Yes. And Phil was wrong.
Building a wattmeter?
John
Negative voltage from Transformerless Capacitive power supply
Jun 22, 2009
76 Replies
"John Larkin">
** No.** Not one tiny little bit.
The OP has just proved himself incapable of explaining his own thoughts.
Same as any other AUTISTIC, CODE SCRIBBLING, LYING
F****IT ALIVE !!
And YOU - you slimy, vile pile of s*it.
..... Phil
No.
JF
Good point. The rectifier diodes need to be moved to the left.
John
=A0 =A0|+ =A0 =A0 |
[C1] =A0[RL1]147R=A0 =A0 | =A0 =A0 =A0|
=A0 =A0| =A0 =A0 =A0|
[C2] =A0[RL2]1470R=A0 =A0 | =A0 =A0 =A0|
--+-> -4.4V
=A0 =A0 =A0|+ =A0 =A0 |
=B5F[C1] =A0[RL1]147R
=A0 =A0 =A0 | =A0 =A0 =A0|
=A0 =A0 =A0|+ =A0 =A0 |
=B5F[C2] =A0[RL2]1470R
=A0 =A0 =A0 | =A0 =A0 =A0|
Interesting, thanks John, I'm just going to have to cob it together and try both configurations. (Oh with a low voltage source of AC) I thought you could just move the current limiting impedance. He'll then need a diode with a higher reverse voltage.
George H.
I was looking back at the OP's comments with your exchange with John Larkin and, in the process, came to exactly the same thoughts that George Herold noted (before he posted them.) It appears that George and I (and John) were quite capable of reading the OP's writing with logical understanding of the range of possibilities allowed by the language used in a situation where you were obviously incapable of that much. John Larkin was not flummoxed in the way you were.
Communication is always a 2-way thing -- the OP could have written more but you could have read into it less, as well. Like all too often happens with you now, you read far more into things that were intended and seem incapable of seeing what actually IS said and expressed, adding your own biases as if they were something others said. That happened between us, as well, and you were on about things I hadn't even imagined meaning in writing. You leap to conclusions, inserting your own assumptions in the process.
I just reviewed some of your posts going back to 2004 and earlier, in a variety of groups. Went back to insults you made about someone you accused of living in Sydney's lower North Shore, for example, with your bigotry about all people living there; to discussions about the
301 opamp, etc. A serious change has happened to you, Phil. You need to figure it out, put it in context, and work on it.Relying upon psychological projection as the only way you can explain to yourself your own failures isn't going to help you, at all.
Jon
Let's just say that in "The Land of Point", you would not be the character that you think you would be.
oc
=A0 =A0 =A0|+ =A0 =A0 |
=B5F[C1] =A0[RL1]147R
=A0 =A0 | =A0 =A0 =A0|
=A0 =A0 =A0| =A0 =A0 =A0|
=B5F[C2] =A0[RL2]1470R
=A0 =A0 | =A0 =A0 =A0|
4=A0 =A0 =A0 =A0|+ =A0 =A0 |
30=B5F[C1] =A0[RL1]147R=A0 =A0 =A0 | =A0 =A0 =A0|
=A0 =A0 =A0 =A0|+ =A0 =A0 |
30=B5F[C2] =A0[RL2]1470R=A0 =A0 =A0 | =A0 =A0 =A0|
Is it OK to reply to myself?
I tried this after hours, but didn't have time to post... (or think) To paraphrase A.B. Pipppard "I'm not quite as smart as some others, and it's helpful to do the experiment to help guide my understanding." "Physicis of Vibration Vol. I".
So the single sided version of the OP's second circuit works just fine. (John F. (and L.) already knows this.) Adding the second side screws's thing's up. All the current flows through the wrong half of the circuit. Moving the diodes gets rid of this. But then there's 'tons' (tons =3D volts) of AC feed through. Intsead of moving the diodes I think a 'better' thing might be to add a second impedance. Make the current for each half of the ciruit flow through different R/ C's and I think everything will work. I can always bread board this tommorrow.
George H.
