OP Amp circuit - Subwoofer circuit simplification. - Page 1
Forum section: electronics / beginners

author: hitech95
author_rank: Regular Contributor
timestamp: March 15, 2016, 09:06:54 pm
content: Hello,Lately I've been watching the theory of op amp (again).The circuit in question isthis, I'm interested to understand the operation and to simplify it.Unfortunately this time I need the circuit.For now I want to focus on the subwoofer section, the two inputs (L and R) pass through two buffers and are then mixed with an inverting adder. (I miss the usage of R14, I believe that it serves for bias currents)At this point there are two second-order filters (Sallen-Key filter), the gain of these is 1. But I do not understand how to calculate the final cut-off frequency. (The first cut is at 282Hz and the second cut is at 374Hz)At this point I have a couple of questions.1) U7 is used for switching the signal in phase, I wonder if it is possible to modify the circuit to make an attenuator with a pot. (Gain <= 1)2) A change that I would like to implement is the ability to choose the cut-off frequency using a potentiometer. Do you have suggestions?3) The buffer in the input stage are needed? Many schemes I've seen online do not have them and do not even have the adder.4) R8 C7, C1, R1, R25 C17 are used to eliminate the DC voltage, right?Thanks, hitech95.

author: DrGeoff
author_rank: Frequent Contributor
timestamp: March 15, 2016, 10:00:00 pm
content: The summing network appears to sum L and R in unequal proportions. Probably not intended, I'd expect R2 and R15 to be the same.The Sallen-Key filters are both LPF types.Cutoff is given as Fc = 1/(2*pi*sqrt(RxCxRyCy))In case 1,Rx=R17=4.7KCx=C13=0.12uFRy=R18=4.7KCy=C15=0.12uFSo Fc = 282HzYou can work out the second one.

author: DrGeoff
author_rank: Frequent Contributor
timestamp: March 15, 2016, 10:14:29 pm
content: Quote from: hitech95 on March 15, 2016, 09:06:54 pmThe first buffer in the subwoofer section is the summing amplifier. When you sum the 2 signals equally you will lose 6dB, so this summing amplifier should have a gain of around 2. In addition it provides a low impedance drive for the filter, so the source impedance becomes negligible in the filter calculations. Without it you would have low signal and have to take into account the source impedance of the passive summing network in the filter calculations as it would no longer be negligible.

author: hitech95
author_rank: Regular Contributor
timestamp: March 15, 2016, 10:25:40 pm
content: Quote from: DrGeoff on March 15, 2016, 10:00:00 pmYes, it is a mistake in the schematic, both resistors have the same value.Quote from: DrGeoff on March 15, 2016, 10:14:29 pmI mean u1 and u8, can you explain me why the adder has a gain of -2?

author: retrolefty
author_rank: Super Contributor
timestamp: March 15, 2016, 10:31:19 pm
content: Maybe you can pick up ideas from this more involved sub-woofer 'controller':http://sound.westhost.com/project48.htm

author: DrGeoff
author_rank: Frequent Contributor
timestamp: March 15, 2016, 10:41:13 pm
content: Quote from: retrolefty on March 15, 2016, 10:31:19 pmThat's a different type of system entirely, relying on a small closed enclosure to push the resonant frequency up above the sub crossover frequency, and then applying a 12db/octave gain stage to compensate for the 12db/octave driver rolloff below resonance.

author: DrGeoff
author_rank: Frequent Contributor
timestamp: March 15, 2016, 10:43:52 pm
content: Quote from: hitech95 on March 15, 2016, 10:25:40 pmU1 and U8 are unity gain buffers used to supply the rest of the circuitry with a known low impedance drive.The circuitry following this are cascaded Sallen-Key HPF, so they should be driven with a low impedance in order for the cutoff frequencies to be easily calculated.

author: hitech95
author_rank: Regular Contributor
timestamp: March 15, 2016, 10:49:04 pm
content: Quote from: DrGeoff on March 15, 2016, 10:43:52 pmOk, so I can remove U1 and U8 and use only the U4. Than I can change the gain of U7 and use a pot for the volume.

author: DrGeoff
author_rank: Frequent Contributor
timestamp: March 15, 2016, 10:53:20 pm
content: Quote from: hitech95 on March 15, 2016, 10:49:04 pmYou could, although it would not be advisable unless you can guarantee that you are driving the circuit with a low impedance.

author: hitech95
author_rank: Regular Contributor
timestamp: March 15, 2016, 10:55:11 pm
content: Quote from: DrGeoff on March 15, 2016, 10:53:20 pmI need only the LPF not the HPF.

author: Audioguru
author_rank: Super Contributor
timestamp: March 15, 2016, 11:57:30 pm
content: If you do not use the highpass filters then your stereo speakers will try to produce subwoofer frequencies with plenty of distortion or maybe exceed their allowed excursion. Also, the frequency response of the entire system will have a boost at the crossover frequency because you will have 3 speakers producing it.Notice that the circuit has the opamps without any power. The LED shows only a single 9V supply but the opamps need a positive and a negative supply.Since the gain of the filter ICs is only 1 then R14 and R24 can be replaced with a wire.Pot R22 adjusts the gain of U7 from 1/6 (0.167) to 6 times.To adjust the cutoff frequency of the lowpass filter you need a pot with 4 sections each with about 15k or 20k and a series resistor in series with each section.There is a calculation somewhere to determine all the parts values for the 4th-order Butterworth Sallen and key filter.

author: danadak
author_rank: Super Contributor
timestamp: March 16, 2016, 02:33:12 am
content: Here is a site with calculators for Sallen Key and other topologies -http://sim.okawa-denshi.jp/en/Fkeisan.htmRegards, Dana.

author: hitech95
author_rank: Regular Contributor
timestamp: March 16, 2016, 07:30:38 am
content: Quote from: Audioguru on March 15, 2016, 11:57:30 pm1) I have the hi-pass filter in the L R speakers.2) I know, i need a dual power supply

author: hitech95
author_rank: Regular Contributor
timestamp: March 16, 2016, 09:48:07 am
content: In that configuration is U1A? I can not understand in a case is inverting, the other I think itcan't workbuffer? because no current flows on RF (R4) when vs- is disconnected