Odd chip.

[Clive Mitchell]

I just cracked open another type of mains powered LED lamp today. This
time it's a GU10 base MR16 sized replacement with 16 white LEDs.

Inside I was expecting to find the capacitive dropper circuit, but
instead there is a 14 pin dil IC with it's number scrubbed off. It has
a small 130K resistor and a 160V 10u electrolytic attached and that's
it!

Just a chip, resistor and electrolytic to drive 16 LEDs directly from
240V AC mains!

What the hell is this chip????

 

[Adam Aglionby]

I just cracked open another type of mains powered LED lamp today. This
time it's a GU10 base MR16 sized replacement with 16 white LEDs.

Inside I was expecting to find the capacitive dropper circuit, but
instead there is a 14 pin dil IC with it's number scrubbed off. It has
a small 130K resistor and a 160V 10u electrolytic attached and that's
it!

Just a chip, resistor and electrolytic to drive 16 LEDs directly from
240V AC mains!

What the hell is this chip????

Not sure but I want one

Any more details on the LED lamp itself?
does it carry genuine looking approvals,CE, TUV, UL etc ?

Only people can think of that do HV LED drivers are Supertex and they aren`t
exactly a single chip and two discreets solution:

http://www.supertex.com/feature_flexfamily2.html

Added sci.electronics.design as crosspost see if anyone has any ideas there.

Adam

 

[Henry]

Harris and Micrel have high-voltage chips.

Adam Aglionby schrieb in Nachricht

 

[Clive Mitchell]

In message

Let's hope it a big time chopper with plenty of heat sink capacity.

It must be a complete rectifier and chopper in a chip, but there's no
heatsink at all. It's just a 14pin DIL chip, an electrolytic, resistor
and the 16 white LEDs in series for it to pass about 20mA through.

It's all on a single small round PCB that has the LEDs on one side and
the chip, resistor and cap on the other. It's glued into a standard
GU10 MR16 glass housing.

I haven't found anything close to this level of simplicity on the 'net
yet. Certainly not for taking 240V AC direct.

 

[Clive Mitchell]

Not sure but I want one

Any more details on the LED lamp itself?
does it carry genuine looking approvals,CE, TUV, UL etc ?

No data on the box other than the misleading text:-

SBO GU10 230VAC 16W

I presume the 16W means 16 White LEDs or the nearest perceived lamp
wattage for intensity (which is pretty good).

The manufacturer appears to be:-

ShenZhen Factory PRC

Only people can think of that do HV LED drivers are Supertex and they aren`t
exactly a single chip and two discreets solution:

I came across their site while searching for more data, but their stuff
seems to be modules.

 

[Victor Roberts]

It must be a complete rectifier and chopper in a chip, but there's no
heatsink at all. It's just a 14pin DIL chip, an electrolytic, resistor
and the 16 white LEDs in series for it to pass about 20mA through.

It's all on a single small round PCB that has the LEDs on one side and
the chip, resistor and cap on the other. It's glued into a standard
GU10 MR16 glass housing.

I haven't found anything close to this level of simplicity on the 'net
yet. Certainly not for taking 240V AC direct.

This is most likely not what you have, but it is a single chip
switching regulator designed to operate from a rectified 240 volt
power line.

http://193.115.250.10/onsemipdfs/MC33363-D.pdf

Have you put a scope on your lamp to confirm that the chip is a
switching regulator and then measured the switching frequency?

 

[Victor Roberts]

In message


It must be a complete rectifier and chopper in a chip, but there's no
heatsink at all. It's just a 14pin DIL chip, an electrolytic, resistor
and the 16 white LEDs in series for it to pass about 20mA through.

It's all on a single small round PCB that has the LEDs on one side and
the chip, resistor and cap on the other. It's glued into a standard
GU10 MR16 glass housing.

I haven't found anything close to this level of simplicity on the 'net
yet. Certainly not for taking 240V AC direct.

Here is another IC from the same company that is listed as a current
regulator and operates directly from a 250 VAC rectified line. (The
switching transistor is rated for 700 volts.)

http://www.onsemi.com/pub/Collateral/NCP1012-D.PDF

 

[Winfield Hill]

Adam Aglionby wrote...

Clive Mitchell wrote ...

Not sure but I want one

Not sure of the chip either, but let me describe one simple way it
might work. The 15 LEDs are surely in series, powered by 10 to 20mA
and taking 25 to 45V to operate, depending on their color.

First we can eliminate a common approach: rectify the 240Vac to get
340Vdc and add a 10mA current source. This would dissipate 3 to 6
watts and can be ruled out because it's too much for a plastic DIP.

But the IC must include an internal bridge rectifier to make dc, and
that the LEDs must operate between ac cycle peaks from the charge on
the 10uF capacitor. From dV/dt = I/C we see the capacitor's voltage
will drop 20V in 10ms from a 20mA current, which isn't a very large
drop. Now imagine a current source that's allowed at least 20V of
operating overhead, this would mean we need to charge the capacitor
to 40V more than the LED operating voltage each cycle. Let's assume
white LEDs with a LED-string voltage of 45V. Each ac cycle the 10uF
capacitor would need to be charged to 85V from its low value of 65V.

