the real world is never perfect) of the input voltage is due to negative feedback. An opamp is an extremely high gain differential amplifier. What does that mean? It means:Vout = (V+ - V-) * almost_infinityWe can solve that:V+ - V- = Vout / infinity.This tells us that if Vout is finite, the input voltage differential must be (nearly) zero. This is a very generic idea, and it applies to pretty much any amplifier using negative feedback whether it is an IC or discrete, uses transistors or valves, or even to things like mechanical governors and hydraulic systems. It is up to you as a designer to make sure that the feedback happens in the right direction (negative feedback, not positive) to make sure that this actually happens. Design it wrong, and the opamp will just saturate and just act as a poor comparator.Most everything you need to know to use opamps can be understood without any reference to transistors, or any of the other elements that make up the opamp. It is a black box. If you want to understand why they work, and why they have certain non-ideal behavior then you need to look inside.

author: kgavionics
author_rank: Regular Contributor
timestamp: November 12, 2012, 03:23:37 am
content: i'm confusing this subject more than before

author: Mechatrommer
author_rank: Super Contributor
timestamp: November 12, 2012, 03:28:42 am
content: Quote from: ejeffrey on November 12, 2012, 02:51:44 amand you need a good recipe for the feedback loop to work. the idea of feedback loop is just that... an "idea".Quote from: ejeffrey on November 12, 2012, 02:51:44 amat least it works! as poor comparator? thats not the worst part. the worst part is it doesnt work at all, it oscillates or even smoke.Quote from: ejeffrey on November 12, 2012, 02:51:44 ami assume by posting such question, the OP is interested at the first bolded quote, and the answer you just gave him in the 2nd bolded quote

author: Gall
author_rank: Frequent Contributor
timestamp: November 12, 2012, 11:33:42 am
content: Let's do some math and see how it works.The opamp has two inputs and one outputs, all Us are relative to ground. The output is a function of two inputs:Uout = f(Uin+, Uin-)Consider it is a differential amplifier. Now the output is the function of difference only:Uout = f(Uin+ - Uin-)This function is generally not linear but it is monotonic in the working range of the opamp. It is also xero if the difference is zero. Let's approximate it with a polynomial (Taylor series):Uout = a1 * (Uin+ - Uin-) + a2 * (Uin+ - Uin-)^2 + a3 * (Uin+ - Uin-)^3 + ... + ak * (Uin+ - Uin-)^k + ...Since the non-linearities are small, we can neglect everything above the 1st order. (This is not essential mut makes our math much simplier). ThenUout = a1 * (Uin+ - Uin-).a1 here is the gain, let's denote it as k.Uout = k * (Uin+ - Uin-)A good opamp has very large k, about 10^7 or more. Let's assume an ideal opamp having infinite k. ThenUout = lim[k->infinity] k * (Uin+ - Uin-)There are two possible cases:1. (Uin+ - Uin-) is nonzero.2. (Uin+ - Uin-) is nearly zero.In first case we get infinite voltage at the output. In practice that means saturation and the output tied to the power rail. Any small output voltage may be caused only by (almost) equal Uin+ and Uin-.Result:the opamp may output voltage without overload and saturation if and only if the voltages at its inputs are (almost) equal.If we want the opamp to work like an amplifier, not like logical gate or trigger, we want its inputs to be at equal potentials. To achieve this we use negative feedback.Consider the Uin- to be connected to Uout through an 1:n voltage divider. Now we get:1. Uin+ > Uin-.Then Uout = infinity, Then Uin- = infinity/n = infinity. This is impossible if Uin+ is not infinity (that is, within operating range).2. Uin+ < Uin-.Then Uout = -infinity. Then Uin- = -infinity/n = -infinity. This is impossible if Uin+ is not -infinity.3. Uin+ = Uin-.Then Uin- = Uout / n, or Uout = n * Uin-. But since Uin+ = Uin-, we get Uout = n * Uin+. This is a non-inverting amplifier.

