Frequency response analysis for closed loop amplifier

[Jenifer_gao]

frequency response

Hi All:

When I looked into the active filter design, I find an interesting phenomena. The frequency response will depend on the external components. An example can be shown in the attachment of a first order low pass active filter. In this example the transfer function is:

H(s) = -(R1/R2)[1/(1+sCR1)]

so ωo = 1/CR1, which is independent of the amplifier itself, and this circuit has a single pole. My understanding is no matter how many poles the amplifier has, it won't contribute to the frequency response performance, only the external component will affect it.
But it doesn't make a sense if the feedback is a pure resistor network, because I know the frequency of this structure is subjected to the amplifier.
If anybody can show me the right way to do the frequency response analysis for the closed loop amplifier. Thanks in advance.

J

 

[noiseless]

Re: frequency response

in what you said the LPF depends on the external components, there is an assumption that the OPamp bandwidth is large enough. If the LPF pole is compatiable to the poles of the OPamp, your transfunction will not work

 

[VVV]

Re: frequency response

The open loop gain of the opamp is so high that generally the gain of the circuit, determined by the external components is much lower. Thus, the characteristic will be like the blue line. As long as this is below the red line, the internal pole will have no effect, even though it is generally at very low frequencies, about 10Hz.

But you cannot push the filter pole too far out to the right either, since then your internal characteristic would take effect. In other words, the red line is like an overall LIMIT. You cannot get a gain higher than that, only lower. That applies to any frequency, so that means that when you try to intersect the sloped portion, it will limit the response, making it look like it has a pole there.

So as long as you stay below the red line the circuit will behave as if the opamp were an ideal one and results will match your calculations.

 

[Jenifer_gao]

Re: frequency response

Thanks VVV and noiseless. Your explaination are very clear and understandable. I would like to discusss it with your guys in further detail.

According to two books at my hand, I find there are two ways to derive the transfer function of the closed loop amplifier. One is to treat the system as two blocks, as shown in figure 1, amplifier and the feedback network; another is using the small signal analysis method, as shown in figure 2.

For the first method:
H(s) = A(s)/(1+A(s)F(s))
where A(s) is the open loop transfer function of the amplifier, F(s) is the feedback factor, which is also a function of the frequency. By knowing the A(s) and F(s), H(s) can be derived. From this equation we can find that the frequency response of this system actually depends on the A(s).

For the second method:
H(s) = -Ceq[C1/(C1+Cin)](Gm-G2s)Ro/{Ro(CLCeq + CLC2 + CeqC2)s+GmRoC2}
(from Razavi's book on page 437)

we can't see clearly the effect of the amplifier on the final frequency response

If anybody can tell me which method is better or more accuracy?
Thanks.

 

[VVV]

Re: frequency response

The frequency response does depend on the characteristic of the amplifier. See my previous post, for example.

But, the transfer function is H(s)=A(s)/(1+A(s)B(s))

For large A(s), which is usually the case within the frequency of interest, you can see that the transfer function depends little on the opamp. But if you approach the point where A is not very large, you will see that effect even more pronounced.

In the problem you are referring to (I am not familiar with Razavi's book), the inverting input sees a star configuration, that is, Cin is connected betweeen the node and ground. You need to get rid of it, since it does not "conform" to the formulas you are familiar with. In all the formulas you only care about input and feedback impedances. You can always use transformations like this to make the circuit look like something you can easily analyze.

Now, it may not be obvious, but the second formula does depend on the opamp characteristic, since Gm depends on frequency.

 

[fantaci]

frequency response

I think the fig1 and fig2 is using different method. The fig1 see the amp as an ideal amp and this amp`s gain does not depend on load of amp. But the fig2 see the amp output as current rather than voltage. that is, the voltage gain is shifting depending on load, but the gm of this amp does not change. I think both way is right, it depends. if your amp has a comparably very low output impedance, you could use the method in fig1 and does not introduce much error.