Oscilloscope AC/DC Rectifier Problem

[Matt21]

Homework Statement

In a AC to DC rectifier the DC signal was Vdc= (6.3+0.4)V and AC signal was Vac=(0.25+0.04)mV. What is the error on the quality factor of the rectifier?

Homework Equations

σQ=√((σr^2)/(Vdc^2 )+(Vr^2 σdc^2)/(Vdc^4 ))

The Attempt at a Solution

I know this is the correct formula and I also know that Vac = Vr. I have tried using this formula in many ways including plugging in the data as it is and converting the Vac data to volts instead of millivolts but it still gives me the wrong answer. For example, when I do that I got error on σQ = 0.000007. Any help as soon as possible would be much appreciated.

 

[gneill]

It looks to me like your Quality Factor is what others call the Ripple Factor of a rectifier circuit, being the ratio of the AC ripple voltage to the DC output voltage while the circuit is connected to a load.

There are a couple of things to check. First, does your definition of Q use the peak-to-peak AC voltage or the RMS AC voltage, and second, is the given data for the AC voltage peak-to-peak, peak, or RMS? If it's supposedly data read off of an oscilloscope then it's likely either peak or peak-to-peak. So you may have a conversion factor to apply in order to bring your data in line with the definition of Quality you're applying.

Converting from mV to V so that all the units match is correct.

 

[NascentOxygen]

the DC signal was Vdc= (6.3+0.4)V

I'm wondering what to make of this? Why did they not say V_dc = 6.7V?

Any ideas?