Re: stumped by WPC power driverboard

I completely agree on your theory,
But then how can Marvin measure 12.7 V there?
http://pinrepair.com/wpc/tstbrig2.jpg

Thats what I am most puzzled about.
What is wrong? Our theory or the measurement from Marvin.
He uses a normal DMM (fluke 83) and I a True RMS (fluke 87)

thanx for thinking this over :)

On 2005-09-12 23:02:44 +0200, "martin" <martin.reynolds@xxxxxxxxx> said:

measurements and meters can be a bit off. Most meters get the AC RMS
right, but can strugge with the DC ripple. As you saw on your scope,
the DC side has a peak of 10V (I'll explain that in a moment) and a
ripple of 2V. Your meter did OK by approximating that as 9V. If you
disconnect the 5V loads, you'll see the voltage float up closer to that
12V number as the ripple will go away and the diode forwad voltage drop
will reduce.

OK so heres the maths:

9V a.c.  = 9* (root 2) peak to peak = 9*1.414 = 12.726V

But that's peak to peak. To get to DC, we have to rectify the AC with
diodes.

Under load, a typical diode has a voltage drop of 1.2V. there are two
diodes in the circuit at any time, so the measured output voltage peak,
under load, would be 12.73 - (2*1.2) = 10.33V.

If there is a minimal load, the Vf of the diodes drops to maybe 300mV.
So the capacitor can get back up to 12V.



.

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