# What should be the V specs of a step down transformer for a 10amp car battery charger

I was looking to create a 10amp manual charger with variable voltage and current option for flooded lead acid battery (normally used in cars).

I only know some basic electronics and usually make solutions by integrating off the shelf modules.

For this project I have to first buy or order a custom built 10amp step down transformer but I was bit confused what should be the secondary output voltage for this transformer

To charge a FLA battery we usually need up to 14.2V and to equalize we need 15V that is maximum output we need from a battery charger

Therefore my question is should I look for a transformer with exactly 15V secondary output? or higher than this and why? because some online battery charging circuits are using 18V transformers.

Thanks,

• You need a transformer that will still supply 15v when delivering the 10A you want. Commented Oct 6, 2018 at 19:26
• @solar mike thanks for your comment, So If a transformer specs says 15v 10amp then I should buy it? Commented Oct 6, 2018 at 19:32
• 14.2 should be adequate to equalize the cells. You need a voltage regulator from a bridge from a step-down XFMR that supplies 14Vrms or 1.4x this with no load peak voltage and THERMALLY COOLED for V drop @ 10A . Windings may drop 10% V at rated current. THis is not the most efficient charger and must be voltage limited to 14.2V and current limited by Ohm’s Law or active current limit. 10Ax3V drop=30W for example for 140W peak charge, so XFMR must be rated at least 25% higher than this as well due to pulse peak/average current crest factor. Commented Oct 6, 2018 at 19:35
• @Tony EE rocketscientist thanks for your comment, So this means I need 12amp Transformer If I am looking to charge at 10A current rate right? Commented Oct 6, 2018 at 19:43
• Do you plan to have any electronics, such as a current limit or voltage regulator, between the transformer and the battery? How a about a rectifier? If so, you need to specify those before you start choosing a transformer. Commented Oct 6, 2018 at 20:08

Most battery chargers that are a simple transformer and bridge rectifier DO NOT fully charge a battery. With a simple setup such as this the variations in line voltage would (and do) change the voltage output.

You need to:

1. Decide the highest voltage you want to charge the battery to. I would recommend you select no higher than 13.8V.
2. Select the bridge rectifier you want to use (capable of 10A) and find out the Vf of the Diodes in it. You will lose 2 * Vf, so if your diodes have a Vf of 0.8V @10A for example you will lose 1.6V across the bridge rectifier. To get 10A RMS into the battery of course requires much higher peak current flow, so you will need to select a bridge that can support at least 25-30A peak. (You can read about conduction angles in bridge rectifiers if you like)
3. Using your charge voltage and diode drop you can now understand the PEAK voltage to be delivered by your transformer. In my example here that would be 13.8 + 1.6 = 15.4V (ignoring any other losses that exist in cables or transformer). Converted to RMS that means you need a transformer with approximately 11VAC.

That is a very non-standard transformer and does not take into account the terminal voltage drop at high current or any other losses.
How successful would this be as a battery charger? At best it would be hit and miss. This type of ultra simple charger is also easily overloaded since there is no method to detect over-current. Connecting a flat battery may result in much more than 10A RMS which may overheat and damage your transformer.
In addition you really should have a battery charger that is short circuit proof and this certainly is not. A short on this would blow the primary fuse (assuming you are sensible and have one).

• Answer might be late, but this might still help someone else: If you have a transformer with secondary coil with center tap, you can use the center as Common to battery, and the other two put Anode on both (so only TWO diodes). Than the diode loss is 1*Vf, so you get transformer with PEAK 13.8 + 0.8 = 14.6V. Multiply that with 0.707 giving you RMS transformer 2 x 10.3AV. You can use 2 x 12AV, the voltage will drop under load. Better solution is to replace the two diodes with two thyristors, and control their gates with a microcontroller to build a better charger. Commented Jun 2, 2019 at 16:35

You need to figure out what will be between the transformer and the battery first.

Transformers provide AC. To charge a battery, you need DC. The method you use to convert AC to DC will affect the output voltage.

Start at the battery and work backwards. Don't worry about the specifications on the transformer until you've worked out everything else.