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This may sound a nonsense question for some of you but I was trying to visualize how an SMPS initializes. Below is a block-diagram taken from a particular SMPS datasheet: http://docs-europe.electrocomponents.com/webdocs/0e77/0900766b80e777c4.pdf

enter image description here

And before posing my question, I would like to make the assumption that all the electronic circuitry in an SMPS is DC powered by the SMPS’s DC output itself. So I assume there is no internal battery or such sort of thing.

If the above statement is true, here is my confusion as follows:

Let’s say the SMPS is switched on, and we started to observe what is happening inside the circuit at an extremely slow motion. My logic says that first the AC input power is delivered to the rectifiers and then rectified to a DC voltage. So far so good.

But just after that moment, the electronic circuitry should start chopping the DC voltage with PWM to provide a high frequency AC for the transformer. But for that to happen the electronic circuitry should already been supplied with a DC voltage which were not created yet at the output.

How initially the electronic circuitry gets DC voltage is very peculiar and confusing for me.

Does anyone have an idea?

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  • \$\begingroup\$ uhm, rectifying (and filtering) the incoming AC is kinda creating the DC needed to run the control part of your diagram... \$\endgroup\$
    – PlasmaHH
    Apr 20, 2016 at 13:43
  • \$\begingroup\$ so you are saying the electronics gets the DC from that point? But it is very high voltage DC for the electronics isnt it? \$\endgroup\$
    – user16307
    Apr 20, 2016 at 13:44
  • \$\begingroup\$ that depends entirely on your smps construction \$\endgroup\$
    – PlasmaHH
    Apr 20, 2016 at 13:47
  • \$\begingroup\$ Some have linear regulators to startup, others use a feed from another controlled supply. \$\endgroup\$
    – DKNguyen
    May 6, 2019 at 15:35

2 Answers 2

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before posing my question, I would like to make the assumption that all the electronic circuitry in an SMPS is DC powered by the SMPS’s DC output itself

An SMPS uses quite often a tertiary winding to obtain power for itself. The initial switching must have begun for this to work though so there is a bleed resistor from the raw dc that provides power at start up: -

enter image description here

Note the tertiary winding and the resistor feeding pin "V".

Very low power devices steal a bit of power down the fly-back back-emf protection devices or through the primary winding.

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The start-up sequence is simple: when you plug the power supply in the ac outlet, the mains is rectified by the diode bridge and the so-called bulk capacitor. The dc voltage you have at this moment equals \$V_{in,rms}\sqrt2\$ if we neglect the diodes drops. If you power the converter from a 100-V rms input then you have around 141 V over the bulk capacitor. A high-value resistor is then connected from the high-voltage (HV) dc rail and feeds the integrated circuit (the controller) \$V_{cc}\$ pin. A capacitor is also connected from that pin to ground. Because of the bias provided by the resistor connected to the HV rail, the \$V_{cc}\$ pin voltage starts rising. Neglecting the current absorbed by the controller as \$V_{cc}\$ goes up, all the start-up current feeds the \$V_{cc}\$ capacitor.

As some point, the \$V_{cc}\$ voltage is high enough (e.g. 15 V) and the controller decides that it can start operating. The controller delivers driving pulses to the power MOSFET and its consumption suddenly increases on the \$V_{cc}\$ pin. Because the current delivered by the start-up resistor is purposely kept small (to limit power dissipation), the \$V_{cc}\$ capacitor remains alone and supplies the controller: the \$V_{cc}\$ pin voltage drops as the controller absorbs current. The drop amplitude depends on the capacitor size and the controller consumption. As the power MOSFET switches, an auxiliary winding connected to the \$V_{cc}\$ pin now feeds the \$V_{cc}\$ capacitor. That auxiliary winding voltage is a scaled image of \$V_{out}\$ the converter output voltage. As \$V_{out}\$ is rising up, so is the rectified auxiliary voltage and the voltage at the \$V_{cc}\$ pin will soon be enough to self-supply the controller. When the output voltage is regulated, the auxiliary voltage also stabilizes.

Please note that the start-up resistor is still connected to the HV rail after the power supply has started. This creates extra dissipation and affects efficiency. To go around that, some manufacturers have access to a high-voltage technology and build an internal high-voltage current source which feeds the \$V_{cc}\$ capacitor at power up and then turns it off once the power supply operates.

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    \$\begingroup\$ Regarding your last paragraph, how does this high voltage present at the Vcc pin of the controller chip not damage it? Since it is now drawing its current from the auxiliary winding, the current through the startup resistor must be minimal, which means the voltage drop in it is minimal, which means a huge voltage must be present on Vcc. \$\endgroup\$
    – Pouria P
    Feb 9, 2022 at 13:06
  • \$\begingroup\$ The auxiliary winding is coupled to the power winding in a flyback converter. So even if you don't draw current from the \$V_{cc}\$ pin, the voltage is maintained at a reasonable level, below the maximum rating of the controller. Variations however exist linked to the leakage inductance for instance at high current or too low a voltage in no-load conditions. This is the designer work to make sure the voltage does not run away or sag in these conditions. \$\endgroup\$ Feb 9, 2022 at 17:16
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    \$\begingroup\$ I didn't quite understand that. There is a diode after the auxiliary winding so no current can flow into it. \$\endgroup\$
    – Pouria P
    Feb 9, 2022 at 21:00

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