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I'm having a hard time explaining to myself why the output voltage of a buck regulator drops at the load when the load pulls more current than the max limit of the buck regulator.

Can someone help me understand this by assuming let's say I have a 3.3V buck regulator circuit that is rated for up to 2A. What happens to the buck topology components' voltages/currents when the load tries to pull 3A instead that causes the voltage output of regulator to drop?

Does same thing happen if I use a linear regulator instead, meaning output voltage drop during overloading?

Thanks, Dan

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  • \$\begingroup\$ Everything drops in output voltage when the current gets too high. If it did not you would have a perpetual motion machine that produces energy from nothing. \$\endgroup\$
    – DKNguyen
    Commented Apr 27, 2020 at 18:34
  • \$\begingroup\$ If you pull paper too hard it rips. Something gives either catastrophically or benignly. What happens when you car wheels cannot supply the torque to drive up a hill? \$\endgroup\$
    – Andy aka
    Commented Apr 27, 2020 at 18:36
  • \$\begingroup\$ Clearly explained it guys. Thanks. \$\endgroup\$
    – Dan_6262
    Commented Apr 27, 2020 at 18:45

2 Answers 2

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Does same thing happen if I use a linear regulator instead, meaning output voltage drop during overloading?

Everything drops in output voltage when the current gets too high. If it did not you would have a perpetual motion machine that produces energy from nothing.

Can someone help me understand this by assuming let's say I have a 3.3V buck regulator circuit that is rated for up to 2A. What happens to the buck topology components' voltages/currents when the load tries to pull 3A instead that causes the voltage output of regulator to drop?

  • Inductors saturate when the current gets too high and they stop doing their inductive thing
  • Transistor switches saturate and can't pass more current to the output.
  • Any and all resistances in the current path will produce increasing voltage drops as current increases to the point where they output voltage as dropped intolerably low.
  • The capacitor is too small to supply enough charge to "tide things over" for the frequency at which the converter is operating at; The frequency being how often that capacitor is topped up.
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  • \$\begingroup\$ That perfectly explains it, thanks DKNguyen. \$\endgroup\$
    – Dan_6262
    Commented Apr 27, 2020 at 18:44
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mainly the inductor can only store so much energy, for a given input voltage and a given ON_TIME.

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