How does this VCF work? - Page 1
Forum section: electronics / projects

author: Dajgoro
author_rank: Frequent Contributor
timestamp: March 14, 2016, 05:55:16 pm
content: Hi.I recently asked about the voltage controlled filter used in the minimoog, but now I'm again stuck with another schematic.Looking for ways to build a voltage controlled filter I found this schematic:http://www.birthofasynth.com/Thomas_Henry/pdf/VCF-1/Sheet_0002.pdfWhich is from this page:http://www.birthofasynth.com/Thomas_Henry/Pages/VCF-1.htmlI do understand the basics of the transconductance opamp and how its supposed to work,but I just can't figure out how does it work?Signal seems to come to the output of opamps, stages go from bandpass to lowpass, what is going on there?

author: Audioguru
author_rank: Super Contributor
timestamp: March 14, 2016, 08:26:33 pm
content: The schematic has two antique CA3080 V to A amplifiers and their control pins say "from collector of Q2" but there is no Q2.

author: Andy Watson
author_rank: Super Contributor
timestamp: March 14, 2016, 08:40:28 pm
content: Acquaint yourself with state variable filters and then the example given in the LM13700A datasheet. The OTAs replace the two integrating functions in the SV-filter.Edit: Don Lancaster's book, "Active Filter Cookbook", is a good place to start.

author: Dajgoro
author_rank: Frequent Contributor
timestamp: March 14, 2016, 09:26:28 pm
content: Quote from: Andy Watson on March 14, 2016, 08:40:28 pmYes, page 17 and 18:http://www.ti.com/lit/ds/symlink/lm13700.pdfIts displayed nicely there and works exactly as I would expect it, but I still don't understand that schematic.For example:Why is the highpass output before the 3080?How is the bandpass generated if there is no lowpass input to it?How is the lowpass generated from the bandpass signal?To me it looks as if the signals are somehow subtracted one from the other.Edit: The Q2 signal is generated on another schematic showed in the second link.

author: Andy Watson
author_rank: Super Contributor
timestamp: March 14, 2016, 10:14:14 pm
content: Quote from: Dajgoro on March 14, 2016, 09:26:28 pmIt's the derivative!QuoteThere is a lowpass:-QuoteConsider the frequency response of an integrator - you should find it is proportional to \$\frac{1}{j \omega}\$ - i.e. low-pass falling at a constant 20dB per decade.QuoteYes.I'll assume that you can figure-out the integrator section and why it gives a low pass response.The derivative will provide the high-pass response, but it is a little harder to see where it comes from. Take a step back and consider (another) integrator - its output is the integral of its input signal - or another way of looking at this is to say that its input must be the derivative of its output. To achieve this state the integrator is put into the negative feedback loop of an op-amp such that the integrator's output is made to resemble the original signal (this is where your subtraction occurs). The output of the op-amp, i.e. the input to the integrator, must be the derivative of the signal.Now that you have these "states" it is a simple matter of picking-off the signal that you require (low pas, band pass etc.). An advantage of building the filter from "state" responses is that the Q parameter can be uncoupled from frequency parameters.Edit: Stumbled on this, better, explanation of how a state variable filter works:http://www.electronics-tutorials.ws/filter/state-variable-filter.htmlThe integrators with the RC time constants are replaced by your OTAs configured as integrator with variable "R".

author: Dajgoro
author_rank: Frequent Contributor
timestamp: March 14, 2016, 11:19:49 pm
content: I think a start would be to understand this first:https://upload.wikimedia.org/wikipedia/commons/thumb/4/4b/BiquadFilter1.svg/640px-BiquadFilter1.svg.pngThat is the exact layout concept that started the confusion, that kind of configuration.Wiki says:QuoteAnd I found this too:http://sound.westhost.com/articles/state-variable.htmI do now see what is the concept, but I have to do more reading before I understand why that works as it does.

author: Dajgoro
author_rank: Frequent Contributor
timestamp: March 15, 2016, 02:46:56 pm
content: I built a simple version of the circuit on a breadboard with a LM324@12V and using 6V as reference generated with another opamp.I added an extra capacitor to the input for ac coupling and biasing with 4.7k at 6V.I did not use exact values from the schematic, but some close to them that I found lying on the desk, that should only change the frequency response as far I understand.Here is the page where I found the schematic:http://sound.westhost.com/articles/state-variable.htmHere is the result I got:The upper trace is always the summing point.200mV per division and input signal was 30Hz + 1kHz generated with a PC.The circuit works as predicted, but I am still trying to figure out how those signals interact exactly.How does this process of summing and derivating acting as a filter?I went through a lot of text on the web on this topic, but there is no explanation in simple words why this effect happens.*Edit, I realized the reference probe was at the wrong position, images updated.

author: Andy Watson
author_rank: Super Contributor
timestamp: March 15, 2016, 03:27:52 pm
content: Quote from: Dajgoro on March 15, 2016, 02:46:56 pmDoesn't that chip have some strange output configuration? I.e. it is not a clean, linear output stage - it might account for the fuzz on the last trace.QuoteAre you happy with the integrator acting as a low-pass filter?A differentiator will operate as a high-pass filter. The trick is in turning an integrator into a differentiator. As I said above, this is done by putting the integrator in the feedback loop of an op-amp.Consider an op-amp with a negative feedback ration of \$\beta\$. Assuming the op-amp is perfect, its gain \$\frac{V_{out}}{V_{in}}\$ will be \$\frac{1}{\beta}\$. Now put the integrator response into feedback: \$\frac{V_{out}}{V_{in}} = \frac{1}{\frac{1}{j\omega}} = j \omega \$, which is the response of a differentiator/high-pass filter.The same op-amp-integrator trick is used in analogue computers to reduce the noise problem associated with direct differentiators.

author: Dajgoro
author_rank: Frequent Contributor
timestamp: March 15, 2016, 03:54:41 pm
content: QuoteI had it lying on the desk and since its 4 opamps in one package I just took it.Datasheet says:-Conventional Op Amp CircuitsSounds good enough and the circuit works. As for the spikes, those are really small, I had to set the scope to 20mV div to see them.Might also be interference from the environment, since this is built on a breadboard and uses PC sound card as signal generator.QuoteYes.I see that as the low frequencies being dampened by the feedback capacitor so the opamp has to gain them in order to keep the + and - inputs in balance.QuoteLooking from a pure math perspective, that should only change the phase.But I can see how low frequencies could be considered DC-like and thus filtered.QuoteI see.I think I now see the why it works when looking at the pictures after reading your post.Now the question is how does the state filter with bandpass work exactly.Can you explain step by step what is going in the schematic below like you did for this example?