To make tester I decide to use computer as platform to control, store and manage test data.
It’s much simpler than to use microprocessor, especially when I have old 486 black/white laptop collecting dust.
For measure I need A/D converter. After searching I found that ADC0804 is ideal:
http://www.sahincnc.com/omersahin/project/recorder/adc/ppadc.htmlhttp://www.quasarelectronics.com/3112-8-bit-pc-data-logger-mark-ii.htmhttp://electricly.com/analog-to-digital-adc-0804/It’s cheap, easy to interface with PC, minimal number of components and to measure up to 15V I need only set resistors to drop voltage 3 times to be in range of 0-5V.
With this and 2 relays I can have full control of starting/ending charger and test load.
For charging I decide to use my regular charger, not to build my own. So I place relay in parallel with AC line voltage for device. With this I can disconnect/connect charger from mains power of electric network without doing any modification to it.
However I find that charger have lamp on it which is lit if charger is connected to battery, newer chargers have voltmeter. To completely remove charger from battery I need relay to disconnect both AC mains and DC charging lines.
The real problem is with constant load for testing.
It’s known that battery is self-discharging after some time. So I need to decide how much current I need to apply to test.
If I apply high current, i.e. 5A (~60W) on battery of say 50Ah, theoretically light bulb will stay lit for 10 hours. In practice usually reaches much less because of internal resistance and heating. For info about all that you can see this page:
http://www.windsun.com/Batteries/Battery_FAQ.htmI can’t use very light load to do testing because self discharge comes in play and result can’t be accurate.
Standard test uses load for 24 hour discharge, because most of time batteries is used under high loads in short time. To calculate what load is needed for 24 hour test of 70Ah battery, divide 70Ah with 24h and get 2,9167A (~36W light bulb or 4,2 ohm resistor).
For smaller batteries of 45Ah you need 1,875A (~23W).
Now, the big question is do I need higher load for 24 hour testing or lower for 100+ hours? After additional research I conclude that higher load actually make some minor damage to battery and any possible repair done by sulfator may be nullified with testing.
Destructive testing isn’t something what I have in mind, so I decide to use load in range from 0,5% to 1% of battery Ah value. My battery is lead-acid 12V 70Ah, so load for it will be from 0,7A to 0,35A.
Next problem is how to construct constant current load, most definitely with transistors or/and with other IC for current regulation. For load I chose to use halogen light bulb of 12V 5W, because load resistors are big and radiated heat is the same.
Load form this halogen bulb at 13V is 0,44A and all schematic I found for loads higher than 0,3A need heat sink for transistor. That’s starting to become big device and I want something tiny...
After conducting experiments with current regulation I concluded that is easier to use halogen bulb directly without regulator.
Testing different voltage and reading currents, I got this results:
Voltage | Current (A) |
13 | 0,440 |
12,9 | 0,439 |
12,8 | 0,437 |
12,4 | 0,430 |
12,35 | 0,429 |
11,9 | 0,420 |
11,8 | 0,419 |
11,5 | 0,415 |
These aren’t linear results, because light bulb doesn’t work linear as resistor. However in testing segment from 12,6V to 11,8V, current can be recalculated from voltage using constant. From my measured readings I recalculate that constant for calculation is 0,01/6 (0,00166...).
Formula for calculating current is now simple:
I=I13 – (13V - Ur)*10*CWhere
I13 is Initial current measured at 13V,
Ur is voltage for which current is calculating and
C is constant for used halogen light bulb – 0,00166667.
Now I can construct schematic for my device and start experimental work.
Here is schematic: