Shades Schematic questions(?)

Hello friends,

I am fairly new to the synth DIY world and have been really enjoying the concepts and joy of soldering.

I have been studying the modern Shades schematic and have questions.

1: is it possible to achieve success with different op amps than the ones used in the schematic? for example, could i swap in a TL072 or LM324? I can’t seem to find a reliable source for the designated pampas.

2: upon breadboarding the top third of the schematic, I notice strange jumps in voltage. when i turn the potentiometer fully in a single direction and meter the output 1k resister, the voltage seems to jump from negative 10v up to zero and then reverts back to negative. It reminds me of using a comparator module in my synth. However shades is obviously not a comparator. What might i possibly have done wrong and how can i manage achieving the smooth DC - to + ? Similar outcome when i switch the DPDT into attenuation mode; Only then the voltage dances around +10

Any insight or help you kind people might be willing to share is greatly appreciated. Thanks

  1. Depends on your definition of success! Noise, distortion and offset will be different. If your goal is to build a circuit that works on your breadboard, you can use a TL07x, that’s good enough.

Any op-amp can be used for the LED driver in place of the LM324.

  1. Not the expected behavior! Probably a feedback resistor is missing somewhere, or you mixed up the V+ and V- pins of the op-amp. Or maybe you forgot to connect the power supply?
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Thanks, My idea of success is a “mostly” functional, smoke-free reaction to turning the power on. This first eight months of learning has produced quite a lot of “magic smoke.”

this might be the case, although I quadruple check my pins and even went so far as disassembling reassembling. One thing I just thought of though, perhaps the op amps themselves are bad/fried. or even fake. Now that i’m aware the tl07x’s are sufficient, I’ll swap them out, run through your suggestions and report back with results. Thanks so much for the quick response by the way!

Turns out the issue was my power supply after all. Thanks for your help. My sanity has been restored.

I do have a follow up question;

The 39k resistor coming off of the Lm324 LED drivers; Is this in place to keep too much voltage from going to ground? I have a two-color LED hooked up that nicely responds to the V+/- but the lights are very dim.

Would it be possible to buffer the voltage before the LED driver and use maybe a ~2.2k resistor on the LED’s as a way to boost the brightness, or would that create an undesirable outcome? I understand it would take another LM324 or similar, but that is about the extent of my understanding.

The 39k resistor sets the amount of current that will flow through the LED in response to the voltage being monitored. You can adjust it to a lower value if your LEDs are not high-intensity.

Voltage does not go anywhere.

I’m not sure what you mean by “buffer”, but you don’t need to do anything besides adjusting the resistor value.

i might be a little lost, conceptually. I’ll get back to my studies before I bombard this forum with a hundred beginner questions.

a voltage buffer came to mind because when I lowered the LED resistor value to 2k I read a pretty big voltage drop at the output’s 1k resistor. +/- 5v in attenuverter mode. Maybe this indicates additional problems in my circuit assembly, or is this an expected behavior when using a much lower value resistor?

Meanwhile I’ll swap the R for something a bit closer to the 39k value to see if that restores the DC at the output.

This is not the expected behavior. Other modules use the exact same circuit to drive regular (not high-intensity) LEDs, thus with much smaller resistors (for example the first version of Shades:

Don’t approach this like black magic :slight_smile: The circuit works for a reason!

At V+ is the voltage you want to monitor visually. Unless it saturates, the op-amp will adjust its output for V+ to match V-. Thus the current flowing through the resistor is V- / R = V+ / R. No current flows through the op-amp inputs. You can thus conclude that the current flowing through the LED is also equal to V+ / R.

This circuit effectively takes care of providing the right voltage at the LED anode (or cathode for the opposite color) so that the current flowing through the LED is proportional to the voltage to monitor.