• Support PF! Buy your school textbooks, materials and every day products Here!

Transfer function of an ideal OP amp

  • #1

Homework Statement



I have to find the transfer function of a linearly and ideally operating OP Amp.

http://i.imgur.com/u3f6E.jpg

Homework Equations





The Attempt at a Solution



I know that the current through the capacitor, R1, and R2 is the same, so i=C(dv/dt) = Vin/R1 = (VIN-V0)/R2, where do I go from here? Is the transfer function just V0/VIN in terms of R1, R2, C and dv/dt? When I equate the second and third equations I get V0/VIN = 1-(R2/R1) but i'm not sure if this is right.

For the second part, I assume I just find a value of R1 that gives a value of A = 20-& in the equation VIN=A(gain) V0
 

Answers and Replies

  • #2
berkeman
Mentor
56,841
6,818

Homework Statement



I have to find the transfer function of a linearly and ideally operating OP Amp.

http://i.imgur.com/u3f6E.jpg

Homework Equations





The Attempt at a Solution



I know that the current through the capacitor, R1, and R2 is the same, so i=C(dv/dt) = Vin/R1 = (VIN-V0)/R2, where do I go from here? Is the transfer function just V0/VIN in terms of R1, R2, C and dv/dt? When I equate the second and third equations I get V0/VIN = 1-(R2/R1) but i'm not sure if this is right.

For the second part, I assume I just find a value of R1 that gives a value of A = 20-& in the equation VIN=A(gain) V0
Welcome to the PF.

You are correct that the same AC current flows through C, R1 and R2. But the current is not Vin/R1. Vin is across both R1 and the capacitor, so that must show up in your equations.
 
  • #3
Welcome to the PF.

You are correct that the same AC current flows through C, R1 and R2. But the current is not Vin/R1. Vin is across both R1 and the capacitor, so that must show up in your equations.


Sorry I didn't realise that, thanks.

So via KCL I can assume that Cdv/dt + Vin/R1 = (VIN-V0)/R2.

This results in a fairly awkward equation if I want to solve for V0(w)/VIN(w):

VIN(1/R2-1/R1) - Cdv/dt = V0/R2

I'm not sure where to go from here, I have no initial or final values for v(t) so i can't really do anything with the dv/dt.

thanks in advance.
 
  • #4
berkeman
Mentor
56,841
6,818
Sorry I didn't realise that, thanks.

So via KCL I can assume that Cdv/dt + Vin/R1 = (VIN-V0)/R2.

This results in a fairly awkward equation if I want to solve for V0(w)/VIN(w):

VIN(1/R2-1/R1) - Cdv/dt = V0/R2

I'm not sure where to go from here, I have no initial or final values for v(t) so i can't really do anything with the dv/dt.

thanks in advance.
No no no no no no no. :smile:

Vin is NOT across R1. It is across both the cap and resistor. So the LHS of your first equation above needs to change...
 
  • #5
Welcome to the PF.

You are correct that the same AC current flows through C, R1 and R2. But the current is not Vin/R1. Vin is across both R1 and the capacitor, so that must show up in your equations.
No no no no no no no. :smile:

Vin is NOT across R1. It is across both the cap and resistor. So the LHS of your first equation above needs to change...


Wow, i'm not on the ball at all today.

Right, so i'll let the voltage across the capacitor be Vc.

via KCL I have:

I(cap) + I(R1) = I(R2)

Cdv/dt + (v0-vc)/r1 = (v0-vIN)/r2


And the RHS is the same

This leads to:

Cdv/dt = V0(1/R2 - 1/R1) + Vin(1/R1 - 1/R2)

thanks in advance.
 
Last edited:
  • #6
berkeman
Mentor
56,841
6,818
via KCL I have:

I(cap) + I(R1) = I(R2)
No. The current through the cap equals the current through each resistor. They are all in series (assuming the ideal opamp needs no input bias current).
 
  • #7
Sorry I meant equals, I had it in the first post I was just getting confused.

