Negative voltage from Transformerless Capacitive power supply

On Tue, 23 Jun 2009 02:53:32 -0700 (PDT), Centron System Solutions wrote:

--- I don't think so.

Here it is in ASCII: (view in Courier)

470R 680nF ACHOT>--[F1]--[R1]--[C3]--+---+---[D3>]---+-->>--+-> +4.4V | |K |+ | | [D1] 330µF[C1] [RL1]147R | | | | ACNEUT>-------------------|---+-----------+-->>--+-> GND | |K | | | [D2] 330µF[C2] [RL2]1470R | | | | +---+---[>--+-> -4.4V

note that when the output of C3 goes positive, the junction of D2 and D4 does too.

Now, since the anode of D2 goes to neutral, it'll conduct, with the result being that the output of C3 will only be one diode drop above GND!

the same thing happens when it goes negative, with D1 steering the signal to GND.

Here:

Version 4 SHEET 1 880 680 WIRE 160 96 96 96 WIRE 304 96 240 96 WIRE 400 96 368 96 WIRE 448 96 400 96 WIRE 560 96 448 96 WIRE 656 96 624 96 WIRE 784 96 656 96 WIRE 96 128 96 96 WIRE 784 128 784 96 WIRE 448 144 448 96 WIRE 656 144 656 96 WIRE 96 240 96 208 WIRE 448 240 448 208 WIRE 448 240 96 240 WIRE 656 240 656 208 WIRE 784 240 784 208 WIRE 784 240 656 240 WIRE 448 256 448 240 WIRE 656 256 656 240 WIRE 656 256 448 256 WIRE 784 272 784 240 WIRE 448 288 448 256 WIRE 656 288 656 256 WIRE 96 320 96 240 WIRE 400 384 400 96 WIRE 448 384 448 352 WIRE 448 384 400 384 WIRE 560 384 448 384 WIRE 656 384 656 352 WIRE 656 384 624 384 WIRE 784 384 784 352 WIRE 784 384 656 384 FLAG 96 320 0 SYMBOL voltage 96 112 R0 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V1 SYMATTR Value SINE(0 170 60) SYMBOL res 256 80 R90 WINDOW 0 0 56 VBottom 0 WINDOW 3 32 56 VTop 0 SYMATTR InstName R1 SYMATTR Value 470 SYMBOL cap 368 80 R90 WINDOW 0 0 32 VBottom 0 WINDOW 3 32 32 VTop 0 SYMATTR InstName C1 SYMATTR Value 6.8e-7 SYMBOL zener 464 208 R180 WINDOW 0 -43 36 Left 0 WINDOW 3 -133 1 Left 0 SYMATTR InstName D1 SYMATTR Value BZX84C6V2L SYMATTR Description Diode SYMATTR Type diode SYMBOL diode 560 112 R270 WINDOW 0 32 32 VTop 0 WINDOW 3 0 32 VBottom 0 SYMATTR InstName D4 SYMATTR Value MURS120 SYMBOL cap 640 144 R0 SYMATTR InstName C2 SYMATTR Value 330e-6 SYMBOL cap 640 288 R0 SYMATTR InstName C3 SYMATTR Value 330e-6 SYMBOL res 768 112 R0 SYMATTR InstName R2 SYMATTR Value 147 SYMBOL res 768 256 R0 SYMATTR InstName R3 SYMATTR Value 1470 SYMBOL zener 464 352 R180 WINDOW 0 -38 31 Left 0 WINDOW 3 -139 68 Left 0 SYMATTR InstName D2 SYMATTR Value BZX84C6V2L SYMATTR Description Diode SYMATTR Type diode SYMBOL diode 624 368 R90 WINDOW 0 0 32 VBottom 0 WINDOW 3 32 32 VTop 0 SYMATTR InstName D3 SYMATTR Value MURS120 TEXT 112 272 Left 0 !.tran 1 uic

This is the obvious fix:

470R 680nF ACHOT>--[F1]--[R1]--[C3]--+-[D3>]-+---------+-->>--+-> +4.4V | |K |+ | | [D1] 330µF[C1] [RL1]147R | | | | ACNEUT>-------------------|-------+---------+-->>--+-> GND | |K |+ | | [D2] 330µF[C2] [RL2]1470R | | | | +-[>--+-> -4.4V

But then, even after you fix the problem there's a low outout voltage and high ripple on the positive supply:

