preamp schematic help

jenks · 03-19-2008, 03:14 PM

Hello, I've been trying for weeks to simulate a preamp circuit found on R G Keens' page; geofex.com. It consists of two LC filters that cause a peak or a notch in the frequency response of the input signal (peak/notch depends on where you set the potentiometer - see attached multisim screenshot) at frequencies determined by the L, R & C values chosen in the LC filter sections.

However, when i simulate the circuit it only increases the gain of the whole signal across its' whole frequency response - as opposed to being frequency specific! Grr!

If anyone out there is willing to take a look at my schemtic and point out any errors I'd really appreciate it!

Thanks.

cbarrow7625 · 03-19-2008, 04:33 PM

Well, 500uF and 1mF are incredibly large capacitors to be using in a solid-state filter circuit like this one.

I would expect maybe .5uF or 5uF but not 500uF!

How did you come about using those values? Can you post a link to the original RG schematic? That would help alot.

jenks · 03-19-2008, 04:44 PM

Hi C barrow, no values were quoted for the R, L and C's in the Lc filter section. i just put in values in order to get the circuit affecting audible frequencies. I'll try changing those to see if it makes a difference... here's the link btw. You'll have to scroll down...

http://www.geofex.com/article_folders/eqs/paramet.htm

Cheers, Jenks.

cbarrow7625 · 03-19-2008, 04:55 PM

Thanks. OK, hopefully this will help you....For a passive notch filter like this one, a small capacitor & larger inductor will give you a higher "Q". A higher "Q" basically means that you are targeting in on a narrower frequency band than with a wide "Q" (see RG's notes on "Q" in the section above the one you cited for good graphic representations of this).

Using a VERY LARGE 500uF capacitor, what you have done is create a VERY LOW "Q" circuit. VERY low Q = affects all frequencies, not just the one you want.

Try dividing your capacitor values by 100 (5uF) and multiplying your inductor values by 100. THis should give you the same center frequency but a much higher "Q". The series resistor will also reduce the Q but in a different way. Any time I use a notch filter like these I always try it without a series resistor in place. THat way I get to see exactly what the Q ofthe filter is and how deep a notch it is making.

Once you are happy with your component values, then start adding resistance back in to tame the depth of the notch.

RG gave you the formula to calculate the needed values as well: The resonant frequency may be calculated as Fr=1/(2*pi*SQRT(L*C)). That along with my hint about how the cap size affects this equation should help you zone in on the proper values pretty quickly.

Let me know what you find.

jenks · 03-19-2008, 05:56 PM

Thanks for all your help C!

I can't believe my circuit was working the whole time! It's just that the values chosen were so inappropriate that I was looking at gain at a set frequency but with too large a Q, so that on the bode plot it didn't seem frequency specific at all!

thanks again!

PS I may be in touch in very soon with more problems as you clearly know what you are talking about in these matters!

Jenks.

jenks · 03-19-2008, 08:10 PM

Hi C,

I've been trying to get the gyrator that R G has added to work onto my circuit for last couple of hours unto no avail!

From looking at the math on R G's geofx page I think a ridiculously low capacitance is required at C2 & C4. I calculated that the capacitance needed to be around 5 x 10^-12 (femto!) farad. Did i get this right? If so then I need to think of a way of getting such low a capacitance! If not then I must have wired something wrong, but what (?) - I'm not exactly sure!

Anyway, any advice you have on the matter will be hugely appreciated.

cbarrow7625 · 03-19-2008, 10:47 PM

C2 & C4 on your sketch should really be pretty large! The input & output caps of the cirucuit are supposed to pass the signal without any low end rolloff. I would expect to see something from 2-22uF inthose places.

Try just putting a 22uF in your simulation. I think that should work.

Make sure to rember all of your zeros & decimal places when calculating values & frequencies.

jenks · 03-19-2008, 11:23 PM

I've tried those values but it still doesn't work! hmm... Not at all sure why

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03-19-2008, 04:33 PM	#2
cbarrow7625 Senior Member Join Date: Mar 2007 Location: Philadelphia suburbs Posts: 366	Well, 500uF and 1mF are incredibly large capacitors to be using in a solid-state filter circuit like this one. I would expect maybe .5uF or 5uF but not 500uF! How did you come about using those values? Can you post a link to the original RG schematic? That would help alot.

03-19-2008, 04:44 PM	#3
jenks Junior Member Join Date: Mar 2008 Posts: 12	Hi C barrow, no values were quoted for the R, L and C's in the Lc filter section. i just put in values in order to get the circuit affecting audible frequencies. I'll try changing those to see if it makes a difference... here's the link btw. You'll have to scroll down... http://www.geofex.com/article_folders/eqs/paramet.htm Cheers, Jenks.

03-19-2008, 04:55 PM	#4
cbarrow7625 Senior Member Join Date: Mar 2007 Location: Philadelphia suburbs Posts: 366	Thanks. OK, hopefully this will help you....For a passive notch filter like this one, a small capacitor & larger inductor will give you a higher "Q". A higher "Q" basically means that you are targeting in on a narrower frequency band than with a wide "Q" (see RG's notes on "Q" in the section above the one you cited for good graphic representations of this). Using a VERY LARGE 500uF capacitor, what you have done is create a VERY LOW "Q" circuit. VERY low Q = affects all frequencies, not just the one you want. Try dividing your capacitor values by 100 (5uF) and multiplying your inductor values by 100. THis should give you the same center frequency but a much higher "Q". The series resistor will also reduce the Q but in a different way. Any time I use a notch filter like these I always try it without a series resistor in place. THat way I get to see exactly what the Q ofthe filter is and how deep a notch it is making. Once you are happy with your component values, then start adding resistance back in to tame the depth of the notch. RG gave you the formula to calculate the needed values as well: The resonant frequency may be calculated as Fr=1/(2piSQRT(L*C)). That along with my hint about how the cap size affects this equation should help you zone in on the proper values pretty quickly. Let me know what you find.

03-19-2008, 05:56 PM	#5
jenks Junior Member Join Date: Mar 2008 Posts: 12	Thanks for all your help C! I can't believe my circuit was working the whole time! It's just that the values chosen were so inappropriate that I was looking at gain at a set frequency but with too large a Q, so that on the bode plot it didn't seem frequency specific at all! thanks again! PS I may be in touch in very soon with more problems as you clearly know what you are talking about in these matters! Jenks.

03-19-2008, 10:47 PM	#7
cbarrow7625 Senior Member Join Date: Mar 2007 Location: Philadelphia suburbs Posts: 366	C2 & C4 on your sketch should really be pretty large! The input & output caps of the cirucuit are supposed to pass the signal without any low end rolloff. I would expect to see something from 2-22uF inthose places. Try just putting a 22uF in your simulation. I think that should work. Make sure to rember all of your zeros & decimal places when calculating values & frequencies.

03-19-2008, 11:23 PM	#8
jenks Junior Member Join Date: Mar 2008 Posts: 12	I've tried those values but it still doesn't work! hmm... Not at all sure why

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