Simple electrical DC conversion 3VDC -->1.5VDC question

[AZ Vic]

Hey all,

I need to convert a plug in to run a couple airaetors (sp?) they currently run on 2 D batts in parallel. The lowest voltage cigarette lighter plugin I could find put out 3VDC.

I know there is a simple way to do this with parts from the local Radio Shack... but I'll be damned if I remember.

anyone?

Thanks in advance,

Vic

 

[UNGN]

If an Aerator uses an electric motor, you can wire two aerators in series, provide 3 volts to it and each aerator will get about 1.5 volts, give or take.

This is how 6 volt speed is achieved on single 12 Volt battery power wheels.

 

[AZ Vic]

Indeed, they are electric motors - I thought that might be the way to go - I just did not want to chance burning them up... I needed somone else to say it would do that too

Thanks!

Vic

I still would like to know how to knock it down - If I remember right, it takes an inline resistor... there is a formula that will tell you what flavor to use...

 

[myclone]

Do an internet search on the LM317 adjustable voltage regulator and the schematics for using it.

Also, there are some useful circuits that do just what youre wanting to using the LM317 on the Bowdens Hobby Circuits site (I dont have the addy handy but google will find it Im sure.)

 

[Drac0nic]

as an EET major I feel obligated to give my .02 here. What you're talking about is Ohm's law, Voltage=Current*resistance. There are a few ways to figure out the resistance, if you've got a VOM you can hook it in series with the batteries and get the current, or you can hook a resistor in series with it and use the voltage drop (something like a 1 ohm resistor) and use the fact that Voltage across the resistor/Resistance=current. You then take the 1.5V drop, divide by your current and that gives your resistor value. Your Wattage value is Voltage squared/the resistance. Oh yeah, if this is too confusing just act baffled, it's what I'd do.

 

[AZ Vic]

Originally posted by Drac0nic
as an EET major I feel obligated to give my .02 here. What you're talking about is Ohm's law, Voltage=Current*resistance. There are a few ways to figure out the resistance, if you've got a VOM you can hook it in series with the batteries and get the current, or you can hook a resistor in series with it and use the voltage drop (something like a 1 ohm resistor) and use the fact that Voltage across the resistor/Resistance=current. You then take the 1.5V drop, divide by your current and that gives your resistor value. Your Wattage value is Voltage squared/the resistance. Oh yeah, if this is too confusing just act baffled, it's what I'd do.

OK, I have a Fluke 87 III, so I can just measure DC current.

DC current = 412 mA w/ new batts This will be times 2 so the total current will be 824mA.

What I have right now is a plug in that reduces the truck's voltage from 12VDC to 3VDC (3VDC is the lowest that I could find)

Help me, I'm rusty here...

V=IR .... So, 1.5VDC(desired voltage) = .824*(resistor needed)

1.5/.824=(resistor needed) ? = 1.82 ohms - Is that right?

Or is (desired voltage) actually (desired voltage drop)? In this case it is coincidence that they are the same...

so, if the figures were....

12 VDC and I need 1.5VDC I would end up with 10.5/.824 = 12.74 ohm...?

As for the wattage it is easier to use P=(IV) so, .824*1.5 = 1.236 watts

Am I right? Do I get a cookie?

Vic

I wonder if I should use 13.4VDC and 3.5VDC for the equations? It's just a little motor... I don't imagine a little one way or the other would matter much.

Can you confirm wiring the two motors in series will provide roughly 1.5VDC each with a 3VDC supply?

 

[AZ Vic]

Originally posted by myclone
Do an internet search on the LM317 adjustable voltage regulator and the schematics for using it.

Also, there are some useful circuits that do just what youre wanting to using the LM317 on the Bowdens Hobby Circuits site (I dont have the addy handy but google will find it Im sure.)

I found that and it would work perfectly, but seems a little complicated for what I need to do.

Thanks for the lead though! If all else fails I'll build it up...

Vic

 

[myclone]

Originally posted by AZ Vic
I found that and it would work perfectly, but seems a little complicated for what I need to do.

Thanks for the lead though! If all else fails I'll build it up...

Vic

If I understand what youre trying to do then using the 317 reg should only require the reg itself, one fixed resistor, and a potentiometer. Prolly less than $5 worth of parts and the potentiometer is so you can adjust the output voltage. You really dont need a potentiometer and could use a fix resistor but I like to use a pot so if I need to change the output later for some reason its just a matter of adjusting the pot to what I need rather than unsoldering and installing another resistor to raise/lower the output.

BTW, wiring the motors in series would in fact provide each with 1.5v if they are supplied with 3v as long as both motors are the same. They are prolly a little different due to manufacturing tolerances but IMO it would be negligable. Just remember if one motor shorts then the full current draw will go through the other motor as well as the full 3v which will destroy the second motor rather quickly. BTW, using a regulator like what I mentioned would keep all that from happening should one motor fail.

Good luck