In my case the owner was no longer the owner because they gave me the money - IMO at the time. If they felt they had legal precedent to collect the money, they wouldn't have needed to try to get me to sign a promissory note which would have been an admission of debt.
I read in sci.electronics.design that Frank Bemelman wrote (in ) about '9V 1A schematic needed', on Mon, 21 Mar 2005:
Well, not thieves under English law, since, as it happens, 1968. In both cases, the owner took no effective steps to re-possess. I'm quite sure that if they had, the pay-back would have occurred.
So you and Jim Thompson are both confessed thieves. Jim in particular, because $400 in 1968 is not what I would call pocket change.
"John Woodgate" schreef in bericht news:gLcis+ESWvPCFw+ snipped-for-privacy@jmwa.demon.co.uk...
Yes, that is my definition of an effective step too. Anyway, I found it remarkable to read this 'aneckdote' from a person that insinuated the other day that someone purchased a few opamps "at his EMPLOYERS expense".
"simulation"
simulation
power
light on
Mr.Brasfield must be joking. Now tell me who said Irms doesn't have a limit? Its obvious to anyone that in any finite power circuit a current has a limit. Were we dealing with infinite power? No.
current
Mr Brasfield. I'm changing the Capacitance which is not going to hold everything constant. I guess you know that I inst = V inst * jwC => I change C => I change the instantaneous current I inst. Idc is the mean of that .. so Idc HAS TO change. All very simple.
increases.
Mr. Brasfield . You can actually see it using common sense but as you seem so stubborn here goes. the definition of Idc is assuming f = 60Hz Idc = 1/T* Integral( Ipeak*Sin 2*PI*60*t) dt the definition of Irms is Irms = sqrt (1/T Integral ( Ipeak^2 * Sin^ 2*PI*60*t) dt) Carry out the integration and tell me if you don't get my result.
I just defined it for you above. I never said the Irms = Idc but that Irms = constant * Idc.
Who said I was holding I rms constant?
What strictest sense is that you pesuod-intellectual trash (PIT from now on out)?
No s*it , PIT?
No s*it, PIT? Too dumb to figure out that Idc is the load current yet? And you apparently don't even know what transformer regulation is...
-which you didn't do, PIT- your Perl program is a joke- and you're too damned dumb to know that your little grade school crap computation becomes invalid because of ridiculously high harmonic content.
Here's where your stupid beyond words , PIT- can't have your large conduction angle with constant Idc and infinite capacitance. You have so many ratios and normalizations in there it has made you confused ,PIT. This also explains why you hesitated to link us to your bullshit "analysis". You're quite the PIT, PIT.
No it doesn't PIT- your original context is a bunch of pseudo-intellectual trash. You may impress others with your cut-and-paste of elementary integral table junk- but as usual with mediocre little pretenders like you- you can't make sense of it.
Not from you he won't- you're too damned to get cause and effect straight, and it all you can do to RMS and mean straight formula- but since you have no understanding, you completely miss quite a few vital points.
Don't you need to write another Perl routine that produces a bs table that you can only index into for maximum Irms rating of transformer and no other value? Yeah- that's useful retard and PIT.
>....
I have never seen the instantaneous value of a signal multiplied by its frequency like that. It may be simple to you, but your formula appears to be taken from a private language of your own. It mystifies me.
Perhaps we can simplify this by showing that Idc HAS TO NOT change when the load current stays constant. I will start with assumed conditions and points of agreement.
The average value of current taken out of the cap is constant, is called Iload, and line frequency is constant. The average value of current into the cap is called 'Idc'.
We are assuming an equilibrium where the peak and valley voltages on the cap do not change from cycle to cycle. In other words, repetitive waveforms.
Then we can say that, on a per cycle basis, the same amount of charge is removed from the cap (going to the regulator/load and now named 'Qcapout') as is added to the cap (and now named 'Qcapin'.)
If we call the time separating charging cycles 'Tcy', then Qcapin / Tcy == Qcapout / Tcy
Qcapin / Tcy is the same value as Idc.
Qcapout / Tcy is the same value as Iload.
Idc == Iload
Since Iload is constant, so is Idc.
Notice that you can change the C value all you like without changing a bit of the above analysis. Note also that it pertains whether the charging current is a sinusoid fragment or not.
My common sense about the circuit operation and analysis I have done are what led me to differ with you. If believing my own carefully done analysis until it is shown to be wrong is 'stubborn', then count me in.
I have done that very integration. Part of the problem here is that the limits of integration that you need to use to get a number out of it are changing under the stated assumption that C is changing. I do not see that incorporated into your result or thinking, yet. When you do that, you will find that Idc and Irms are no longer proportional. Another issue is that under the assumption that capacitance is changing (and that it matters), the currents in question are not sinusoid fragments as your expressions claim.
Yes, your 'constant' is my 'K where K is a constant'..
You did, if you followed your own logic.
Given these conditions, and I quote: If you hold everything else constant, (including the DC current taken from the cap by the regulator/load), then you claim the following: I'm changing the Capacitance which is not going to hold everything constant. Idc *is* constant as established above. If we take as true your assertion (and I quote): Idc is directly proportional to Irms where Idc refers to that constant charge per cycle, then Irms cannot change without violating your assertion of proportionality.
I've posted a link in this thread to an analysis I've done. (under subject "Re: 9V 1A schematic needed [link]") Below is an extract from it that may get this cleared up. You can lift and use the definite integral expressions by plugging in your equivalent variable names.
The model behind this analysis employs an ideal transformer having an output resistance Ro and open circuit peak output voltage Vp, rectifier diodes that turn on at a fixed forward bias and have no resistance [1], and an infinite output filter capacitance [2]. An independent variable, Vd, represents the difference between the filter capacitor voltage limited by R and what it would be if R was 0 Ohms. The transformer open circuit output is assumed to be sinusoidal. The analysis is for full wave rectification, but can be easily adjusted for half wave.
[1. Diode resistance can be folded into the transformer output resistance. The actual P/N junction current/voltage characteristic contributes negligible error because it only affects the current flow when that current is relatively close to zero anyway.] [2. The infinite filter capacitor represents the limiting (and worst) case for RMS/average current ratio. The situation can improve for lesser capacitors as the ripple becomes an appreciable fraction of Vp.]Get the rectifier turn-on level: Vx := Vp - Vd Substitute an angle, a, for the phase velocity: a := 2 pi Fp t (where Fp is line frequency and t is time) The voltage charging the cap thru Ro: v(a) = max(0, Vp cos(a) - Vx) Derive (half of) the conduction angle g = acos(Vx/Vp) The squared charging voltage during conduction: v(a)^2 = Vp^2 cos(a)^2 - 2 Vp Vx cos(a) + Vx^2 Mean charging voltage over a half cycle: v_ = 1/pi Integral[from -g to +g] v(a) d(a) = 1/pi (2 Vp sqrt(1 - (Vx/Vp)^2) - 2 Vx g) Mean squared charging voltage over a half cycle: vsq_ = 1/pi Integral[from -g to +g] (Vp^2 cos(a)^2 - 2 Vp Vx cos(a) + Vx^2) d(a) = 1/pi (Vp^2 g + Vp^2 sin(2 g) / 2 - 4 Vp Vx sqrt(1 - (Vx/Vp)^2) + 2 Vx^2 g)
Note that the integrals for finding means are evaluated only within the conduction angle and the interval over which the mean is computed is a half cycle (pi radians).
These identities are were substituted: sin(+acos(x)) = +sqrt(1-x^2) cos(x)^2 = 1/2 + cos(2 x)/2
The mean and RMS output current will be: I_ = v_ / Ro irms = sqrt(vsq_) / Ro
[snip]
YHBT. YHL. HAND.
--Mac
What a total crock of s*it- you are simply regurgitating what was told you days ago- and something that you missed with your long-winded pseudo-intellectual analysis in Perl- the rectifier Idc=Iload. But now being the fake you are, you pretend to educate someone else about it. And your so-called analysis above shows that you don't know your ass from a hole in the ground. Real engineers know that the capacitor is a "charge balance" circuit element- in steady state the time variation of charge must balance to zero. Other people know this is as capacitors block DC voltage. Because you don't understand these simple physical facts, you have to resort to more of your artificial "analysis"- you are a true p.o.s.
Your "analysis" is no analysis at all- you are a clueless little pedant and pseudo-intellectual.
Makes about as much sense as that mess you posted to rec.audio.
This is just a bunch of elementary crap you pasted from other sites and a handbook with very little comprehension. Your garbage analysis is not even close enough to be called a first-order approximation.
[snip extract]
I recall that post also, but not from anyone in this subthread. To which person and anecdote are you referring?
Have something to add? Share your thoughts — no account required.
Ask the community — no account required