May 13, 2011
Cracking The Marketing Code
Intel Announces New Transistors and Cadence Snatches Up Another EDA Start-Up
This week I try to crack the marketing code of Intel’s new tri-gate transistor announcement and will dig through some of the details to figure out how this new transistor technology will shape the future of semiconductor design. I also report on Cadence’s new acquisition of Altos Design Automation and investigate how the newest rumblings at Cadence will affect the future of EDA360.
Did you know that video game creation is now considered art? Check out this week’s Fish Fry to find out how I plan on reshaping the world of virtual conferences to include artful video game features.
If you like the idea of this new series be sure to drop a comment in the box below. I appreciate all of your comments so far, and we will be working to enhance the Fish Fry each week - as long as you're watching.
Fish Fry Links - May 13, 2011
Posted on May 13, 2011 at 10:12 PMWell this is only for bragging rights since I don't have an iPhone and am more inclined in the Android direction.
The peak to peak voltage of a sine wave is twice the square root of 2 of the rms voltage. So a 120 V sine wave would be 339.36 p-pV. If your waveform is other than a sine wave your results could be different.
Posted on May 16, 2011 at 7:37 PMSteve has the right formula - he's just using a rounded version of square root of 2. Without the rounding it would be 339.41 p-pV. No iphone here either - cool app idea though
Posted on May 18, 2011 at 1:14 AMAdam is correct that I used a rounded, from memory version of the square root of 2. I should have properly rounded of the answer to 339 p-pV given that was the precision of the stated problem.
Posted on May 18, 2011 at 12:53 PMAll this discussion of the square root of two seems really irrational to me. However, to be nit-picky in a different way, I would like to point out that I think Amelia said:
"...the peak voltage for a sine wave whose RMS voltage is 120."
Absent a DC component, I think the "peak" voltage would be half of the "peak-to-peak" voltage, wouldn't it?
Yeah, I know. I had a guy like me back in EE316 too...
Posted on May 18, 2011 at 3:06 PMV_p=((2)^(1/2))*V_rms
Embarrassing - about forgetting to square
Added Later: As mentioned by kevin - in the absence of any DC components.
Posted on May 18, 2011 at 4:33 PMKevin, Yes if you want to be nit picky in that way the correct answer would be 170 Vpeak, or -170 Vpeak if you prefer .