# Battery Test.

Discussion in 'The Projects Forum' started by Leek1001, Sep 13, 2011.

1. ### Leek1001 Thread Starter New Member

Sep 13, 2011
9
0
Good morning all,

At first, I'm not so good in the English language so excuus me for that.

My Question:
I want to test 2 serie battery's (12V+12V=) 24V 7Ah (C20 test) lead acid VLRA type.
They are used to power a motor (24VDC, maxload = 40ADC) (maxload only occurs only for a few seconds (20-30sec)).
After doing this they will be recharged.

How could I test these battery's ?
I want to know how the behave.

What do I want to accomplish is:
A emplacement of a battery connectect to a Resistive Loadbank (to simulate the motor) and I want to read de battery by checking the voltage with an ADC (microcontroller) and send this to a PC for analysing.

What I thought so far:
A lead acid battery cannot read out correctly when there is an load attached to this battery, so I want to try to switch off the load and then read it out.
The resistive loadbank is not a complete simulation of a motor but would it be enough?
P = U*I = 24VDC * 40ADC = 960Watt, so i need to dissipate a lot of heat.
If I want to withdraw 40A out of the battery I need to use a 0.6ohm 1kW resistor (or a lot of resistors in parallel), but when the load is attached to the battery, the OCV (Open Circuit Voltage) drops, so I won't withdraw 40A from the battery anymore. how to solve?

I've checked these topic's:
Electronic Loads A constant current withdraw.
Power MOSFET is core of regulated-dc electronic load

Points:
1) How to withdraw 40A from a lead acid battery (for only a few seconds 20 -30 sec)
3) The resistive loadbank is not a complete simulation of a motor but would it be enough?
4) How to withdraw a constant current from a lead acid battery?

Any thoughts, comments and help would be greatly appreciated.

Greets,
Leek1001.

2. ### wayneh Expert

Sep 9, 2010
11,897
2,845
What do you really want to know? The performance of lead-acid batteries is a well known thing. Note that voltage at the terminals can take quite a while to equilibrate. It will rise a little after a large discharge and fall a bit after charging. This is diffusion within the battery, and temperature plays a role. So you really need to know what question you want answered.

I'd consider using a constant current control into a load made of headlights. They're made to dissipate the heat and will give a great visual indication of what is happening.

Last edited: Sep 13, 2011
3. ### Leek1001 Thread Starter New Member

Sep 13, 2011
9
0
@wayneh.

Well the preformance isn't documentated at a discharge current of 40A.
So, I've absoluutly no idea how the battery will behave.
- How long can he handle this current.
- How long can he handle other currents.
- What happends if I give the battery a chargeboost for like 10min and then decharge again?
This is what I wan't to find out.

So I was thinking to create a constant current control with max 40A, but I've no clue how to begin.

Greets,
Leek

4. ### wayneh Expert

Sep 9, 2010
11,897
2,845
Here's an example of a simple constant current supply, just to show you the general concepts that many such circuits use.

The op-amp LM358 watches the small voltage across the low-ohms shunt that is in series with your load. Ohms law teaches that the voltage across that shunt will be proportional to the current through it. The op-amp compares that voltage to the voltage reference on its other input, set by the variable resistor, and adjusts the op-amp output to bring the two voltages in line (that's the magic of what op-amps do). That output in turn controls the current through the load via the darlington arrangement of the two transistors. The feedback loop is completed.

For your load size, you'd probably be better off to omit the darlington BJTs and use MOSFETS in parallel instead. They don't require the base current that the BJTs need. You'll need a big heat sink and fan to keep them all cool and fuses to protect things. Be sure to post your plans here for review by the pros here. 40 amps is not trivial and you need to consider safety first.

5. ### vrainom Member

Sep 8, 2011
109
19
Generally this type of batteries have an internal resistance of about 30 milliohms which means at 40A it would dissipate around 50 watts.

The 7AH are rated over 20 hrs (that is 350 milliamps steady for 20 hrs) but assuming an average 20A consumption it will last for under 10 minutes, 30 A = under 5 minutes, 40 A = ???? Profit!!!

If you want to measure the battery under load I think it's normal to consider a battery discharged at around 10.8v (1.8v per cell) Under heavy load you could go down as low as 9.6v (1.6v per cell), but be sure to charge it asap.

But try to locate the datasheet for the battery brand you have to get the specifics.

6. ### Smoke_Maker Active Member

Sep 24, 2007
126
15
Leek,

I understand what what you want to do, I do battery testing like what you want.

First find out what battery you have and go on the Internet and find out what the rating is, it will read something like 10 amps for 60 minutes this is how they rate the battery. 10 amps for 60 minutes = 10AH

Next discharge the battery at the battery maker discharge rate with a stop clock and compare it to the manufactures specifications.

You need a amp meter and a voltmeter, you can use a packaging banding strap (flat steel strap to hold box on a wood shipping pallet). Start with 4 to 6 feet and then trim to correct current draw.

7. ### wayneh Expert

Sep 9, 2010
11,897
2,845
That would be my approach, since usually you want to know if the battery is still performing, but it isn't what the OP said he's interested in, per #3:
"- How long can he handle this (40A) current.
- How long can he handle other currents.
- What happends if I give the battery a chargeboost for like 10min and then decharge again?
This is what I wan't to find out."

But he can run any test he wants if he has the right tools. Packing strap as a constant current control is pretty clever! How hot does it get?

