Negative voltage from Transformerless Capacitive power supply

Hi Denis

Here goes:

For the sake of clarity, I will assume the following circuit is just powered from 5V with respect to 0V. It will also work if you are hanging a 0V line

Connect a capacitor from an output pin of the micro to the anode of a 1N4148 diode. Connect the diode cathode to ground. The capacitor needs to have its

• plate connected to the micro and its - plate to the diode.

Then connect the cathode of another 1N4148 to the junction of the anode of the first 1N4148 and the capacitor. Connect a capacitor from the anode of the second 1N4148 to ground. The second capacitor needs to have its + plate connected to ground. Hell, if you understand this, I suggest brain surgery...

Here is how it works: Micro output goes postive and charges the first capacitor (the first 1N4148 conducts current to ground). When the micro output goes low, the capacitor -ve plate follow this transition and goes negative. This forces current to flow in the second diode and charges the second capacitor negatively. Thus the second capacitor has to be wired with its +ve plate to ground.

Cap values can be electrolytics at about 1uF to 10uF. YOU MUST keep the duty cycle of the micro port symmetrical (roughly). if not, the capacitors will not charge or discharge and your output will sulk.

Cheap too!

Let me know what you think

One way would be to replace the zener with two series back-to-back zeners, then use one diode from the zener-capacitor junction to give you +V and one (the other way 'round) from the same point to give you

-V.

You may not like the arrangement wrt the N line if you are switching a triac with the micro, so you could also put another zener in series with the one in the original circuit and another diode to give you a more negative voltage wrt N than in the original circuit (eg. ~= -5V, ~=-10V).

Once you have the +5V you can simply use a standard switched capactitor inverter (e.g. 7660 type) to generate the -5V with respect to the circuit GND. But that's probably a fairly expensive and not very elegant solution compared to a discrete part solution others have mentioned.

Dave.

How much current do you need at -5v? A simple diode charge pump from a PIC pin may be sufficient. Another alternative is to level-shift to avoid the need for a -ve rail. Or use a rail splitter from a 10V supply.

Power integrations do some nice chips for transformerless PSUs in the power region where caps start to become rather big - LNK302/4/6

"Centron System Solutions"

The positive end of this stack connects to the active while the negative end returns to the neutral via a 400 volt series diode and the X cap etc. Gives -5 and -10 volts wrt active.

Another 400 volt diode passes current from the X cap etc direct to active when the polarity reverses.direction.

The may or may not be the same as Spehro's second idea.

BTW

I sure hope that " input wave" of yours arrives optically isolated or similar.

..... Phil

Or connect the neutral to the middle of the stack and rectify into both ends. Same number of parts.

Seems safer to float the circuitry on neutral, instead of hot, unless there's a reason to be on the hot side.

John

If it isn't it soon will be.

"John Larkin"

He did NOT want + and - rails around active or neutral.

...... Phil

"Spehro Pefhany" >>

Not if it's temp probe you stick in the mouth ....

...... Phil

He seemed to be following the Micron appnote, which floats on the hot side. I suggested that is's safer to be on the low side unless there's a reason otherwise. That's reasonable, isn't it?

John

After the morgue truck leaves..

I think I had the same problem with the circuit shown in the app note. First I assume that (L) stands for live and (N) for neutral on the AC power line. Then why not put the current limiting resistor and / or capacitor on the live side? At least then the +/- 5V is only flopping around the neutral line rather than the 120 VAC on the live side.

George h.

Yeah, less drama when you get forgetful and clip your scope probe ground to V-.

John

"John Larkin" "Phil Allison"

He did NOT want + and - rails around active or neutral.

.... Phil

Are you sure? He didn't explicitly say that. He did say he was confused about how to generate + and - 5 volts.

John

"George Herold"

"The capacitive design for this type of PSU requires the +5V to be at mains Live with circuit GND at 5 Volts less than this level. Its hard to see how to create a circuit that would produce -5V with respect to circuit GND."

He could just be talking about the app note circuit, and not the one he's building.

- you stupid F*****AD.

Lets' face it guys...

Many mains switching arangementt REQUIRE the switch to be in the active line.

Piss OFF.

..... Phil

Hi Guys thanks for all your input

And yes it is time to clarify. Although its only after all your helpful input that Im in a position to clarify.

The microcontroller and OP-amp in this design do two things.

(a) The OP-amp measures voltage across a sense resistor (0.47 ohm) passing mains current. And feeds its output to the A/D in the microcontroller. Importantantly the voltage across the sense resistor swings + and - about mains Neutral.

(b) The microcontroller controls a TRIAC wired on the low voltage side (Neutral) of the mains connection.

So ideally I want +5V (+4V to +5V), and -5V (-4V to -5V) relative to mains neutral to power my OP-amp. And + 5V for the microcontroller.

From your input here I have created a quick schematic and visible at:

Can you have a look please.

Its designed to give me 33mA with say 30mA on +4V and 3mA on -4V. I have added component values.

Im unsure of best location for Diodes D3 and D4. Should they be before the zeeners?

Is this circuit going to work?

Thanks again for all the help.

Denis _______________________________

On Tue, 23 Jun 2009 20:48:43 +1000 "Phil Allison" shrieked in Message id: :

How do you even type with a straitjacket on? Or do you just peck at the keys with your long, pointy nose?

On Jun 23, 2:53=A0am, Centron System Solutions wrote: [....]

I only took a quick look but I have the following comments:

The rectifier diodes need to come before the Zeners. Zeners conduct in the forwards direction but they don't spread the current around well and can thus be damaged more easily than the rectifiers.

I worry about what happens when the power is connected right at the peak of the sine wave. The resistor in the input needs to be able to withstand the surge.

You don't show any EMI protection etc. This needs to be part of the considerations. Imagine what happens for things like a nearby electric drill adding chatter. You also should defend against high voltage spikes.

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)

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:

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