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How does this circuit start up?
How does flux exist in a transformer?Does winding a choke inductor produce a transformer?How are these RC values calculated in this buck converter?How does this wall-wart switcher work?Modifying 12V PSU, how do I identify the +12V Return Sense?How to Model a Transformer in LTSpiceI need help to identify this old SMD deviceSMPS voltage drop in 230v ac to 5v dc flyback
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$begingroup$
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 connect 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?
power-supply transformer switch-mode-power-supply vcc
asked 8 hours ago
DIP4DIP4
184 bronze badges
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add a comment
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$begingroup$
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 connect 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?
power-supply transformer switch-mode-power-supply vcc
asked 8 hours ago
DIP4DIP4
184 bronze badges
New contributor
DIP4 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Check section High Voltage Start up on page 5 of the link you provided.
$endgroup$
– Huisman
8 hours ago
$begingroup$
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 ...".
$endgroup$
– DIP4
8 hours ago
add a comment
|
$begingroup$
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 connect 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?
power-supply transformer switch-mode-power-supply vcc
asked 8 hours ago
DIP4DIP4
184 bronze badges
New contributor
DIP4 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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 connect 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?
power-supply transformer switch-mode-power-supply vcc
power-supply transformer switch-mode-power-supply vcc
asked 8 hours ago
DIP4DIP4
184 bronze badges
New contributor
DIP4 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
DIP4DIP4
184 bronze badges
New contributor
DIP4 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 7 hours ago
DIP4
asked 8 hours ago
DIP4DIP4
184 bronze badges
New contributor
DIP4 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
DIP4DIP4
184 bronze badges
asked 8 hours ago
DIP4DIP4
184 bronze badges
184 bronze badges
New contributor
DIP4 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
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Check out our Code of Conduct.
$begingroup$
Check section High Voltage Start up on page 5 of the link you provided.
$endgroup$
– Huisman
8 hours ago
$begingroup$
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 ...".
$endgroup$
– DIP4
8 hours ago
add a comment
|
$begingroup$
Check section High Voltage Start up on page 5 of the link you provided.
$endgroup$
– Huisman
8 hours ago
$begingroup$
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 ...".
$endgroup$
– DIP4
8 hours ago
$begingroup$
Check section High Voltage Start up on page 5 of the link you provided.
$endgroup$
– Huisman
8 hours ago
$begingroup$
Check section High Voltage Start up on page 5 of the link you provided.
$endgroup$
– Huisman
8 hours ago
$begingroup$
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 ...".
$endgroup$
– DIP4
8 hours ago
$begingroup$
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 ...".
$endgroup$
– DIP4
8 hours ago
add a comment
|
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered 8 hours ago
Dave Tweed♦Dave Tweed
136k11 gold badges174 silver badges298 bronze badges
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add a comment
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$begingroup$
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.
answered 8 hours ago
G36G36
6,0491 gold badge5 silver badges11 bronze badges
$endgroup$
$begingroup$
+1 for referring to the datasheet... The datasheet saves all! :)
$endgroup$
– KingDuken
8 hours ago
add a comment
|
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2 Answers
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2 Answers
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$begingroup$
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.
answered 8 hours ago
Dave Tweed♦Dave Tweed
136k11 gold badges174 silver badges298 bronze badges
$endgroup$
add a comment
|
$begingroup$
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.
answered 8 hours ago
Dave Tweed♦Dave Tweed
136k11 gold badges174 silver badges298 bronze badges
$endgroup$
add a comment
|
$begingroup$
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.
answered 8 hours ago
Dave Tweed♦Dave Tweed
136k11 gold badges174 silver badges298 bronze badges
$endgroup$
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.
answered 8 hours ago
Dave Tweed♦Dave Tweed
136k11 gold badges174 silver badges298 bronze badges
edited 8 hours ago
answered 8 hours ago
Dave Tweed♦Dave Tweed
136k11 gold badges174 silver badges298 bronze badges
answered 8 hours ago
Dave Tweed♦Dave Tweed
136k11 gold badges174 silver badges298 bronze badges
answered 8 hours ago
Dave Tweed♦Dave Tweed
136k11 gold badges174 silver badges298 bronze badges
136k11 gold badges174 silver badges298 bronze badges
add a comment
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add a comment
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$begingroup$
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.
answered 8 hours ago
G36G36
6,0491 gold badge5 silver badges11 bronze badges
$endgroup$
$begingroup$
+1 for referring to the datasheet... The datasheet saves all! :)
$endgroup$
– KingDuken
8 hours ago
add a comment
|
$begingroup$
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.
answered 8 hours ago
G36G36
6,0491 gold badge5 silver badges11 bronze badges
$endgroup$
$begingroup$
+1 for referring to the datasheet... The datasheet saves all! :)
$endgroup$
– KingDuken
8 hours ago
add a comment
|
$begingroup$
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.
answered 8 hours ago
G36G36
6,0491 gold badge5 silver badges11 bronze badges
$endgroup$
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.
answered 8 hours ago
G36G36
6,0491 gold badge5 silver badges11 bronze badges
answered 8 hours ago
G36G36
6,0491 gold badge5 silver badges11 bronze badges
answered 8 hours ago
G36G36
6,0491 gold badge5 silver badges11 bronze badges
answered 8 hours ago
G36G36
6,0491 gold badge5 silver badges11 bronze badges
6,0491 gold badge5 silver badges11 bronze badges
$begingroup$
+1 for referring to the datasheet... The datasheet saves all! :)
$endgroup$
– KingDuken
8 hours ago
add a comment
|
$begingroup$
+1 for referring to the datasheet... The datasheet saves all! :)
$endgroup$
– KingDuken
8 hours ago
$begingroup$
+1 for referring to the datasheet... The datasheet saves all! :)
$endgroup$
– KingDuken
8 hours ago
$begingroup$
+1 for referring to the datasheet... The datasheet saves all! :)
$endgroup$
– KingDuken
8 hours ago
add a comment
|
DIP4 is a new contributor. Be nice, and check out our Code of Conduct.
DIP4 is a new contributor. Be nice, and check out our Code of Conduct.
DIP4 is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Check section High Voltage Start up on page 5 of the link you provided.
$endgroup$
– Huisman
8 hours ago
$begingroup$
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 ...".
$endgroup$
– DIP4
8 hours ago