0
\$\begingroup\$

This question is a follow up to a previous question found here.

As a follow up question: for a half-wave rectifier, what is the "AC" voltage and how do you measure it? Additionally, What is the DC voltage and how do you measure it? From another answer, the DC voltage is the "offset" between the AC and DC signals when viewed in AC and then DC mode. How do I then measure AC signal? Go into AC mode and measure from the peak to GND? If I do it that way, the those values are: VDC= 2.9V and VAC=6.0V (from reference to peak). Does that seem correct? Finally, is it incorrect to talk about the voltage in this way?

\$\endgroup\$
1
\$\begingroup\$

I'm going to call the voltage that you're measuring on the scope \$V_{in}\$.

When you set your oscilloscope to AC coupling, the voltage that you see on the screen is: $$ V_{in,AC} = V_{in} - \text{average}(V_{in}) $$ so turning on AC coupling effectively just shifts the signal down so it's centered at 0 volts.

If you're looking for the actual value of \$V_{in}\$, AC coupling isn't a great way to do it - there's no way to tell what the average value was before we removed it.

You asked a lot of questions, so I'm not sure if I've answered everything.

\$\endgroup\$
  • \$\begingroup\$ So - if my signal is composed of both AC and DC, can I call the "long time average of Vin" the DC voltage? \$\endgroup\$ – metropolis Jun 4 '15 at 19:40
  • \$\begingroup\$ Yeah. A lot of the time, you use an oscilloscope to measure periodic signals; then, the "long time average" is really just the average over one period. \$\endgroup\$ – Greg d'Eon Jun 4 '15 at 19:53
0
\$\begingroup\$

How to measure AC: No, not from peak to GND but from top peak to bottom peak. Then you would have a peak-to-peak AC voltage. In AC mode GND will be the long time average of the AC voltage.

No it is correct to talk about a voltage this way: x Volts DC + y Volts AC You just split the constant (DC) and time-varying (AC) part which makes it easier to talk about and do calculations.

\$\endgroup\$
0
\$\begingroup\$

Without specifying units of measurment, it is assumed that RMS values are implied. Firtly, the RMS value of a full wave rectified sinewave (assuming perfect diodes) is Vpk/1.4142.

The "1.4142" part is the square root of 2 in case you were wondering.

For a half wave rectified square wave, the RMS value is Vpk/2

But, to measure (or determine) only the AC (RMS) part of that waveform requires the use of pythagoras.

Noting that a full wave rectified signal has an average of 0.637x Vpk, you can subtract this from the full wave signal like this: -

Vac (RMS) = \$\sqrt{(\dfrac{Vpk}{\sqrt2})^2-(0.637\cdot Vpk)^2}\$

If Vpk is 100 volts Vac (RMS) = \$\sqrt{5000 - 4052.8}\$ = 30.78 volts

I leave the half wave example to someone else.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.