Maybe the books say that there is no load on the rectifier.Most textbooks start with a statement like
'The capacitor charges up to the peak value of the input voltage.'
If this were simply true then the capacitor would thenceforth always be equal or greater than the input voltage. Therefore no more curent could flow into the capacitor. Therefore no current could flow out of the capacitor into the load.
And no, the waveform before the capacitor is not a ripple waveform.
You need to have V in the numerator of the right side of the equation.Not sure I understand your comments get device info?
My original is not a trick question, by the way.
Most texts describe in some detail the voltage waveforms in a transformer/rectifier/reservoir capacitor lineup. Many offer some version of the formula
ΔV = 1/2fRC or ΔV = 1/4fRC
But few address the much more difficult issue of the current waveforms, or the plain fact that the thing would not work without the ripple. Or to put it another way - it is impossible to smooth out all the ripple. The greater the smoothing the greater the peak current.
OK, I totally agree that the cap is charged by a series of very short duration, very high current pulses. In fact, it is possible to make the cap so large that the peak repetitive current spec of the rectifier is exceeded. All the coulombs that are removed from the cap by the load during the time the rectifier is reverse biased must be replaced during the time it is forward biased.Yes Ron the delta is relative so to get an absolute value you need to multiply the factor by the average output voltage, which for most purposes can be taken equal to the capacitor peak voltage.
The different formulae refer to half wave (2) and full wave (4) rectification. The ripple on full wave being half that of half wave.
But I am trying to move away from the voltage equations to get people to acknowledge that the capacitor is charged by a series of very short duration very high current pulses, not by a continuous waveform as with the voltage.
I see what you mean. I was thinking peak-to-peak ripple, you were thinking peak deviation from average. This makes sense for delta, although ripple, in my experience, is almost always specified as p-p.I suppose that depends upon your Δ doesn't it?
The factor of 2 comes in because to a first approximation the average voltage is halfway between the peak and minimum of the ripple so
Vav = Vpk - ΔV or Vav = Vmin + ΔV
Ripple = 2Δ
Which is particularly useful if the rectifier is powering a voltage regulator, because they look like a constant current load (if the regulator load resistance is constant).That's why I said 'some variation on the formula'.
The most useful one I've seen actually includes current
Vav = Vpk - I/4Cf (for full wave)
A simulation with waveforms wouldn't be too difficult.I agree with you Loosewire a picture is often worth a thousand words.
However creating and placing said picture on AAC involves me in considerably more effort than spewing out a few words so I only do it if I am convinced of the interest.
You raise an interesting point by saying this. For the theoretical case usually described in the textbooks, this is true. But the impedance of the source driving the rectifier is crucial. Looking through several texts of mine, I only found one that includes the source impedance in the analysis.The greater the smoothing the greater the peak current.