# Design an integrator

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## Homework Statement

Design an integrator to integrate a ##2 kHz## square wave with peak to peak amplitude ##0.5V##.

Choose ##R_F = 150 k##, ##C_I = 10 \mu F##.

Choose ##C_F## such that ##R_F >> \frac{1}{\omega C_F}##.

Choose ##R_I## such that ##R_IC_F << 0.5 ms## and ##\frac{1}{\omega C_I} << R_I##.

Comment on the phase shift. ## The Attempt at a Solution

I am having difficulty with this for some reason.

I derived the equation for the square wave (in volts) to be:

##v_i(t) = 0.25, \space 0 \leq t \leq \frac{1}{4000} s##
##v_i(t) = -0.25, \space \frac{1}{4000} s \leq t \leq \frac{1}{2000} s##

Now here's where I had trouble. I don't know how to meet the specifications which have been prescribed.

I tried deriving a transfer function (which was an absolute mess) using:

$$\frac{v_o}{v_i} = - \frac{Z_2}{Z_1} = - \frac{Z_{C_F} || R_F}{R_S + R_I + Z_{C_I}}$$

After massaging the expression a bit, I got:

$$\frac{v_o}{v_i} = - \frac{ \frac{R_F}{C_F} }{ (\frac{R_S}{C_F} + \frac{R_I}{C_F} + \frac{R_F}{C_I}) + j( \omega(R_F R_S + R_F R_I) - \frac{1}{\omega C_F C_I} ) }$$

Which I believe to be incorrect, because when I substitute ##\omega = \omega_0 = \frac{1}{R_F C_F}## into the gain, the magnitude of the denominator is not ##\sqrt{2}## due to the ##- \frac{R_F}{C_I}## term.

I can't see what I'm doing wrong in terms of the transfer function, and I'm not sure it's the right way.

It would be much appreciated if someone could assist me with this.

Thank you.

I thought of a completely different approach. I was so hungry earlier my brain went on vacation, but I believe that this question is more specific than I originally thought.

I didn't originally figure that ##f = 2 kHz \Rightarrow \omega = 4000 \pi \frac{rad}{s}##

This allows the choice of ##C_F## such that ##R_F >> \frac{1}{\omega C_F} \Rightarrow C_F >> \frac{1}{\omega R_F} = 5.31 \times 10^{-10} F##.

So choosing ##C_F >> 5.31 \times 10^{-10} F = 0.531 nF## would be fine, say ##C_F = 1 \mu F## since I'm sure I can find a capacitor like that.

Now I need to choose ##R_I << \frac{0.5 \times 10^{-3} s}{1 \times 10^{-6} F} = 500 \Omega## and ##R_I >> \frac{1}{(4000 \pi)(10 \times 10^{-6})} = 7.96 \Omega##.

I notice that the choice of ##C_F## was very important, otherwise the resistance ##R_I## would have a very large upper bound.

I figure an average of the lower and upper bound would be appropriate, so choose ##R_I = \frac{500 + 7.96}{2} = 253.98 = 254 \Omega##.

Does this make more sense?

Now here's where I had trouble. I don't know how to meet the specifications which have been prescribed.
With the specified squarewave as input, is there a prescribed amplitude for the triangular wave it must produce at the output?

It's the current through Rin that charges C during each half-period, so you can readily determine how many volts the output will ramp up during that 0.25 msec interval. (Your examiner may not be content with the output being mere fractions of a μV.)

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Which I believe to be incorrect, because when I substitute ##\omega = \omega_0 = \frac{1}{R_F C_F}## into the gain, the magnitude of the denominator is not ##\sqrt{2}## due to the ##- \frac{R_F}{C_I}## term.

I can't see what I'm doing wrong in terms of the transfer function, and I'm not sure it's the right way.
It would be much appreciated if someone could assist me with this.
.

