为什么运算放大器电路需要反馈?


17

我知道,为了使运算放大器正常工作,需要从输出到反相或同相输入(取决于外部电路)的直流反馈环路。

使用运算放大器时,直流反馈的目的是什么?为什么有必要,没有它会有什么影响?



37
这是电阻制造商财团的阴谋。
奥林·拉斯罗普

1
因为它出奇地好。大多数工程师没有这种经验,但是:实际使用节点分析时没有理想的OpAmp假设。将其视为有限增益放大器。您会看到将获得相似的结果,当您假设增益是无限的时,您将获得理想的运算放大器。
Cyber​​Men 2012年

@OlinLathrop他们为什么不禁止电压跟随器?
德米特里·格里戈里耶夫

Answers:


19

理想的运算放大器具有无限的增益。它会放大+和-引脚之间的电压差。当然,实际上,这种收益不是无限的,而是仍然很大。

运算放大器的输出(在某种程度上也是输入)受电源的限制,我们的输出不能超过电源的投入。

如果我们简单地将信号放入运算放大器而没有反馈,则将其乘以无穷大并获得二进制输出(它将在电源轨处饱和)

因此,我们需要某种控制增益的方法。这就是反馈的作用。

反馈(DC和AC)从输入中获取放大输出的一部分,因此,增益受反馈网络的约束更大,这是可预测的,而受大规模(和不可预测的)开环增益的约束则小得多。

即使在仅交流的电路中,我们仍然需要在直流(零赫兹)下工作的反馈,否则增益将仅为直流信号的开环增益。尽管受到限制,但交流信号会被直流开环增益淹没。


没有反馈,运算放大器就可以用作比较器,因此输出并非完全没有意义。
starblue'5

并非所有运算放大器都能与比较器一起使用,对于比较器,您应该只使用它。许多比较器都不能很好地发挥作用。这有点像说电阻器就像保险丝一样工作。是的,但是确实不是一个好主意。(尽管我知道至少有一种设计在这里!)
杰森·摩根

....也许我应该包括一些运算放大器在驶入铁轨或超出其CM范围超载时会做非常奇怪的事情。
杰森·摩根

您仍然可以这样做,编辑答案:还建议您通过编辑(文字左下方的按钮)来改善您的信息
clabacchio

@JasonMorgan:问题不仅仅在于共模范围。如果两个输入之间的电压差变得太大,即使两个输入都在器件可以处理的范围内,某些运算放大器的性能也会很奇怪。
2013年

22

您已经知道运算放大器具有很高的开环放大率,通常为10万倍。让我们看一下最简单的反馈情况:

在此处输入图片说明

运算放大器会放大V 之间的差异: V+V

VOUT=100000×(V+V)

现在V = V O U T,然后V+=VINV=VOUT

VOUT=100000×(VINVOUT)

或者,重新排列:

VOUT=100000100000+1×VIN

一样好

VOUT=VIN

× 1放大器,主要用于获得高输入阻抗和低输出阻抗。

×VOUTVIN

编辑
现在,通过在反馈仅使用输出电压的一小部分,我们可以控制放大。

在此处输入图片说明

再次

VOUT=100000×(V+V)

V+=VINV=R1R1+R2×VOUT

VOUT=100000×(VINR1R1+R2×VOUT)

Or:

VOUT=100000×VINR1R1+R2×100000+1

The term "1" can be ignored, so that

VOUT=R1+R2R1×VIN

Notice that in both the voltage follower and this non-inverting amplifier the actual amplification factor of the opamp cancels provided it is high enough (>> 1).


5

The ideal op-amp has infinite gain, and this is of little use in analog electronics. The feedback is used to limit the gain of the circuit. You can find many examples in the wiki article.

Consider the simple feedback loop:

enter image description here

Vout=AVx

Vf=FVout

Vx=VinVf=VinFVout

Vout=AVinAFVout

Av=VoutVin=A1+AF

In the case of the op-amp, its gain defines A: it will be a quite nasty function, because these amplifier are made for just giving brutal gain, and won't have a nice linear function. Luckily, if you look at Av, if A is big enough it will cancel the 1 and itself leaving 1/F to determine the gain.

