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What is the role of the BJT and capacitor C2 in the capacitive power supply circuit shown below?

Capacitive Power Supply Circuit

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  • \$\begingroup\$ The BJT is a current amplifier. The capacitor stores charge. What is the context? Is it taken directly from an exam you are taking in school? What have you tried? \$\endgroup\$
    – JYelton
    May 24 '19 at 19:36
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They are implementing a "charge pump" that increases the available output current. It's easier to see if you redraw the circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

When "Line" goes negative, C2 charges through D1 and D4. Q1 is cut off, as are D2 and D3.

When "Line" goes positive, it not only supplies current through D3 to the load, it also switches on Q1. This pulls the negative end of C2 up to the "Neutral" voltage, allowing it to discharge through D2 and supply additional current to the load.


Note that the same increase in available current could have been achieved with an ordinary 4-diode bridge rectifier. However, this circuit has the advantage that the negative side of the regulator remains directly tied to neutral.

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  • \$\begingroup\$ How does switching on of the transistor pulls -ve of capacitor to neutral? The collector itself is at neutral, so how does the collector current flow actually? \$\endgroup\$
    – G-aura-V
    May 25 '19 at 2:04
  • \$\begingroup\$ @JuneStar_2918: The collector current flows through the emitter into the negative (bottom) end of the capacitor. At the same time, current flows out of the positive end of the capacitor, through D2, to the load. This current flow discharges C2. Q1 is saturated, so its Vce drop is very small. You can think of it as a switch that shorts neutral directly to the negative end of the capacitor during the positive half-cycles. \$\endgroup\$
    – Dave Tweed
    May 25 '19 at 2:14
  • \$\begingroup\$ I can't actually figure out where the collector current comes from. \$\endgroup\$
    – G-aura-V
    May 25 '19 at 2:35
  • \$\begingroup\$ @JuneStar_2918: To complete the loop, the current flowing out of the positive end of C2 flows through the load, and back to the collector via the neutral line. \$\endgroup\$
    – Dave Tweed
    May 25 '19 at 4:02
  • \$\begingroup\$ @Dave Tweed: How is the negative side of the regulator not tied to neutral if we use a normal four diode bridge rectifier considering that this is a non isolated power supply. Also what is the advantage of the negative side of regulator directly tied to neutral. \$\endgroup\$
    – CS Thakur
    May 25 '19 at 19:38

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