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A boost converter is a circuit that uses power from a voltage source, to act as another voltage source with a greater voltage. To conserve power, it draws more current from the input than it provides to the output. A boost converter looks like this:

(image from Wikipedia)

The general idea is to charge the inductor up (with the switch closed) then discharge it through the load, in series with the power supply. The capacitor is for smoothing the output voltage, and the diode allows the capacitor and load to retain a voltage while the switch is on (it prevents the capacitor from discharging through the switch).

 

The following circuit uses power from a current source, to act as another current source with a greater current. To conserve power, it "draws" more voltage across the input than it provides across the output.

(shown in LTspice IV) Here Q1 and I2 represent the switching element - I2 is a pulsed current source.

The general idea is to charge the capacitor up (with the switch open) then discharge it through the load, in parallel with the power supply. The inductor is for smoothing the output current, and the diode allows current to continue circulating through the inductor and load while the switch is off (it prevents the inductor from discharging through the switch, and damaging the switch).

What is this circuit called?

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  • \$\begingroup\$ Wild guess: current-fed buck converter. \$\endgroup\$ – Nick Alexeev Feb 8 '16 at 1:51
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That's a buck converter. I'm not quite sure if this question is in jest, but, to be more precise, Nick Alexeev is correct: it's a current-fed buck converter.

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  • \$\begingroup\$ ... So it is. No, it was not a trick question. Interesting that the only changes require to make it work from a constant-current source is the addition of that capacitor on the left. \$\endgroup\$ – immibis Feb 8 '16 at 4:11
  • \$\begingroup\$ @immibis, That current source (I2) is a switching current source \$\endgroup\$ – Dave Feb 8 '16 at 5:19
  • \$\begingroup\$ Indeed. I already said that I2 represents the controller, which is mostly irrelevant (and so I abstracted it as a pulsed current source). \$\endgroup\$ – immibis Feb 8 '16 at 6:10
  • \$\begingroup\$ The interesting facts are that the input is a current source, and that the output also behaves as a current source. \$\endgroup\$ – immibis Feb 8 '16 at 6:12
  • \$\begingroup\$ @immibis, Ah, I see what you're saying. It's not that it needs a current source per say so much as a source of current. I.E. You could replace that current source I1 with a voltage source and then that diagram would still be true. It's displayed as a current source to show that no matter what voltage is produced by I1, it will always be lower on the load and the power transfer characteristic is how you determine current draw. Also, that input capacitor is used in every switching design I've ever seen, including boost, so I don't know why you found that interesting. \$\endgroup\$ – Dave Feb 8 '16 at 9:46

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