7.2v battery pack to power a 4.8v motor

[Shawn]

I'm hoping I can use a resistor to lower the voltage so I don't burn out the motor, but I don't know what resistor to use. Any knowledgeable people here willing to help out?

 

[the DRIZZLE]

How large a motor are we talking about here? What's the application?

 

[Shawn]

portable wet vac/carpet cleaner

 

[ussfletcher]

You could connect it directly and get some extra power... for a time.

 

[Paperdoc]

In theory this can be calculated IF you know the CURRENT (amps) the motor pulls when connected properly to a 4.8V supply. BUT I will say up front, it's a bad idea and will comment a little on why later.

You want to drop from 7.2 to 4.8 volts, a drop of 2.4 volts. Ohm's law says V=IR. So, to drop 2.4 volts the value of R, the resistor, must be 2.4/I, where I is the amperes flowing in the circuit. Now, suppose the vac motor is pulling 4 amps (a pretty modest current); then the R value is 0.6 ohms. That is not an easy one to get. But hold on, there's another factor to consider. The resistor will be dropping 2.4 volts with a current of 4 amps, so it will be converting that much power into heat that must be removed to avoid overheating the resistor. How much? Well, P=VI; that is, power (heat) is 2.4 x 4 = 9.6 watts. You need a resistor of 0.60 ohms able to handle over 10 watts power.

Now, even supposing you can do that, here's where the complications start. The voltage dropped across the resistor (and hence, the remaining voltage actually available to the motor) is IR. Any time the motor needs extra current (higher I), the resistor will drop even more voltage than before, thus reducing the motor's supply voltage just exactly when it needs all that voltage and current! When? Well, at start-up, for one time. And any time it needs to do extra work. Think about how often a vacuum cleaner motor sounds like it is straining to work hard. To make matters worse, a motor is an inductive load. Its effective "resistance" comes about because it generates internally a "back-emf" that opposes the supplied voltage, and this depends on the motor speed. So a motor that runs into heavier load slows down slightly, reducing the back-emf and thus experiencing a higher driving voltage, causing more current to flow through it and allowing it to do extra work and then speed back up. But if you have a resistor in the voltage supply line, that higher current flowing through the motor will result in MORE voltage drop across the resistor, and LESS voltage available to the motor under these conditions. In certain cases, the motor thus deprived of the required voltage will slow down even more, trying to use more current to speed up, but it won't - it may just STALL!

Why not use the proper battery in the first place? Or a 7.2V motor?

 

[LTC8K6]

Remove a cell from the pack...

 

[Shawn]

bummer. I have the battery hooked directly to the motor but it starts to overheat after a couple of minutes. I don't know how long it will last. the previous battery pack had a shorted cell and it fried the circuitry, so I was going to use an RC battery pack wired directly to the motor and charge it with an RC charger. I can't find any 4.8v batteries cheap and I have no way of charging them. guess I'll just have to trash the vacuum. damn things are expensive to replace.

 

[disappoint]

You can use silicon diodes that drop the voltage 0.7 volts each. 3 of them will get you down to 5.1 volts. 4 of them will get you down to 4.4 volts. Observe proper polarity and current capacity of the diodes. They will also prevent back emf.

 

[Shawn]

You can use silicon diodes that drop the voltage 0.7 volts each. 3 of them will get you down to 5.1 volts. 4 of them will get you down to 4.4 volts. Observe proper polarity and current capacity of the diodes. They will also prevent back emf.

Would this work: http://www.radioshack.com/product/in...ductId=2062575 ?

 

[PottedMeat]

Would this work: http://www.radioshack.com/product/in...ductId=2062575 ?

they'll probably fry with a rated current limit of ~1.5A continuous, plus the estimated drop of 0.7v is low - the chart says ~1.1V at 1.5A

motors draw a crapload of current

edit: you could try these 6A diodes, but i can't tell the forward voltage drop because they don't give a part #
http://www.radioshack.com/product/index.jsp?productId=2062591

 

[Shawn]

Looks like this is a bust then. This: http://www.alibaba.com/product-gs/378273570/4_8v_dc_motor_high_rpm.html is a similar motor. Can draw up to 62A.

 

[guyver01]

by this time, you coulda run down to Radio Shack and picked up the right size battery

 

[Shawn]

The only 4.8v batteries they have are for phones. I don't think they have enough power for a vacuum cleaner.

 

[Jeff7]

You can use silicon diodes that drop the voltage 0.7 volts each. 3 of them will get you down to 5.1 volts. 4 of them will get you down to 4.4 volts. Observe proper polarity and current capacity of the diodes. They will also prevent back emf.

This will still have the same issue of heat generation.

Take the 0.7V drop on each diode, and multiply it by the current that would be flowing through it. That's how many watts of heat each diode will have to dissipate.

Another option would be a DC-DC converter, but to provide enough current for the motor, it might be $20-$30 for the converter.
Is the motor labeled as to its current draw?



(There's also the effect of this on battery life - all of that heat generation is going to be coming out of the battery, not doing anything useful.)


Edit:

Looks like this is a bust then. This: http://www.alibaba.com/product-gs/378273570/4_8v_dc_motor_high_rpm.html is a similar motor. Can draw up to 62A.

2.5A when it's running. (That 62A would be if the motor is stopped dead.)
Sounds like this is definitely something that's best left running directly off a properly-sized NiMH or NiCad battery.
Batteryspace.com has all kinds of 4.8V battery packs. Be warned, most of their stuff is Powerizer brand, and I've not had good experiences with them, specifically issues with lower-than-rated capacity, low output voltage, and premature cell die-off. Link.
Have a look at the existing pack too to see if it's NiCad or NiMH. I think there are some slight differences in how the two are to be charged, though some "chargers" I've seen are little more than a resistor in series with the output of a bridge rectifier + capacitor, which keeps the charging current low. (This doesn't perform any kind of charge status monitoring, it just keeps some voltage across the battery terminals at all times.)