In the case of the non-inverting amplifier, the block F is a voltage divider, so it will be something like 1/X. This will set the gain of the amplifier to X.

In the case of real op-amps, A won't be infinite, but big enough to allow cancelling it in the DC gain equation. And the advantages of feedback are even more, like increasing bandwith, linearity, S/N ratio and more. For instance, in a closed loop the gain is determined only by the inverse of the feedback gain, provided that the op-amp gain is big enough.

Actually, one resistor only is not that useful as a feedback, as it behaves the same as a short circuit. A voltage divider to ground makes it behave like a fixed ratio multiplier of the same factor (for the same reason mentioned above).


1
Thanks, I understand that a feedback is primarily needed to control the gain of the amplifier, so whatever the feedback gain, the amplifier gain will be equal to its inverse. Is that correct?
user1083734

1
And do I understand correctly that the single resistor between output and input is not effective because it will not alter/divide up Vout and so the amplifier gain will be the same as its open loop gain, without any feedback. I am not sure on this last point.
user1083734

@user1083734 it's right: if you understand how the op-amp works, and what is the transfer function of the feedback circuit, you are a step closer to understand the whole circuit
clabacchio

Is the feedback transfer function the same as the transfer function of the whole circuit? I can calculate the latter, but do not know how to calculate the former.
user1083734

2

The purpose of DC feedback is to define what you want the op-amp to do, i.e. what its output voltage will be. Without it, the output will rise or fall until it hits the power rails.

This can be useful, and there is a large market for op-amps specialized to work this way, called "comparators".

A comparator is simple: if the + input is greater than the - input, the output is +Vcc. Otherwise, the output is −Vee. The schematic symbol is the same as an op-amp, and they can even with sufficient effort be coaxed into working in both roles, but in practice, the two types are highly specialized, and such efforts are not really worth it.

With the DC feedback path, an op-amp can be stable at some point other than "output hard against the rails", and the circuit is generally designed to find that point.

Rather than thinking about it statically, think about an op-amp as an integrator. Whenever its + input is greater than its − input, an op-amp's output will RISE, rapidly. This rise should being the inputs closer together, finally stopping when they are equal. Likewise, + input less than − input will cause the output to fall. The feedback is generally to the − input because that's the simplest way to make a circuit that works this way.


1
"This rise should being the inputs closer together, finally stopping when they are equal." You don't explain why that happens.
stevenvh

1

A typical power supply error amplifier has no DC feedback path:

Sipex app note  - error amplifier

I can assure you, however, that this amplifier works quite well.

Visualize this error amplifier controlling a buck converter. Vcomp would be used to control the duty cycle of a switch, which controls current flow through an inductor and controls Vout. As Vcomp increases, so does the duty cycle, which causes Vout to increase and Vcomp to decrease. The compensation network will increase or decrease Vcomp in a controlled manner, to force Vout to match Vref (as closely as the opamp will allow).

[ Of course, the power train is providing some semblance of DC feedback, but I digress :) ]


1
I think you are overcomplicating things trying to find an exception to OP's answer, especially because he's asking about feedback (try to abstract from him mentioning a resistor) and your circuit actually HAS feedback, but only for AC signals.
clabacchio

2
The circuit depends on DC feedback also. It's just not shown in the circuit. The circuit shown is not the complete amplifier. Vcomp controls the duty cycle of a switch which then controls Vout, and this is effectively a DC feedback path. There has to be DC feedback, otherwise what will stabilize the amplifier? The AC local feedback will not do that.
Kaz

@Kaz I guess Olin is the only person allowed to have some fun here.
Adam Lawrence

-4

DC feedback in op-amp uses due to stability, also op-amp gain is too high so we use feedback to have a specific gain in output


3
"DC feeback in opamp uses due to stability" makes no sense, at least in English.
Olin Lathrop
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