Switching Operation of ICL8201 (Buck Controller)

Hi,
I am working on this Infineon IC ICL8201, which is an AC/DC Buck Controller with PFC for LED Lamps.
Here is a ref design for the IC.



I quote from the datasheet, "art 4.2.3. In it's normal regulation operation, power MOSFET is turned on by valley detection through DRAIN pin, is turned off by constant on time."

I mean HOW? This MOSFET switching part is not clear to me.
What is the function of Diode D2 here?

Please do explain it, I am kinda newbie here! Thanks.

** Datasheet here -
View attachment Infineon-ICL8201-DS-v02_01-EN.pdf

there is a timer which switches the mosfet off after that time has elapsed.
The drain pin looks at the voltage there and when it is low it turns the fet on again.
The advantage is that you are always in DCM but not too deeply.
Also you can know your average led current is peak/2, so you choose the right sense resistor, and you get th exact current you want.
When the inductor has discharged completely, the diode goes off, and the voltage swings down naturally, as there is an LC cct there, partky involving parasitics.

treez said:

Also you can know your average led current is peak/2, so you choose the right sense resistor, and you get th exact current you want.

Click to expand...

So you were saying sense resistor is the only way I can change the O/P current rating of the entire circuit?

When the inductor has discharged completely, the diode goes off, and the voltage swings down naturally, as there is an LC cct there, partky involving parasitics.

Click to expand...

You meant diode D2 from the circuit which is connected between external power mosfet gate and Vcc pin, right?

So you were saying sense resistor is the only way I can change the O/P current rating of the entire circuit?

Click to expand...

yes

You meant diode D2 from the circuit which is connected between external power mosfet gate and Vcc pin, right?

Click to expand...

No i meant the D1, the power diode of the buck.

The LC "swinging" of the drain voltage is a common thing in SMPS, and in this case, they are making use of it..to know when to turn the fet back on. That way they stay in Boundary conduction mode, right at the inbetween of ccm and dcm

treez said:

No i meant the D1, the power diode of the buck.

Click to expand...

I almost had an idea about diode D1, but thanks for clearing this up more!

But the diode "D2" is making all the confusion, I still have no idea why it's there and how it's working!

Moreover, in previous reply u said "timer circuit which switches the mosfet off"; now from a simple point of view, without gate voltage (LOW level input voltage) mosfet goes off, right?
And this IC had it's only connection with mosfet gate via diode D2 which is directly connected to PIN Vcc.
So maybe is there any possibility that IC sending pulse or something via PIN Vcc and Diode D2; that causing gate voltage to go up which will eventually trigger zener diode ZD1 (as Vzener = 12V) and discharge the capacitor C3 through it, which will result in a LOW level input in mosfet gate?

diode d2 just helps keep the fet q1 on all the time.
Q1 is not the switching fet....Q1 is just the the FET that allows the chip to start up without needing a high voltage start up circuit inside the chip, as that takes up much room in there.
The switching fet is the small one you see in the IC.....that fet q1 is just on all the time. Some people call it a common base kind of configuration.

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sorry , let ,me adjust my words, fet q1 is only on when its VGS is >5v...SO obviously its not ON when the diode D1 conducts.

Ahh all this time I was thinking wrong (lack of understanding u may say)!! Thanks man.

c4 looks like it has a joint purpose of being the resonating cap for the buck inductor to resonate with after d1 has turned off...but also it cunningly acts as a charge pump kind of thing to keep the IC supplied.

The cct relies on the resonance between the buck inductor and c4 to resonate down when d1 turns off, and then the low voltage there is detected (by the drain pin), and that way the IC knows that the inductor has discharged, and so it knows that it must turn the fet on again.

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incidentally, d3 is there to stop charge getting sucked out of the vcc capacitor when the source of q1 flys upwards...as you know, when the drain of q1 flys upwards, the source goes with it because they are connected by a capacitor.

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d2 blocks the discharge of c3, so that the fet q1 is kept on...you dont want c3 to supply the ic , because then c3 would discharge....also, you dont really want chip power to come through R1A r2a and r2c, (as they woudl overpower and heat up) so you have the diode d2 to block current going thru those three resistors to the chip.

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actually some of the supply current for the chip will come through q1 just after the internal fet turns off.

treez said:

sorry , let ,me adjust my words, fet q1 is only on when its VGS is >5v...SO obviously its not ON when the diode D1 conducts.

Click to expand...

Now from basic buck operation, Diode D1 conducts only when mosfet switch goes off, right? Thats the question if this is "THE switching FET" than how it's Vgs leveling up or down keeping the switching operation going? Not to forget, there is a capacitor C3 always charged up (as I find no discharging path) via 3 series resistor, which is clamped by a zener.

ok i know what you mean cuzz thats where i was going wrong at first sight.....well....what is situation for fet turn on or off.?...it is VGS>5V........so you are thinking fet can only go off if the gate voltage goes down....but also, it can also go off if the source voltage goes up! ...above the gate voltage, and that is what is happening here.

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d2 also allows excess vcc voltage to get burned up in the zener...since the charge pump is a bit 'hit and miss' at supplying the chip

here is an ltspice simualtion of the flyback version of your converter. (ltspice is free)

@treez It took me a while to understand the whole operation, but a big thanks to you!

i must admit it looks a cunning chip, supplying itself efficiently by that charge pump, and using the ringing of the drain node to detect when the inductor has just discharged and its ready to turn on again...without needing an extra winding to do that, just a plain inductor. The external transistor stands off the high voltage, meaning the chip can be cheaply made from low voltage process, and the chip is less likely to overheat.

if suppose I am connecting 18 LED's in series and current is 150mA, So in the above reference design only change I need to do is changing the sense resistor.

Led Voltage is 3.3V

Please Help