I have this multicopter. it is a flying robot, with one huge, powerful LiPo underneath.
I say it again, it is huge, and it can pull a car with all the amps and burst-c's that I dont understand.
Building the thing was terribly nerve wrecking - though it mostly was made from other peoples semi-premade parts. Because; I have extremely delicate 3Volts electronics side by side next to stuff that needs cables fat as a thumb, because of all the juice (in lack of better words) they have to let through.
Now, my question is this:
I just purchased a new tiny video transmitter (and a matching receiver). Cool! The specs on this transmitter, that is to be placed on the flying thing says that in terms of Volts, it is fine within the range of what the huge brick of a powerhouse LiPo can deliver.
But I also know that this brick of a LiPo is so strong on the Amp-side, that I could use it to welder.
Does this matter? Is it OK to hook up a tiny thing right up to something that can kill an elephant in Amps, if the Volts match? Or are some volts more heavy than other - will my transmitter fry?
Good to go
To the best of my knowledge, “amps” (current) can only be taken, not given. Your camera should take only what it needs and that’s it.
I would like someone to check me on this one though.
Getting close to flying, Frits?
That’s pretty much that Chris
That’s pretty much that Chris
Thanks CTC :)I thought
I thought that was how it was… but stilll it seams scary
Yes, I am fully flying, have been so for a while - and I will post about it… once I know more
Thanks. I think I’ll shoot a
Thanks. I think I’ll shoot a video of it first time I hook it up though, just in case it blows, might at least get a good video for the money
I will side with Chris here.
Current is pulled not pushed. You can put motors and LEDs on the same circuit. LEDs pull in the neighborhood of 20mA. Whereas, motors can pull 1000’s of mA, depending on the motor and driver circuit you have set up.
You almost always put a resistor inline with a LED to limit the voltage. Even though it is called a current limiting resistor.
Maybe too much information, but, you calculate the resistor value with the formula:
resistor in ohms = (supply voltage - LED required voltage)/LED current in Amps
The supply voltage is whatever you are using to drive the LED, 5 volts for instance. The required voltage is listed on the packaging or in the datasheet for the LED, red LEDs are around 2.5 volts or less. The LED current is also listed on the packaging or with the datasheet, typical LEDs as I said pull 20mA. So, resistor = (5 - 2.5)/.020 = 2.5/.02 = 125 ohms.
My point was that the resistor only drops the voltage. Current draw is what the component will pull all by itself. You could power a SHR with a car battery, as long as you made sure to include the proper voltage regulator.
Not a problem…but…
Watch for huge inductive spikes if you are switching those motors on and off. Don’t run any power lines parallel to the sensitive equipment power lines without some kind of conditioning. The more parallel the run, the worse the problem. Doesn’t matter how good your regulation to 5 volts for the logic is, kick off and stop those motors close to the sensitive stuff will not blow them, but may cause intermittent problems.
Good point, Bangstick.
Good point, Bangstick. You’re huge battery won’t push too much current through your camera, but noise and spikes on the power line are something to watch out for.
The good news is that you can use caps to help filter noise and spikes. Diodes on your motor drivers can prevent problems with back EMF from the motors. As Bangstick says, separate your motor power from signal lines (inductive noise falls off quickly with distance). If you have to cross a noisy line with a sensitive one, it is best to do so at a right angle.
This is the spirit what this
This is the spirit what this website is made for
But don’t be scared…Chris is right…current is not coming from alone…it’s just taken by the device. You can lit a LED with a huge lead acid truck battery and it will not kill the LED but you have to reduce the voltage.
I’m getting so old, I had totallly forgot about crossing at right angles to minimize inductive interference.
Thanks for that.