Design Op-Amp Bandpass Filter for 500 & 1500 rad/s

[hogrampage]

Homework Statement

Select component values to design for corner frequencies at 500 rad/s and 1500 rad/s.

Homework Equations

KCL, transfer function

The Attempt at a Solution

Gain:

[1] G = 1 + $\frac{R_{A}}{R_{B}}$

Voltage division:

[2] V₂ = V₁$\frac{R_{2}}{R_{2}+1/sC_{2}}$

KCL:

[3] 0 = $\frac{V_{1}-V_{i}}{R_{1}}$ + $\frac{V_{1}-V_{2}}{1/sC_{2}}$ + $\frac{V_{1}-V_{o}}{R_{3}}$

I am stuck at this point and don't know what to do next. There is a similar problem in the book and it said to substitute equations [1] and [2] into [3].

I sort of know what to do once I find $\frac{V_{o}}{V_{i}}$, but I just don't know how to get there from where I am.

Thanks!

 

[aralbrec]

Equations 2 and 3 are two equations in three unknowns (Vo/Vi is one unknown). The third equation is [1] the gain formula which relates V2 to Vo. The gain formula gives gain from the + terminal of the amp to the output but you could also relate V2 to Vo just by using the resistor divider Ra/Rb. So this equation let's you replace V2 by an expression in Vo and equation [2] will let you replace V1 by an expression in Vo. Then equation [3] will only contain Vo and Vi.

But honestly it seems like you are throwing equations at it, particularly the gain equation, when it's much simpler not to memorize any equations at all. I'd be looking at removing as many unknowns as possible at the outset. So I would have found V2 in terms of Vo by the resistor divider Ra/Rb. This voltage in Vo would then appear at the + terminal of the op amp. Then KCL at the + terminal would find V1 as a function of Vo and the final KCL at V1 would involve only Vi and Vo.

 

[hogrampage]

I looked up the circuit on Google and found this: http://en.wikipedia.org/wiki/Sallen–Key_topology#Application:_Bandpass_filter

That is the same circuit and the transfer function is in the image attached. According to the page R₂ = 2R₁ and C₁ = C₂.

Would be a good idea to pick C = 1F or and R₁ = 1kΩ, and then use scaling to find the standard values to use?

 

[rude man]

I looked up the circuit on Google and found this: http://en.wikipedia.org/wiki/Sallen–Key_topology#Application:_Bandpass_filter

Would be a good idea to pick C = 1F or and R₁ = 1kΩ, and then use scaling to find the standard values to use?

For 500 rad/s you're looking at capacitors on the order of C= 1/ωR with R=10K to 100K.

 

[hogrampage]

No matter what values I try, none work :| (I am always ~30-50Hz away on one or both of them). R_A and R_B don't have much effect, but I notice R₃ changes the bandwidth (or maybe it's something else, but it makes graph wider/smaller depending on value, as well as the center frequency). I am not sure where I am going wrong.

 

[rude man]

No matter what values I try, none work :| (I am always ~30-50Hz away on one or both of them). R_A and R_B don't have much effect, but I notice R₃ changes the bandwidth (or maybe it's something else, but it makes graph wider/smaller depending on value, as well as the center frequency). I am not sure where I am going wrong.

Go to your link. First, pick your mid-band gain A. That defines R_a and R_b. You might pick A = 1 since nobody told you what it should be.

Then: what is f₀, given your corner frequencies f₁ and f₂? Then, how is Q expressed as a function of f₀, f₁ and f₂? This part is standard textbook stuff you should have covered by now.

Once you know f₀ and Q, use the formulas in your link to determine all the R's and C's. Note that you have only 2 formulas, so you need to pick some of the R's and/or C's arbitrarily to solve for the remaining two components.

(Your instructor might have wanted you to determine the transfer function yourself).

 

[hogrampage]

If I set A to 1, then G = 1.5, so I set R_B = 10kΩ resulting in R_A = 5kΩ.

f₀ = $\sqrt{\frac{500}{2\pi}\frac{1500}{2\pi}}$ = ~138Hz

Q = $\frac{f_{0}}{f_{2}-f_{1}}$ = ~0.86

(2$\pi$f₀)² = $\frac{R_{3}+R_{1}}{C_{1}C_{2}R_{1}R_{2}R_{3}}$

R₃ = $\frac{R_{1}}{(2\pi f_{0})^{2}(C_{1}C_{2}R_{1}R_{2})}$

Now, arbitrary values:
R₁ = 10kΩ
R₂ = 20kΩ (2R₁)
C₁ = C₂ = 0.1μF

R₃ = ~6.6kΩ

The simulation is still giving the wrong corner frequencies (this simulation shows ~103Hz and ~310Hz). I am using the universal op-amp 2 in LTSpice IV. Either my circuit is wrong or I calculated something wrong. An image of the circuit I made in LTSpice IV is attached.

 

[aralbrec]

Does 'AC 1 0' mean a 1V amplitude ac signal? If so, try changing V2 and V3 to +/-10 rather than 1. You could be clipping the output.

 

[hogrampage]

I tried a different schematic (attached) and same issue. I am so confused, I have gone over my calculations numerous times and I get the same answers, but it just doesn't give the correct simulation.

 

[aralbrec]

R₃ = $\frac{R_{1}}{(2\pi f_{0})^{2}(C_{1}C_{2}R_{1}R_{2})}$

Now, arbitrary values:
R₁ = 10kΩ
R₂ = 20kΩ
C₁ = C₂ = 0.1μF

R₃ = ~6.6kΩ

You can't set them all to arbitrary values; Q is a function of these values too. You have to find component values that satisfy both equations for Q and foIf that's too annoying, I have a simpler design procedure that doesn't let you control the max gain:

1. G=2 (it must be 2 because this causes a term to zero in the Q equation)
2. all C the same
3. R1=R3=1/(w_oQC)
4. R2=2Q/(w_oC)

Max gain is 2Q²

(p116 ISBN 9783527407668)

 

[hogrampage]

EDIT: See my last reply.

 

[rude man]

The simulation is still wrong, where am I going wrong?

Check you op amp power supplies. They look backwards to me in polarity, and the voltages (+/-1V) look too low.

 

[hogrampage]

I figured it out while in the lab today:

Q = ~0.86
G = ~1.8
C₁ = C₂ = 0.47μF
R₁ = R₃ = 2.4kΩ
R₂ = 4.7kΩ
R_A = 12kΩ
R_B = 10kΩ

Thank you for all of your help!

 

[rude man]

Check you op amp power supplies. They look backwards to me in polarity, and the voltages (+/-1V) look too low.

Also make sure you're not overdriving the input. Does your output look like a nice clean sine wave?