Low cost mains power supply

[Nomen Nescio]

In most DIY stores you can find these cheap Chinese digital timers to
turn you lamps or appliances on at specified time. The thing I'm
wondering is, how do they power the digital circuitry from
110VAC in a device costing only $5-$10? Are they using a transformer or
are they directly powering the digital circuitry from the AC voltage?

[Robert Wessel]

For a small amount of power (and efficiency in the single digit
percentage range), a rectifying bridge, a zener for regulation, a
current limiting resistor and a capacitor or two is all you need. A
switch and fuse are probably a good idea too.

Isolation is key for safety on something like that.

[Pete]

They don't use a transformer. They usually use an X2 rated capacitor to
drop the voltage, then some rectification and regulation. This might be
as simple as a single series diode, a resistor, a zener, and a smoothing
capacitor.

Or, there are small IC's around that basically only need the mains in
via an X2 capacitor on one side, and deliver regulated 5v out at the
other side.

Peter

[Stef]

In comp.arch.embedded,

Pete <pje...@yahoo.com> wrote:
> Nomen Nescio wrote:
>> In most DIY stores you can find these cheap Chinese digital timers to
>> turn you lamps or appliances on at specified time. The thing I'm
>> wondering is, how do they power the digital circuitry from
>> 110VAC in a device costing only $5-$10? Are they using a transformer or
>> are they directly powering the digital circuitry from the AC voltage?
>
> They don't use a transformer. They usually use an X2 rated capacitor to
> drop the voltage, then some rectification and regulation. This might be
> as simple as a single series diode, a resistor, a zener, and a smoothing
> capacitor.

Google for [capacitor dropper] and you will find lot's of examples.
Efficiency is not that bad, power factor is terrible.

And keep in mind that this type of circuit is not isolated, so your 'low
voltage' side is live AC!

--
Stef (remove caps, dashes and .invalid from e-mail address to reply by mail)

Even a hawk is an eagle among crows.

[Jasen Betts]

I've got one here it has a capacitive dropper and a tiny 4mAh NiCd cell (I
assume NiCd because the charge rate printed on it is C/10) the AC switch
is a 48V relay and there's a TO92 device which probably switches it.
image in ABSE

--
⚂⚃ 100% natural

--- news://freenews.netfront.net/ - complaints: ne...@netfront.net ---

[Robert Macy]

I one time needed a super efficient 5Vdc [allowed to be directly
connected to the mains] power supply but had NO room for the size of
caps needed for the normal drop stage. So, I designed up a small
circuit that connectd the mains to the rectifier diode starting at
zero crossover and DISCONNECTING when the mains input went above
around 7V and started to take the 5 Vdc output voltage up to 5.1, or
so. thus I had efficiency AND regulation.

[John S]

On 4/9/2013 9:36 AM, Robert Macy wrote:

> I one time needed a super efficient 5Vdc [allowed to be directly
> connected to the mains] power supply but had NO room for the size of
> caps needed for the normal drop stage. So, I designed up a small
> circuit that connectd the mains to the rectifier diode starting at
> zero crossover and DISCONNECTING when the mains input went above
> around 7V and started to take the 5 Vdc output voltage up to 5.1, or
> so. thus I had efficiency AND regulation.
>

The mains voltage rises to about 7V input (5V output) in about 100us and
to about 7.1V input (5.1V output)in 111us. So you have about 11us to
charge your storage capacitor. Then, the storage cap must hold up the 5v
for about 16ms until the next rise (this assumes 170V peak, 60Hz,
half-wave).

I must be missing something very important and I would appreciate your
help in understanding what it might be.

Many thanks,
John

[Oliver Betz]

John S <Sop...@invalid.org> schrieb:

>> caps needed for the normal drop stage. So, I designed up a small
>> circuit that connectd the mains to the rectifier diode starting at
>> zero crossover and DISCONNECTING when the mains input went above
>> around 7V and started to take the 5 Vdc output voltage up to 5.1, or
>> so. thus I had efficiency AND regulation.
>>
>
>The mains voltage rises to about 7V input (5V output) in about 100us and
>to about 7.1V input (5.1V output)in 111us. So you have about 11us to

I consider it useful to mention mains voltage (and frequency) when
presenting such calculations.

>charge your storage capacitor. Then, the storage cap must hold up the 5v

I also think that "7.1V" is somewhat tight.

The Harris HV-2405E used this principle, of course with wider
thresholds.

Oliver
--
Oliver Betz, Munich http://oliverbetz.de/

[Arlet Ottens]

On 04/09/2013 07:56 PM, Oliver Betz wrote:
> John S <Sop...@invalid.org> schrieb:
>
>>> caps needed for the normal drop stage. So, I designed up a small
>>> circuit that connectd the mains to the rectifier diode starting at
>>> zero crossover and DISCONNECTING when the mains input went above
>>> around 7V and started to take the 5 Vdc output voltage up to 5.1, or
>>> so. thus I had efficiency AND regulation.
>>
>> The mains voltage rises to about 7V input (5V output) in about 100us and
>> to about 7.1V input (5.1V output)in 111us. So you have about 11us to
>
> I consider it useful to mention mains voltage (and frequency) when
> presenting such calculations.

They were mentioned (170V peak, 60Hz, half-wave) in the part that you
snipped.

>
>> charge your storage capacitor. Then, the storage cap must hold up the 5v
>
> I also think that "7.1V" is somewhat tight.
>
> The Harris HV-2405E used this principle, of course with wider
> thresholds.

And of course, it all depends on the current consumption. With very low
currents used, the cap would only discharge slowly.

[Oliver Betz]

Arlet Ottens wrote:

>> I consider it useful to mention mains voltage (and frequency) when
>> presenting such calculations.
>
>They were mentioned (170V peak, 60Hz, half-wave) in the part that you
>snipped.

correct, sorry!

[rickman]

On 4/9/2013 3:17 AM, Stef wrote:
> In comp.arch.embedded,
> Pete<pje...@yahoo.com> wrote:
>> Nomen Nescio wrote:
>>> In most DIY stores you can find these cheap Chinese digital timers to
>>> turn you lamps or appliances on at specified time. The thing I'm
>>> wondering is, how do they power the digital circuitry from
>>> 110VAC in a device costing only $5-$10? Are they using a transformer or
>>> are they directly powering the digital circuitry from the AC voltage?
>>
>> They don't use a transformer. They usually use an X2 rated capacitor to
>> drop the voltage, then some rectification and regulation. This might be
>> as simple as a single series diode, a resistor, a zener, and a smoothing
>> capacitor.
>
> Google for [capacitor dropper] and you will find lot's of examples.
> Efficiency is not that bad, power factor is terrible.
>
> And keep in mind that this type of circuit is not isolated, so your 'low
> voltage' side is live AC!

It may not be isolated, but it can be current limited and so not lethal.

--

Rick