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Puse charging a capacitor.

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selwyn.mileham

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Can someone please explain what happens if an electolytic capacitor is pulse charged with a square wave pulse.
Example 1M in series 1uf cap is charged with a pulse train of Ton =10ms and Toff =990ms.
Will the capacitor eventually charge to Vp if there is no discharge path? Vp = pulse peak volts.
 
Theoretically, if the cap has no leakage, it will eventually get charged. But the current is going to be pretty small with that big resistor, so it's going to take a while. The time constant is 1 s and the duty cycle is 1%, so if you assume it will take n time constants to charge up the cap, it will take around 100*n s.
 
If the pulse is going between 0V and Vp then the cap will charge up to the average value of the pulse (Vp x duty-cycle of pulse) or Vp *10/1000 = 0.01Vp.

If the pulse is actually an open circuit when "off" then it will charge to Vp.

To get it to charge to Vp if the pulse goes to 0V, you will need to add a diode to prevent the cap from discharging.
 
Capacitor voltage at any given time is determined by the equation in the attached. Where t is the pulse duration, Vc(t) is the voltage of the capacitor at time, t (end of the pulse); Vi is the initial voltage of the capacitor just before the pulse; Vf is the voltage of the capacitor at infinite time which in this case is peak of the pulse.
So the question of what the capacitor voltage will be at the end of the pulse now depends on undefined Vi (capacitor's voltage just before the pulse)
 

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I believe the op is talking about applying a train of pulses, not one pulse. Thus the voltage will to to the average value of the pulse train (see post #3).
 
I believe the op is talking about applying a train of pulses, not one pulse. Thus the voltage will to to the average value of the pulse train (see post #3).

understood! Using the equation i posted earlier, after the 6th pulse it charges to 0.992Vp. This is above the average voltage of the pulse. So, on the long run the capacitor tends to charge to Vp.
 
I would say that the basic equation will only give an approximation to the voltage level.

We should consider the effect of the capacitive reactance of the capacitor when applying a 10mSec pulse, ie 100Hz and its harmonics
 
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understood! Using the equation i posted earlier, after the 6th pulse it charges to 0.992Vp. This is above the average voltage of the pulse. So, on the long run the capacitor tends to charge to Vp.
And as I posted, it will not charge to Vp unless the signal is rectified or is open circuit when off. If the signal goes to 0V when off, then the cap will exponentially charge only to the average value as determined by the pulse duty cycle.
 
The OP is asking
Will the capacitor eventually charge to Vp if there is no discharge path?
The answer is 'yes'.
 
pulse duration alone is not enough to determine the frequency but time for a complete cycle (i.e, Ton + Toff). In this case, it is 1000ms, i.e, 1s and f = 1Hz.

So that implies a capacitor will have no effect in reducing electrical noise on a power rail, when used for decoupling.???
 
i didn't say that but pointing at the fact that the frequency here is 1Hz not 100
Hz

Its a 10mSec square wave type pulse, so it is made up of the fundamental and a series of decreasing amplitude odd sine wave harmonics, the low 990mSec period is zero

Look thru this link.
https://en.wikipedia.org/wiki/Square_wave
 
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