# What resistor to simulate a 2CR5 under load (homemade battery tester)?

Discussion in 'Digital Photography' started by Jack Blake, Jan 1, 2005.

1. ### Jack BlakeGuest

Hello, and Happy New Year to everybody.

My battery tester uses a (too weak) 10mA load for its 6V test.

I'm guessing that a 2CR5 Photo Lithium battery should be tested at

Any electronics geniuses out there? What type/size resistor should I
connect to the battery, while testing with my battery meter, to
simulate a Canon SLR's load, and provide me with an estimation of the
usability of my 2CR5 (and other 6V lithium) batteries?

I also have a DMM.

Thanks for all advise and tips.

Jack Blake, Jan 1, 2005

2. ### Charles SchulerGuest

40 Ohms (or about) will work. 2-W is a good power rating.

Charles Schuler, Jan 1, 2005

3. ### kahoutGuest

Assuming:
2) the battery has a 6 volt potential

Then R = V / I, so that R = 6 /1.5 =4
Then W = V x A. so W = 6 x 1.5 = 9

You would need a 4 ohm, 9 watt minimum resister load.

If the load is a factor of ten lower (150 mA), then the resistance would be
40 ohms at 1 watt.

Kevin

kahout, Jan 1, 2005
4. ### Bob WilliamsGuest

Digicams are really tough on batteries. I would guess that a camera with
the LCD screen on, would probably draw 1,000mA.
So, I would connect a 6 Ohm, 5-10 Watt resistor across the battery
terminals while measuring the voltage. I'd leave the load on for 5-6
seconds and see how much the voltage drops in that time.
For a fresh Lithium Battery, I'd expect the voltage drop to be no more
Bob Williams

Bob Williams, Jan 2, 2005
5. ### Al DykesGuest

Google for "2CR5 specifications" and you'll find manufacturer's pages
that have electrical specs for the battereies that include
voltage/current/time graphs.

Al Dykes, Jan 2, 2005
6. ### wayneGuest

I find a light bulb works pretty good at the right voltage and wattage as
you can see the drain. so a 6V 10 watt should work or an equivalent 12V
should work fine.

Wayne

wayne, Jan 2, 2005
7. ### GuestGuest

( R = E/I ) = ( 6V /.150A ) = 40 OHMS

( P = I * E ) = ( .150A * 6V ) = .9W

A 40 ohm, 2 watt resistor will provide the correct load.

THO,
get a;

6v, 100ma bulb
and a
6v 150 ma bulb

That way, you'll have a quick "visual"
re the state of your battery.

<rj>

Guest, Jan 2, 2005
8. ### secheeseGuest

A light bulb initially provides close to a dead short across your
battery because the tungsten starts off relatively cold. I wouldn't
do this to my camera battery. I think using a resistor as others have
suggested, is the better plan.

secheese, Jan 2, 2005