I am new to this field so spare me if I am asking something wrong.

I have a battery of around 48 volts and a dc motor of 2hp. So I was thinking is there a way to increase the torque generated by the motor when I need it and increase the speed of the motor when needed by changing the voltage being transfered to it by installing a transformer.

Please explain it to me and if I am wrong, it will be very helpful if you could provide some useful reference as I have to enter an electric cart making competition in few months thanks.

  • \$\begingroup\$ The "transformer" you need is called a gearbox \$\endgroup\$ Dec 6, 2015 at 12:18
  • \$\begingroup\$ I know about the gearbox, I just disnt want to use it \$\endgroup\$
    – Deiknymi
    Dec 6, 2015 at 15:51
  • \$\begingroup\$ Then you are stuck with the torque constant Kt of the motor you have. More torque = more current = more wasted power and more heat. \$\endgroup\$ Dec 6, 2015 at 15:54
  • \$\begingroup\$ I think there's some confusion between "increase maximum torque above normal rated value" (which you can't do without gearing it down) and "control torque from zero to normal rated maximum" (which you can do with a speed controller) \$\endgroup\$
    – pjc50
    Dec 7, 2015 at 14:58

2 Answers 2


The torque (and speed) of a DC motor are both controllable.

The torque is controllable up to the maximum rated torque of the motor, and is controlled by varying the armature (and if a shunt wound motor, the field) current. Maximum sustainable torque is achieved at full rated current on the armature (and field). Most DC motors, especially small ones such as your 2HP example can easily handle up to 150% armature current overload for a brief period (say, about 30 seconds).

The speed is controllable from 0 to maximum rated speed by varying the armature voltage from 0 to base rated voltage (48VDC in your case). For a shunt wound motor, the speed can be increased by reducing the field excitation, but above base speed, the motor will lose torque. Essentially, it goes into a constant power mode.

A transformer doesn't work on DC, so you can't use that for any control. A small motor like your example one should work fine driven off of a PWM controller (if you have a DC supply), or an SCR Phase angle controller (if you have an AC supply).

You didn't give any specs on the motor besides rated HP and voltage, so I can't determine the actual speed or torque curves. If you are stuck with that motor, a 'gear' reducer is probably going to be necessary in addition to a PWM or SCR controller. This doesn't have to be a gearbox, it could just be a belted pulley system between the motor output shaft and the driven load.


With the same voltage and the same voltage, we will have the same current.
And the same current generates the same power.

But: you can spend more time to get more torque.
Or spend less time to get faster but less strong rotation.

This is done with gearbox as Brian said. It contains some physics.

Power is conserved in a gearbox. so:


I don't want to prove this equations, just accept it:

$$\frac{\omega _{out}}{\omega _{in}} =\frac{r_{in}}{r_{out}} =\frac{n_{in}}{n_{out}} =\frac{\tau _{in}}{\tau _{out}}$$

where omega is the angular speed. r is the radius and n is the number of teeth on a gear.also tau is torque.

you can calculate what you want.

If you want more torque, attach a small gear with a few teeth to the motor shaft. and connect a big gear with many teeth to the small gear.

the big gear will rotate at a low speed but it has much torque. though it will take much time for it to perform a full revolution.

  • \$\begingroup\$ I don't want to use gearbox \$\endgroup\$
    – Deiknymi
    Dec 6, 2015 at 15:52
  • \$\begingroup\$ I said. you can't increase power with the same voltage and motor. change your motor or increase voltage. there is a third way which is opening your motor and cleaning it. it may increase the motor efficiency. \$\endgroup\$
    – AHB
    Dec 6, 2015 at 15:57

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