I'm building a simple project that consists of a 9v LED, a 5v Gas Sensor and a 3.3v micro controller. All the devices have low current draw, so I'm not really worried about hooking them up in parallel.

I was thinking that the 9v LED could be directly connected to a battery. In parallel with that I would like a 5v and 3.3v voltage regulator, where both of the voltage regulators would power the other two components.

I've looked at this thread: Power Supply Design - Multiple Voltage Regulators but got some mixed signals regarding hooking up the voltage regulators in parallel or series, and was wondering if anyone could give me a definitive answer on best practice? Additionally, I've seen some articles that encourage bypass capacitors to help smooth out the DC signal, because I'm connecting directly to a battery I can ignore that, correct?

The LED is a MegaBrite (https://www.sparkfun.com/products/10236)

The gas sensor is a MQ-3 (https://www.sparkfun.com/products/8880)

and the microcontroller is an MSP430 (breadboard/launchpad version)


simulate this circuit – Schematic created using CircuitLab

Thanks for your help!

  • \$\begingroup\$ I think most LED's would be happy running on 5V; but will the microcontroller talk to either LED and/or sensor? If so, don't forget level shifting 5V -> 3.3V -> 9V. \$\endgroup\$
    – Ron J.
    Jul 2, 2013 at 20:49
  • \$\begingroup\$ Some more details about the LED, gas sensor and uC would be nice. The gas sensors I know are power hungry, they would draw a 9V battery in no time (if the battery was able to sustain them at all). Most uC's can run on 5V. \$\endgroup\$ Jul 2, 2013 at 21:02
  • \$\begingroup\$ I've updated the description above, you can see the parts now. I'd like to power it with one power source. If I need more batteries, or an alternative, that's fine. I'd rather switch batteries than use multiple types of batteries though. I doubt it would need to work more than four hours or so at a time. \$\endgroup\$ Jul 2, 2013 at 21:17
  • \$\begingroup\$ The LED will run from 5.5V - any voltage above this is wasted energy unless you know that the brightness increases. According to the spec 5.5V to 9V is fine and I'd definitely use a switcher for making the power rails run more efficiently. \$\endgroup\$
    – Andy aka
    Jul 3, 2013 at 8:13
  • \$\begingroup\$ Is this the sort of voltage switcher you were talking about? robotshop.com/… Where I would just need one to get me to 3.3v? \$\endgroup\$ Jul 3, 2013 at 13:10

1 Answer 1


You are wasting quite a bit of power by using linear regulators to drop 9 V to 5 V and particularly to 3.3 V. There is also no need to run the LED from 9 V. There is no such thing as a "9 V LED". There are some packaged devices that contain a LED and resistor so that the whole thing is intended to run from some particular volage, like 9 V, 12 V, 24 V, or whatever, but then you don't have just a LED anymore.

It would be useful to know the current requirements of the 5 V circuitry, the 3.3 V circuitry, and the LED. I'm guessing that the sensor probably doesn't take much current. In that case, I'd probably use a linear regulator to make 5 V as you suggest, but a switcher to make the 3.3 V from the 9 V. The sensor runs from 5 V and everything else, including the LED, runs from 3.3 V. That will be a lot more efficient. Assuming this is a normal red, yellow, or green LED, it can be run from 3.3 V easily with the right resistor in series.

  • \$\begingroup\$ Well, the LED I have contains three groupings of LEDs, a 32-bit register and a controller to drive it all. The minimum voltage of that setup is 5.5v. I tried to get it to work with 5v, but I wasn't able to get it to behave as well. Do you see serious concerns with going from 6v to 5v then to 3.3v? I could do that instead. I liked the size of a 9v battery though. \$\endgroup\$ Jul 3, 2013 at 13:05
  • \$\begingroup\$ Also, I plan on having the 5v sensor switched. Do I need to have anything in place to help control extra current that may exist in the 9v and 3.3v line when the sensor is disconnected? \$\endgroup\$ Jul 3, 2013 at 13:25

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