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This question already has an answer here:

Can someone explain me the working of this boost converter and the function of individual components. Please. I am unable to understand.

Why are the transistors connected to the MOSFET. What purpose do they serve?

Can someone explain me one cycle of the output current path, please?

Boost Converter

The data sheet for IC1 can be found here

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marked as duplicate by PeterJ, Dave Tweed Jan 24 at 12:43

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • \$\begingroup\$ From the data sheet you linked: "The driver IC2 is ON Semiconductor low cost dual NPN/PNP transistor BC846BPD. Its NPN transistor is connected as a super diode for charging the gate capacitance. The PNP transistor works as an emitter follower for discharging the gate capacitor. This configuration assures sharp driving edge between 50 − 100 ns as well as it limits power consumption of R7/R8 divider down to 50 mW" \$\endgroup\$ – Colin Jan 24 at 12:05
  • \$\begingroup\$ Yes I read that in the datasheet but unable to understand the working for one cycle. how does the output current flow and the working of the IC. please. I am really confused \$\endgroup\$ – Electronic_Maniac Jan 24 at 12:09
  • \$\begingroup\$ Why do we need a voltage divider of R7/R8? And why do we have the transistor set for charging the gate capacitance of the MOSFET? MOSFET requires only gate voltage to turn on and off , right? why do we care about current? why the resistor divider and how does the current flow at the output? please explain. i am confused \$\endgroup\$ – Electronic_Maniac Jan 24 at 12:12
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    \$\begingroup\$ If you have additional information related to your original question, use the "edit" button to add the information there. Do not create a duplicate question. \$\endgroup\$ – Dave Tweed Jan 24 at 12:43
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    \$\begingroup\$ This isn't a duplicate. It is about the same circuit, but this is "how does it work" and the other is "how to estimate the needed input current for a specific output voltage and current." \$\endgroup\$ – JRE Jan 24 at 12:50
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There are two parts to the cycle for a boost converter, when the SWE output is high, the FET is 'on' and current is able to flow through the inductor, and mosfet to ground, this builds a magnetic field around the inductor.

When the SWE output goes low the FET is 'off', current can no longer flow through the inductor to ground, the magnetic field collapses and a voltage is induced which forward biases the diode D1 and charges the capacitors.

R5 and R4 create a potential divider which is used to set the voltage at COMP, the IC will then decide whether to have a longer or shorter on time (duty cycle) to create the correct output voltage. It creates a feedback loop.

Q1 has gate capacitance, the PNP transistor is used to quickly discharge that to achieve a sharp turn off, the NPN is configured as a diode.

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  • \$\begingroup\$ Thank you. This really helped. But what's the use of resistor divider R7 and R8? And why do we need two transistors in this fashion? \$\endgroup\$ – Electronic_Maniac Jan 24 at 12:44
  • \$\begingroup\$ The converter IC has an NPN output which SWE is the emitter of, it's unable to sink current, R8 provides a path for the base current for the PNP to turn it on and discharge Q1. The NPN part of the dual transistor acts as a diode and prevents a positve voltage appearing at SWE when it's low. \$\endgroup\$ – Colin Jan 24 at 12:55
  • \$\begingroup\$ Thank yyou very much. In the datasheet of NCV3063, how to calculate the Vswce? It is not mentioned in the datasheet. Please help \$\endgroup\$ – Electronic_Maniac Jan 25 at 9:59

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