Optimal Load Value for Battery Tester Simulating Vehicle Start-Up Conditions

Hello
I need a simple tester to check the dry condition of the batteries.
The tester is to simulate the start of the car if the battery voltage does not drop below 10V.
Construction: a piece of wire on the section of which resistors or other components giving an artificial load will be attached.

The problem is choosing the right load value. I have browsed a few topics with inrush current and some say that the starter takes 100-150A from the battery, and others that 300-350A.
Another problem is that at the moment of start-up, these values jump significantly because many factors of work change, and with such an artificial load, the current is constant.

What value will be the most optimal for such a test?
On the one hand, I want to get a strong load so that the measurement is reliable, and on the other hand, I don't want to finish the battery.
I will add that the measurement length would take about 5 seconds for the multimeter to have a moment to stabilize the values.

Hello
Original testers available for purchase load 100A or 125A
greetings

This is already concrete, thank you

It should be realized that to load 350A you have to take into account the power of 4.2kW.

For example, if you load a copper wire of an appropriate length, its temperature will change for a few seconds, which will cause the resistance to change as well.

Maybe I would focus on 100W resistors.
For example like this:

Such a resistor can be loaded 5x with a time of 5 seconds, which gives us 500W / piece.
http://www.tme.eu/pl/details/ax100wr-0r47/rezystory-100w/te-connectivity/5-1625999-3/

Counting on ...


470mOhm, it gives it at:
= 10V, current: 21A, power 212W
= 12.6V, current: 27A, power 340W

220mOhm, it gives it at:
= 10V, current: 45A, power 454W
= 12.6V, current: 57A, power 718W
Here, the activation time is max. 3.5 seconds.

Connect the load with a car relay, each resistor separately - for example a time relay.

Probably wind a few turns of steel wire cooled with a PC 12V fan.

It will be cheaper to buy a new one (100 A) for up to PLN 100, or a stimulant (professional equipment 150/300 A) for about PLN 250.

I would like to make a 100 ? resistor with a power of 0.1 W, but the cost is about 14 cents and in the store they are 10 cents. Advise on how to go below 10 cents, I need two such resistors.

zasilaczen wrote:

I have browsed a few topics with inrush current and some say that the starter takes 100-150A from the battery, and others that 300-350A.

Both are right. It depends on the engine model and temperature at start-up.

zasilaczen wrote:

I will add that the measurement length would take about 5 seconds for the multimeter to have a moment to stabilize the values.

Longer than normal boot time? 3 seconds is enough to stabilize the measurement result, unless you want to check if the voltage above 10V is maintained for at least 5s.

zasilaczen wrote:

What value will be the most optimal for such a test?

The value will depend on the rated inrush current of the battery under test.
Do you want to test "so-so" or according to the standard? The standard specifies the method of performing the measurement.

So yes gentlemen.
I will aim at 125A, i.e. how well I counted 1500W.
On resistors, it would definitely work fine, but the cost of three is almost PLN 100.
The scale of my needs is also not such that I should immediately obtain a professional tester.
I am more inclined towards something "Home Made". I have workshop facilities, so there is nothing to prevent me from doing something like this myself, I just need your substantive help to make it all have hands and feet.

The idea with wire is nice, but I'm afraid it will have to be a lot to get this load, and I need something more handy.

The idea with sheet metal is equally interesting, but I am not an electronics engineer and I do not know how 60m? is up to 125A

I really like the idea of glow plugs, small, handy and power-hungry, this could be it 5 seconds of connect should be enough. Describe something more about it and how you have built the structure

@vodiczka Tester is only to exclude batteries where the voltage drops below 10V. For more accurate measurements, I would already invest in a testing device.

To test the condition of the battery, it is not necessary to load it with electricity. You have to measure its internal resistance. Then you can easily determine the starting current and the condition of the battery very accurately. A healthy battery is approximately 3 to 5 mOhm which gives a starting current of approximately 350A.

Good quality testers measure the voltage difference with no load and with a load of about 10A and from this they determine the internal resistance and the possible starting current. Even better instruments, e.g. Hioki, measure the battery resistance for 1kHz AC current and do not require any current load (but the cost is considerable).

It should also be remembered that the "starting" current given on the battery is the maximum permissible current, which in a sufficiently short time (specified in the standard which is written opposite) allows to charge the battery without damaging it. Otherwise, the lead connectors between the cells will burn out or the lead plates will bend.

I do not see any sense with the construction. You have ready both load and newer electronic ones. Cheap as dirt, so why bother and think about it.
The cheapest electronics are $ 20 with a $ 30 resistor. So what will you build below PLN 100?

http://www.jula.pl/catalog/motoryzacja/akceso...akumulatorow/tester-akumulatora-604178/#tab04

On the well-known Yato auction site, you can buy for less than PLN 100, I do not see any sense in DIY in this case.

Do not read the voltage during the test with the built-in voltmeter - it lowers the measurement.
Measure with a separate multimeter directly at the battery poles.
However, the built-in resistor actually has the value of 0.12 ?, i.e. a load of about 100 A.
This applies to my tester similar to this one: