Totally new to this guys, so sorry if my question are dumb. Anyways I have a small 3v motor I want to turn at max speed/torque whenever it is turned on. I am using a 8.4v power supply to power the motor. This is controlled by a 5v pulse that turns on a MOSFET which allows power to the motor. The power supply is 6 AA's so they slowly will reduce in voltage. So what I want to know is should I use a LM317T voltage regulator and regulate it at 3V or just simply put in a resistor to controll the voltage to 3V and not worry about the batteries slowly becoming weaker.? I measured the motor and it measurs 1Ω so it would be a fairly small resistor. I'm trying to keep this small but I do want peak performance, so any input would be much appreciated.
On a side note, what kind of difference would I see from a 6v motor? Thanks for any help. Also the load on the motor will be fairly constant.

 

If you want the maximum performance of your motor drive it with a LM317 if the current it needs is less than 1.5 Amps, which it is i think.

If you use a 6V motor driving the same load, then its output speed will be higher and thus the output power will increase.

 

I forgot to mention that a motor, when starting, draws more current than when running. You said your motor resistance is 1R, thus with 3V applied to it it will try to draw 3 amps from the LM317. However, the LM317 does not allow it and it will limit the current down to few mA and your motor will not have enough torque to start. One solution to this is to use a transistor to give the motor an initial push to start and then switch it off or use a voltage regulator which can supply more than 3 amps (if the starting current of your motor is really 3 amps).

 

Mik3 Thanks for all your help. Do you know of another voltage regulator that would work? If not, what if I use two in parallel so the current is split between them, will that work?

 

The LM138 is suitable for you. It can supply 5 Amps continuously and 12 Amps for 0.5ms, thus its good for driving small motors like yours.

 

Thanks Mik3, one moe quicky. Will this be able to handle the current if I bump up to the 6V motor?

 

Depends on the power rating of the motor.

 

Afraid I disagree with Mik3 here. An LM317 has a minimum voltage drop of 1.7; nearly 3v in current limiter mode. Using it as a voltage regulator throws away a whole battery, minimum.

Do a Google search on "chopper driver", which is basically what you need.

Since you're limited to a few volts, MOSFETs (unless logic level) are just about out of the question, unless you're using logic-level MOSFETs.

You need to regulate the current through the motor, but you haven't given all of the specifications of the motor. You've mentioned 3v, and you say you've measured about 1 Ohm through it - if so, that's 3A current, which is quite a bit. Do you have any more data for the motor?

Chopper drivers are great, but aren't particularly simple to set up - especially at such low voltages.

 

Perhaps one of these? http://www.v-infinity.com/pdffiles/V78XX-1000 Series.pdf

 

Thanks for all your help guys. Well sorry guys, sound like I need to find out some specs on my motor. I'm not sure where to get the specs, I know its a small hobby motor and its caled a 1.5-3v motor.
Sgt Wookie, I belive your right that it doesn't need 3 amps, but again need to find out motor specs. As for my mosfet I'm using a IRF510. Basically like an on/off switch, i think its called enhancement mode. But this gets gated by a seprate 5v power supply. As for the LM317 being used in current limiter mode, i dont know anything about that. I just was trying to regulate the voltage to the motor so it was always at peak performance. Maybe that is the same. But with battery pack at 8.4v and 7.2v as the batteries are about dead, that still gives me >4v room to power the regulator correct? Am I looking at this wrong?
On a side note, Sgt wookie, if your were or are in the corps, I would like to thank you for your service to this country on this veterans day. Your sacrafice is appreciated.

 

An LM317 has a minimum voltage drop of 1.7; nearly 3v in current limiter mode. Using it as a voltage regulator throws away a whole battery, minimum.

Whats the problem? He can drive it with a battery less.

 

I tried to use two D cells to the motor and get a good current reading. Nothing happened. I switched over to AA's and used like 5 to get it to spin and it read OL. My meter tops out at 300mA. So I thought it was >300mA. I then decided I would try to get an exact value for the current for the motor. I placed a 1 ohm resistor in series with the motor and got a voltage across the resistor at 240mV. That's 240mA using ohms law. What gives? What am I missing here? Does anyone know a good way to get motor data/specs. This simple problem is definitely putting me in my place.

 

Whats the problem? He can drive it with a battery less.

A DC-DC converter circuit would be a far better choice. Linear regulators are simple to implement, but they sure waste a lot of power.

 

I tried to use two D cells to the motor and get a good current reading. Nothing happened. I switched over to AA's and used like 5 to get it to spin and it read OL. My meter tops out at 300mA. So I thought it was >300mA. I then decided I would try to get an exact value for the current for the motor. I placed a 1 ohm resistor in series with the motor and got a voltage across the resistor at 240mV. That's 240mA using ohms law. What gives? What am I missing here? Does anyone know a good way to get motor data/specs. This simple problem is definitely putting me in my place.

Your motor probably has a very low resistance itself; perhaps less than an Ohm.

At the same time, your AA batteries have internal resistance. As the batteries become depleted, the internal resistance increases.

When you put five 1.5v AA cells in series, simple math says that you should get 7.5v across the cells. However, when you place a load across the cells, the internal resistance of the cells comes into play; the heavier the load (less resistance) the less voltage will be present across the cells. That's why cells get hot when they're being rapidly discharged; power is being expended in the internal resistance. If the heat builds up faster than it can dissipate, the cells may rupture forcefully (explode).

 

A DC-DC converter circuit would be a far better choice. Linear regulators are simple to implement, but they sure waste a lot of power.

I know that but for him its a lot easier to do it with a linear regulator as i guess from his knowledge.

 

I'll look into using a DC- Dc convertor circuit. I do have a low level of knowledge but half the fun with this project is learning how all this works, and doing the research. Do you guys have a good website I can learn how to properly use a DC-DC convertor?

 

I've attached a schematic of what i think I want and should be doing now. The diode is a zener with a reverse voltage of 3v. (I know i didn't draw it quite right.) Will this give me the desired result? Also is resistor R1 required to drop the additional voltage so the rest is regulated at 3v? How would I go about figuring out the required specs of the components? Thanks ahead of time for anyone still putting up with me.

 

You dont need the inductor and the capacitor in the circuit. This is a DC chopper power supply. A simple PWM motor control circuit is shown here:

http://www.discovercircuits.com/H-Corner/PWM.htm

Here the 555 is the control circuit. If you want you can drive the MOS with another control circuit, like a uC, rather the 555 shown here.

 

Here is a very interesting link:

http://www.nxp.com/acrobat_download/applicationnotes/APPCHP3.pdf

 

You measured a resistance of 1 ohm for the motor. But is the 1 ohm including the 0.4 ohm resistance of the meter's leads? If it is then the motor's true resistance is 0.6 ohms.

A motor draws more current when it works hard. It works the hardest when it starts and when it is stalled. Its current then is 3V/0.6 ohms= 5A.
If your AA batteries are Ni-Cad or Ni-MH and their connections are good then they can supply 5A for a few seconds to start the motor.
When the motor is running then its current will be much less than 5A depending on how much work it is doing.