Best way to slow down my 12v motor

Hi All,

I have been doing a bit of reading on the net about ways to slow a 12v motor down and have a few different options.

I have 2 12v blower fans for a boat heater and I want to be able to slow them to half their speed (so I have a high/low setting etc).

The motors have a draw raiting of 6Amps and are fused with a 10A fuse, so they are not drawing horrible amounts.

Would a simple resistor be the best way to slow these fans down, or should I be looking for a full on MOSFET type controller?

Do you know at what voltage the fans run at half-speed?

Resistor is the most simple and probably the cheapest but it will be bulky.

Any type of solid state device, if a plain regulator will have to dissipate the same power - don't see any advantage.

PWM method gives the most control.

Would it be possible to use the heater coil as a 'resistor' ?

Is it DC or AC?

It seems that using a MOSFET driven by a PWM voltage could vary the speed of the 12V DC motor (I assumed it is DC since it is on a boat).
The main idea of this configuration is to decide on the PWM frequency. It should be chosen be low enough to minimize the MOSFET power dissipation that occurs mainly at transitions, and be high enough to give a smooth rotation.

I have just been investigating a problem with my VW Passat heater fan that stopped working and can pass on the following facts. the fan has a 4 speed control, full power is a direct connection, the other speeds are via a 100 W resistor with taps, the next speed down is via a really low value resistor - .1 ohms?, right down to the lowest speed which is via a 2.2 ohm resistor. So unless you can make your own resistors out of fire element wire, a resistor dropper is impractical.
I would use a PNP series pass darlington transistor ( E to +12V, C to fan, B to control), this way the 12 A would only drop .2V (12 X .2 = 2.4 W) across this transistor when its switched on. Connect a diode across the output (cathode to fan, anode to E), this is to absorb any switching transients from the motor brushes. the control input (the base) must now be connected to earth via transistor that can handle over 10 mA (put a 1K current limiting resistor in series with this lead) or switched off. using a 555 IC set up to give a variable mark/space ratio would do this. So build it as a 1KHZ oscillator and by switching the resistors, you will be able to change the mark/space ratio and hence the mean voltage across the fan and hence its speed.
Frank

chuckey said:

I would use a PNP series pass darlington transistor ( E to +12V, C to fan, B to control), this way the 12 A would only drop .2V (12 X .2 = 2.4 W) across this transistor when its switched on

Click to expand...

I think a PNP [or NPN] transistor would drop .7V and using a darlington would mean that a maximum of 12v - 1.4V = 10.6V is given to the motor, as the second transistor [in the Darlington] would also have a voltage drop.

A PWM using the 555 is here **broken link removed** and here https://www.electronics-lab.com/projects/motor_light/005/ as examples.

No you missed the point. I am taking the output from the collector of a PNP with its emitter taken to the input +12 V, so when its saturated, E = +12V, B = 11.2V, C = 11.8V . This would give a dissipation of Collector, 12 X .2, base .6 X 1 = 3W. When its off E =+12V, B = > +11.5V, C = 0V. If this transistor is replaced by a darlington, the numbers are different and I hope are better, Nope you end up with a voltage drop of 1.4V @ 12A - absolutely rubbish. So use a discrete darlington with the collector of the first transistor returned to earth. in this case the transistor would have to dissipate 12 W (12v @ Ib output transistor ~ 1A ), so best to put a 10 ohm 10 w resistor in series with it.
Frank

http://www.analog.com/library/analogdialogue/archives/41-05/ldo.html
http://microblog.routed.net/wp-content/uploads/2007/06/voltage-regulators.pdf
http://www.onsemi.com/pub_link/Collateral/SR004AN-D.PDF