Active Filter Design

[audioguru]

now your resistor values are far too low for an opamp to drive. A few opamps can drive a 600 ohm load but most cannot drive less than 2k ohms.

 

[boozi]

Will work on it later this week (got exams on Wed + Thurs). I'll get back to you with more updates!

Audioguru -> That's one of the problems..... If I want that resistor's value to increase, I have to decrease
the capacitors, but that causes the other resistors to increase by a huge factor :\

 

[CRISespecialist]

good day my friend,,i suggest that all the couplres like resistor,capacitors must be experemented by changing values one by one,sence you are stell studying,write also your observation.and make sure you have ground in circuit to prevent humming.
CRIS...

 

[audioguru]

Audioguru -> That's one of the problems..... If I want that resistor's value to increase, I have to decrease
the capacitors, but that causes the other resistors to increase by a huge factor :\

But some of your resistor values are a dead short and will not allow the circuit to work.

 

[boozi]

CRISespecialist -> Thanks for the tip. That's what I've been doing with my first design (the one I made a month ago), and it worked out
ok, but I had to modify it again...

Audioguru -> That's the problem. If I want to increase that resistor's value, I'll have to increase the other ones as well (in the order of
mega).

JimB -> I've looked more into the stuff you told me about. The problem with cascading a low-pass with a high-pass filter is that it won't
work for a hiqh-Q factor (mine has to be 80). The book suggested that I should use the bandpass filter shown in Fig 5.16. I did
some more research into that design and it turns out that in order to have one thing (High-Q factor), you have to give up another (a gain
of 1 at the desired center frequency), meaning that I cannot change them independently.

For now, I'm thinking of using the LC filter I designed last semester. I'll try and add an op-amp to it to see what kind of results I get.

 

[boozi]

So I've been looking at the design I created last semester, and one of the things we couldn't figure out was why there was a gain (about 8) in the LC filter.... Does anyone have any thoughts on that? The circuit contains only inductors, capacitors, and one resistor (to match the load).

 

[audioguru]

I have never seen an opamp used at a frequency as high as 80MHz with such a Q as high 80.
Now you are talking about an LC filter without posting its schematic so we don't know causes its problem.

 

[boozi]

Sorry about that. Here are the schematic & simulation!

 

[audioguru]

It is normal for a high Q filter to have voltage gain.

 

[boozi]

Can you please elaborate more on that? What's exactly the reason for the gain?

 

[audioguru]

The high Q makes the impedance higher. The output power must equal the input power so the voltage is increased.

 

[boozi]

Ok, thanks for the clarification. When we actually built the PCB last semester, we tried it out, and in the end, the output signal was massively attenuated... Is there any chance you know how to add the noise factor in PSPice? I would like to try and simulate the design with the additional noise.

 

[audioguru]

I have never used PSPice and never added noise to LTspice IV that I use.

 

[boozi]

I have never used PSPice and never added noise to LTspice IV that I use.

Ok, thanks anyways. I'll see what I can do about my issues.

 

[boozi]

The high Q makes the impedance higher. The output power must equal the input power so the voltage is increased.

I just thought about it a little.... If the high Q makes the impedance higher, how come the gain is not higher than 1 in the "smaller" circuit I posted? The one with only one inductor, one capacitor, and one resistor.

 

[audioguru]

I just thought about it a little.... If the high Q makes the impedance higher, how come the gain is not higher than 1 in the "smaller" circuit I posted? The one with only one inductor, one capacitor, and one resistor.

I can't find your "smaller" circuit. A parallel LC circuit is a high impedance at its resonant frequency which causes the voltage to increase if it is not loaded down too much. Maybe you have a low value load resistance which caused its peak to be suppressed.

 

[boozi]

I can't find your "smaller" circuit. A parallel LC circuit is a high impedance at its resonant frequency which causes the voltage to increase if it is not loaded down too much. Maybe you have a low value load resistance which caused its peak to be suppressed.

My bad. I thought that I included it with my original circuit. I'll upload the schematic/simulation when I get back home!

 

[MRCecil]

I can't find your "smaller" circuit. A parallel LC circuit is a high impedance at its resonant frequency which causes the voltage to increase if it is not loaded down too much. Maybe you have a low value load resistance which caused its peak to be suppressed.

Would you care to give a referent as you failed to do in you post #31? Regarding the former are you speaking of parallel insofar as the tank itself or the I/O, and to the latter are you speaking to I/O, shunt or series? Without the references, one is left with a vacuum as to specifics.

 

[audioguru]

Would you care to give a referent as you failed to do in you post #31?

The OP showed the graph of a peaked voltage so obviously it is from a parallel tuned LC circuit. A series tuned LC circuit produces a voltage null (not a peak) at resonance.

If the parallel tuned LC circuit is loaded down by a fairly low resistance load then it does not produce a voltage peak because at resonance it is a high impedance.

 

[boozi]

Here's the smaller circuit. So that's what you meant by having a small resistance?