Electronic – Charging SLA using step down converter

batteriessolar cell

I have a 10w solar panel which outputs about 17V 0.5A at full sunlight. I use a small buck converter to charge my SLA 7.2Ah battery. I have currently set the buck converter output voltage to 14v and in the day time it charges the battery and at night I use the battery to power a 10w led (which uses about 9.5W with the fan on). Battery charges at about 0.5A when there is full sun light.

My questions are,

  1. 14V is the best voltage to charge the battery on CV or I should change the voltage of buck converter ?

  2. Is it okay to charge a SLA battery using this method even if I used a power supply to charge it? (power supply ——buck converter——battery)

  3. Is it okay to leave the battery charging at 14v for longer periods of time in case if I dont use the battery at night to power the light ?

Best Answer

There are two main battery usage modes

  • In cyclic use where you take substantial energy every or most days
    it is OK to charge to 14 to 15 volts.

    This is how you usually use the battery.
    .

  • For standby or float use with no regular deep discharge
    charge to 13.5 - 13.8 volts.
    This applies when the battery remains fully or almost full charged with no daily deep discharge.

Below I'll use 13.8/14.5 as the target for float/cyclic mode charging.


14 to 15 Volt is acceptable for cyclic use.

If you are floating the battery for some time without discharge then typically 13.5 - 13.8 float Voltage is recommended.


If you can get 17V x 0.5A the a buck converter should get 14V at 17/14 * 0.5A x Z/100
where z is % efficiency. If you are getting 0.5A in and 0.5A out as you state (which may just have been an approximation) then Z = 82% - and a piece of wire would work as well.

A reasonable buck converter should be able to achieve 90% efficiency giving about 0.55A at 14V out for about 17V at 0.5A in.

At lower battery voltages the buck converter is probably worth using.

PV panels will provide more current when loaded beyond they maximum power point (Wmp) and it may be that by connecting the panel directly to the battery (via a diode) you would get more current than by using the buck converter. This is not certain but is easily tried. The advantage of the buck converter is that it limits Vbattery_max to a preset voltage (eg here = say 13.8V/14.5V, while a direct diode connection from PV panel to battery will overcharge the battery if no means is provided to limit voltage rise.

For best use of the available energy you can run the charger in a CC/CV mode where it provides as much current as possible (CC mode nominally) until the battery reaches say 13.8V/14.5V and then changes to CV mode and hold the voltage at 13.8V/14.5V and the battery takes "what it wants".

An easyish way to run the buck converter in CC mode is to set the output voltage just high enough above the battery voltage that the battery accepts all available charge, until the battery reaches 13.8V (for float mode) or 14.5 - 15 volts for cyclic mode.
A simple way to do this is to use a target voltage a diode drop above vbattery. So eg if Vbattery = 12V Vtarget = 12V + VF_diode = say 12V + 0.6V = 12.6V. As long as the battery accepts all available current at that voltage the converter is operating at near optimum Vout. When Vbattery reaches 13.8V (or 14.5V) the target voltage is clamped at 13.8V/14.5 .

______________

Example of loading a panel below MPP to get higher current.

This shows that for the example panel - which is NOT the user's panel - direct connection of panel to battery may be more efficient than using a 90% efficient buck converter!. Actual panel in use and converter performance need to be checked.

The diagram below is a curve from an existing panel with Vmp ~= 35V and Imp ~= 8.6V.
I have added X axis labelling for an ~= Vmp=17V panel amd Y axis can be scaled proportionately if desired. Your panel may not have exactly the same V/I/Power curves but these are typical.

The dashed curves are power out at 3 insolation levels. Say 100%, 75% and 50% of full sun - not important.
Blue vertical lines drop from the peak of eac power curve (Wmp) to the relevant VI curve touching then at current-at-peak-power = Imp.
The horizontal red lines to the Y axis allow the current Imp to be read.
Units are 1-10 but can be adapted to the existing panel if desired.
The green vertical line is a load of ~= 12V. (maybe 12.3).
The orange lines show the panel current at 12V. Y axis current figures are rough as the scaale has been hacked about somewhat in editing.

It can be seen that for 100% insolation Imp ~= 8.6A but I12V ~= 9.2A.
At 75% insolation figures are 7.1A / 7.6A
At 50% insolation figures are 5.5A / 5.8A.

So: 100% 9.2/8.6 = +7%
75% 7.6/7.1 = +8.5% 50% 5.8/5.5 = + 5.5%

SO without any controller, current to a 12V load would be increased beyond Imp by 5.5% y 8.5% above Imp by just supplying the battery directly by the PV panel.

With a buck converter Icharge = Vpv/Vbattery x Ipv x %_buck_efficiency/100

If a good efficiency buck converter was used (say 90% efficient) then at 12V battery

100% insolation Vpv=14V, Ipv = 8.6, Icharge (90%)= 9.03A = lower than direct connection.

50% insolation Vpv = 13.75V, Ipv = 5.5A, Icharge = 5.7A = lower than direct connection.

How this compares to the panel under consideration needs to be checked.
At much higher Vpanel:Vbattery ratios a 90% buck converter is easily superior. A 95% efficient buck converter should have a slight 'edge'

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IMPORTANT

The Imp line must be drawn as shown here.
The apparently flat top of the curve is misleading.

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