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# 18-20V battery capacity measured on 15V capacity meter

#### SentinelAeon

##### Member
I am using a battery capacity tester as seen on image. It will test batteries that are max 15V and up to 3A current is allowed. I have some 18V and 20V battery packs that i would like to test. So i was wondering ... could i use a resistor in series with the battery to reduce the voltage at the meter, so that the voltage will be lower than 15V ? And then use the value of capacitor to figure out how much energy went into the capacity meter and how much was lost as heat on a resistor ?

What value resistor would i use and how i would calculate the capacity then ?

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You certainly can’t use a capacitor. I think the fact that you suggested a capacitor means that you need better understanding of what a capacitor does.

A zener diode should work. A zener diode in series will reduce the voltage by the voltage rating of the zener. You need to make sure that the power rating (in Watts) of the zener is large enough for the discharge current.

For instance, if you have an 18 V battery, and you use a 4.7 V zener, the voltage across the discharge resistor will be about 13.3 V. If you use a 50 Ohm resistor, the current will be around 0.27 A. The heat dissipated in the resistor will be about 3.5 W and the heat dissipated in the zener diode will be about 1.25 W.

I suggest that you use a resistor rated at 10 W or more and a zener rated at 3 W or more as components run near their maximum power rating get really hot. Using a fan helps a lot.

The zener won’t affect the Ah capacity measured.

I'm sorry, it was a typo, i meant a resistor ofcourse ! Resistor in series with the batteries. I am asking because i have lots of 50 and 100W resistors at home and thought i could use them so i dont have to buy anything. And then just calculate how much is lost on resistor and how much goes into a capacity meter. It doesn't have to be very precise, so +- 15% is perfectly fine, i mostly measure to compare different packs as in which is better.

The problem with a resistor is that when the discharging isn’t running, so before and after the test, the tester will see the full battery voltage and my be damaged or may not work.

And, if you put a resistor in series with the tester, you'll only get a maximum of 20V/100 ohms = 0.2 amps load (or 4 watts).

You can do the math a 50 ohm resistor.

Note that this is the absolute maximum, of course. Because the load tester has its own resistors in series.

You could put a bunch of 50 ohm resistors in parallel before the tester but then, as mentioned above, you run into the risk of applying nearly 20 volts to the input when the device is set to a very low load as the voltage across the load of the device and the load of the resistor pack will be determined by ohm's law.

I will be honest and say that i have never used a zener and much like current, i choose the path of least resistance. But i will take your advice and buy a zener and try it out.

So if i understand it correctly, the voltage rating of the zener is how much the source voltage will be reduced. The resistor is connected directly to the capacity meter device, i will adjust resistor rating to get 3A at the start of the test.

Also if i understand it correctly, the amount of heat dissipated on the zener will be (Vin - Vout) * current ? So if i shoot for 3A current and lower the voltage from 18V to 15V, then i will need a 9W zener and/or cool it with heatsink/fan ?

Is there anything else i need to know about ?

These load testers are posted on AMAzon as a 2-pack. Just put both in series with your battery pack. Total current will be the same - voltage through each will be half. Assuming you bought the 2-pack...

I’ve got one of those load testers and they use an external resistor to set the current. The load testers turn the current on and off at the start and end of the test so putting two in series won’t work as they will turn off at different times.

So if i understand it correctly, the voltage rating of the zener is how much the source voltage will be reduced. The resistor is connected directly to the capacity meter device, i will adjust resistor rating to get 3A at the start of the test.

Also if i understand it correctly, the amount of heat dissipated on the zener will be (Vin - Vout) * current ? So if i shoot for 3A current and lower the voltage from 18V to 15V, then i will need a 9W zener and/or cool it with heatsink/fan ?

Is there anything else i need to know about ?
That is correct, but I would say the heat is the zener voltage times the current.

If you have a 3 V zener and discharging at 3 A, then 9 W will be dissipated and the zener should be rated at 9 W or more. The larger the rating, the easier it will be to keep from overheating.

I’ve got one of those load testers and they use an external resistor to set the current. The load testers turn the current on and off at the start and end of the test so putting two in series won’t work as they will turn off at different times.
Then add a 100 ohm resistor in parallel across each as well and you can easily analyze the circuit.

I did a bit of digging and compared to step-down, zener diode's advantage is that it will give me the correct mAh rating, including the part getting used on resistor, unlike step-down where i will just get the amps coming out of the step-down but not the ~10% getting wasted as heat on step-down.

Could someone give me a scheme on how to connect everything, zener diode and resistor (resistor that acompanies zener diode, not the resistor thats connected to capacity meter)

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