Not true. RMS voltage would be correct for a resistive load where the current varies with the voltage, but not for a constant load current as is the case here. The regulator input is a constant current as determined by the load current and is independent of the input voltage. Thus the input power is linear function of voltage, or V*I where V is the instantaneous voltage and I is the load current. If V is varying then the average dissipated power is the average value of V times I.
Hello there,
Yes that is correct Carl.
Here are some numbers using some carefully chosen waveforms:
First with a constant current which would be like the voltage
regulator and a triangular wave with a dc offset:
v(t)=1+t*1
i(t)=1
T=1
vrms=Rms(v(t),t,0,T,100)=1.528
irms=Rms(i(t),t,0,T,100)=1
vavg=Avg(v(t),t,0,T,100)=1.5
iavg=Avg(i(t),t,0,T,100)=1
p(t)=v(t)*i(t)
P_True=(1/T)*Integrate(p(t),t,0,T,1000)=1.5
Pa=vavg*iavg=1.5
Pr=vrms*irms=1.528
Now with a current that is also triangular:
v(t)=1+t*1
i(t)=t
T=1
vrms=Rms(v(t),t,0,T,100)=1.528
irms=Rms(i(t),t,0,T,100)=0.5774
vavg=Avg(v(t),t,0,T,100)=1.5
iavg=Avg(i(t),t,0,T,100)=0.5
p(t)=v(t)*i(t)
P_True=(1/T)*Integrate(p(t),t,0,T,1000)=0.8333
Pa=vavg*iavg=0.75
Pr=vrms*irms=0.882
So in the case of constant current the average voltage works out better
than the rms value for the constant current case, but when the current
varies (for this example) the rms value worked better.
When in doubt, use the integral form
It's still a good idea however to use a larger capacitor and a better matching
input voltage for the regulator because the lowest dip in the ripple has to be
at least as high as a smooth input dc voltage would have to be, so that means
that the power dissipation for a rippling wave input will be higher than when
using a smooth input, providing of course you have control over the input
voltage itself as well as the capacitance.