batteries
I am planning to make a battery operated device. I am using spring contacts for the battery. Since the device will be dropped, I am planning to put in a capacitor. I just don't know what type (ceramic or electrolytic type) and value.
Can you please let me know how to compute the value? There will be three NIHM batteries (1.2V each). Maximum load is 500mA.
The total discharge caused by your device will be 20uA * (8 years). The issue is you need to convert to mAh to match units.
.02 mA * 8 years * 365.25 days/year * 24 hours/day = 1402.56 mAh are required for this
Your battery is only 240mAh capacity, so it will only last a fraction of this time.
You can approach this with algebra also.
240mAh / .02 mA = 12000 hours = 500 days of lifetime from your battery
This is actually still not perfect. You need to factor in the self discharge rate of your battery. This will be specific to the battery you have chosen and can be very high, possibly dwarfing your 20uA limit and placing you in the mA range.
Let me know if there is something you need more information on.
If you power off 9V then consider using some of the 3W LEDs in series: -
Typically red leds need about 2.5V across them so wiring three in series will need at least 7.5V then there should be some current limiting circuit that might take 0.5V.
Typically for blue or green leds they need 3.4V across them so maybe only series together two LEDs.
You might just about get enough brightness with the NCP5623 chip this way but if not you'll need a current limit circuit and mosfets. The NTD80N02 looks OK but you need to limit current into the LEDs.
Best Answer
I suggest using double spring loaded contacts, so that the batteries can float a bit without losing contact.
Otherwise, measure the disconnect time for many drops, and then calculate the capacitor based on this time.
Let's say you find out that the disconnection does not last for more than 50ms. Let's assume 100ms to be on the safe side. Let's also assume that your system can tolerate a voltage drop to 3.0V.
$$ Q_{max} = I_{max} * t_{max} = 500mA * 100ms = 50mC \\ C_{min} = \frac{Q_{max}}{V_{drop,max}} = \frac{50mC}{3.6V - 3.0V} = 83333\mu F $$
Simulation
The voltage source is the batteries, the switch simulates the drop event at \$t=0\$. Observe in the time domain simulation how the capacitor voltage drops linearly, reaching 3.0V after 100ms.
simulate this circuit – Schematic created using CircuitLab
Conclusion
Some of the assumptions in this calculation lead to unfeasible capacitor values. Maybe the time the contacts break typically is much shorter. If not, reconsider whether your device will use 500mA during a drop condition.
If you still get to such large capacitor values, consider soldering in the batteries. Or use double spring floating contacts, as suggested above.