VRLA in-series current limitting resistor - Page 1
Forum section: electronics / projects

author: cte7ds
author_rank: Contributor
timestamp: November 27, 2012, 11:49:47 am
content: Wikipedia claims the open-circuit voltage for a fully discharged lead acid battery is 11.7V. Using the standard 13.8V float/standby supply voltage, the maximum voltage drop across battery and charger would be (13.8V-11.7V) = 2.1V.So if we put a 1-ohm resistor in series with the charger and the battery, the absolute maximum power dissipation of the resistor would be (2.1V^2)/1 = 4.41W. Now the rule of thumb when using power resistors I believe is that you should use a resistor with 1.5-2 times bigger wattage value. So a 1-ohm 10W resistor would be the safe bet. However, since the load will be decreasing over time then maybe a 5-Watt rated resistor would be enough in this case? It's half the size and easier to find which makes it tempting.

author: Lightages
author_rank: Unknown
timestamp: November 27, 2012, 11:53:59 am
content: What are you trying to accomplish? A minimalist trickle charger? A float maintenance charge?What AH rating is the battery?

author: cte7ds
author_rank: Contributor
timestamp: November 27, 2012, 12:11:54 pm
content: Quote from: Lightages on November 27, 2012, 11:53:59 amFloat/standby service with a very simple design. 7-30Ah range. I was thinking if I get a bigger battery I can simply add more resistors in parallel to increase the current.

author: Lightages
author_rank: Unknown
timestamp: November 27, 2012, 01:12:59 pm
content: Well actually, float maintenance is not usually current limited. The charging of a lead acid battery is usually as follows:1. Dump current into the battery until it reaches a specific voltage, usually around 14.5-14.72. Hold that voltage at a steady voltage until the charging current drops to around 2% of the battery capacity.3. Drop the voltage to the maintenance charge of around 13.8 and let the battery take what it will.So a couple of things. Size your resistor, if you really must use one, to not limit the current to less than the 2% of the capacity of the battery. Please read this:http://batteryuniversity.com/learn/article/charging_the_lead_acid_batteryIf using resistors and you cannot find the wattage you want, remember that you can parallel resistors to get higher wattage. Just keep in mind that parallel resistors need to be calculated for correct resistance.

author: cte7ds
author_rank: Contributor
timestamp: November 27, 2012, 01:44:24 pm
content: Quote from: Lightages on November 27, 2012, 01:12:59 pmCurrent limiting resistor is required when the initial charge current to an empty battery would exceed the capability of the power supply. The multi step charging is more complicated than a simple constant-voltage supply. I've often seen standby systems described where rectifier/charger/voltage supply is directly in parallel with the batteries. I guess simplicity increases reliability. Fast charging speeds that multi step charging offer aren't often necessary.QuoteQuestion is, are 1-ohm 5W resistors adequate or do I have to go 10W?

author: Lightages
author_rank: Unknown
timestamp: November 27, 2012, 01:50:58 pm
content: I would go the 10W route. The 5W ones will heat up to their limit and if you have them in a warm climate they may fail prematurely.

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: November 27, 2012, 02:43:01 pm
content: For intermittent service I wouldn't worry so much about being near the rating. But, and it's a big but, be sure to look at the manufacturers operating curves. They'll have info on correct heat sinking- what size pads and traces for smaller resistors, what size heat sinking and temperature rise for large ones. That 5W rating may not be at all realistic in the actual installation, so you might have to go with the 10W anyway. I use a lot of Dale RH series in the aluminum housing because they have very good data on exactly what the part is and isn't capable of.