Could someone help me understand a couple parts of this circuit?
i’m assuming both the resonance and volume potentiometers are where I have the two arrows pointing on the left of the circuit? What does it mean when the output has an arrow back to the middle of the variable resistor. Does that mean the output feeds back in? if so what pins on the potentiometer are they? Would it be input on pin1, output pin 3 and output going back to pin 2?
For the Freq potentiometer on the right side. Is that +12v on pin 1, 100k resistor on pin 2 and -12v on pin 3?
Where pin4 and 5 on the V2044 chip have a 820PF cap between them it looks like to get that value i would need three different caps together. How would the caps be wired to get that value, in series?
Hi there
1- so yes they are the potentiometers that you will be playing with once you are done.
The arrow “back” means that the potentiometer is used as a variable resistor and not as a voltage divider. So yes pin2 and 3 have to be connected together.
2-you need to add a 4k7 resistor to pin 3 and connect it to -12v.
3- capacitor have the opposite rule. They add up in parallel.
Conventionally when using a pot as a variable resistor the wiper is connected to one of the ends; I don’t think that’s exactly necessary, you could leave that end unconnected, but that’s how it’s usually done.
If you number the pins as 1 = counterclockwise limit and 3 = clockwise then usually the wiper (pin 2) connects to 3, but it depends whether the circuit makes more sense with resistance increasing or decreasing as you turn clockwise. Here, from the datasheet, “The Q Control (pin 2) is a current input summing node. Minimum resonance occurs at zero current.” so you want counterclockwise to give you minimum current, or maximum resistance. Therefore wire the wiper to potentiometer pin 1.
When used as a voltage divider usually pin 1 connects to the low voltage and pin 3 to the high. But again it depends on what makes sense in the circuit. Here you do want higher CV as you turn clockwise so pin 1 is -12 V (via a 4.7k in series) and pin 3 is +12 V.
The V2044 datasheet doesn’t say what any of the caps do but shows 560 pF on pins 4–5 and 6.8 nF for the other three. But the SSM2044 datasheet shows 820 pF and 10 nF like here. My guess is it affects the frequency range, and that a precise value is unimportant, so just a 680 pF might work fine. Otherwise use caps in parallel (not series) to add their values. 680+100 I bet would be close enough. (Or of course just buy an 820 pF cap.)
Thank you Analogoutput!, So with the res pot the wiper connects to pin 1 which has +12v and pin 3 is the output that goes to the 30k resistor?
And what about the volume pot across the opamp, Would that be pin 1 from the opamp output with the wiper connected to pin 1 and pin 3 going back into the inverted input?
The V2044 output is current, and that current goes through the feedback resistors whose left end is at virtual ground, so the voltage at the right (output) end is current times resistance. So for higher volume you need higher resistance. That means the wiper needs to connect to pot pin 1 (counterclockwise).
And whether it’s pin 1+wiper connected to the op amp output and pin 3 to the - input or vice versa doesn’t matter: It’s acting as a variable 2-pin resistor, and a resistor works either way around. All that matters is which end the wiper connects to. Likewise on the resonance pot, connect wiper to pin 1, and which way around it is doesn’t matter.
When it’s a voltage divider, it does matter which end connects to where.
