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Wonder if the way I connect LM4040 is causing this? [SOLVED]
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Author Wonder if the way I connect LM4040 is causing this? [SOLVED]
batchas
When you look at this part of the circuit. do you think the problem I have is coming from the way I send the +10V coming from the LM4040 to both pots SAMPLE-SEL-POT and PITCH-POT?
I am asking because when POT 1 from SAMPLE SEL is turned clockwise and I turn POT 2 from PITCH clockwise, POT 2 is influencing what comes out at the PROBE from SAMPLE SEL (adds like minus 3mV to pot 1 output).
So before looking further in the circuit (analog GND, DGND) I am wondering if LM4040 which connects to both pots is the issue?
Is there a way the signal is flowing back at pin 3 of each pot?
Should I have added some diodes between >>> +10V from LM4040 output and +10 >>> at pin 3 of each pot?

Circuit is here
m.o
First thing I'd look at:
What values are the pots? 100k? 10k?
And what is the worst case resistance to ground for both circuits in paralell?
(Too low resistance will draw too much current, which the lm4040 can't deliver).
flts
1) Values of R6 and the pots aren't specified, you should check the datasheet of the LM4040 and make sure the operating current will stay on the specified range, and possibly adjust R6 / pot values accordingly.

2) Look at "Reverse Breakdown Voltage Change with Operating Current Change" specification in the datasheet. With 10V versions A and B, if the operating current changes in the normal range of 1mA to 15mA, the reverse breakdown voltage change you are seeing is completely normal (in 1 mA ≤ IR ≤ 15 mA and 25 degrees celsius temperature, typical change is 8mV and maximum 12mV).

So if the current you draw from the LM4040 changes even in the specified range, you are looking at 8-12mV of change in the output voltage, and if you're out of the range, the specifications don't hold anymore.

I'm a complete amateur with this type of stuff so I'd guess someone more knowledgeable can provide the "correct" answer, but I suppose the obvious options are to try to keep the current draw in the circuit as constant as possible, buffer the LM4040 with a (precision?) opamp, if you actually want to draw more current than what it gives, or just assume the output voltage may change a few mV and plan the rest of the circuit accordingly...
Synthiq
If you see 3mV change at PROBE, the change at +10V is higher and depend on the SAMPLE_ATT setting and value but can be as high as 6mV. The change in load current when you change POT 2 must be less than 10V/100kohm but is probably less than half of this. Thus we can calculate the LM4040 output resistance as delta_U/delta_I of at least 3mV/100uA which is 30 ohms but can be higher than 100 ohm, far above the specified 0.7 ohm at 1mA. So my guess is that the current in LM4040 is way too low for it to work correctly. If you use 100kohm pots R6 has to be less than 10kohm to even meet the minimum spec for operating current but 3.6kohm to meet the conditions for the specified 0.7ohm output resistance. For 10kohm pots, the values would be 2.2kohm and 1.5kohm respectively. These values doen't take the tolerances of the pot resistance into account so they may have to be lower.
batchas
Sorry I forgot to mention R6 value d'oh!
Here is a better overview, putting the two PCBs circuits on one.

R6 is 10k.
The pots are 50k.

Circuit iwith values s here
batchas
Thanx a lot for all the explanations, I understand.
Makes very much sense thumbs up
batchas
Changed 10k to 5k (parallel 10k on proto): it works great It's peanut butter jelly time!



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