12V to 7.2V DC converter circuit?

How much current do you need?

What do you want to power with those 7.2 V for several minutes?

By the way, here's an idea that may become part of the solution.

A 7805 voltage regulator chip with a green LED inserted into its ground connection (between the ground pin and circuit ground) will give 7.1 volts.

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

Hmmm, I would check with Radio Shack, they may have one for you already built for less than it would cost you to do it.

Luhan

I grabbed this from web forum.

I 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 Shaft diameter 2,3 mm Weight 73 g

Thanks for all the input so far...I moved this answer to the .basics group as well.

-joe

I'd prefer using a cellular phone loader for the cigarette lighter. They are cheap and use a mc34063 dc/dc converter. With the aid of the datasheet you will easy find the resistor voltage divider that sets the output voltage. If the current you need is up to 0.5A that will be the cheapest solution.

Since you want to run a motor, you probably don't need tight voltage regulation. If you want dirt simple you could just use a bunch of diodes in series. Each diode will drop a fixed voltage. The voltage drop does depend on current but resides in a pretty narrow range. The nominal figure you see usually cited 0.7 volts for a conventional silicon diode, but with substantial current you may see a bit more. If you use chunky enough diodes you might not even need a heatsink.

Try a 7806 regulator with 2 series 1n4148 diodes between the GND pin and 0V FB

He needs several amps. would require a pass transistor. g20, you can get better regulation with a three terminal regulator like Bear suggested but you also need a power transistor in the circuit to carry the several amps of current you mentioned in one of your previous posts. Google "linear voltage regulator pass transistor circuit" or such, if you can't find anything or have more questions just ask us, one of us will post an ASCII diagram of a circuit.

Or maybe look at a data sheet:

Cheers! Rich

Yup. According to this:

it draws 6 amps at maximum efficiency - which implies the possibility of even more.

I wonder if he really needs a regulator. He may have a fixed load on the motor, in which case he could use a high wattage series resistance. At no load (1 amp), he needs 5 ohms at 10 watts (safety margin) which yields a range of 8.8V (engine running) to 7V (engine off). At maximum efficiency (6 amps) he needs 1 ohm at ~50 watts (safety margin) to yield 7.8 engine running to 6V engine off. The chart on the url show a voltage range for the motor of 6 - 9, so those values work.

From the chart at the url mentioned, the motor equivalent resistance computes to 1.2 ohms. (7.2V/6amp at max efficiency) That 1.2 ohm figure was used in computing the needed resistors.

Ed

Ed, Will the resistors be able supply the startup 25A to the electric motor? I would imagine the cigarette outlet can handle it...since it can handle a portable air compressor which must have a far larger motor than what I'm dealing with.

-joe

Resistors tailored for operation under normal loading will not allow it to draw anywhere near enough power under startup or for that matter any heavy load. Motors spin up ex_treme_ly slowly driven from a current source (a resistor is a current limiter not a voltage regulator).

Where do you get "the startup 25A to the electric motor"?

This is a tiny 3 oz motor, and it is highly unlikely that it will draw 25 amps.

But lets say, for the sake of discussion, that it does require 25 amps to start. You won't get that with any resistance I mentioned. You asked about this on another newsgroup, and I answered there. It is simple to provide the higher current, bypassing the resistor, momentarily (at startup). See the other post.

Ed

I saw the specs for a *similar motor* that said the stall current drain was 21A. I'm not saying that website is correct, but maybe it's a start. Yes, I did see your bypass circuit using the cap...and that looks good.

-joe