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I came upon this sentence in a paper concerned with dielectric elastomer actuators (DEA):

Due to the capacitive behavior of a DE, a converter with current source functionality should, therefore, preferably be used to adjust the voltage vp, and by this the electrostatic pressure, in (1).

Why is a current source preferable compared to a voltage source?

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    \$\begingroup\$ with a current source, you can control the time exposed to the current, and know exactly what charge has been added to the total energy. \$\endgroup\$ Oct 9, 2019 at 10:04

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There will be more to it than the basic reason, but -

A capacitor's voltage can not be changed instantaneously, but the current into it and out of it can.
If you have a capacitor that you wish to charge to say 100 volts and can charge at up to 1A from the supply available.

  • If you apply a 100V power supply to the capacitor when it is fully discharged (so at 0V across the capacitor), then as the voltage CANNOT change instantaneously the supply will experience a hard short circuit. 1A (the supply max) will flow into the capacitor and the voltage will rise, but the power supply will in most cases "not be happy.

  • If you apply a 1A constant current source to the same discharged capacitor, it will again charge at 1A and its voltage will increase, but as the supply is designed to operate in this manner there will be no problem. The capacitor will charge until some preset voltage limit is reached - usually the charging supply in this scenario will be set to 1A constant current with a 100V maximum voltage limit.


Inductors have the opposite constraint - they WILL accept a voltage step but the current they conduct cannot change instantaneously. To "charge" an inductor to 100V across it and 1A flowing in it you could apply a 100V step voltage and the current will increase until the desired 1A current is reached.

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  • \$\begingroup\$ Also the capacitor equation (Q = CV) where V is capacitance voltage says us that a current source supplying constant charge (Q) per second increases the capacitor voltage in a constant speed whereas when you apply a constant voltage the charging of the capacitor follows a curvature slowing in time. \$\endgroup\$
    – Rehin
    Oct 13, 2019 at 9:51
  • \$\begingroup\$ @Rehin - that is true if you use constant I or V sources. Usually for most paid charging you use ~~= constant energy = a power supply rated at a given wattage and adapting its output as V rises - usually using a switchmode power supply. \$\endgroup\$
    – Russell McMahon
    Oct 13, 2019 at 9:54
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The typical DEA operating voltage is in the kV. While the end goal is to get a specified voltage across the device, just slamming a voltage source across the DEA will lead to a poorly controlled peak current that could damage the device or the driving circuit.

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    \$\begingroup\$ True - especially so given the refusal of a capacitor to instantaneously budge from zero for an infintesimal period and its then insistence on ramping up it's V. \$\endgroup\$
    – Russell McMahon
    Oct 13, 2019 at 9:55

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