The rectifier diodes only see about twice the zener voltage, 10 volts or so. One BAV99 maybe. And Zetex makes a dual zener in SOT23.
line --------r-----c------+---ak----+------+-----+5 | | | | k | | zen a c | | | | | | neut---------------------(----------+------+-----com | | | | k | | zen a c | | | | | | +---ka----+------+------5
John
e
doc
e=A0 =A0 =A0|+ =A0 =A0 |
=B5F[C1] =A0[RL1]147R
=A0 =A0 =A0 | =A0 =A0 =A0|
=A0 =A0 =A0| =A0 =A0 =A0|
=B5F[C2] =A0[RL2]1470R
=A0 =A0 =A0 | =A0 =A0 =A0|
D4
e=A0 =A0 =A0 =A0|+ =A0 =A0 |
330=B5F[C1] =A0[RL1]147R=A0 =A0 =A0 =A0 | =A0 =A0 =A0|
=A0 =A0 =A0 =A0|+ =A0 =A0 |
330=B5F[C2] =A0[RL2]1470R=A0 =A0 =A0 =A0 | =A0 =A0 =A0|
r=A0 =A0 =A0|
=A0 =A0 =A0|
=A0 =A0c
=A0 =A0 =A0|
=A0 =A0 =A0|
=A0 =A0 =A0|
=A0 =A0 =A0|
=A0 =A0c
=A0 =A0 =A0|
=A0 =A0 =A0|
>
> John- Hide quoted text -
>
> - Show quoted text -
Thanks for the ascii image John. I found that this,
line ----+----r-----c---------------+--ak--+-----+5
| | |
| k |
| zen a c
| | |
| | |
neut-----(--------------------------+------+-----com
| | |
| k |
| zen a c
| | |
| | |
+----r----c----------------+-ka---+----------5
Works just fine. Whereas moving the diodes gives a lot of ripple on
the output.
To the OP, I would suggest buidling some test protoype circuits and
try them first with low voltage AC.
George H.
Speaking of points, you couldn't prove a point if you molded it into a Lawn Jart© and pierced the skull of a cocker spaniel from 30 yards, fucko.
Sorry, dumbfuck, but the word "jart" is not copyrighted, nor is it a trademark. So much for your pathetic attempt to take yet another jab at me.
How is that for a point, you absolutely witless twit?
[snip]
I found that this,
line ----+----r-----c---------------+--ak--+-----+5 | | | | k | | zen a c | | | | | | neut-----(--------------------------+------+-----com | | | | k | | zen a c | | | | | | +----r----c----------------+-ka---+----------5 Works just fine. To the OP, I would suggest buidling some test protoype circuits and try them first with low voltage AC.
George H.
Simple and effective. Just what the OP needs.
Wastes twice as much power as necessary.
John
.
=A0| =A0 =A0 =A0|
=A0k =A0 =A0 =A0|
=A0 =A0 =A0c
=A0| =A0 =A0 =A0|
=A0| =A0 =A0 =A0|
=A0| =A0 =A0 =A0|
=A0k =A0 =A0 =A0|
=A0 =A0 =A0c
=A0| =A0 =A0 =A0|
=A0| =A0 =A0 =A0|
Yup, but the design is hardly power efficient either way. Do you have some way to deal with the large ripple when the diodes are moved?
George H.
Do you mean take away the two R's in series with the current-limiting caps and replace them with a single R in the line... or is there something else I'm missing?
Yup, but the design is hardly power efficient either way. Do you have some way to deal with the large ripple when the diodes are moved?
George H. Would this work?
,------------+---, | | | /\\ | Rload k k zener | | l a a = | i / \\ | | n--R--C--< >--------+---+--neut e \\ / | | k k = | a a | Rload \\/ zener | | | | | '------------+---'
I don't see why there would be more ripple. The droop voltage between recharge events is V = I*T/C, same no matter where the diodes are, where I is the load current and T is one line cycle. There would be more ripple only if the zener curve isn't flat, which is not an issue for zeners above 5 volts or so. Use bandgaps for flatter curves.
Besides, caps are cheap.
John
I did an appliance controller chip design (Emerson Electric) with dual shunt regulators (on-chip) in place of the zeners. Worked just ducky.
...Jim Thompson
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC\'s and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
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Gourmet Puzzles:
What part of the fish are the "sticks"?
Likewise where are the chicken "fingers" located?
Maybe you could find the chicken fingers in the Pope's nose?
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