You can now see how the circuit likely works: use a say 400V MOSFET
to connect the capacitor-charging circuit only long enough to get it
up to 45V higher than the LED string, 85V in this case, and charge
fast enough to insure the ac line voltage hasn't gone much above 85V
when the charging is finished and the FET is shutoff. E.g., for a
150mA charging current, it would take only 1.3ms to charge the cap.

In this fashion, most of the circuitry in the IC works with voltages
under 90V, and only the rectifier and FET portions need high-voltage
ratings. The power dissipated in the chip should also be quite low.
I imaging this IC can work with any color LED, with various numbers
of LEDs in series, and at other ac line voltages, without adjustment.
I don't know the 130k resistor's purpose, perhaps to program the LED
current?

Clive, I suggest that you carefully use a multimeter to measure the
average dc voltage across the capacitor to give us some guidance.
You can also place the meter in series with the LEDs to measure the
average current, and then change the value of the 130k resistor to
see its effect.

Thanks,
- Win

whill_at_picovolt-dot-com

 

[Andrew Gabriel]

You can now see how the circuit likely works: use a say 400V MOSFET
to connect the capacitor-charging circuit only long enough to get it
up to 45V higher than the LED string, 85V in this case, and charge
fast enough to insure the ac line voltage hasn't gone much above 85V
when the charging is finished and the FET is shutoff. E.g., for a
150mA charging current, it would take only 1.3ms to charge the cap.

This is a similar idea to a circuit I built for running a US
120V 1050W frying pan off a 240V UK mains supply. I used a
pair of MOSFETs to chop out the middle of each half mains
cycle so the frying pan only got 120V RMS.

The interesting bit was working out the phase angle where I
had to switch the MOSFETs off and then back on again. Given
any particular phase angle, it was quite easy to work out the
RMS voltage, but working back the other way from a desired
RMS voltage to the phase angle left me with an unsolvable
piece of calculus (at least, unsolvable to me;-). Worked
backwards using iteration, and then verified by comparing
the intensity of 100W lamp on my circuit verses a genuine
120V AC supply.

OK, the power factor of the circuit is only 0.5, but at least
it's drawing power only from the portion of the sine wave which
most electronic devices don't use, rather than at the peaks.

 

[Winfield Hill]

Andrew Gabriel wrote...

This is a similar idea to a circuit I built for running a US
120V 1050W frying pan off a 240V UK mains supply. I used a
pair of MOSFETs to chop out the middle of each half mains
cycle so the frying pan only got 120V RMS.

The interesting bit was working out the phase angle where I
had to switch the MOSFETs off and then back on again. Given
any particular phase angle, it was quite easy to work out the
RMS voltage, but working back the other way from a desired
RMS voltage to the phase angle left me with an unsolvable
piece of calculus (at least, unsolvable to me;-). Worked
backwards using iteration, and then verified by comparing
the intensity of 100W lamp on my circuit verses a genuine
120V AC supply.

OK, the power factor of the circuit is only 0.5, but at least
it's drawing power only from the portion of the sine wave which
most electronic devices don't use, rather than at the peaks.

I was going to jump on you, saying such schemes may be OK at
20mA, etc., but for the ac-line's sake are better avoided at
high currents. However, your high current draw ignoring the
ac-peak region seems not so awfully bad. But still bad. :>)

Thanks,
- Win

whill_at_picovolt-dot-com

 

[Tony Williams]

But the IC must include an internal bridge rectifier to make dc,
and that the LEDs must operate between ac cycle peaks from the
charge on the 10uF capacitor. From dV/dt = I/C we see the
capacitor's voltage will drop 20V in 10ms from a 20mA current,
which isn't a very large drop. Now imagine a current source
that's allowed at least 20V of operating overhead, this would
mean we need to charge the capacitor to 40V more than the LED
operating voltage each cycle. Let's assume white LEDs with a
LED-string voltage of 45V. Each ac cycle the 10uF capacitor
would need to be charged to 85V from its low value of 65V.

A possible way of doing that would be with a Triac (or something)
that connects the fw bridge to the top of the cap at every
zero-crossing, with the Triac being shut off when the cap voltage
reaches 85V.

Continuing to use your numbers, the cap would start to take
current when the fw rectified sine reaches 45V, (at 11 deg),
and end at 65V, (which is at about 15 deg). Assuming that a
sine is a reasonable straight line in this region, then the
Volts/Second onto the cap is defined, and that automatically
defines it's current. About 0.44A for about 0.45mS.

Using the same numbers, a 65-85V charge of a 10uF cap is about
0.015J and 100 charges/sec would be 1.5 Watts into the LEDs.

If the 130k resistor sets the Vpk on the cap, then the value
of the cap sets the joules per half-cycle (more or less).
Quite a flexible and tidy little circuit.

 

[Tony Williams]

............. About 0.44A for about 0.45mS.

Try................... 0.86A for about 0.23mS.