# Why is the output of a boost converter (shown on many waveforms) square like across a real load, when the current is triangular?

I am trying to learn about boost converters, but I don't understand why the images of the voltage output always seems to be like a square wave PWM.

I understand that 1) (assuming perfect inductors) the current wave form on the charge state of the inductor would lead to a linear increase in current flow and constant voltage across. And that 2) The moment the switch is open, the voltage at Vl would be Vi-Vo. However its then part right after, from looking at the current flow at this point..

It looks asif the curent decreases at a linear rate meaning from my understanding it must mean the voltage across the inductor is constant (agreeing with the first image). However i dont see how this agrees with kitchoffs law on how the voltage in a loop equals 0. If the inductor and supply voltage is assumed to be constant, but there is a linear decrease in current, then if the load is resistive surley this means the output voltage changes proportionaly (V = IR), but that would lead to a non 0 loop voltage?

Apologies is all this is wrong as its my first time tring to understand this.

• You seem to have misinterpreted the plots; they aren't saying Vo is square, they're saying Va(t) is square with the high level labeled as Vo. Vo is ~constant, as a large enough C is used to make change in Vo much smaller than Vi or Vo-Vi. Commented Sep 21, 2023 at 15:03
• Thanksyou, so how will the voltages at Vo and Va be diferent ( asuming diode was ideal)? Commented Sep 21, 2023 at 15:16

• I think you mean $V_L(t)$ don't you? Enacted by this: $V_L(t)$ <-- the output capacitor (so necessary) keeps the inductor voltage constant during the output charge phase i.e. it is big enough so that changes in $V_L(t)$ are small enough to ignore. Commented Sep 21, 2023 at 17:22
• Yes. and Ahh i see (i think), so without the capacitor but with a load, the voltage $V_L(t)$ would also exponetially decay just like normally discharging over a resistive load? Commented Sep 21, 2023 at 18:12