According to this Cdv/dt = (V0 - VIN)/R1 = (V0 - VR1)/ R2

where VR1 = (R1)Cdv/dt
 
  • #8
berkeman
Mentor
56,841
6,818
Sorry I meant equals, I had it in the first post I was just getting confused.

According to this Cdv/dt = (V0 - VIN)/R1 = (V0 - VR1)/ R2

where VR1 = (R1)Cdv/dt
Two things. First, use the "virtual ground" property at the - input node of the opamp. Since the + terminal is connected to ground, what is the voltage always going to be at the - input (due to the negative feedback and infinite gain of the ideal opamp)? That helps simplify the equation for the current through R2.

Second, you really do need to combine the cap and R1 together as the LHS of an equation. When you try to say things individually about them, you make errors (because the voltage at the point between them would need to be an additional variable if you want to treat them separately. Use the vitural ground property of the - input node to write an equation for the voltage Vin across both C and R1 and the current that flows through both of them.
 
  • #9
Two things. First, use the "virtual ground" property at the - input node of the opamp. Since the + terminal is connected to ground, what is the voltage always going to be at the - input (due to the negative feedback and infinite gain of the ideal opamp)? That helps simplify the equation for the current through R2.

Second, you really do need to combine the cap and R1 together as the LHS of an equation. When you try to say things individually about them, you make errors (because the voltage at the point between them would need to be an additional variable if you want to treat them separately. Use the vitural ground property of the - input node to write an equation for the voltage Vin across both C and R1 and the current that flows through both of them.
If I remember correctly it's V input = V output = 0 in an ideal OP Amp so the potential at the node above the - op amp terminal is 0. This means that the current through I2 is

V0 - 0/R2 = V0/R2.

The Capacitor and resistor in series are a basic RC circuit, and the voltage across them is the same

Vc = Vin[1 -exp[-t/RC]]where R is R1 and C is the capacitance.

I know that this is the voltage across the capacitor but I'm not sure how exactly to combine the resistor and capacitors voltages and even if I do would dividing by the resistance R1 suffice to find the total current flowing into the node?

Regardless, I now have

Cdv/dt = (Vin-0)/R1 = (V0-0)/R2

Cdv/dt = Vin/R1 = V0/R2

So I can conclude that V0/VIN = R1/R2.
 
Last edited:
  • #10
berkeman
Mentor
56,841
6,818
If I remember correctly it's V input = V output = 0 in an ideal OP Amp so the potential at the node above the - op amp terminal is 0. This means that the current through I2 is

V0 - 0/R2 = V0/R2.

The Capacitor and resistor in series are a basic RC circuit, and the voltage across them is the same

Vc = Vin[1 -exp[-t/RC]]where R is R1 and C is the capacitance.

I know that this is the voltage across the capacitor but I'm not sure how exactly to combine the resistor and capacitors voltages and even if I do would dividing by the resistance R1 suffice to find the total current flowing into the node?

Regardless, I now have

Cdv/dt = (Vin-0)/R1 = (V0-0)/R2

Cdv/dt = Vin/R1 = V0/R2

So I can conclude that V0/VIN = R1/R2.
You have the output equation correct.

The input equation is not right still. It's probably best to devine a new voltage at the node between the cap and R1. That way you can write the Cdv/dt using the voltage just across the capacitor. The voltage across the resistor is then related to Vin and Vc.
 
  • #11
You have the output equation correct.

The input equation is not right still. It's probably best to devine a new voltage at the node between the cap and R1. That way you can write the Cdv/dt using the voltage just across the capacitor. The voltage across the resistor is then related to Vin and Vc.

Thanks.

I'll call the node potential between the capacitor and the resistor E.

