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I can't understand the circled symbol and didn't manage to google it. It looks like a variable resistance and a capacitor. What does it mean?

enter image description here

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    \$\begingroup\$ What do you mean by "what does it mean"? You've interpreted it correctly. \$\endgroup\$ – duskwuff Sep 15 at 5:10
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    \$\begingroup\$ Yeah that is pretty strange and I've never seen that before... But I think your interpretation is correct. Why it's placed there, I'm not entirely sure. Also accidentally deleted my comment. Yes, variable resistors have three terminals :) \$\endgroup\$ – KingDuken Sep 15 at 5:12
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    \$\begingroup\$ It’s a potentiometer, not a variable resistor. \$\endgroup\$ – Chu Sep 15 at 6:57
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    \$\begingroup\$ @pipe if you aren't personally up for finding an answer, just don't respond. "just google it" is very rarely an appropriate response. \$\endgroup\$ – dn3s Sep 16 at 5:00
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    \$\begingroup\$ @dn3s pipe's comment was addressed to another comment, not to the question. "Just Google it" seems like a reasonable response, given that follow-up questions shouldn't be being asked in comments anyway. \$\endgroup\$ – David Richerby Sep 16 at 9:49
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It's not one symbol.

It's just a capacitor connected to the wiper terminal of a potentiometer.

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    \$\begingroup\$ Thank you Photon! So the "out" signal is found at the wiper of a second potentiometer? \$\endgroup\$ – Anna Sep 15 at 5:12
  • \$\begingroup\$ @Anna, yes, that's correct. \$\endgroup\$ – The Photon Sep 15 at 5:13
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Just to explain the potentiometer, since you have your answer now... if you look at one physically, you'll see a carbon track, from the left terminal to the right one. That's a resistance. The middle terminal connects to the wiper, the copper slider that contacts the track. Turning the shaft moves the wiper, varying the resistor.

You seemed to understand that already. In some cases you'd use just 2 terminals as a variable resistor. But it's often more useful to use all three. If you connect, say, 5V to one side, gnd to the other, then the wiper will give a voltage between 5V and gnd, variable.

If you connect one end to gnd, and the other to a signal, the wiper will give a voltage partway between the signal and gnd. As in a volume control.

Often electrical circuits are controlled by a particular voltage somewhere that you want to control. You need the 3 terminals of a potentiometer, aka "pot", to do that. A simple variable resistance would, by itself, just limit the current coming through. That's not always useful. In most circuits, a pot is used with all 3 terminals.

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As others say - it's a capacitor on a pot wiper.

Q1 is an inverter with a gain without the mystery pot of about 38 x (V+ - 1) [for reasons related to transistor physics] ~= 300 if V+ = 9V, but with Vc_Q1 amost at ground.

Q2 emitter provides a buffered inverted input signal with is fed back via the left hand 100k pot to stabilise gain. The position of the pot wiper alters the frequency response of the RC feedback network in a probably undesigned and 'interesting' manner.

  • Wiper far left = a sub 1 Hz low pass filter for feedback PLUS a large cap on the input. so signal probably low. Pot wiper far right - the Q2 emitter follower drives the cap but it again probably clamps the voltage enough to prevent feedback so you get high gain overall plus high gain from Q2 so massive output clipping.

  • As you slide pot right to left you probably get increasing modification of signal, reduction in overall gain but change in frequency response.

  • The circuit appears to require enough signal to drive Q1 into conduction, so on very low signals it probably produces no output.

I started to say what that change would be but decided "it's complex" :-). It would sound very bad (or very good with the right ears on).

As a 'bonus' the circuit acts overall as a frequency response modified Schmitt trigger. I won't even start to try to suggest what happens with pot variatioj - but simulation would be interesting.

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    \$\begingroup\$ Not sure about that... I read it as wiper far left, lowpass determined by the source impedance feeding the thing and the end resistance of the pot, far right you get a high frequency boost by bypassing the emitter resistor of Q2 (Zero determined by pot end and wiper resistance). DC wise there is feedback via the pot track (never brilliant), so the thing will auto bias. I don't see the Schmitt action (no common emitter load resistor and all the feedback is negative). I think it is a tone control from a guitar amp or similar. \$\endgroup\$ – Dan Mills Sep 16 at 10:22
  • \$\begingroup\$ @DanMills I agree re fuzz box and DC feedback bias. || When pot is at lh end 22 uF shunts down input signal from 1uF . Looks like it settles down with about 0.6V across the 2k2. eg Q1 off = Q2 on so V2k2 = 2.2/12.2 x 9 ~= 1.5V =- so Q1 on so it would oscillate but 22 uF and LPF MAY slug it enough. It will then settle into 0.6V across 2K2 so Vb Q2 ~= 1.2V so V across 22k = 9-1.2 = 7.8V so gain Q1 ~= 38.4 x 7.8 = ... and . Simulation easier ;-) \$\endgroup\$ – Russell McMahon Sep 16 at 13:55
  • \$\begingroup\$ Thank you for these explanations Russell and @DanMills. Enlightening. Indeed I tried to build a fuzz face. It's the first pedal I've tried, and as the total amateur I am, I picked the circult from this list that used transistor symbols I recognised. \$\endgroup\$ – Anna Sep 21 at 9:02

protected by W5VO Sep 16 at 17:41

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