Hi,

I currently have a crude 24V battery load tester that I use to check the capacity of large lead acid batteries (in order of 200ahr). Its a big mother that has 3x 3ohm 1000 watt resistors in it. Normally I pull about 30amps during the load test.

I'd like to replace it with something physically smaller and lighter and was thinking of using some MOSFETS in replace of the wire wound resistors.

After doing a few calculations I'm finding that although the MOSFETS can handle the current (eg 6x mosfets drawing 5 amps each) the power they would dissipate is quite small eg 6watts each (i need around 1000w) using them purely as a resistive load.

Instead I would like to use the mosfets to switch an inductor/ coil at a certain frequency generating an inductive load.

How easy would it be to build my own coil?

Any other way to build this load tester that I haven't thought of?

Cheers
Squib

 

That would be more like 120W each (24V x 5A)
There would be no point in doing this because you want to waste energy not be efficient by building a switching regulator..... Now there is an idea; build a boost inverter that drains/load-tests one set of batteries while charging another set.

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Hi,

I don't want to get too fancy charging other batteries and the like. Your right 120W, what I meant was that the mosfets when passing 5amps would loose 6w in heat....

So what I was thinking was loosing the remainder (eg 114W) through a coil wired in series with the mosfet so that the reactance of the coil would indeed dissipate the power.

Of course, this all depends on how big the coils are going to be.... and in the above scenario, there would be 5 of them.

The benifit of this design would be not as much heat generated as most the power is being lossed as emf in the coils.

Squib

 

I think you are missing the point that if you want to create a load that allows you to run 30 amps during a test that the power will have to end up as heat unless you plan to store it (such as the battery charger). A coil or inductor will store some energy but only for a moment. Any losses will still end up as heat.

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stevez

 

Last time I made a 12 and 24 volt battery tester I used car head lights for the load.
In the case of 24 volts use two 12 volt bulbs. Car lights are available and 100 watt resistors are a little hard to find on Sunday afternoon.

 

To supplement the car headlights you could add tail light bulbs to add smaller increments of load.

Don't dismiss a modest quantity of resistors in parallel and series - if you can obtain them via surplus, ham flea market, etc. Arrangement depends on resistance and power handling capability.

The electric radiator cooling fan from a car might also serve as a load though you might need to employ several. The energy from the motor shaft is converted to "motion" by the fan blades and ultimately ends up as heat - however it is more diffused. A water pump or hydraulic pump might also be considered.

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stevez

 

I would keep it simple and use the resistors set up as it is. ( KISS principle )

The lamp option as ronsimpson suggests is a good alternative and you get some extra light as well as heat.

Considering the 200 Ah 24 Volts accu's you are testing, these are very heavy anyway and you need reasonably sized testleads ( 6 mm² ) and alligator clips anyway to run 30 Ampères testload.

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