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### Topic: 05/Mood Cue - Capacitor across potentiometer (Read 2201 times)previous topic - next topic

#### joews

##### Feb 05, 2013, 09:56 pm
Hi, electronics newbie here.

I understand that the capacitor in parallel with the servo provides "top-up" voltage when the servo starts moving to prevent the current across the rest of the circuit dropping, but what is the function of the second capacitor in parallel with the potentiometer?

Thanks!

#1
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#### Raybrite

#2
##### May 31, 2015, 10:58 am
Actually , if you follow the wires around the schematic diagram you will see that they are actually connected in parallel. You may be able to take one of them out with no consequences if I am right. Otherwise it would have to do with the capacitance value needed across the servo. You have 200 ufd across the motor now.
I haven't played with these circuits in years.
Hope this helps.
Ken

#### ctcarton

#3
##### Jun 04, 2015, 05:43 pmLast Edit: Jun 04, 2015, 06:52 pm by ctcarton
The physical positioning of the capacitors right next to the components they decouple is actually a very important, as is having two separate capacitors. The book doesn't explain why because it's a bit too advanced for a beginner. I will try to explain why concisely.

Your understanding of the purpose is actually reversed.  The capacitor actually provides a 'top-up' current to the device it is connected over in order to protect the rest of the system from a drop in the supply voltage

Devices such as the servo motor require relatively large amounts of current when they are engaged. Switching very quickly from a low current to a high current (or vice-versa) isn't something the power lines are good at doing and causes the entire power line to act like an long inductor as it resists the change for a short time. During this time the power line, because it acts as an inductor, drops voltage itself. That voltage drop is of course visible to other components. (Edit: The Arduino board has a voltage regulator that tries to mitigate this effect, but sometimes we need to help it out.)

To deal with this effect we put a decoupling capacitor in parallel, and physically close, to the motor. Capacitors are capable of fast current changes. It will charge up when the motor is not running, and when the motor starts up and demands an instant current, the capacitor can supply that current instead of forcing the power lines to do so. The current on the lines can rise much slower minimizing the inductance generated. To be most effective the capacitor should be as close to the motor as is reasonably possible. And we put a separate one around the pot as well because it's function can also cause fast current changes which, although not as significant as those caused by the motor, might still have visible effects.

So that is overall gist of what is going on.  I hope this explanation was helpful.

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