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

5V below a (mains) 5V 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

--
Bill Naylor
www.electronworks.co.uk
Electronic Kits for Education and Fun

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.

--
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Check out my Electronics Engineering Video Blog & Podcast:
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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"

**Connect two 5.1 volt zeners in series and put an electro across each.

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"

** But not consistent with what the OP asked for.

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"

** The OP wanted rails that were BOTH negative wrt the AC line - ie the ground to be at -5.

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.

** Now read the rest of his damn post, instead of seleting one bit

- you stupid FUCKHEAD.

Lets' face it guys...

** Go straight to hell - you POMPOUS TURD.

** It is not crazy to want the processor supply linked to the active - you ignorant ass.

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: :

[...] crap snipped.

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.