Hi, I'd like to understand what the highlighted part of this circuit does. This is supposed to be an active low pass filter with a cutoff that can be controlled with a voltage.

I believe that the part I highlighted works as voltage controlled capacitance. If that's the case, can someone explain how this works please?

 

The components circled in yellow do not constitute a low pass filter.
It is the TL071 opamp that is configured as a low pass filter hence that is why it is an active filter.
The gain of the amplifier circuit is determined by R2 / R1.
But R2 has frequency dependent components C1 and C2 across it. Hence the gain is now a function of frequency.
As frequency increases, the combined impedance across R2 drops and hence the gain falls. Thus you have a low pass filter.

 

I doubt that the 1n4148 has much in the way of variable capacitance especially compared to the 100 nF capacitors C1 and C2 connected to it. More likely to my mind is that the 1n4148 acts as a variable resistance depending on the current
through it. As the current increases the resistance decreases and the signal through C1 and C2 is shunted to ground.
This lowers the feedback at higher frequencies and would tend to raise the frequency response.

 

The components circled in yellow do not constitute a low pass filter.
It is the TL071 opamp that is configured as a low pass filter hence that is why it is an active filter.
The gain of the amplifier circuit is determined by R2 / R1.
But R2 has frequency dependent components C1 and C2 across it. Hence the gain is now a function of frequency.
As frequency increases, the combined impedance across R2 drops and hence the gain falls. Thus you have a low pass filter.

Yes, I do understand this. I said " I believe that the part I highlighted works as voltage controlled capacitance. "
The whole circuit is a voltage controlled LPF. I am interested in the highlighted part which I believe is a way of changing the capacitance using an input voltage. I want to understand why this arrangement using 2 capacitors allows the capacitance to be modified using a voltage source.

 

I doubt that the 1n4148 has much in the way of variable capacitance especially compared to the 100 nF capacitors C1 and C2 connected to it. More likely to my mind is that the 1n4148 acts as a variable resistance depending on the current
through it. As the current increases the resistance decreases and the signal through C1 and C2 is shunted to ground.
This lowers the feedback at higher frequencies and would tend to raise the frequency response.

I'm not sure that the diode is used this way since the input "VC" would usually be of an unknown, but high, impedence. Current would not be controlled on this input, but voltage would be, and it is supposed to change the cutoff frequency of the filter by changing the capacitance of C1 and C2

 

Current would not be controlled on this input, but voltage would be

The voltage on the VC input feeds the diode via a 1k resistor as does the voltage from the cutoff pot thus those two voltages result in an adjustable diode current. A controlled current flows in that forward biased diode. The magnitude of the current changes the impedance of the diode.

The circuit around the opamp is a low pass filter with higher frequencies being fed back to the inverting input - negative feedback. When the diode impedance is low - higher current in the diode - some of that feedback signal is shunted to ground through the diode and less feedback means higher gain and the reverse when the diode impedance is high - low diode current.

 

The voltage on the VC input feeds the diode via a 1k resistor as does the voltage from the cutoff pot thus those two voltages result in an adjustable diode current. A controlled current flows in that forward biased diode. The magnitude of the current changes the impedance of the diode.

The circuit around the opamp is a low pass filter with higher frequencies being fed back to the inverting input - negative feedback. When the diode impedance is low - higher current in the diode - some of that feedback signal is shunted to ground through the diode and less feedback means higher gain and the reverse when the diode impedance is high - low diode current.

Hmmm, ok. I'll have to read up about that. Thank you.
So what does this 2 capacitor arrangement do then? In a LPF, I usually see a capacitor and a variable resistance in parallel on the feedback. But this circuit is adding a variable resistance (the diode, according to what you are saying) between 2 capacitors.

 

Michael is correct. D1 is acting as a current-controlled resistance. Back in the day, this was called a "variolosser:

https://encyclopedia2.thefreedictionary.com/variolosser

Here is a patent that mentions it: https://patents.google.com/patent/US4405903A/en

I first came across the term in gear from CBS Labs back when they built stuff like the Volumax and Audimax - automatic audio signal processing gear for TV and radio *way* before DSP was invented. The Volumax used two diodes in series, presumably for a wider control range and reduced distortion.

Nothing in the circuit changes the value of any capacitor. The only variables are the pot RES and the *effective* impedance of D1. C1, C2, and D1 form a "T" network, in this case a high-pass filter. This is one-half of a common filter and oscillator circuit called a "Twin-T" network. Here is an introduction on that:

https://allaboutelectronics93882548.wordpress.com/twin-t-oscillator/

As with the common opamp feedback network with a single R and C in parallel, your circuit has a high-pass filter (the t network) in a negative feedback loop, which forms a lowpass filter at the opamp output.

ak

 

@AnalogKid , thank you. So it's called a twin-t network. The link you provided looks like a very good explanation. I'll read that later today.

Was this a common way to control the cutoff of a filter in the analog days? I believe we could also achieve the same thing by using only 1 capacitor on the feedback but replacing R2 by a jfet. If that's the case, then what would be the advantages of one method over the other.

 

To be clear, what you have is not a twin-T because there is only one T.

Not super-common, but only because continuously-variable remote control of a filter's cutoff frequency was not a very common requirement. But in that applications space, it was a known tool in the tool box.

A J-FET needs a negative bias voltage, which often is not available in a single-supply system. In an audio system it is possible to drive a charge pump with the audio signal to create a low-power negative bias voltage that varies with the audio signal level, but there are a loooooooot of other non-audio circuits and systems that need a variable filter. Nowhere in any of your posts have you said what the circuit is filtering, and why.

ak

 

Yes, this would be for audio. Negative voltage is not issue. I'm using an ATX power supply so I have access to a good selection of voltages.

I was under the impression that to create a variable filter, one would only need to adjust either the capactor or the resistance on the feedback of the opamp. This design that I posted above got me confused. I thought this would be a way to create a variable capacitor (instead of using a varactor). But if the diode is only used as a variable resistor, then I wonder why this wasn't used in the place of R2 instead. Because the typical design, in audio applications, would be to replace R2 by a pot. So if it needs to be voltage controlled, it only makes sense to use a voltage controlled resistance (vactrol, jfet, or this way with the diode that I didn't know about until now) instead of the pot.

 

Varying R2 changes *both* the passband gain and the filter corner frequency. Not good.

Also, varying R2 with a remotely-supplied DC control voltage is not simple. The voltage at the source is varies with the instantaneous value of the audio signal because the opamp inverting input is a virtual ground, not the real system GND. The gate voltage must track this in order to present a constant resistance at all signal voltages. If D1 were replaced with a FET, the source would be at GND. This makes Vgs independent of the audio signal amplitude going through the FET.

ak

 

The same question has been asked here Electro-music -Why does this filter have two capacitors? And other queries. and the creator of the schematic is here Most simple VCF on earth.
Contains a fair bit of information and alternative ideas about simple Voltage controlled filters

I simulated it in Falstad for the heck of it Simple VCF.

Edit: I used 12V instead of 9V as in the original schematic(OP probably meant it to be a guitar/bass pedal) since that is the eurorack standard voltage, seems to work on +/-15V systems as well...