8. ### Leek1001 Thread Starter New Member

Sep 13, 2011
9
0
@wayneh:
Thanks for the constant current supply, I'll see how to turn this into the right circuit.

@vrainom:
Thanks, I've send an email to the manufacture.

@Smoke_Maker:
12V 7Ah (C20 test).
What do you mean with the packaging banding strap ?
Could this be used as a load?

Greets,
Leek

Sep 24, 2007
126
15

10. ### Smoke_Maker Active Member

Sep 24, 2007
126
15
Go to some where they ship large stuff on wood platforms, they use this strapping to hold items to the wood platform. The strap is about 19mm wide and .76mm thick.

You can buy large load resistors but this packing strap works very well and is most times free.

11. ### Leek1001 Thread Starter New Member

Sep 13, 2011
9
0
@Smoke_Maker
Thanks for the tip.

Only the assigment is a bit changed.
because: Now I need to use a microcontroller to change the resister values (switching between different resistor values) to change the load while testing.

So:
Option 1: discharge 40A
switch.
Option 2: discharge 10A
etc.

If the OCV (Open Circuit Voltage) drops, the current will change also, so the option then is, to put (or take away) an other resistor to the resister value selected by the microcontroller while running the test.

What do you guys think about that?
Is that achievable?

Greets,
Leek.

12. ### Smoke_Maker Active Member

Sep 24, 2007
126
15
It will be easy if you use a micro controller.

Using a microcontroller you will be able to very the load to keep it constant.

13. ### wayneh Expert

Sep 9, 2010
11,897
2,845
In the circuit I gave you, you need only to control the reference voltage to get any current you want at the output. So you
could use a MC but you could also just use a switch setup with a few resistor dividers to give you the voltages you need.

If you're also doing data acquisition under computer control, you might look into LabJack. I love my U3HV.

14. ### Smoke_Maker Active Member

Sep 24, 2007
126
15
Wayneh:

Thats only a 15 amp power transistor what do you suggest for higher amps, he wants up to 40 amp testing?

15. ### wayneh Expert

Sep 9, 2010
11,897
2,845
You'd want a few MOSFETs in parallel on a heat sink. The controller sets the voltage seen by all the gates. The mosfets are not fully on and therefore offer a resistance and will burn off a significant amount of heat, so that's why you need a few and need them on a sink. That's true with BJTs also, but they require a base current of about 10% of the collector current. The MOSFETs need only a voltage.

I've never built such a beast so that's about as far as I can take you.

Smoke_Maker likes this.
16. ### Leek1001 Thread Starter New Member

Sep 13, 2011
9
0
Yeah but the assingment is to use resistors.

I've been calculating and I think to it like this:
creating a resistor of 4.8ohm at 120W (2.7ohm 100W5 % + 1.0ohm 50W 1% + 1.0ohm 50W 1%)
when putting 24 of these resistors in parallel then u get 0.6ohm min 960W 40A.

17. ### wayneh Expert

Sep 9, 2010
11,897
2,845
You must use ONLY resistors? Then you just need a switch from one resistor arrangement to another.

Even with the MOSFETs to control current, it still makes sense to let the resistors burn off a majority of the power. But the MOSFETs (or any transistor) cannot control the current without themselves offering some resistance and thus burning a portion of the total power.

18. ### Smoke_Maker Active Member

Sep 24, 2007
126
15

So I take it you want to build the load first and then do the recording to the computer next ??

To build your load you will need 2k watts of resistors, the reason why is that if you run the resistor near max capacity they will fail quickly.

For resistors you could get some high wattage resistors and series/parallel them to get .6 ohms. You can use .3 + .3 ohms in series to get to .6 and parallel them to get 2k watts or you can use 2 of these, see the link below.

http://www.meci.com/electronics/parts/resistors/braking.html

19. ### Leek1001 Thread Starter New Member

Sep 13, 2011
9
0

A resistor parallel circuits:

| | | |
[] [] [] []
| | | |
5A 5A 10A 20A

With this I could create:
5A
10A
15A
20A
25A
30A
35A
40A

Now I need switchs to switch the resistor on or off the ciruit.
5A*24VDC = 120W -->
R = U/I --> R = 24/5 = 4.8ohm 120W -->
creating this by:
1ohm 50W + 1ohm 50W + 2.7ohm 100W.

current that will flow like this:
24/(1ohm 50W + 1ohm 50W + 2.7ohm 100W) = 5A

1ohm 50W * 5A = 5V --> P=I^2*R = 5^2*1 = 25W
1ohm 50W * 5A = 5V --> P=I^2*R = 5^2*1 = 25W
2.7ohm 100W * 5A = 13.5V --> P=I^2*R = 5^2*2.7 = 67.5W
Total W = 25+25+67.5 = 117.5W (almost the 120W calculated with 4.8ohm)

Am I right with this calculation?
(I dropped the constant current story.)

20. ### Smoke_Maker Active Member

Sep 24, 2007
126
15
That should work. You should get some resistors and experiment with them before you make a big buy, see how hot they get and also see what the amp draw is.

As you change the load on the battery the voltage will drop some so you need to experiment to see what the next resistance value should be.

Once again, be careful with with manufactures specifications, lets say that the 1 ohm resistor is listed as 25 watts, that means that the resistor will continue to work and live a long life at a certain temperature like 25 C. when you go outside that temperature you shorten it's life and it changes resistance.

Experiment, Experiment, Experiment it's the best way to learn.