## Trying to understanding the Rungler in depth

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ketem13
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### Trying to understanding the Rungler in depth

I try to understand the part of the circuit marked in purple -
opamp U6A is acting as a comparator? is comparing between the output of the shift register that connected to the inverting pin via R17, but also the same output is going to the non inverting pin via R16? The same R16 is also connected to the collector of T1. in every High voltage of P1 the connected to the gate of T1 the signal appearing at the collector with be pushed to ground? what is the role of R43? pull down resistor?

The output of the 4021 Q8 is the most significant bit of the 8 bit shift register right?
If I have a logic level HIGH at the input of the 4021 (pin 11), on the first clock the output at Q6 will be HIGH while the output on Q7 and Q8 will be LOW?
next clock will shift the output from Q6 to Q7 and the output from Q7 to Q8? on the same new clock the new signal that is at the input will be present at Q6?

guest
Super Deluxe Wiggler
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### Re: Trying to understanding the Rungler in depth

the opamp looks like an optional inverter, but the output is labeled XOR. this sort of looks like a random sequence generator, where you XOR two or more outputs of the shifted sequence and then reinsert them back to the beginning. lets assume both the 4021 and opamp are running at +/-9V rails. if the transistor is on, then the non-inverting input is at 0V, so +/-9V digital signals from the 4021 are just going to get inverted at a ratio of 1/2 (15/30k), so you get +/-4.5V out (which is odd, since those arent valid logic levels for the 4021). if the transistor is off, the voltage at the noninverting pin is 2/3 whatever the input is, and then the output is inverted from there. so for 9V, there is 6V at the noninverting pin and inverting pin (due to feedback). so the inverting pathway has 9V in, and 6V at the inverting pin, so 3V drop across the 30k. this then gets converted to a 1.5V drop across the 15k, which is subtracted from the 6V at the inverting pin, so 4.5V at the output. so with the transistor not active, you get a positive "gain", with the output being +/-4.5V for a +/-9V input. again, this is odd as this doesnt match the input requirements of the 4021. if the 15k was 30k and the 62k wasnt there, it would be a standard optional inverter and give +/-9V outputs.
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Navs
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### Re: Trying to understanding the Rungler in depth

I realize you are looking at a specific part of the Rungler, but maybe the general description provided by Rob Hordijk will help too:

viewtopic.php?t=92049

There is another, longer thread at electro-music.com with a similar title.

ketem13
Ultra Wiggler
Posts: 808
Joined: Tue Jan 16, 2018 2:02 pm

### Re: Trying to understanding the Rungler in depth

guest wrote:
Thu Jul 22, 2021 12:56 am
the opamp looks like an optional inverter, but the output is labeled XOR.
XOR meaning the output Is HIGH only if on of the inputs are LOW and the second is HIGH, otherwise if both input are at the same level (either LOW or HIGH) the output is LOW ?
What does it mean optional in that context?
guest wrote:
Thu Jul 22, 2021 12:56 am
this sort of looks like a random sequence generator, where you XOR two or more outputs of the shifted sequence and then reinsert them back to the beginning.
guest wrote:
Thu Jul 22, 2021 12:56 am
lets assume both the 4021 and opamp are running at +/-9V rails. if the transistor is on, then the non-inverting input is at 0V,
Is that because the transistor pull the signal via the collector to grpund?

guest wrote:
Thu Jul 22, 2021 12:56 am
so +/-9V digital signals from the 4021 are just going to get inverted at a ratio of 1/2 (15/30k), so you get +/-4.5V out (which is odd, since those arent valid logic levels for the 4021).
they get inverted in the previous stage at the opamp input, right?
And just to be sure - we identify the signal as digital because it is either a +9V or -9V. two distinct levels? a pulse wave for that matter could be called digital?
guest wrote:
Thu Jul 22, 2021 12:56 am
if the transistor is off, the voltage at the noninverting pin is 2/3 whatever the input is, and then the output is inverted from there.
so for 9V, there is 6V at the noninverting pin and inverting pin (due to feedback).
for 9V there is 6V at the noninverting pin and also 6V at the inverting pin(input)? isn't it 9V at the inverting pin?

guest wrote:
Thu Jul 22, 2021 12:56 am
so the inverting pathway has 9V in, and 6V at the inverting pin, so 3V drop across the 30k.
why is 6V the the inverting pin?
guest wrote:
Thu Jul 22, 2021 12:56 am
this then gets converted to a 1.5V drop across the 15k, which is subtracted from the 6V at the inverting pin, so 4.5V at the output. so with the transistor not active, you get a positive "gain", with the output being +/-4.5V for a +/-9V input. again, this is odd as this doesnt match the input requirements of the 4021. if the 15k was 30k and the 62k wasnt there, it would be a standard optional inverter and give +/-9V outputs.

elektrouwe
Common Wiggler
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Location: wild south of Germany

### Re: Trying to understanding the Rungler in depth

you can also explain the XOR circuit from an "modular synth" view: it's like a common "attenuverter" with fixed gain and selectable + or - sign.
Or you can view it as a "ringmodulator" where the digital carrier applied to the NPN switches polarity of the inverting signal path.
There is nothing special about the Rungler, you could use any kind of discrete,gate , or opamp circuit that provides controlled inversion == XOR.

guest
Super Deluxe Wiggler
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Joined: Mon Aug 19, 2013 11:49 am

### Re: Trying to understanding the Rungler in depth

"optional inverter" is the name for an opamp setup like the one above, where you can either have a positive gain, or a negative gain, depending upon whether the transistor is on. so there is the "option" to invert. so if you take the transistor input as one of your inputs, then the output of the opamp does digital XOR as youve described.

when analyzing opamp circuits, the first thing i look at is if there is a signal pathway from the output back to the inverting pin. if so, then there is negative feedback and you can assume the noninverting pin is at the same voltage as the inverting pin. i then figure out what voltage is at the noninverting pin. in this case, is a voltage divider to 9V that produces 6V. so this is also the voltage at the inverting pin.
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