2
\$\begingroup\$

I am interested in adapting this Jameco kit schematic to my own needs. I have run into an issue though. I am not sure how to lower the voltage requirements. This circuit is built for 12V, I would like to power it with between 3.3 and 5 volts, specifically from a lithium-ion battery.

Schematic

My question is specifically about the various op-amps used in the circuit.

The MC33204P Datasheet says that the minimum supply voltage is 1.8V (+/- 0.9V) and the maximum is 12V(+/- 6V). I realize that this means that I can use a 3.3V supply, but how much do I have to change in terms of the resistors and capacitors, if anything at all?

The MC34072P Datasheet does not mention a minimum supply voltage as far as I can see, although I may be mistaken. One of the graphs showing supply current vs supply voltage, starts at 1V. Will this op-amp work as intended with the resistors of the original schematic? Will is work at all? Should it be replaced?

I do not plan on following the schematic past the transistors.

\$\endgroup\$
4
  • 1
    \$\begingroup\$ The important property for MC33204P is its rail-to-rail input and output capability, I'm unsure why another type was used for the fifth opamp. R27/28 set the threshold for all the other opamps and R22/23 set the input gain. You may want to experiment with those values to get a fit for your purpose at lower power supply voltage. \$\endgroup\$
    – jippie
    Aug 13, 2014 at 6:12
  • 1
    \$\begingroup\$ You know, there are plenty of other color organ schematics out there that don't require opamps or virtual grounds. But aside from the led stage at the end needing to be switched to a parallel setup, you can just reduce the voltage. Response might be a bit clippy tho \$\endgroup\$
    – Passerby
    Aug 13, 2014 at 16:40
  • 1
    \$\begingroup\$ Just a couple of things to consider in addition to what has already been mentioned: 1)Reducing the number of LEDs or put all 4 in parallel (Not recommended as one LED may end up drawing all the current) for the outputs as you won't have enough voltage to cover two ~3V (depending on what LEDs you use) drops 2)Reducing the gain on the input amplifier so that it doesn't saturate and the signal does not saturate the following op amps (using the lower supply voltage) 3)The ~0.7V drop across the 1N4002 diodes may cause some issues with lower voltages as well (transistors may not turn on properly) \$\endgroup\$
    – Luc
    Aug 13, 2014 at 17:55
  • 1
    \$\begingroup\$ The data sheet you link to is 11 years old. You can find the latest ones here. MC34072 supply voltage 3.0 to 44V onsemi.com/PowerSolutions/product.do?id=MC34072 MC33204 supply voltage 1.8V to 12V onsemi.com/PowerSolutions/product.do?id=MC33204 \$\endgroup\$
    – Matt B.
    Aug 13, 2014 at 22:41

1 Answer 1

2
\$\begingroup\$

The "5th" opamp is a virtual ground so that you other amps have a DC offset. This is done because you're using a single supply (just 'positive' volts). No matter what input voltage you pick, this circuit will take half and use it as the negative rail for the op-amps.

In that sense, this circuit is robust enough to handle the voltage change well. However, the LED's in series will probably need to have their current limiting resistors changed so that they get enough juice from your lower supply. I think that's all you'd really need to change. This circuit just lights up according to the frequency of the input right? I'm just looking at this at a glance.

\$\endgroup\$
1
  • \$\begingroup\$ Yes, the input is bandpassed to three light channels according to frequency/intensity. \$\endgroup\$
    – Passerby
    Aug 13, 2014 at 16:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.