Can a small capacitor bank be used to power a robot with stable voltage?

AI Thread Summary
A small capacitor bank charged with 24 volts can be used to power a robot, but it is not ideal for providing stable voltage to motors that require 9 to 12 volts. Capacitors are better suited for short bursts of energy rather than sustained power, as their voltage drops rapidly when current is drawn. While supercapacitors can store energy for brief periods, batteries are recommended for longer durations due to their ability to maintain stable voltage and hold more energy. The discussion emphasizes that using capacitors as a primary power source for a robot is impractical, and batteries are the preferred choice for reliable operation. Ultimately, for effective motor operation, understanding both voltage and current requirements is crucial.
14shreyas
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hi there,
i want to know that how can i use a small capacitor bank which can be charged with 24 volts and gives out the same, please help.
i want it to move motors for a robot like 9v to 12v each
 
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You can use current limiting resisters in series to charge the capacitors.
But why do you want to use capacitors to move robot motors, why not battery? Capacitors can't hold much energy.
 
i want to use it for a purpose because i am running the robot on a generator i built myself, so the genrator will create electricity and charge the capacitor bank so i can use it afterwards. there is one more thing after this but i want to keep it a secret! ; )
 
You might use a supercapacitor to store little energy available over a short duration, but to supply power for over 10s or 100s batteries are better.
 
The thing about capacitors is that they are not ideal for using as power supplies. Their use is for filters, coupling and decoupling mostly. In digital systems they are used to power memory cells, but we are talking very, very short times (much less than a second).

A capacitor has a the relationship

i_{c} = C \frac{dv_{c}}{dt}

Here, the current i_{c} is the current running into the capacitor.
Rearranging the equation gives you

\frac{dv_{c}}{dt} = i_{c}/C

As you can see, if the capacitor is used to supply a system (robot) the voltage will decrease. It is impossible to use a capacitor to power something (led, robot, resistor, whatever), for times longer than a few miliseconds or microseconds without the voltage changing rapidly.

So you can't charge a capacitor to 24 volts and then use it to provide a circuit with a stable 24 volts. The more current you use, the quicker the voltage goes down.

In conclusion: A capacitor is not use able as a power supply. You need to use batteries for this. Batteries can hold much more energy also, and they provide a stable voltage at their nodes.

Remember also, that if you want a power supply to run a motor, it is not enough to know the voltage it needs, you must also know something about how much current the motor needs to operate in the region you want it in.

IF you wanted to use capacitors to power your robot, you would need a very, very big capacitance. But again, the capacitor would not be able to provide a stable 24 volts. As soon as you begin tapping energy from it, the voltage drops.
 
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