Zondrina, as I can see you are allowed to neglect the influenmce of the input coupling capacitor (1/wCi<<Ri).
That means: Setting Ci to infinite in your transfer function, you get - as expected - the first-order lowpass function.
There was nothing wrong in your calculation - but you didnt consider that the coupling capacitor Ci disturbs the classical 1st-order lowpass function (and the corresponding expectation regarding -3dB)

Comment: Please note that it is very uncommon to have in numerator and denominator expressions like R/C.
It is better and more logical to have a numerator showing the maximum gain (in this case: RF/Rin with Rin=Rs+Ri)

Zondrina, as I can see you are allowed to neglect the influenmce of the input coupling capacitor (1/wCi<<Ri).
That means: Setting Ci to infinite in your transfer function, you get - as expected - the first-order lowpass function.
There was nothing wrong in your calculation - but you didnt consider that the coupling capacitor Ci disturbs the classical 1st-order lowpass function (and the corresponding expectation regarding -3dB)

Comment: Please note that it is very uncommon to have in numerator and denominator expressions like R/C.
It is better and more logical to have a numerator showing the maximum gain (in this case: RF/Rin with Rin=Rs+Ri)

There is a problem with your theory. The Input capacitor ##C_I## is intended to be ##10 \mu F##, so I can't possibly allow it to be infinite.

I think because I'm designing the circuit to handle a specific signal, then I can take the frequency of the signal as the network frequency.

Hi Zondrina,
in case you are not allowed to neglect the influence of Ci (app. 10 ohms at f=2kHz) you have, in principle, a bandpass response (rather than a first order lowpass).
Hence, your SQRT(2)-test is not applicable.
For evaluating the transfer function, I suggest to create a "normalized" form:
Numerator N(jw) and denominator D(jw) as polynominals in (jw).
More than that, D(jw) should have the form D(jw)=(1 + A(jw) - Bw²).
This is the classical bandpass expression which enables you identification of the pole frequency (center frequency) as well as the Q value (bandwidth).

Zondrina - my last response concerns the (theoretical) transfer function only.
Do you have further questions regarding the integrating operation?

I am having difficulty with this for some reason.
I'm quite sure that you are not expected to use the transfer function. You should apply the design guidelines in your first post, viz., RC time-constants. The design process looks like it should take you all of 10 minutes, as half the element values have already been decided upon.

There are a couple of additional considerations not mentioned, but which you have probably covered in lectures. One is that input and feedback resistors in general OP-AMP circuits should be kept within the range roughly 1kΩ .. 10MΩ so that OP-AMP limitations don't come to the fore.

The other I alluded to by asking about the expected amplitude of the triangular wave when the squarewave is integrated. Presumably we want it to be of a usable amplitude, and not mere microvolts. By the same token, we don't want to drive the OP-AMP into saturation!

When faced with choosing R or C so that its value is ">>" some other, you may find there is not so much leeway that you can go for a factor as wild as x1000. Maybe start with x50 and see how that jives with other considerations. If all looks good, perhaps try x100 or x200.

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I decided the ##C_F## I calculated was much too large. I decided to let ##C_F = 1 nF##, so the resistance ##R_I## was between ##8 \Omega << R_I << 5k \Omega##.

I then chose ##R_I = 2.5 k\Omega##, and it worked quite nicely.

I decided the ##C_F## I calculated was much too large. I decided to let ##C_F = 1 nF##, so the resistance ##R_I## was between ##8 \Omega << R_I << 5k \Omega##.

I then chose ##R_I = 2.5 k\Omega##, and it worked quite nicely.
You built and tested this? Or simulated it?

I'm wondering how you answered "Comment on the phase shift."?

Zondrina, as I can see you are allowed to neglect the influenmce of the input coupling capacitor (1/wCi<<Ri).
That means: Setting Ci to infinite in your transfer function, you get - as expected - the first-order lowpass function.
There was nothing wrong in your calculation - but you didn`t consider that the coupling capacitor Ci disturbs the classical 1st-order lowpass function (and the corresponding expectation regarding -3dB)

Comment: Please note that it is very uncommon to have in numerator and denominator expressions like R/C.
It is better and more logical to have a numerator showing the maximum gain (in this case: RF/Rin with Rin=Rs+Ri)

I understand this more now, though it wasn't relevant to the problem. For ##\omega >> \omega_0##, ##Z_C ≈ 0##. So the capacitor acts like a wire.

As for the phase, ##\phi = 90^o##.

So the capacitor acts like a wire.
As for the phase, ##\phi = 90^o##.

A piece of wire with 90 deg phase shift?
Just the opposite: Because 1/wCf<<Rf the feedback path acts nearly as an ideal capacitor.