Reducing amperage while maintaining voltage?

Discussion in 'General Electronics Chat' started by AznGothic, Feb 17, 2008.

Feb 17, 2008
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I'm not sure if this has been discussed already or not. I did a search and the closest thing I found was a thread talking about reducing 12V to 6V with a 5amp current. In my case I have a 6VDC battery that is putting out 15A but I want to reduce it to about 7A while keeping the 6VDC. I'm actually fairly new at all of this. I guess I'll get into what I'm trying to do....

I have a high RPM DC motor that runs off of 4 D cell batteries. I actually run this motor quite often and constantly buying 4 D cell batteries is getting a bit costly. I initially tried getting 4 rechargeable D batteries but couldn't find any in my area. I went to the near by city and found rechargeable D cell's but no charger. So then I decided to try and hook up this motor to a AC/DC converter. I was using a 2A converter that allowed you to select between 1.5V to 12V. The converter barely spun the motor at all. I then tested the motor with the 4D batteries with a multimeter and found out that it was using about 4.5A at peak spin with an initial draw of 7A to get it started. So I set out to buy a 6V R/C Car battery figuring it'd give me the amperage I needed and it came with a charger for a fairly cheap price. Only problem now is when I hooked it up, the motor spun extremely fast and started burning up (got really hot and started putting off a smell). I tested the R/C battery to find it's putting out 15A opposed to the 7.5A the D batteries put out (7.5A is a single D battery, with 4 tied in a series it put out 4.5-5A). How can I reduce the 15A to around 5-7A and keep the 6V output?

Also I don't know if I understand this completely but if I am, it's not exactly the voltage and amperage that matter but the power (wattage) that matters. So 6VDC at 7A is 42Watts, 6V at 15A is 90Watts. Would it be safe to say, even if I lost some voltage while reducing the amperage, as long as I still get 42Watts it'll work just fine?

I apologize for the extreme length of my post...

2. Audioguru New Member

Dec 20, 2007
9,411
896
Your 6V throw-away battery voltage dropped when the motor was running so the current also dropped. Since power equals voltage times current then the power to the motor was much lower than you thought.
The RC battery voltage did not drop much so its voltage and current were much higher. Then the power to the motor was too high.

The internal resistance of the throw-away battery reduced the voltage and reduced the current going to the motor. Add a resistor in series with the RC battery so it works the same.

You need to find out the resistance of the motor to be able to calculate the value of the resistor.
If you measure the voltage and current when the throw-away battery is powering the motor then the calculation is easy.

3. nomurphy AAC Fanatic!

Aug 8, 2005
567
12
You don't control the current, per se, the motor takes what it needs to operate per the voltage applied. It appears your supplying to much voltage (and the D batteries sagged).

If you want to spend some more money on batteries for an experiment, go by 12 more of the D batteries. Connect 3 in series to get 4.5V, and then do this 3 more times with the remaining batteries. Connect these four sets of batteries in parallel, this will provide 4x the current @ 4.5V and help prevent battery sag.

Now see how the motor runs with just 4.5V, but with more current provided as the motor needs. If it still runs too fast, remove the middle battery from each chain so that you have 2 batteries (3V) x 4 sets.

A series resistor with the RC battery will also create a voltage drop based on the amperage the motor is drawing. If the motor is drawing 15A, then a 0.1 ohm 50W (1.5V * 15A) resistor would drop 1.5V. So, the voltage to the motor would be 4.5V. However, these numbers will vary with the VA of the motor.

Without going into more complicated circuitry, getting a large rheostat may do what you need (but it's just another form of series resistance).

4. mik3 Senior Member

Feb 4, 2008
4,846
63
If the motor is rated at 6 VDC then your new battery its ok, but if it is rated at a lower voltage and you apply to it 6VDC then you will probably burn it.

However, if the motor id rated at 6 VDC and you apply 6 VDC (the normal working voltage) and you dont have the proper load on its axis then it will spin very fast.

Feb 17, 2008
11
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Thank you for the responses. I feel like I'm understanding a lot more of what's going on. So basically because my D batteries tested at around 7amps but when the motor was applied I was getting 4amps then I was also not getting 6V and instead less voltage was used?

Also to Nomurphy, what is a rheostat? I was wondering if I'd just be able to buy a variable resistor like those found on a computer case fan or computer cpu fan that lets you control how fast the fan spins and apply that to my RC car battery with my motor?

Also I'll take your advice and go buy some cheap D batteries, tie them first in a series and then group the series into parallel and see what happens. Could you also go into a little bit of detail on the formula or equation you used to find out I need a 0.1ohm 50W resistor?

6. nomurphy AAC Fanatic!

Aug 8, 2005
567
12
I think mik3 makes a good point, using the 6V battery try putting a load on the motor first. Otherwise, it could free-run out of control.

If the motor is drawing 15A, then a 0.1 ohm 50W (1.5V * 15A) resistor would drop 1.5V.

E = I*R
1.5V = 15A * 0.1 ohm

W = I*E
22.5W = 15A * 1.5V

for heat dissipation, double 22.5W = ~50W

A rheostat is a wire-wound pot, usually pertaining to higher power requirements (such as a toy train or slot-car speed controller). As you "turn-up" the speed controller, the resistance decreases and more voltage is applied to the motor.

Using a fixed resistor in series with a motor really isn't a good idea, because the motor is dynamic (speed/voltage/current varies).