[Amps] Calculating ripple % ?
Jim Thomson
jim.thom at telus.net
Sat Sep 22 11:18:29 EDT 2018
There appears to be 2 ways to calculate ripple %. There also appears to be no consensus as to which method is correct.
One method uses the peak to peak value of AC ripple, as seen on a scope.....then divide by loaded B+ (then multiply that result by 100). This is with a full bore load on the B+ supply, like key down.
The other method uses the RMS value of the AC ripple, then divide by loaded B+ value. (than again, multiply that result by 100). Again, with a full bore load.
Problem is, most simple AC meters..and also cheaper DMMs will not measure RMS AC voltage. These lower cost type meters assume the AC is a pure sine wave. On a simple B+ supply, with
full wave rectification and a C filter, the AC component looks more like a saw tooth wave form.
Then I see there is various formulae used to find the RMS value of the typ saw tooth wave form...with varying results.
These days, with something like my Fluke 87, it will read true rms, so the saw tooth wave form is a moot point. A true RMS DMM, like a fluke 87 etc, will easily read the rms component.
But regardless of which method is used, the calculated ripple % difference between each method is huge.
EG: 8.75% using the P-P method equates to just 3.1% if using the RMS method.
So which method is... the real deal ?? Which one should we be using ? I suspect the P-P method is in vogue by some folks..since back in the day, it was..and still is, easy to measure. The formulae used to find the rms value of a saw tooth
wave form are plenty, varied....and all over the map..hence the use of a simple scope to measure P-P. Im inclined to believe the RMS method is probably the real deal.... but I may well be wrong.
Jim VE7RF
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