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In message <l8vmqj+1p9b1hj@...>, dated Thu, 19 Dec 2013, sawreyrw@... writes: I mean the 500 mA, not the 400 A in the original simulation. And the 'DC component' doesn't go away

John, That current is correct. It is due to the start up transient and the DC component will eventually go to 0. Whether the magnetizing inductance is too low depends on the VA rating of the

In message <l8vjeg+tmtpa2@...>, dated Thu, 19 Dec 2013, smalcolmbrown@... writes: I didn't realise that by 'lopsided' you meant that the waveform appeared to have a large DC

Thank you sawreyrw That really solved that one. I am glad that it is a simple fix :) Thank you John For confirming what I thought was correct. :) Thank you Jerry Lee Marcel For also confirming another

In message <l8vch8+10ue8vj@...>, dated Thu, 19 Dec 2013, smalcolmbrown@... writes: There should not be a 'lop-sided' waveform anywhere. Whatever you do with adding and subtracting

Suusi, What is happening is the voltage across the primary is not 0 at time=0, so you have a huge primary current at time=0. Due to the long time constant it takes a long time for the primary

Sri, just a test, i don't see my messages. Rudi --- Ce courrier électronique ne contient aucun virus ou logiciel malveillant parce que la protection avast! Antivirus est active. http://www.avast.com

What do you mean "lopsided waveform"? The traces I see are perfectly correct. Indeed you can't take what happens at time-zero for granted, it could never happen in real life. Nature hates

The question is is the lopsided waveform between ~A and ground an accurate representation of a winding connected across a across any two of the phases? I was under the impression that the output was

This problem in the topic is long solved, but . . . The last windows I ran and maintained was windows 98. If one reinstalled or updated like that, junk seemed to collect, such that such that one

for aurora_viii, I did, and the administrator merged my complaint to an existing similar complaints (a lot, not just my) on their website. That means, it is not a scam, it is a real issue. ... How I

Martin, Not really, I could not access to the group anymore through my old yahoo account. But I could activate my group's membership through my existing sbcglobal account ... Thanks for advices guys.

I do not understand how a 1uH can be too large. according to the formula provided in the datasheet (for an output current ripple of ~40): l=(vout/(f*ripple curent))*(1-(vout/vin))

Hello Business Kid, You wrote: “... that was a great way to clog up a registry with junk, and end up with random files on your drive.” I’m not interested in Unix problems for many years. But: 1/

I cannot understad the functionality of this statements. Nothing from help, could someone explain that? Many thanks, Enrico

I am uploading 2 HEMT models which I am using for the benefit of others. The ATF34143:

Many thanks Helmut. Your answer helped me a lot.

Hello Ahmed, You could use .step and .MEASURE commands to get the frequency versus voltage. The results of MEASURE will be saved in the errorlog-file(.log). Then right mouse click in this file to plot

+1 for phishing scam. Send a copy of the email to yahoo help, subject 'phishing email' or 'spoof email' No Internet set-up needs a 'phone number that I've heard of, and no don't give out your 'phone

Hello all, What is the theory behind selecting the right max time step for FFT analysis? I always assumed that the max time step should be selected depending on the highest frequency resolution one