Cdv/dt = (E - 0)/R1 = V0/R2

Edit: Sorry it's very long winded.
Cdv/dt = E/R1 = V0/R2

You then said to relate the voltage across the resistor (E) to Vin and Vc. I'm not sure at how to do this but i'd guess that


Vin - E = Vc
Vc - Vr1 = E

E = Vin - Vc = Vc - VR1

2Vc = Vin+VR1

but VR1 = E-0 = E

2Vc = Vin + E

but from Vin-E=Vc we get E = Vin - Vc

so

2Vc = Vin + Vin - Vc
3Vc = 2Vin

Vc = 2/3 Vin

E = Vin - Vc = Vin-2/3Vin = 1/3 Vin

Therefore


Cdv/dt = E/R1 = V0/R2

(1/3Vin)/R1 = V0/R2

Giving:

V0/Vin = R2/3R1

Edit: Sorry it's very long winded
 
Last edited:
  • #12
berkeman
Mentor
56,841
6,818
I'm sorry, I didn't track your last post at all.

If you call the voltage between the cap and R1 "Ve", then the v(t) across the capacitor is Vin(t)-Ve(t). And the voltage across R1 is Ve(t)-0V. Does that change your equations?
 
  • #13
I'm sorry, I didn't track your last post at all.

If you call the voltage between the cap and R1 "Ve", then the v(t) across the capacitor is Vin(t)-Ve(t). And the voltage across R1 is Ve(t)-0V. Does that change your equations?


Vin(t) - Ve(t) = Vc(t) (eq1)

Ve(t) - 0 = Vr1(t) (eq2)

Another equation that should work is


Vc(t) - Vr1 = Ve(t) (eq3)

using eqs 2 and 3

Ve(t) = Vr1(t)
and
Ve(t) = Vc(t) - Vr(1)

therefore

Vr1(t) = Vc(t) - Vr1(t)
Vc(t) = 2Vr1(t) (eq4)

Does up to here make sense?

from eqs 1 and 3

Ve(t) = Vin(t) - Vc(t) = Vc(t) - Vr1(t)

from eq 4, Vc(t) - Vr1(t) = 2vr1(t) - vr1(t) = vr1(t)

giving:


Vin(t) - Vc(t) = Vr1(t)

since vc(t) = 2vr1(t)

vin(t) = 3vr1(t)

Vin(t)/3 = vr1(t)

and

Vc(t) = 2/3 Vin(t)


ALSO V(E) = Vin(t) - Vc(t)
Vin(t) - 2/3 Vin(t)
Ve(t)= 1/3 Vin(t)

So going back to the original equations:

Ve(t)/R1 = V0/R2

Giving V0(t)/Vin(t) = R2/3R1


Yes it yielded the same equations.



is up to here correct?
 
Last edited:
  • #14
bump.
 
  • #15
berkeman
Mentor
56,841
6,818
Vin(t) - Ve(t) = Vc(t) (eq1)

Ve(t) - 0 = Vr1(t) (eq2)

Another equation that should work is


Vc(t) - Vr1 = Ve(t) (eq3)

using eqs 2 and 3

Ve(t) = Vr1(t)
and
Ve(t) = Vc(t) - Vr(1)

therefore

Vr1(t) = Vc(t) - Vr1(t)
Vc(t) = 2Vr1(t) (eq4)

Does up to here make sense?

from eqs 1 and 3

Ve(t) = Vin(t) - Vc(t) = Vc(t) - Vr1(t)

from eq 4, Vc(t) - Vr1(t) = 2vr1(t) - vr1(t) = vr1(t)

giving:


Vin(t) - Vc(t) = Vr1(t)

since vc(t) = 2vr1(t)

vin(t) = 3vr1(t)

Vin(t)/3 = vr1(t)

and

Vc(t) = 2/3 Vin(t)


ALSO V(E) = Vin(t) - Vc(t)
Vin(t) - 2/3 Vin(t)
Ve(t)= 1/3 Vin(t)

So going back to the original equations:

Ve(t)/R1 = V0/R2

Giving V0(t)/Vin(t) = R2/3R1


Yes it yielded the same equations.



is up to here correct?
Equation 3 looks wrong to me. The voltages across the capacitor and R1 add up to Vin...
 

Related Threads on Transfer function of an ideal OP amp

  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
1
Views
10K
Replies
32
Views
4K
Replies
6
Views
1K
  • Last Post
Replies
11
Views
642
  • Last Post
Replies
12
Views
3K
Replies
1
Views
626
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
8
Views
938
Top