# Transfer function of OpAmp

Discussion in 'Homework Help' started by Emil Skovgaard, Dec 28, 2015.

1. ### Emil Skovgaard Thread Starter Member

Jun 6, 2015
37
1
Hi,

I have this exercise where i shall show that the transfer function for the OpAmp is as shown (se the attached files for the transfer function and the circuit).

I know, that when working in the s-domain, R, C and L changes, or at least C = 1/sC and L = sL. Then i gotta work with Laplace to get it from the t-domain to s-domain, but i simply can't see how i should add the R, C and L, and in what order, as i don't see how 1/R1*C occurs, for example.

Hope to get some help,

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2. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
This "1/R 1*C" has nothing to do with how the individual component are connected.
Your transfer function is shown is standard form and this requires a lot of math work. So just be pure math you have 1/R 1*C in your equation.
Try it yourself
$H(s)=-\frac{Z2}{R_1} =- \frac{\frac{(R_2+s*L)*(\frac{1}{s*C})}{(R_2+s*L)+(\frac{1}{s*C})}}{R_1} = -\frac{(R_2+s*L)*(\frac{1}{s*C})}{R_1*((R_2+s*L)+(\frac{1}{s*C}))}$

3. ### MrAl Well-Known Member

Jun 17, 2014
2,433
490
Hello,

If i understand you correctly you just want to know how 1/R1C comes about separated from the rest of the equation?

To find out, calculate the full form transfer function as Jony had shown for example, then multiply that by R1 and by C, then separate numerator N and denominator D, then divide both N and D by C and by L, then expand both into individual terms. You can then reconstruct the final form by using the expanded N as the new numerator, the expanded D as the new denominator, then multiply that by 1/R1C and you'll have the same form as shown in your diagram of the equation.

4. ### Emil Skovgaard Thread Starter Member

Jun 6, 2015
37
1
Thanks for the help both of you, but i think i must have explained my question badly, my apologies..

With attention to the OpAmp circuit, i shall explain that the transfer function is as the H(s) expression (both files are the attached ones).
But it is unclear for me how to show that. I can see the transfer function is similar to the voltage divider formula, but how the components insert in the formula is unclear for me.
Hope that is a better explanation?

5. ### sailorjoe Member

Jun 4, 2013
361
63
Emil, I think you have everything you need.
Jony130 showed you how to derive the transfer function from the circuit topology (arrangement).
You have the transfer function format you want to target.
Now it's just algebra.
Can you handle it from here?

6. ### Emil Skovgaard Thread Starter Member

Jun 6, 2015
37
1
I think i'm getting a better understanding. Correct me if im wrong. When working with transfer functions for OpAmp's the general formula is something like H(s) = Z2/Z1. In this case, as there is nothing more than R1, it's just called R1, and not Z2.

Then the voltage runs through both (R2 +L, and C) i see thats where the (R2 + sL) expression comes from, but then why * 1/sC. Is that because they sit in parallel ?

I really appreciate the help!

7. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
Yes, exactly. We simply know that the voltage gain for inverting amplifier is Vout/Vin = - R2/R1.

Notice that R2 is connected in series with a XL. And capacitor is in parallel with (R2 + XL). So, the equivalent impedance is equal to:
Z2 = (R2 + XL)||Xc; The equation for parallel connection is
https://en.wikipedia.org/wiki/Series_and_parallel_circuits#Resistors_2
1/Z = 1/(Z1) + 1/Z2 ... 1/Zn But if we have two component connect is parallel we can use this equation:
Z = (Z1 * Z2)/(Z1+Z2)

And this is why in post 2 you can see this:

Z2 = (R2 + XL)||Xc = ((R2+XL)*Xc)/((R2+XL)+Xc)

8. ### Emil Skovgaard Thread Starter Member

Jun 6, 2015
37
1
In the Z = (Z1*Z2)/(Z1+Z2), since we don't have any Z1 it just equals zero, and therefore it ends up being Z2/Z2 ?

9. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
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This equation is used when we want to find a equivalent impedance of a two impedance connected in parallel.
In our example Z1 = (R2 + XL) and Z2 = Xc

10. ### Emil Skovgaard Thread Starter Member

Jun 6, 2015
37
1
Of course. So when using math on the expression, i end up having the shown expression. Thanks for the clarification!

In the addition to this, i'm asked to determine the gain H(jw) when w is going to infinity, and when w is 0.
As it is an inverting OpAmp, it guess the formula is Vout/Vin = - R2/R1. Is it just inserting the two Ω-values? But how do i come around the infinity and 0? If i multiply by 0, the whole expression would just be 0, and by infinity, it would never stop growing?

11. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
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For ω = 0 we have Xc = ∞ and XL = 0Ω . So the voltage gain is ??

But for ω = ∞ we have Xc = 0Ω and XL = ∞, ans the gain is ?

12. ### Emil Skovgaard Thread Starter Member

Jun 6, 2015
37
1
Sorry, i don't understand the inputs. When ω = 0, why do Xc = ∞, and XL = 0Ω. Do we insert the ω-value all the places where x is? cause x = s = jω ? Do the resistor values stay the same, we don't do anything about them? Can you elaborate, please?

13. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
One quick question. Do you know what Xc and XL is ??

tsan likes this.
14. ### Emil Skovgaard Thread Starter Member

Jun 6, 2015
37
1
When you ask that way i'm not sure. I supposed it was just another letter for s, as it would fit with sL, and 1/sC when you rename the components.. but i might be wrong?

15. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
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So, what sL and 1/sC is and what ω (omega) is ?? And do you ever herd about capacitance reactance/inductive reactance?
Also, is this some kind of a homework?

16. ### Emil Skovgaard Thread Starter Member

Jun 6, 2015
37
1
Yes, i know what sL, 1/sC and ω is, and i have heard of capacitance, reactance and inductance.

No, it's not homework. It's supplementary material for those who wan't, and i just wanna get a better understanding

17. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
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If you don't understand the basics, you can do it in a hard way. Simply s = 0. And plug this number into your transfer function.
s--->0
$H(s->0)=-\frac{1}{R1 C1} * \frac{0 + \frac{R2}{L}}{0^2 +\frac{R2}{L}*0 + \frac{1}{C1 L}} =-\frac{1}{R1 C1} * \frac{\frac{R2}{L}}{\frac{1}{C1 L}}=- \frac{1}{R1 C1} *\frac{R2}{L}*\frac{C1*L}{1} = - \frac{R2}{R1}$

But this result should be obvious for you, without even knowing the transfer function. These are really basic stuff. How can you study s-domain analysis without knowing the basics.
For ω = 0 it should be clear for you that capacitance reactance is ∞ and the inductive reactance is 0Ω.
Which means that capacitors become open circuits and inductors become short circuits. And this is why we have -R2/R1 for ω = 0

18. ### Emil Skovgaard Thread Starter Member

Jun 6, 2015
37
1
I know, and apologies for missing 'spots' in my knowledge and basics about circuits - it is a very speedy course i'm attending, so hard to get hang of everything, so therefore. But i'm trying my best, and thanks for taking the time

So by going the hard way around, it would just be inserting s = 0, and s = ∞? But in the example you show, where you put s = 0, and end up with -R2/R1, is that considered the gain when s = 0?

I just refreshed reactance ( http://www.allaboutcircuits.com/textbook/alternating-current/chpt-5/review-of-r-x-and-z/ ). But i'm not sure what it has to do with the calculation of the gain?

19. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Yes, gain for s = 0.
This give us a "tool" that we can use to predict the gain by inspection for ω = 0 and ω = ∞ without even knowing the transfer function.

20. ### Emil Skovgaard Thread Starter Member

Jun 6, 2015
37
1
And the solve for s = ∞ the same way, and i will have the gain for both conditions?

So you mean predicting the gain, just by looking at what components the circuit contains? If so, i'm not familiar how to do that - what may i search for to read about such, if that is possible?