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I need some help in figuring out what size capacitor I need, and if my idea will even work.

I have a small remotely-controlled helicopter, and a very small camera. I am trying to record video while flying the helicopter around. So far, cutting down the body of the helicopter and taking the PCB out of the camera and mounting it to the helicopter has worked well. However, the usable flight time is reduced from 10 minutes or so to 1 minute. I noticed that the battery on the camera and the battery on the helicopter are the same type and voltage. What I would like to do is power the helicopter and the camera from the same battery to save on weight.

My original plan was to simply cut the camera battery out of the picture and tie the camera directly do the helicopter. However, I thought this would cause problems, as the voltage while charging would likely fry the camera.

Then, I noticed the decorative LED on the front of the helicopter. The voltage (3.7V or so, depending on battery charge) is exactly what is needed to power the camera. The LED is also off while the helicopter is turned off and/or charging. Now, this helicopter alternates the polarity going to the LED twice a second, as it is a bi-color LED. My current plan is to cut out the LED, rectify the voltage with 4 diodes so I have straight DC, and use this to power the camera. (Barring any current issues of course. I plan to measure the current load of the LED vs. the camera before doing this, but I expect they will be similar, given the amount of time the camera can run off of its own battery [1 hour+] of similar capacity to that of the helicopter.)

I am going to experiment with this plan tonight, but I just realized that there is probably a very short time where the LED driver outputs 0V in between its polarity switch. I'm thinking that sticking a capacitor in there will help with the problem, but I have no idea how to calculate what size I need.

Can you tell me what I need to do to calculate the size cap that would likely keep things running smoothly? Again, the voltage output of the LED driver and the voltage requirement of the camera is 3.7V, but I'm guessing will be anywhere from 4.2 to 3.6 during operation. I don't yet know the current requirements, but plan to figure those out tonight.

Thank you for your time.

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2 Answers 2

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I would guess the camera takes 100-150 mA (180 mAh 3.7V battery, ~1 h runtime), and your LED will only be driven with about 20 mA, plus there should be a resistor in series with it.

I'd just power them both off the same battery if weight is a major issue, and if you're concerned about the charging voltage damaging the camera, add some cutoff switch (or just unplug it) while charging.

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    \$\begingroup\$ Thanks Nick. I tried some things tonight. The LED was blinking on its own, and the power from the helicopter was not alternating as I originally suspected. I soldered the power that originally was going to the LED right to the camera and tried to power everything up. Sure enough, it wasn't enough to run the camera. Prior to camera power-on, the voltage at the camera's input was 4.17V, and after attempted camera power-on, it was 1.7V. I am going to follow your suggestion next. Thanks! \$\endgroup\$
    – Brad
    Commented Oct 11, 2010 at 22:48
  • \$\begingroup\$ I've tried connecting the camera to the battery, but the camera dies after about 40 seconds. The camera shuts itself off if the battery voltage dips below 3.6V. I think that the voltage drop while the main propeller motors spin up is a bit too much for it. After the camera dies, the helicopter is fine flying around for several more minutes. I think this project is dead. \$\endgroup\$
    – Brad
    Commented Oct 12, 2010 at 2:54
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I think that just putting a 2.2mF capacitor across the battery terminals will work well enough to smooth voltage. I presume the voltage dips when the ESC draws large current pulses for the motor.

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