Capacitor Charging time?

Other Hardware General Electronics
1

there are two scenarios in first,

if i use 7 volt motor but with 1000 uf capacitor on power line,

and in second if i use 7 volt motor with 470 uf capacitor on power life,

which scenario will take longer to charge again after motor uses the current?

If i am using a motor which consumes 1A but needs a power spike of 2A at start does the capacitor will be discharged completely after it ?

what if i start my motor slowly with PWM and does not let it spike to 2A will capacitor be discharged after it too ?

2

You really haven't provided enough information.

If you read the following, you will more likely be able to help us, to help you:

http://forum.arduino.cc/index.php?topic=149015.0

3

[u]Capacitor charge/discharge[/u].

No simple-easy answers...

if i use 7 volt motor but with 1000 uf capacitor on power line,

and in second if i use 7 volt motor with 470 uf capacitor on power [line]

which scenario will take longer to charge again after motor uses the current?

A larger capacitor charges and discharges more slowly (with the same series resistance).

If they are both connected to the same power source at the same time, parallel devices always share the same voltage so they will charge at the same rate. And together you have 1470uF so it's slower than either one alone.

In general, if you are connected to a power supply you can't calculate the charge rate because you don't know the effective series resistance of the power supply or the current capabilities/characteristics of the power supply.

If i am using a motor which consumes 1A but needs a power spike of 2A at start does the capacitor will be discharged completely after it ?

If a power supply is connected it won't completely discharge because the power supply is continuously supplying voltage. If the power supply can supply the required current it won't discharge at all. If there is no power supply and if you know the load resistance you can calculate the discharge rate. But, a motor is not a pure resistance.

And theoretically, a capacitor never "fully discharges" (and it never "fully charges"). If you have a known resistance (or known current) you can calculate when it's 90% charged/discharged or 99.999% charged/discharged.

what if i start my motor slowly with PWM and does not let it spike to 2A will capacitor be discharged after it too ?

You'll still get spikes but the overall discharge rate will depend on the average current.

But again, with a power supply attached it's not easy to make a calculation because some of the current will come from the power supply and with PWM the capacitor will be re-charged between current pulses.

4

Mark248:
there are two scenarios in first,

if i use 7 volt motor but with 1000 uf capacitor on power line,

and in second if i use 7 volt motor with 470 uf capacitor on power life,

which scenario will take longer to charge again after motor uses the current?

charge = capacitance x voltage.

So more capacitance, more charge needed, so either more current or more time (or both)

If i am using a motor which consumes 1A but needs a power spike of 2A at start does the capacitor will be discharged completely after it ?

Not enough information. If the power supply can supply 2A why should the capacitor discharge?

what if i start my motor slowly with PWM and does not let it spike to 2A will capacitor be discharged after it too ?

Learn the basic capacitor equations:

Q = CV
dQ/dt = C dV/dt = I

[ thus dV/dt = I/C ]

Q is charge on the capacitor in coulombs
I is current (as is rate of change of charge, dQ/dt) in amps
V is voltage on the capacitor in volts
C is capacitance in farads

dV/dt is rate of change of voltage on the capacitor.

In your situation if the power supply is good enough, the capacitor is really only reducing
high frequency noise on the supply wiring.

If the supply is struggling, 1mF (1000µF) is unlikely to be nearly enough to make any practical
difference. At 1A drain a 1mF cap has 1000V/s droop, ie droop from 7V to 0V in just 0.007s