- Mon Nov 23, 2015 5:13 pm
Whats the reason for 12v/8NiCd's over 6v/4NiCd's?
Twice the storage capacity (if you use a buck converter). If you have 800mA/h batteries @ 1.2V.
4 cells x 1.2V = 4.8V x 0.8A = 3.84 VA (Watts).
8 cells x 1.2V = 9.6V x 0.8A = 7.68 VA (Watts).
If you use a voltage regulator to supply 3.3V, the current drawn from the batteries is the same as the load and lost as heat from the regulator. So with a load of 100mA.
4 cells: 4.8V-3.3V = 1.5V x 100mA = 0.15 Watts as heat. (Useable power 3.3V x 0.8A = 2.64 Watts).
8 cells: 9.6V-3.3V = 6.3V x 100mA = 0.63 Watts as heat. (Useable power 3.3V x 0.8A = 2.64 Watts).
So no advantage.
With a buck converter: it works like a transformer so the current drawn from the batteries is less than the load. Again with 100mA @ 3.3V.
4 cells: 3.3V x 100mA = 0.33 Watts + 10% (losses) = 0.37W / 4.8V = 77mA from batteries.
(Useable power 4.8V x 0.8A = 3.84 Watts – 10% (losses) = 3.46 Watts).
8 cells: 3.3V x 100mA = 0.33 Watts + 10% (losses) = 0.37W / 9.6V = 39mA from batteries.
(Useable power 9.6V x 0.8A = 7.68 Watts – 10% (losses) = 6.91 Watts).
The end result is you only take half as much power from each cell, they last twice as long (depending on other things) and you only need half the current from your solar panel to charge them. The last bit is a big plus as this means they fall in to trickle charging rather than full charging which is much easier. 50mA @ 12V would charge 8 cells where you would need 100mA 6V to charge 4 cells both having supplied the same power. It doesn’t make any difference to the size of the solar cells as 12V 50mA would be the same physical size as 6V 100mA. You just have more overhead for none-sunny days where charging doesn’t take place and charge at a lesser C rate which is better for the batteries.
As for the parts. It depends a lot on how much power you are going to use and how sunny it is.
For the application above a diode to stop the batteries discharging through the solar cell and an LM317 constant current source would be fine up to a charging current of “cell capacity” / 10. So 0.8Ah / 10 = 80mA.http://theparanoidtroll.com/2011/01/05/ ... oad-lm317/
The discharge limiter can be in the form of a Zener diode and pull down resistor connected to the buck converter enable pin, such that when the battery voltage drops below the Zener value the enable pin is no longer held high, switching off the converter until the batteries charge again. This is quite useful as it self-resets when the batteries charge.
The buck converter, anything with an enable pin that suits you application. Whether you are sleeping or not and reliability of solar charging dictates how important the quiescent current is . http://www.ebay.co.uk/itm/DC-DC-Power-S ... SwgQ9VjSCk
I'll try to sketch it up & post tomorrow.