Version 4 SHEET 1 964 680 WIRE 160 96 96 96 WIRE 304 96 240 96 WIRE 400 96 368 96 WIRE 496 96 400 96 WIRE 608 96 560 96 WIRE 688 96 608 96 WIRE 816 96 688 96 WIRE 96 128 96 96 WIRE 816 128 816 96 WIRE 608 144 608 96 WIRE 688 144 688 96 WIRE 96 240 96 208 WIRE 608 240 608 208 WIRE 608 240 96 240 WIRE 688 240 688 208 WIRE 816 240 816 208 WIRE 816 240 688 240 WIRE 608 256 608 240 WIRE 688 256 688 240 WIRE 688 256 608 256 WIRE 816 272 816 240 WIRE 608 288 608 256 WIRE 688 288 688 256 WIRE 96 320 96 240 WIRE 400 384 400 96 WIRE 496 384 400 384 WIRE 608 384 608 352 WIRE 608 384 560 384 WIRE 688 384 688 352 WIRE 688 384 608 384 WIRE 816 384 816 352 WIRE 816 384 688 384 FLAG 96 320 0 SYMBOL voltage 96 112 R0 WINDOW 3 24 104 Invisible 0 WINDOW 123 0 0 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName V1 SYMATTR Value SINE(0 170 60) SYMBOL res 256 80 R90 WINDOW 0 0 56 VBottom 0 WINDOW 3 32 56 VTop 0 SYMATTR InstName R1 SYMATTR Value 470 SYMBOL cap 368 80 R90 WINDOW 0 0 32 VBottom 0 WINDOW 3 32 32 VTop 0 SYMATTR InstName C1 SYMATTR Value 6.8e-7 SYMBOL diode 496 112 R270 WINDOW 0 76 29 VTop 0 WINDOW 3 66 26 VBottom 0 SYMATTR InstName D4 SYMATTR Value MURS120 SYMBOL cap 672 144 R0 SYMATTR InstName C2 SYMATTR Value 330e-6 SYMBOL cap 672 288 R0 SYMATTR InstName C3 SYMATTR Value 330e-6 SYMBOL res 800 112 R0 SYMATTR InstName R2 SYMATTR Value 147 SYMBOL res 800 256 R0 SYMATTR InstName R3 SYMATTR Value 1470 SYMBOL zener 624 352 R180 WINDOW 0 46 30 Left 0 WINDOW 3 35 69 Left 0 SYMATTR InstName D2 SYMATTR Value 1N750 SYMATTR Description Diode SYMATTR Type diode SYMBOL diode 560 368 R90 WINDOW 0 70 33 VBottom 0 WINDOW 3 72 34 VTop 0 SYMATTR InstName D3 SYMATTR Value MURS120 SYMBOL zener 624 208 R180 WINDOW 0 46 30 Left 0 WINDOW 3 35 69 Left 0 SYMATTR InstName D1 SYMATTR Value 1N750 SYMATTR Description Diode SYMATTR Type diode TEXT 112 272 Left 0 !.tran 1 uic

So, what to do???

More later... :-)

JF

Reply to
John Fields
Loading thread data ...

Yes. And Phil was wrong.

Building a wattmeter?

John

Reply to
John Larkin

"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

FUCKWIT ALIVE !!

And YOU - you slimy, vile pile of shit.

..... Phil

Reply to
Phil Allison

--
No.

JF
Reply to
John Fields

Good point. The rectifier diodes need to be moved to the left.

John

Reply to
John Larkin

=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.

Reply to
George Herold

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

Reply to
Jon Kirwan

Let's just say that in "The Land of Point", you would not be the character that you think you would be.

Reply to
TutAmongUs

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.

Reply to
George Herold

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

Reply to
John Larkin

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.
Reply to
George Herold

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.

Reply to
Bill Palmer

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?

Reply to
TutAmongUs
[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.

Reply to
Michael Robinson

Wastes twice as much power as necessary.

John

Reply to
John Larkin

.

=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.

Reply to
George Herold

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?

Reply to
Michael Robinson

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 | | | | | '------------+---'

Reply to
Michael Robinson

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

Reply to
John Larkin

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  |             |
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  |
| E-mail Icon at http://www.analog-innovations.com |    1962     |
             
Gourmet Puzzles:

        What part of the fish are the "sticks"?

        Likewise where are the chicken "fingers" located?
Reply to
Jim Thompson

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.