# Transformer current and a NiMH battery pack

Discussion in 'Electronic Basics' started by [email protected], Feb 17, 2005.

1. ### Guest

A while back I built an adjustable power supply with a 12V 1.2A power
supply, a full wave bridge rectifier, and an LM117 adjustable
quick charge a small 5-cell 6V NiMH battery pack I use in a R/C car.

If I connect the battery pack to my adjustable supply (set @ 6V) what
current will the power supply put out? Will the power supply put out
the 1.2A rating of the transformer? I do plan to carefully time the
charge so that I do not overcharge the pack.

2. ### Larry BrasfieldGuest

You built a power supply by using a power supply? Hmmm.
Power supplies, unless designed with current limiting to protect
themselves, will put out enough current to damage themselves.
The question you want to ask is what current can be taken
from the supply without degrading its reliability to a level
worse than you would like or need.
It is a rare power supply using linear regulation that can
deliver (for very long) a DC current equal to the RMS
current rating of the transformer. A typical ratio is half
as much DC current as RMS tranformer rating, due to
the cresting factor in the transformer current waveform.
I doubt that will help. Your charge times are likely
going to be longer than the thermal time constants
of the components that would be over stressed.

3. ### Guest

You built a power supply by using a power supply? Hmmm

I meant to say transformer.

4. ### Rheilly PhoullGuest

Well I guess you havent got too much to lose by connecting up to the battery
with the volts set low and then increasing until the charge rate or the max
output of the supply is reached (current). At that point you should be
checking the heat generated on the heat sink of the supply and of the
battery. This would be a check to see if it would be possible to do what you
want. If successful then you will have to consider how to make it work for
practical purposes, ie when the battery is discharged it would be easy to
overload the supply if you set the volts too high. Probably some form of
current limiting could be used and constructed easily if it's not in your
supply anyway.

5. ### BillGuest

Well, you can't set it at 6V or else it will never charge because a 5
cell NiMH has 1.45V/cell under charge.
Therefore your pack voltage under charge will be
5 x 1.45 = 7.25VDC
You don't say what the capacity of your pack is but IIRC from my days
in R/C racing they used 1.4AHr packs so that is what I'll base all the
following calculations on.
If you wanted a slow charge you could charge at C/10 rate which would
be 140mA.
If you set your power supply to 12VDC and then hooked a current limit
resistor between the power supply and the battery you would calculate
as follows
R = (12-7.25)/.14
which equals 33.9 ohms.
Figuring your power dissipated in the current limit resistor
P = .14x.14x33.9
which equals .664 Watts
so for a C/10 charge you would be fine with a 33 ohm, 1 watt resistor.
To summarize set the power supply to 12VDC and put a 33 ohm, 1 watt
between the power supply and the battery pack.
It should take about 15 hours to get your pack charged completely at
this rate but at C/10 you will not cause any damage if it goes longer.
If you want a faster charge C/5 (.28 amps) then half the resistor
value to 16 ohms and double the power rating to 2 watts. This should
have you charged in about 7 to 8 hours.
Not knowing anything about the internals of your power it's hard to
say whether either of these current rates would cause damage to your
supply but I doubt that .28 amps out would cause any grief unless you
have some kind of low current fuses or really low power devices in
your internal power supply circuity. Really it's not possible to know