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I am working on a color organ and want to use a design similar to this. I understand the gist of the schematic, but I have a few questions about it and how it works.

First: I understand that it is trying to tell me how to hook up my input audio signal, but I don't understand what it is telling me.

Second: This is describing a potentiometer connection, right? If so, do you know why having a potentiometer here is advantageous over having a static voltage divider?

Third: The active band pass filter used is slightly different than the band pass filters I see on google, so what does the resistor that is highlighted do?

Fourth: I want to increase the number of band pass filters so I can control more LEDs at different frequencies. To do this, I believe I would have to have a higher Q value and smaller band widths per filter, correct? How do I increase the Q of my filter so it is a smaller band pass?

Fifth: What is the point of this bit of circuit? Can't the op amp output go straight to the transistor?

Finally: They don't have transistor models marked. Since they are powering lots of LEDs, would you suggest a power transistor? Or maybe a darlington pair for better control?

Thanks in advance for your help! I really appreciate it!

EDIT: Quick followup question: Should I assume the VCC in the red squared part of the schematic is 12V as well?

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  • \$\begingroup\$ First off the thing that is confusing me about this is that it appears to show the audio input jack wired backward... the signal (tip & ring) are grounded and the ground (sleeve) is wired to the signal input. So for this to work it has to be battery operated, and there doesn't seem to be a good reason for this. \$\endgroup\$ – Daniel Dec 14 '15 at 4:04
  • \$\begingroup\$ @Daniel -- I agree, the jacks are totally ass-backwards. I built two of these (the schematic is from a Jameco kit) so I could have stereo light organs and the only way to make that work right was a traditional common sleeve ground. Otherwise it worked great. \$\endgroup\$ – tcrosley Dec 14 '15 at 4:22
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First: it's saying the audio input should be a 3-pin (stereo) jack, probably 3.5mm. However, they've connected two of the pins together, so you might as well use a mono (2-pin) jack. "JACKPTH" is probably their name for a plated-through-hole jack, i.e. one with pins that goes through the PCB instead of being surface-mount.

Second: R24, R25 and R26 are effectively volume controls for each of the three channels. They allow you to adjust the sensitivity of the circuit to deal with different incoming sound levels in each band. Say if you have very bass-heavy sound, you might want to turn down one of the pots so that you don't just get the bass LEDs lit constantly while the treble ones never come on or vice-versa.

Third: Consider the R1/R7 junction as the input to the bandpass filter, and consider the input of the filter as seen by R1. You've got a voltage divider formed by R1 and the parallel combination of R7 and the capacitors. Without R7, that voltage divider would form an RC low-pass filter but with R7 present, the bottom leg of the divider is largely real (dominated by R7).

Stated alternatively, R7 makes the input impedance of the bandpass filter be mostly-real. That means that the only filtering effect you get is as-designed and not from the interaction between the filter block and the output impedance of the block that supplies it (R1 and R24).

The Google schematic you post has a very different input arrangement to the bandpass, which is not purely capacitive.

Fourth: You need a narrower band for each filter, yes. I would suggest googling up the formulae for a chosen bandpass filter topology and plugging in the values (of centre frequency and Q) that you want, and deriving component values from that. And do some experimenting in a circuit simulator.

Fifth: D1+C11 is an envelope detector. It produces a signal that tracks the average volume of the output of the bandpass filter, with all the audio-frequency signal removed. I am surprised by the lack of a base resistor though!

Finally: two chains of LEDs drawing 10mA to maybe 50mA is not a heavy load. I would go with BC548 or anything similar. Doesn't really matter as long as you don't exceed the max collector current of the device. Note the resistors in series with the LEDs have interesting values, which will depend upon the specific models of LEDs chosen, i.e. their intended operating current and their forward voltage.

Vcc: seems reasonable to me.

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  • \$\begingroup\$ The transistors in the original kit are 2N3904s. Parts list on the page I linked to. \$\endgroup\$ – tcrosley Dec 14 '15 at 4:44
  • \$\begingroup\$ Thank you so much for your detailed answer! I think that gives me a good idea of what I'm working with. As for the LEDs, I think the different resistor values are there because they are lighting different color LEDs. Also, I was thinking of a more robust transistor because eventually I want to power more than this, maybe a string of LEDs or some EL wire. But as long as I know the current draw, I should be able to pick the right transistor. \$\endgroup\$ – Gigaxalus Dec 14 '15 at 4:51
  • \$\begingroup\$ Yes, different colour LEDs have different forward-voltages. A blue LED is about 3.6V while a red might be 1.8V or less, which means that there is less voltage across the resistors in series with the blue LEDs (12-2*3.6=4.8V) than the red (12-2*1.8=8.4V), which in turn means you need a lower resistance to get the same current. And since our eyes are less sensitive to blue, you probably need a higher current through them to get similar apparent brightness, hence the very very different values. \$\endgroup\$ – William Brodie-Tyrrell Dec 15 '15 at 1:18
  • \$\begingroup\$ If you want to drive a heavier load, I suggest using a darlington, e.g. TIP120. You could merely install a larger transistor like a TIP31, it will have lower gain and may change the behaviour of the envelope detector by drawing more current from C11. Nothing that can't be overcome with a little adjustment (probably a reduction of that 100k resistor or the insertion of a small base resistor), but not a simple replacement. \$\endgroup\$ – William Brodie-Tyrrell Dec 15 '15 at 1:18

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