Design an Integrator: R_F, C_I, C_F, R_I, Phase Shift

  • Thread starter STEMucator
  • Start date
  • Tags
    Design
In summary, the homework statement states that an integrator is needed to integrate a 2 kHz square wave with a peak-to-peak amplitude of 0.5 V. The equation for the square wave is obtained, and it is found that the input capacitor, C, must be 10 µF in order to meet the specifications. It is also found that the resistor, R, must be smaller than 0.5 ms in order to not cause distortion. The phase shift of the square wave is calculated to be -0.25 ms.
  • #1
STEMucator
Homework Helper
2,076
140

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.

Screen Shot 2015-02-01 at 1.56.45 PM.png


Homework Equations

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.
 
Physics news on Phys.org
  • #2
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?
 
  • #3
Zondrina said:
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.)
 
Last edited:
  • #4
Zondrina said:
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 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)
 
  • #5
LvW said:
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)

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.
 
  • #6
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).
 
  • #7
Zondrina - my last response concerns the (theoretical) transfer function only.
Do you have further questions regarding the integrating operation?
 
  • #8
Zondrina said:
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.
 
Last edited:
  • #9
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.
 
  • #10
Zondrina said:
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."?
 
  • #11
LvW said:
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##.
 
  • #12
Zondrina said:
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.
 

What is an integrator and how does it work?

An integrator is an electronic circuit that performs the mathematical operation of integration, which is the accumulation of a signal over time. It uses a combination of resistors and capacitors to achieve this function. The input signal is applied to the capacitor, and the output signal is taken from the resistor. The capacitor charges and discharges over time, creating a smoothed output signal that is proportional to the integral of the input signal.

What is the purpose of R_F in an integrator circuit?

R_F, or feedback resistor, is used to control the gain or amplification of the integrator circuit. It determines how much of the output signal is fed back to the input, and thus, affects the overall output signal. A larger R_F value will result in a higher gain and a smaller R_F value will result in a lower gain.

How do C_I and C_F affect the performance of an integrator?

C_I, or input capacitor, and C_F, or feedback capacitor, are both used to control the integration time constant of the circuit. A larger value for either capacitor will result in a longer integration time and a smaller value will result in a shorter integration time. These capacitors also affect the frequency response of the circuit, with larger values allowing for a wider bandwidth and smaller values resulting in a narrower bandwidth.

What is the purpose of R_I in an integrator circuit?

R_I, or input resistor, is used to limit the current flowing into the input capacitor. This helps to prevent damage to the capacitor and ensures that the circuit operates within its linear range. A larger R_I value will result in a slower charging and discharging of the capacitor, which can affect the overall performance of the integrator.

How does phase shift affect the output of an integrator?

Phase shift is a phenomenon that occurs when signals are processed through electronic circuits. In an integrator, it can introduce a delay or advance in the output signal compared to the input signal. This delay or advance can affect the accuracy of the integrated output. To minimize phase shift, the circuit must be designed with components that have low parasitic capacitance and inductance, and the circuit layout should be carefully considered.

Similar threads

  • Advanced Physics Homework Help
Replies
1
Views
989
  • Engineering and Comp Sci Homework Help
Replies
3
Views
938
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Electrical Engineering
Replies
26
Views
36K
  • Engineering and Comp Sci Homework Help
Replies
30
Views
6K
  • Engineering and Comp Sci Homework Help
Replies
10
Views
3K
  • Advanced Physics Homework Help
Replies
4
Views
2K
Replies
1
Views
504
Replies
3
Views
409
  • Engineering and Comp Sci Homework Help
Replies
2
Views
8K
Back
Top