Electronics Forum

by **Macman** » Thu Aug 27, 2015 11:24 pm

just new to electronic wiring and I'm having trouble with potentiometers. first let me explain what I'm trying to do. trying to control the speed of a 12 v DC motor. what I have built so far. I have 120 v going to a 12dc power converter wired the 12 v to go into a buck converter to get steady 12 v out then run the 12 v to the motor with a pot in line and also a digital volt and amp meter in line. started with a 5k ohm pot and it was just on and off no variance went to a 100 ohm and it got hot and I think burned up. is there a better way to do this I have been thinking DAC or PWM but this is all new to me any help would be great oh. motor is 12 v DC and pulls about 5 amp at full draw.

by **pebe** » Mon Aug 31, 2015 5:50 pm

Using a series pot to control the motor speed is not practical when the motor consumes 5A.
[quote] ......wired the 12 v to go into a buck converter to get steady 12 v out ...[/quote]
You are on the right lines when you use a buck converter because it is a switched mode device and is very efficient. Can the converter you are using be adjusted to give a variable output voltage?

by **Macman** » Tue Sep 01, 2015 2:37 am

[quote="Macman"]Can the converter you are using be adjusted to give a variable output voltage?[/quote]
yes it can but it uses those little pots that have to be adjusted by turning a little screw with a screw driver. I want to adjust speed by turning a knob by hand. was thinking of a PWM controller but never have work with one

by **pebe** » Tue Sep 01, 2015 11:28 am

Your best approach would be to modify the buck converter so that it uses an external pot. That way you won't burn out the pot and you can vary the voltage applied to the motor - and hence its speed. I modified a buck converter in this way to provide a bench variable power supply.

Most of the converters I have seen use a 10k linear pot (marked 103). You can remove it if you have a soldering iron with a tip large enough to melt the solder on two of the tags (centre and end) at once. That enables you to lever up that side of the pot. Repeat that operation for the centre and other tag. After a few times the pot should come free and you can then replace it with a normal pot.

You need a hot iron because the connections are plated through holes and it takes a while for the heat to penetrate to the other side of the board.

If you feel its a bit too fiddly, then you could just attack the pot with sidecutters to destroy it. You could then desolder the contacts one at a time.

I hope that helps.

by **Macman** » Wed Sep 02, 2015 5:45 am

thanks Pebe but being a newbee to electronic circuitry i'm questioning how that is different from placing a pot right after the buck converter and very the voltage to the motor

by **pebe** » Wed Sep 02, 2015 11:40 am

OK. If the motor takes 5A at 12V then it will take approximately 2.5A when it has 6V applied to it. If you put a variable resistor in series with the motor to reduce its supply to 6V, then the resistor has to 'drop' the other 6V. That means the power dissipated in the resistor will be 6V x 2.5A = 15W. You would require an enormous wirewound pot to handle that power - which is why you burned out your pot.

By contrast, the pot in the buck converter is only sampling the output voltage of the converter chip and feeding back a tiny current to it to control that voltage. So the power dissipated in the pot is peanuts compared to putting a pot in series with the motor.

I hope that explains it.

by **Macman** » Wed Sep 02, 2015 4:48 pm

Thanks again Pebe I understand now. what is cool is I have never done much of anything with electricity or electronics and I'm learning allot. a few weeks ago I didn't even know what a buck converter was, now I have a breadboard and I am experimenting with circuits. thanks for taking the time to type an answer Pebe.