AC offline to low-V supply, with a cap

Years ago we had threads about the ugly but seemingly simple scheme to get 12V DC at 5mA, etc, with a film capacitor straight to the AC line. We said, gotta have a series resistor to minimize the startup inrush current.

Does anybody recall a specific instance of this scheme being used commercially?

Paul and I are days away from turning in the manuscript on our new x-Chapters book, and I decided to add this and a few other cheap AC-to-DC-power circuits, for entertainment. So I created one with SPICE, 12V at 5mA, only a few parts, cost under $1.50, and the series resistor sized to limit the worst-case inrush (cap with -170V, new line connection at +170V) to 720 mA, with a 470-ohm 2-watt wirewound.

OK, the numbers look bad. Yes the cap current is 90-deg out of phase with the voltage, but not so for the 470-ohm resistor, swallowing up power. The efficiency is less than 5%. Yes I could reduce the 470, but its maximum pulsed instantaneous power (0.2ms) is already 245 W.

Maybe an NTC inrush limiter, but I don't know if the time scales are right.

--
 Thanks, 
    - Win
Reply to
Winfield Hill
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Half wave or bridge? The efficiency/surge situation is better for a bridge.

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John Larkin         Highland Technology, Inc 
picosecond timing   precision measurement  
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Reply to
John Larkin

onsdag den 6. marts 2019 kl. 22.21.00 UTC+1 skrev Winfield Hill:

a capacitive dropper is used in a crap ton of cheap led lights and then there's death traps like this:

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Reply to
Lasse Langwadt Christensen

The cheapest guys woulda gone for half-wave.

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 Thanks, 
    - Win
Reply to
Winfield Hill

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This Snap Power Night Lite uses a circuit like you describe. This reviewer has a circuit diagram of it around the 7 minute mark.

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Reply to
bloggs.fredbloggs.fred

I think it is done for antiparallel strings of LEDs too.

Those fake Edison lamps. I wonder how they work.

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John Larkin         Highland Technology, Inc 
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Reply to
John Larkin

Gawwk!

No inrush resistor, and no lots of other stuff.

--
 Thanks, 
    - Win
Reply to
Winfield Hill

5% is unusually poor. 170v 470R is 0.36A, why do you need to limit it to this low value? Too much R there. 0.36A x 170v = 61 watts peak in the R, not 245.

If you draw 5mA at 12v, your 470R is dropping 23.5v at 5mA, giving around 3

3% efficiency. Not good due to too much R, but nowhere near 5%.

The problem with NTCs: a) you can reapply power when hot, so you can't use reduced R. Having highe r R at other times is no plus b) they tend to deteriorate. c) they cost more

RC supplies are widespread inside appliances. The chinese make supplies like this with touchable outputs too. They're not legal here but do turn up.

There are some basic problems with these types of supply. a) shock obviously, or when used internally the need for more care over ins ulation since the LV is live. b) Cs on mains like to fail. Look up 'wattless dropper' for plenty of stori es from the valve TV days. Class X2 caps are better but some still fail. Rs taking startup pulses on mains, even when run well below their rated 200v, also have a history of failure, albeit better than that of caps. c) Zeners are needed if you can't guarantee enough load current at all time s, leading in some apps to power waste. d) RC PSUs are only really practical for very low currents.

NT

Reply to
tabbypurr

I have seen a capacitive dropper used in an electricity meter from a reputable manufacturer.

John

Reply to
jrwalliker

Series antiparallel diodes, lots of 'em. No limiter in sight. The sheath around the long skinny PC boards is the phosphor.

Reply to
whit3rd

OK, full bridge reduces cap size and current, discharge resistor cuts max V in half.

--
 Thanks, 
    - Win
Reply to
Winfield Hill

No kidding. Fast edges are going to blow straight through the series C, won't they? Nearby lightning strikes, or someone plugging a nasty load into the same power strip, or who knows what. Worst-case inrush voltage on a power line is a lot higher than the 170V nominal peak.

IMHO this doesn't belong in AoE except as a "Bad circuit," accompanied by an explanation of the various things that might go wrong with it.

-- john, KE5FX

Reply to
John Miles, KE5FX

Voltage straight into the LEDs? Wow.

I was just looking at one, and it does seem to have a little electronics in the base.

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John Larkin         Highland Technology, Inc 
picosecond timing   precision measurement  
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Reply to
John Larkin

this one at least got a varistor,

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Reply to
Lasse Langwadt Christensen

AC outdoor security lights [1] typically use a dropper. The dropper circuit inside of one looks like this:

+----/\/\/\------||------+-------------------+

| 1W 250V _|_ `-+-, | \ / 1N4003 /_\ 1N4749A L o --+-- | N o | _|_ | --+-- \ / 1N4749A | /_\ 1N4003 `-+-, | | | | | | +------------------------+-------------------+

FWIW, Todd Harrison produced a video [2] that shows how to convert a readily available medical grade isolation transformer into a tech iso transformer. In the video Harrison also reviews safety considerations apropos to dropper supplies.

Notes:

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Thank you, 73,

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Don Kuenz KB7RPU 
There was a young lady named Bright Whose speed was far faster than light; 
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Reply to
Don Kuenz

Use a capacitive voltage divider, i.e. add a capacitor also across the load. This will bypass part of the mains spikes.

A few decades ago, when 4000 series CMOS and LCDs became available, digital clocks with capacitive mains power was used.

Reply to
upsidedown

The capacitive dropper is a way to invite trouble, due to its high-pass nature. The more transients and crud is on the mains line, the better they get in.

--

-TV
Reply to
Tauno Voipio

'Ang about. The point of the R in the supply is to limit peak inrush curren t to something the parts can all cope with. There's nothing disreputable ab out that - but the circuit does get used where it shouldn't, and there lies the problem. An EE that can't use RC PSUs is missing a basic widely used s kill.

NT

Reply to
tabbypurr

The R limits them to no problem on mains. But if you plug an RC PSU into a modified square wave invertor, it's toast very quickly, the R burns out.

NT

Reply to
tabbypurr

In a capacitive voltage divider with series capacitor Cs and hence Xs reactance and parallel capacitance Cp and reactance Xp, is used to get

12 Vac, Xs must be 18 Xp for 230 Vac mains, i.e. Cs = Cp/18. With a load resistance Rp = 12 V / 5 mA = 2400 ohm. Using Cp 1,3 uF will give about the same reactance and hence Cs = Cp/18 = 72 nF.

Assuming that up to 1500 V mains transients may be present, the capacitive voltage divider voltage could be 72 V. The source impedance at the load would be about 1 kOhm, thus the current would be about 60 mA during the transient.

Compare this with a series capacitor only with 1500 V 1 kV/us (1 kV/ms) transient.

Of course, all the calculations should be done properly with impedances and not just reactances.

Reply to
upsidedown

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