I have a RC car and I wonder if its startup speed will increase if I mount a capacitor in parallel with the battery. My thinking is that while the car is idling the capacitor will load. When I start the car, the capacitor will improve the internal resistance of the battery. Am I wrong? 1000uF is enough?

# Electronic – Boost starup of a motor with a capacitor

capacitormotorpower

## Best Answer

Interesting question.

We can get a rough estimate. Along with the voltage

U, you need to know the expected currentIof the motor starting up, and how much timedtyou expect it to take to start up. That will tell you how much current the cap has to supply for how long, and consequently, how big the cap should be. Since you don't list the specs of your motor, I found a random motor here: http://www.pololu.com/product/1574It says 6V, 6A stall current, 450mA running current. So for my motor

U=6V. I honestly don't know if motors draw more than the stall current while starting up, but I think the stall current makes a good estimated upper bound for the current you'll need. So we'll sayI=6A.You'll also need to know how long it takes the motor to get up to speed. That will be application dependent, but let's say 1 second to keep it simple.

Now, let's suppose for the moment that the battery has an internal resistance such that it provides an insignificant amount of current during startup, that is, we're starting the motor up on the capacitor alone. This is the absolute worst case scenario (i.e., biggest capacitor scenario); obviously the battery will be providing at least 450mA (again assuming my motor) otherwise your battery isn't using the motor to its fullest once it's started.

Finally, we have to account for the fact that we're discharging a bit. In my limited experience with motors, if you undervolt them, you don't always get a proportional amount of torque out of them. Besides, if the capacitor voltage drops too low, sooner or later the battery (even with its higher internal resistance) is going to overwhelm the capacitor. So let's say that during startup, the capacitor only loses 10% of its voltage; we'll call this proportion

p.So we have:

U=6VI=6Adt=1sp=0.1Now, we'll make the (basically false) assumption that current is constant over time and voltage varies linearly, in order to avoid calculus. I've tried to structure my assumptions to overestimate the size of the capacitor we'd need anyway, so we should be fine doing this. Then, by the capacitance formula,

I=CdU/dt, where the change in voltagedU=pU=600mV. SoC=Idt/(pU)=(6A)(1s)/600mV=10F.Yikes. You could get one of these: https://www.sparkfun.com/products/746, the datasheet says that it can do 3A but is rated at 1750mA (I think those are mA, at any rate). The above advice about thick traces is definitely a good idea, and I think I, at least, would test the internal resistance of that cap by putting it in series with a 1 Ohm resistor and an ammeter on the high current circuit (the 1Ohm resistor is in case I'm wrong about the current). I'm worried about shorts.

At this point I should let you know that I didn't know the answer before I started, so I wasn't sure what I'd get; maybe I shouldn't have shot for an upper bound. I'd recommend you plug your own motor's numbers into this equation and fiddle with it; your 3.6V motor is almost certainly not going to draw 6A, which is good news for your capacitor.

As an "exercise for the reader", you can also get an estimate by computing the amount of energy stored in a capacitor (1/2

CU^2), and then computing the kinetic energy of your car at full speed (1/2mV^2), and then figure out how big a capacitor you need to get the car up to full speed. For that, you'll need to bear in mind that the motor only converts energy at a certain efficiency (60% for a cheap 3.6V motor?), and the capacitor will probably not discharge all the way because of the battery and because the motor may not be effective at low voltage. You'll get a completely different answer, but there's a lot to be learned in comparing these two estimates and the assumptions underneath.