airtronics transmitter and 3 lipo with diodes

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airtronics transmitter and 3 lipo with diodes

Postby amtpdb1 » Mon Jun 18, 2012 2:27 am

Hi:
I just bought a 3 cell lipo transmitter battery from Hobby King. For My Airtronics radio I had to switch the positive and negitive posts in the plug. As the voltage is higher with the lipos then in the nicad pack, I was going to install 4 diodes,2 series sets set parallel to cut the voltage by 1.4 volts. The reason for 2 parallel sets is in case one diode gives up the ghost the other set will continue working. This transmitter has a 9.5 volt alarm and I figure that with the 1.4 volt drop the lipo will still be 10.9 volts which is about 3.6 volts per cell which seems good. I have heard of people doing this, but cannot find any thing now that I am thinking about doing it. Is there any draw backs to doing this? Will the diodes draw current and heat up in the transmitter? Does anyone have any feedback or reasons one way or the other to do this or not do this? Any suggestions?
Thank in advance for any thoughts and or help with this.
Don :?
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Re: airtronics transmitter and 3 lipo with diodes

Postby BillGriffiths100 » Mon Jun 18, 2012 8:25 am

If you're worried use 1N5004 's well overrated at 1 amp that will mean you don't have to double up(ie parallel) , and ensure a good current margin however the current draw will be small anyway.
The diodes do heat up when they are dropping voltage like a resistor, but unlike a resistor they maintain a constant voltage drop irrespective of the applied voltage (within their rating).
Say your outside current was 400Ma (although it will be much less) your drop will be 0.4*1.4 = 0.56 watts . Each one dropping half 0.28w the diodes will hardly get warm.
Note we are not talking an over-current condition.
Last edited by BillGriffiths100 on Fri Jun 14, 2013 3:23 pm, edited 3 times in total.
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Re: airtronics transmitter and 3 lipo with diodes

Postby ergocentric » Mon Jun 18, 2012 4:31 pm

Parallel diodes and transistors:
in an overcurrent condition, the diode with the lowest forward voltage will pass pretty much all the current, and burn out
then the next device, then the next.

Bipolar devices do not share current well.
unconcerned but not indifferent - MAN RAY
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