Re: 12V to 7.2V DC converter circuit?

 

On Wed, 10 May 2006 21:51:52 -0700, g20zoom wrote:


First, I've crossposted this to sci.electronics.basics, and redirected
followups to there, since that's really where this question belongs.

Given that, it's conceptually quite simple - an adjustable regulator,
like an LM317, a series pass element, if you need more than one amp;
a transzorb and a hash choke, a BIG heat sink - since car voltage is
usually more like 13.2, you have to drop 6 volts, which means getting rid
of 6 watts of heat for each amp you draw.

Find out how much current your load takes, and we can be more specific.

What exactly _is_ it that you're powering?

Thanks,
Rich

 

Thanks for moving the post.  Sorry, I didn't know.

I think I need about 3-4 amps.  It's a motor to start a small nitro RC
engine (.15 size)
I guess the motor is about a mabuchi 380 in size.
The starting requires a bit of pulse starting(ie on for 15-20 seconds
off again, on again) so I imagine the initial start loads maybe a
little higher?

I took this from some other posting on the web:
checked the specs for the 380/400 motor too :Specification
Nominal voltage 7,2 V
Operating voltage range 3,6 ... 8,4 V
No-load rpm 16400 min1
No-load current drain 0,5 A
Current drain at max. efficiency 3,3 A
Current drain when stalled 21 A
Max. efficiency without gearbox 72 %
Length of case, excl. shaft 37,8 mm
Diameter 27,7 mm
Free shaft length 13,8 mm=20
Shaft diameter 2,3 mm=20
Weight 73 g

 

Yes, in reality I'm dropping from 12V car battery, which is probably
13.2v like you mentioned
to the equivalent of a 6 nicad cell (C or subC size) battery pack in
series.  the 6-cell is supposed to be 7.2v, but I think after a full
charged it's somewhere around 8.0 to 8.4V.

 

On Thu, 11 May 2006 09:35:04 -0700, g20zoom wrote:


This should be pretty easy:
http://www.national.com/pf/LM/LM338.html

It's good for 5 amps, but you'll need a considerable heat sink. It also
has overload protection; I notice the lock-up current is about 25A; if
your motor stalls you'll need more power. :-)

Good Luck!
Rich

 

g20zoom@gmail.com wrote:

Your best approach is to use NiCds and charge them from
the car or the AC mains if available.  Why force yourself
to start the RC engine near the car?  The charge circuit
using the car or a 12V supply is real simple: one resistor,
a 1N400x diode and an LM317.

               -----
+12 ------Vin|LM317|Vout---+
               -----        |
                Adj        [R] 5 ohm 1/2 watt
                 |          |
                 +----------+--->|---> To NiCd +
                               1N400x

Gnd --------------------------------> To NiCd -

This will provide a charge current of 250 mA, which
will take 14 hours to charge a fully discharged 2.5 aH
C cell.  You can get more sophisticated, but it is
probably not necessary.

But if you must, you can run a comparator across the
NiCd pack. When the voltage rises to full charge, the
comparator switches and drops out a relay, which
disconnects the charge circuit from the source. That
prevents draining the car battery and terminates the
charge to the pack so it can't overcharge. (You can
leave a NiCd on a 14 hour charger for well in excess
of 14 hours with no problems, but not forever.)
If you elect to go with the comparator/relay disconnect
circuit, you could change the value of the resistor to
1.5 ohms, 2 watt.  The charge rate  would than be .83
amps, and charging a fully discharged pack would take
about 3 1/2 hours.

Ed