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I'm trying to understand how CI HT7A6322 starts up.

Relevant part of the datasheet says:

Control circuit power supply. Also provides a charging current during start up due to a high voltage current source connected to SW. For this purpose, a hysteresis comparator monitors the VCC voltage and provides two thresholds:

VCCON: Voltage value (typically 14.5V) at which the device starts switching and turns off the start up current source.

VCCOFF: Voltage value (typically 8V) at which the device stops switching and turns on the start up current source

When the circuit is connected to mains the internal current source is ON state which is connected together with SW pin.

This current source in winding A induces a voltage in winding B powering the VCC pin. Is my explanation correct?

enter image description here

DIP4
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    Check section *High Voltage Start up* on page 5 of the link you provided. – Huisman Sep 19 '19 at 19:45
  • I get it. The trick part is ".. which will charge the VCC pin ... ". I think the correct one should be " ... which will charge **external capacitor** connected in VCC pin ...". – DIP4 Sep 19 '19 at 20:00

2 Answers2

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Normally we are using the start-up resistor. But in this case, there is an internal MOSFET between SW and VCC pin. And this MOSFET will provide a start-up current.

enter image description here

G36
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At the top of the block diagram on page 2, there's a MOSFET that functions as a current source connected between SW and VCC that can withstand the full line voltage. It passes a tiny current (via winding "A" of the transformer) that doesn't cause too much dissipation inside the device. This current is not sufficient for operation of the switching regulator, however.

This current source charges the capacitor attached to VCC until it reaches Vccon, at which point, the switching begins (powered temporarily by the charge on the capacitor) and larger amounts of power can be transferred via the auxiliary winding "B" on the transformer to keep things going continuously.

Dave Tweed
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