3 ms·

Hmmm... makes sense. Is it possible to ELI5 what should be put after the PWM so it can handle low impedancies? From https://www.electronics-tutorials.ws/power/

Hmmm... makes sense. Is it possible to ELI5 what should be put after the PWM so it can handle low impedancies? From https://www.electronics-tutorials.ws/power/switch-mode-power-supply.html https://www.electronics-tutorials.ws/power/switch-mode-power... "The Buck Switching Regulator" looks like an arrangement with a diode, inductor and capacitor is enough. Interesting.

amelius 5y ago
The link you found explains it well, but there is a different perspective if you want to focus on impedances. The idea is to drive the capacitor with a high impedance. This impedance should not be a resistor, because a resistor will limit current (more current means a higher voltage drop in the output), further it will consume power and become hot. Instead, you should look for something that works more like an ideal current source. An ideal current source has an infinite output impedance (because loosely speaking the impedance R=dv/di and di=0 for an ideal current source). A constant current source is a good fit for charging a capacitor since a constant current means a constant dv/dt slope. Now, as your link shows, a switching power supply does not have a constant current source in its main circuit, but it does use an inductor which has similar characteristics since inductors resist changes in current. The governing equation of an inductor is v=L*di/dt, and for the output impedance of the inductor you get loosely speaking R=dv/di =L/dt, which is large for large L and short timescales (dt close to 0), hence the similarity to an ideal current source. Now the driving part of the circuit at the left of the schematics in your link is a voltage source. So to drive an inductor, you use a voltage source. And to drive a capacitor, you use a current source. Inductors can be seen as current sources, and capacitors can be seen as voltage sources. I hope this explains it.

amelius 5y ago

The link you found explains it well, but there is a different perspective if you want to focus on impedances. The idea is to drive the capacitor with a high impedance. This impedance should not be a resistor, because a resistor will limit current (more current means a higher voltage drop in the output), further it will consume power and become hot. Instead, you should look for something that works more like an ideal current source. An ideal current source has an infinite output impedance (because loosely speaking the impedance R=dv/di and di=0 for an ideal current source). A constant current source is a good fit for charging a capacitor since a constant current means a constant dv/dt slope. Now, as your link shows, a switching power supply does not have a constant current source in its main circuit, but it does use an inductor which has similar characteristics since inductors resist changes in current. The governing equation of an inductor is v=L*di/dt, and for the output impedance of the inductor you get loosely speaking R=dv/di =L/dt, which is large for large L and short timescales (dt close to 0), hence the similarity to an ideal current source. Now the driving part of the circuit at the left of the schematics in your link is a voltage source. So to drive an inductor, you use a voltage source. And to drive a capacitor, you use a current source. Inductors can be seen as current sources, and capacitors can be seen as voltage sources. I hope this explains it.