DC Motor overvoltage

[Emil Johnsen]

I want to run a 3.6V DC motor from a 12V battery. The motor will not run
continously, only 0.5sec at a time with several seconds delay before it will
run again.

Can I do this without damaging the motor?

(I'd like to avoid using a regulator or PWM if possible, also the extra
torque from higher voltage would be useful.)

 

[Rich Grise]

Emil Johnsen wrote:

I want to run a 3.6V DC motor from a 12V battery. The motor will not run
continously, only 0.5sec at a time with several seconds delay before it
will run again.

Can I do this without damaging the motor?

(I'd like to avoid using a regulator or PWM if possible, also the extra
torque from higher voltage would be useful.)

If you know the motor current at normal operation (you can measure
this ;-) ), just add a series resistor. Lessee, I think R = E/I,

so your resistor would be (12 - 3.6)/(Imotor) ohms.

Good Luck!
Rich

 

[Jamie]

you can on some.. but i can bet you that the RMP's are really
cranking.!
what you should at least do is pu some diodes in series.
to drop the level.

Emil Johnsen wrote:

I want to run a 3.6V DC motor from a 12V battery. The motor will not run
continously, only 0.5sec at a time with several seconds delay before it will
run again.

Can I do this without damaging the motor?

(I'd like to avoid using a regulator or PWM if possible, also the extra
torque from higher voltage would be useful.)

 

[Emil Johnsen]

If you know the motor current at normal operation (you can measure
this ;-) ), just add a series resistor. Lessee, I think R = E/I,

so your resistor would be (12 - 3.6)/(Imotor) ohms.

The motor draws ~3amp, so I will need a ~30W resistor, which means either a
bulky wirewound resistor or an expensive film resistor. Since it won't run
continously I can probably use a smaller resistor, but I don't know how
small.

 

[Don Kelly]

"Emil Johnsen" <nospam.emil.johnsen@student.uib.no> wrote in message
news:cg34nk$1265$1@toralf.uib.no...

If you know the motor current at normal operation (you can measure
this ;-) ), just add a series resistor. Lessee, I think R = E/I,

so your resistor would be (12 - 3.6)/(Imotor) ohms.


The motor draws ~3amp, so I will need a ~30W resistor, which means either
a
bulky wirewound resistor or an expensive film resistor. Since it won't run
continously I can probably use a smaller resistor, but I don't know how
small.

However, what is important is the current at startup in this operating mode.
That will be higher than at normal load. In addition, the resistor while
cheap will not help at no load or light load- (speed will be high and the
speed will change more rapidly under load.
Also- what kind of motor? shunt, permanent magnet, series? it makes a
difference.
--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer

 

[Emil Johnsen]

However, what is important is the current at startup in this operating
mode.
That will be higher than at normal load. In addition, the resistor while
cheap will not help at no load or light load- (speed will be high and the
speed will change more rapidly under load.
Also- what kind of motor? shunt, permanent magnet, series? it makes a
difference.

I'm guessing series, because this motor comes from en electric screw driver,
but I'm not 100% sure.

I'm going to use the motor as an actuator for a ball valve, obviously with
significant gear reduction. I will leave the motor on for <0.5sec, during
this time it will rotate the valve shaft 90deg and stall as it hits a
mechanical stop. The motor will not see particuarly light loads.

 

[Doug]

Emil Johnsen wrote:

However, what is important is the current at startup in this operating
mode.
That will be higher than at normal load. In addition, the resistor while
cheap will not help at no load or light load- (speed will be high and the
speed will change more rapidly under load.
Also- what kind of motor? shunt, permanent magnet, series? it makes a
difference.

I'm guessing series, because this motor comes from en electric screw
driver, but I'm not 100% sure.

I'm going to use the motor as an actuator for a ball valve, obviously with
significant gear reduction. I will leave the motor on for <0.5sec, during
this time it will rotate the valve shaft 90deg and stall as it hits a
mechanical stop. The motor will not see particuarly light loads.

My guess is that you stand a good chance of doing some kind of mechanical
damage to the motor, particularly in light of the fact you will be
inpacting it on a stop With 3 times the voltage, the starting torque may be
more than the motor can handle.

 

[Don Kelly]

If you are going to use a screwdriver motor -why not keep the gears that go
with it? If it is a series motor, you can limit current with a resistor. I
would suggest that as you are running at 3 times the voltage, use 2 times
the measured locked rotor resistance in series- rate the resistor on the
basis of the current under locked rotor conditions. This will limit the
current to what the motor normally takes in this condition and stall torque
to what it is normally. The speed will drop more under load than it would
normally.
A screwdriver motor is expected to stall so you could reduce the extra
resistance somewhat. Watch for arcing or sparking at the brushes- I think
the windings will take the overcurrent for the time under consideration but
the brushes may be a problem.


--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer
"Emil Johnsen" <nospam.emil.johnsen@student.uib.no> wrote in message
news:cg3ehl$19uo$1@toralf.uib.no...

However, what is important is the current at startup in this operating
mode.
That will be higher than at normal load. In addition, the resistor while
cheap will not help at no load or light load- (speed will be high and
the
speed will change more rapidly under load.
Also- what kind of motor? shunt, permanent magnet, series? it makes a
difference.

I'm guessing series, because this motor comes from en electric screw
driver,
but I'm not 100% sure.

I'm going to use the motor as an actuator for a ball valve, obviously with
significant gear reduction. I will leave the motor on for <0.5sec, during
this time it will rotate the valve shaft 90deg and stall as it hits a
mechanical stop. The motor will not see particuarly light loads.

 

[Terry Pinnell]

"Emil Johnsen" <nospam.emil.johnsen@student.uib.no> wrote:

I want to run a 3.6V DC motor from a 12V battery. The motor will not run
continously, only 0.5sec at a time with several seconds delay before it will
run again.

Can I do this without damaging the motor?

(I'd like to avoid using a regulator or PWM if possible, also the extra
torque from higher voltage would be useful.)

I haven't tried it, but I would have thought the answer was No. And if
it wasn't for your final proviso, I'd have suggested a simple 12V PWM
approach. Apart from probably minimising risk of motor damage, that
would allow you to secure extra torque.

FWIW, that's what I was considering doing for my curtain controller
(subject of recent threads). That too uses a screwdriver motor and
gearbox. Mine was originally powered by 2.4 V (2 x 12 V Nicads). But
that didn't quite cut the mustard. So I increased it to 3.6 V (with a
permanent trickle charge sourced from an existing convenient 12 V DC
supply). In initial tests that seemed to give too violent an action
(especially dramatic when the curtains opened automatically at dawn
<g&gt. But it turned out that the necessary longish wiring to the motor
provided sufficient resistance to tame it.

BTW, I originally tried my 2.4 V motor at progressively higher
voltages from my bench power supply, but quit at 6 V.

Did you also consider a limiting microswitch? I'd be interested in
what stall-detection circuitry you are using.

--
Terry Pinnell
Hobbyist, West Sussex, UK