Back to Basics video: What is RMS Voltage?

[w2aew]

I hope this isn't TOO basic for this group....

Here's a very basic, practical explanation of RMS voltage, what it means, how to calculate it (without using calculus). 

--video removed due to bonehead error in the calculation shown for the square wave--- 

Serves me right for doing a video late at night without thinking it through enough.  I put together a new video on the topic when I have a chance.  Sorry about that folks.   Apparently it wasn't too basic for my overtired brain when I put it together - doh!
 

---UPDATE---
I put together a revised video, correcting the error in the "digital" example that was in the original video.  Hopefully no similar bonehead errors will be in this one!

[Deagle]

In my opinion: Nothings wrong with basic tutorials. At the very least I like to think it reinforces my knowledge. I figure its worth it if there's something slightly new that I can learn from it.


I waste a lot of time going over things I already know, It's not a BAD form of procrastination right?

[Telequipment]

Not to basic at all, it's good to go over things, as we all can drift off, I really like your videos, were you a teacher!!!!

[BillyD]

Good video, very well presented as always. Keep it up + thanks for sharing.

[Telequipment]

Also here's is your Fast edge pulse generator, lovely project, other people should build, there will lots of uses for this, I run mine on 2x 1.5 penlight batteries on a holder on the back, it will be interesting to see how long they last under normal use.

[vk6hdx]

Great video & well presented. Thank you.

[IanB]

I hope this isn't TOO basic for this group....

Here's a very basic, practical explanation of RMS voltage, what it means, how to calculate it (without using calculus). 

Whoa, I got lost at the very beginning of the video.

We had 2 V DC into 100 ohms giving a power dissipated of 40 mW. OK.

Then we switched to a 50% duty cycle on the same DC voltage. So the power is halved, giving 20 mW dissipated.

Now by the power formula we have:

P = V²/R = 0.02 W

Solving for V gives:

V_RMS² = 0.02R = 0.02 x 100 = 2

V_RMS = sqrt(2)

This is not equal to V_PP x duty cycle as indicated at 1m44s of the video?

In general it would seem that:

V_RMS²/R = D x V_PP²/R

=> V_RMS = sqrt(D) x V_PP

where D is the duty cycle.

[kjpye]

IanB has identified one problem with the video, but two other things jumped out at me when I watched it.

1. What is the purpose of the "of the same level" at the end of the definition. I simply don't know what this is meant to mean, and it would be better left out.

2. I had trouble with the calculations of the average value of the squared voltage. The first thing I noticed was the division by 7. Why 7? Just because there are 7 points in the calculation doesn't mean that you should divide by it. What you are really trying to do is calculate the area under the curve without using calculus. The example given splits the first quarter of the sine-wave into 6 sections, so the calculation should approximate the area of each of the six sections, add them up and divide by 6! I was very surprised that the answer came out so accurately. According to my calculator, adding the seven squared values does not give an answer of 3.500 (not even close). A better solution would be to calculate the value at 7.5, 22.5..., 82.5 degrees, add them up and divide by six, but a better approximation would be to approximate each section by a trapezium and work with that. (This gives the standard trapezoidal rule for calculating the area under the curve; basically calculate the seven values as in the video, but give a weighting of 2 to all the values except the first and last, and divide by 12.) This gives a very good approximation.

My other comment is that I would have emphasised that what was happening with the calculations on the sine-wave was that we were calculating the instantaneous power, and then averaging that.

[w2aew]

Well, crap.  That serves me right for throwing a video together late a night without putting too much thought into it. I pulled the video down.  duh!

[kfitch42]

=> V_RMS = sqrt(D) x V_PP

where D is the duty cycle.

I get the same answer working straight from the voltage values. Computing the area under this 'curve' is easy ... just rectangles.
Given duty cycle D and period P.
The signal has squared value Vpp*Vpp from time 0 to D*P and squared value 0 from D*P to P
The mean of the squared values becomes 1/P*(D*P*Vpp*Vpp + (1-D)*P*0) = D*Vpp*Vpp
taking the square root: sqrt(D)*Vpp.

[w2aew]

I hope this isn't TOO basic for this group....

Here's a very basic, practical explanation of RMS voltage, what it means, how to calculate it (without using calculus). 

Whoa, I got lost at the very beginning of the video.

We had 2 V DC into 100 ohms giving a power dissipated of 40 mW. OK.

Then we switched to a 50% duty cycle on the same DC voltage. So the power is halved, giving 20 mW dissipated.

Now by the power formula we have:

P = V²/R = 0.02 W

Solving for V gives:

V_RMS² = 0.02R = 0.02 x 100 = 2

V_RMS = sqrt(2)

This is not equal to V_PP x duty cycle as indicated at 1m44s of the video?

In general it would seem that:

V_RMS²/R = D x V_PP²/R

=> V_RMS = sqrt(D) x V_PP

where D is the duty cycle.

Yeah - bonehead error in the video, sorry about that.  I pulled that video down.  Serves me right for putting together a video late at night, without putting much thought into it. 

I put together a revised video, which hopefully is free of such bonehead mistakes: