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Old 12th October 2012, 02:40 AM   #1
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Default 40hz boost question

Could someone explain how this circuit works to boost 40hz. Its a variable boost and I don't understand how the second op-amp provides gain. I've circled the portion in red. Ok, my brain hurts! Does it pickup its source from ground and pass through 56k into non-inverting then out of the op-amp through 2.7k and 2.2uf then into the variable resistor?
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Old 12th October 2012, 09:27 AM   #2
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OOpppss..... All op-amps are TL072.
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Old 12th October 2012, 10:28 AM   #3
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the op-amp in red is an inductance emulator of 6 something Henries with about 53K in parallel and 2.7K in series with the parallel RL. along with the 20K pot acts as a variable NOTCH at about 40Hz. a really awkward design!
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Old 12th October 2012, 11:19 AM   #4
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OK, thanks much. So I bet there is a better design and will have to research.
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Old 12th October 2012, 12:45 PM   #5
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You can build a common EQ boost/cut circuit and only use the boost (one example).

This circuit uses three filters to increase/decrease the feedback for a range of frequencies. The more feedback, the more cut. Less feedback results in greater output at that band.

The old Rockford circuit is relatively simple but requires a reverse log pot to make it easily adjustable.
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Old 12th October 2012, 08:59 PM   #6
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Thanks... I should have read Elliott Sound Products as there is a nice explanation on this... Parametric and Sub-Woofer Equaliser
The particular crossover I drew this from I've had laying around for about 15 years. I benched it with a sig gen and scope. With 1 volt in @ 40hz and no boost it puts out about 1.2 volts. (This is just one of the channels.) When the boost is turned up to full the output goes to 7 volts with a decent looking sine-wave. I thought, that's a heck of a boost!
Scott

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