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More efficient SMPS IC?

for a long time I wondered how my PSU switches from 240vac down to Xvolts.

So being ignorant as i can be, I just assumed, if you want 12vdc, there was some kind of IC that would deliver voltage only when the AC voltage (at50/60hz) between 1vac and 12vac meaning you only draw current for a small percentage of time..

but if that was true how do you power the IC? and if that method ever was to be implemented could it work? with as much or more efficiency?

excuse my ignorance - thanks.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
I presume the PSU you've got isn't one with a transformer and a rectifier in it. You're probably asking about a switchmode power supply.

Remember that other thread about a step-up DC-DC converter? Well, (and pretty surprisingly) the power supply you're probably talking about is also a DC-DC converter.

But wait (you say) isn't the mains AC?

Well, yes, it is. But the first thing your PSU does is rectifies it to produce something like 400V DC.

Then all it does is use a switchmode DC-DC converter to reduce the voltage.

BUT simple DC-DC converters have a DC path from the input to the output, and this would be very dangerous for a power supply running from the mains. So what is done in this case is that the oscillator switches current not to a simple inductor, but to a transformer. The output of this is rectified and becomes your 12V DC.

So you have an AC to DC to AC to AC to DC converter! Why go through all of this? Two reasons:

1) the transformer used operates at a much higher frequency so it needs fewer turns of wire and is thus smaller and lighter and (most importantly) cheaper.

2) Instead of having to regulate the voltage by dissipating heat, the oscillator can be varied to allow the current to flow into the primary for more or less time, thus controlling the amount of power. Feedback from the rectified output is what is used to control this.

3) (did I say 2 reasons?) Because of the high frequency the output filtering can be done with smaller (and thus lighter and cheaper) capacitors.

Switching power on and off is essentially lossless, so the main dissipation often comes from the rectifier diodes on the output. The switching transistor(s) often have smaller heatsinks.

Power to the IC that controlls this can come from one of several sources.

The first is a separate low power DC power supply.

The second (and this sounds similar to the first) is a very simple low current DC source.

The third is from the output of the regulator itself.

The first one is easy to understand, and it's also somewhat rare.

The second isn't even capable of providing enough power to run the IC, and you might think it's no use at all.

The third just sounds impossible because there is no power to start the regulator.

Well, the second and third are used together. The low current that is leaked from the high voltage DC supply is used to charge a capacitor. When the voltage across the capacitor is high enough there is enough power for the regulator to pulse the DC supply once or twice. As it pulses the DC source, the output becomes active and the power from a special output can charge that capacitor to keep the regulator running. It pulls itself up by its bootstraps.
 
um, ok lol.. thanks for that but.

let's presume 0.7v voltage drop per diode, what if, 180 diodes in series (or however many more it would take) to get the voltage low enough to around 3 - 5c for an IC chip to operate, the IC then is able to power up, to start switching, the 240v gets turned to 240dc then switched down to 12,9,5,3 etc.... through your typical DC > DC switch mode?...

would that work?
 
Probably not

I agree, mains voltage is not consistent across the board or all the time so trying to find the sweet spot number of diodes is a crap shot...

In the US it regularly swings from about 110v to 120v depending on location, usually close to the 110v mark as the electric company isn't going to give you extra for free if they don't have to, but none the less it does swing and is not consistent...
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
If you look at diodes as ideal (with a constant voltage drop) then the mains voltage varying from a peak of 1.414*110 to 1.414*120 would result in a change in the voltage at the other end of the string of diodes from x to x+14 volts.

Worse, since the voltage drop is dependant on current, the voltage would tend to creep upward if the load was small.
 
Not to mention that diodes are no more efficient at dropping a voltage than a resistor is. You can produce a low voltage supply from mains voltages by using a capacitor to drop the voltage, which is more efficient since the capacitor dissapates no power.

Bob
 
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