author: Kremmen
author_rank: Super Contributor
timestamp: November 12, 2012, 12:33:32 pm
content: Quote from: kgavionics on November 12, 2012, 03:23:37 amTo properly un-confuse yourself, please forget initially all about op amp internals and physical details. Before getting there, you need to understand the basic idea of a feedback system. This idea has nothing whatsoever to do with any physical implementation. As already noted a feedback system can be an electronic amplifier, a mechanical device or it can be something totally different. To illustrate, the dynamic predator-prey system formed by owls and lemmings, is a feedback system. Feedback is aconceptnot an implementation.Going to an ideal op amp it has been pointed out that the output is the difference of inputs multiplied by infinite gain. How then can you create a circuit that makes sense from that? The answer of course is (negative)feedback. A simple to understand concept is the voltage follower: feed a signal to positive input, and feed the output back to the negative input (thereby making it negative feedback) and the amp output will follow the positive input precisely.More generally if we note the op amp open loop gain with G and the feedback path gain with H and the closed loop gain of the combined circuit with g, it generally applies thatg = G / (1 + GH)For very large values of G, you can see from the equation that actually the gain is in practice defined by H and H only. Thus if we consider the previous example we get:g = [very large] / ( 1 + [very large] * 1) (the term H is 1 because you feed all of the output signal back to neg input)Further you can easily see that a large number divided by another that is 1 bigger won't make much of a difference, so in practice the answer is 1.What happens if you feed the amp output into a pair of series resistors grounded at the other end and take the signal from the midpoint. If the resistors are identical, the voltage fed back to neg input is ½ the output. Now the g equation becomes large/(1+large*0.5) which is as close to 2 as does not matter.So, for very high forward gain G, it is only the feedback that defines the magnitude and nature (proportional, integrative, derivative) of the output.To understand amp circuits properly, you must understand feedback. Feedback is properly a subset of control theory so look in that direction for more info. Should be plentiful in the interweb.

author: Mechatrommer
author_rank: Super Contributor
timestamp: November 12, 2012, 02:25:40 pm
content: there's nothing precious than an advice from AoE (or from here iirc)..."the opamp will output anything in order to make +ve and -ve input equal"thats all i need to know. further detail behaviours i will look into the datasheet. there is no way i'm going to dig whats inside the opamp physical (unless i have time to waste), i dont care how it works why it works, all i know it got to work. as ejeff said, its a blackbox. so buckle up your ohm and kirchoff law and use just that Rule of Thumb to design your circuit with an opamp.

author: AlfBaz
author_rank: Super Contributor
timestamp: November 12, 2012, 09:38:27 pm
content: Ahh... I was wondering what this was all about. So we are talking about the virtual earth.Lets see if i can describe it in plain english.We have a typical inverting op-amp arrangement with Rin going to the negative input, the Rfeedback going from the negative input to the output. Positve is tied to ground.If both +ve and -ve inputs are at 0V then the differnce is 0V so we have no output.You now apply a positive voltage to Rin, this voltage appears on the -ve input. The differential amplifier sees a difference and amplifies it. Since this is a +ve voltage into the -ve input, the output will go -ve.This -ve voltage is fed back via Rfeedback to the -ve input which currently has a positive voltage on it. Since the two voltages are opposite polarity, the negative ouptut voltage starts bringing the Postive input voltage at the inverting input down toward zero, at which point the diferential amplifier is happy since both of its inputs are at zero.If we were to increase the input voltage further the output voltage will rise proportionally to cancel out the input voltage.Keep in mind that although you asked "what keeps the input at zero?" if the non-inverting input was tied to a voltage other that zero, then this would be the voltage present on the inverting input.The current path is as follows.Since the amps input is very high inpedence, the current cannot flow into the amp, so the input current flows through Rin and Rfeedback into the output (which is -ve) through the amps totem pole bottom transistor to the negative supply rail.

author: kgavionics
author_rank: Regular Contributor
timestamp: November 13, 2012, 03:07:21 am
content: thank you guyswell explained