hand held night vision

Stephen-mx3

A friend and myself have both taken apart a pair of Call of Duty NVGs and taken them down to bare bones (which is quite bare) and have both mounted them in different style housings

My question is directed to the electronically minded out there. The NVGs are powered by 5 AA batteries (standard AAs are approx 600mah each IIRC, and 5 equals 6v) but the battery holder is bleedin massive. SO, i bought a cheapo flea-bay 7.2v 3800mah NiMH battery which is smaller and easier to handle

What kind of resistance (if needed) should I install to stop this from frying the board? Should I just butcher the NiMH battery and remove a cell to drop it down to 6v, or would the current still be too strong?



(Also, watch this space for my Black Ops RC Car project )

Leftyflip

The unit will only take the current needed, so there's no need to worry about the amps.
Ohm's law is your friend - V = i/r or in your case v/i = r.

Stephen-mx3

I hated physics! What's "i" again?

So you reckon it'll be ok ger?

thermo

please kepp discussion within the charter guys

Leftyflip

Stephen-mx3 wrote: »

I hated physics! What's "i" again?

So you reckon it'll be ok ger?

V= Voltage
I = Current
R = Resistance.

It should be, I've been running effects pedals off 12V 15000mah adaptors for months, just using a simple resistor circuit to take it down to 9V. Mind you, you do loose some of the Mah.

Stephen-mx3

ah well, with 3800mah running a small camera and screen it should be fine for a few games

thanks ger!

Stephen-mx3

so just so i have it right, it would be 7.2 divided by 3800, and the answer would be the resistance i need?

Stonewolf

I think it's actually 7.2/3.8 as 3800 is in milliAmps

Stephen-mx3

That is slightly easier to work out!

So resistance needed is approx 2 ohms?

Stonewolf

Well, since I gather you don't want to have no voltage surely it's (7.2-6)/3.8 which would be a bit over .3

Stephen-mx3

<insert facepalm emoticon>

that makes a lot of sense, thanks man

DeBurca

I am sorry to say lads but ye have got it wrong, the formula is right but ye are applying it to the battery and that is no good as it only tells you about the battery and tells you nothing about the load that you are applying it too, the load (IE the NVGs)
You have two choices either modify the battery by removing a cell and thereby reduce the output from 7.2v to 6v or fit a resistor in series with the load that will have a voltage drop across the resistor of 1.2v thereby reducing the remaining voltage to 6v
The only proper way to do this is to find out the internal resistance of the NVGs

There are two ways of doing this, one is to use a multimeter set on OHMS and simply read off the reading but this may not give an accurate reading as it will probably be a dynamitic circuit with a few I.C. and OP AMPS etc and as such the reading will not be reliable plus most multimetres use an internal 9v battery and there is a small chance that it could cause some damage to one or more of the internal components

That leaves us with method two which is to put the same multimetre set on milliamps in series with the battery and take a reading with the NVGs switched on

Once we know the reading then it is a simply calculation using Ohms law (I=E/R) or more simply
VOLTS divided by AMPS equal RESISTENCE measured in OHMS, remember that the multimeter is set to milliamps and you need it in amps.

Remember 100 milliamps is .1 AMP. 200 milliamps is .2 AMP etc

So after the calculation is done and you know the internal resistance of the NVG goggles all you need is a resistor that will cause a voltage drop of 1.2v across it and is equal to 1/5 of the internal resistance of the NVGs

When selecting the resistor make sure that you get a resistor that can handle the wattage, wattage can be calculated by the formula P=VxI or Power measured in WATTS equals AMPS multiplied by VOLTS

I hope this is of some help to you

As for me I would have removed a cell from the battery long ago

Stephen-mx3

well haven't started anything yet apart from mounting the night vision on the holder so I'll just whip a cell out

Sorted, thanks DeBurca

Gaz2010

did ye get anywhere wit this after,good idea all the same

Firekitten

Did you consider that now youve split the two units, the power isnt going to be the same? One unit, isnt going to need the juice of 2...

Stephen-mx3

Jeez, it has been a while since I looked at this.

Going to have another look at it tmw. There actually has been a surprising amount of times that this could have been useful. I might try it with 2 CR123 batteries instead. Rechargeable ones, that's pretty much the smallest I can make the power source!!

It's only one unit dude.