# How does a capacitor reduce voltage drop in a DC motor start?

I know that a capacitor stores charge: C=Q/V

but what i don't understand is how this would reduce the voltage drop caused by high current draw.

My theory is that the capacitor would need to be in parallel with a conductor (minimal resistance) and then connected to the motor.

simulate this circuit – Schematic created using CircuitLab

(Diode is motor)

By doing this, current will only be drawn from the capacitor when the power source can't supply enough.

If this is true why? and if it this isn't true how is capacitor supposed to be set up then? what would happen if the capacitor is in series?

• The cap should be in shunt and only big enough to compensate for transient commutation spikes, not boost conductance of weak batteries or supply, The intent is to absorb RF loop currents but not damage the armature from arc pitting on surface from excess watt-seconds of arc energy. So the cap should be sized according to radiated noise and located directly near motor. cable inductance is also a factor. – Tony Stewart Sunnyskyguy EE75 Nov 23 '16 at 22:03

What you show is just a diode in parallel with a voltage source. There is no capacitor here since it's shorted. Removing the capacitor would change nothing.

It's not clear what you are really asking, but some types of motors have a "startup capacitor" in them. These types of motors run on AC, and don't have any torque when the rotation speed is 0. The capacitor unbalances the motor to cause some torque at 0 speed.

This capacitor would reduce efficiency at normal operating speed, so there is usually a way to switch it out of the circuit. A common means is a centrifugal switch.

If you really mean a capacitor in parallel with the power supply a DC motor is connected to, then that's just a capacitor holding up a supply. There is nothing special about a motor being connected to that supply.

A large capacitor across a supply provides extra charge to the load when the supply voltage drops. This helps the supply look more beefy to the load than it really is, at least in the short term. In effect, the supply/capacitor combination is capable of larger short term current than just the supply alone.

A motor draws a surge of current at startup, so a capacitor can help. However, this motor initial current surge is "long", so in most cases a unrealistically large capacitor would be needed to make a significant difference.

A small capacitor across a motor can help to reduce emissions. The capacitor keeps the voltage more steady, and keeps the high frequency noise current circulating close to the motor. The time over which such a capacitor can make a meaningful difference in holding up the voltage is so small that this only does anything useful at frequencies that can radiate.

• perhaps he meant a DC motor, where e-caps across small power motors get a boost in power at the expense of faster wearout – Tony Stewart Sunnyskyguy EE75 Nov 23 '16 at 23:21
• @Tony: You may be right. I have added to the answer. – Olin Lathrop Nov 24 '16 at 16:08
• I think i understand, the capacitor is in parallel with the supply, so that when the voltage drops (in this situation as a result from the motor's start current) the voltage also drops over the capacitor resulting in the charges being repelled from the negative capacitor pole and continuing the only possible way to the plus terminal, being thru the motor? – Ryan Abbas Nov 24 '16 at 21:15
• @RyanAbbas understand that all Power Sources, caps and motors have an effective series resistance, ESR , often spec'd as DF in caps and DCR in motors. For commutation motor noise, the caps will act as snubbers to the motor. – Tony Stewart Sunnyskyguy EE75 Nov 25 '16 at 6:50

Your voltage source is assumed to have some resistance. The capacitor would provide some of the current during the start-up surge.

simulate this circuit – Schematic created using CircuitLab

So, say it takes 0.5 second for the motor to wind up and it draws 0.5A during the surge and 100mA after (of course it will vary more smoothly than that). Then you would like the voltage to drop less than 1V during the surge. Let's assume all the current is supplied by the capacitor.

So $C = \frac{I_{start}\cdot T_{start}}{\Delta V}$ = 0.25F (250,000uF).

As you can see, you'll need a pretty large capacitor for even a small motor.

Putting it in series would just cause the motor to jog for a bit and then stop when the switch is closed, assuming the capacitor was discharged to begin with. It could only reduce the current available to the motor (again assuming it was discharged). Once the cap is charged there is no more current and the motor stops.

• i don#t understand what the arrows are supposed to mean in the diagram – Ryan Abbas Nov 24 '16 at 9:48

I believe that capacitors are sometimes used to prevent voltage drop when starting a DC motor that is connected to a power supply that is not really suitable. In the diagram below, the power supply is represented by an ideal voltage source, an internal power-supply resistance, and an internal power-supply capacitance. When the motor is switched on, the armature is not turning and the back emf is zero. The motor current is then limited only by the armature resistance and the internal resistance of the power supply. Unless the internal power-supply capacitance is quite large, it will discharge through the armature resistance and much of the supply voltage will be dropped across the internal power-supply resistance. Adding more capacitance will "fix" that problem by supplying current for a longer time before discharging to a significantly lower voltage. If the motor accelerates sufficiently in that length of time, the back emf will build up and prevent further discharge.

If the power supply has a current-limit or overcurrect-trip circuit the capacitor can prevent the power supply from electronically reducing the voltage or shutting off. Since the power supply may have difficulty charging a large capacitor, a charging circuit may be needed.

This scheme is suitable only for small DC motors that have sufficient armature resistance to be started at full voltage without incurring commutator damage. Larger motors require a starting circuit that will limit the current until the motor has reached full speed.