Buck converter (PWM)

WHAT YOU ARE LOOKING AT

A synchronous buck converter, simulated the most literal way possible: step through time, work out which switch is on, integrate the inductor current, integrate the output voltage, repeat. No averaged model, no small-signal approximation — just the two state equations

V_o = (1/C) ∫ (I_L − I_o) dt I_L = (1/L) ∫ V_x dt

written as a loop. The output voltage settles near VDDH·D, and the reason is almost embarrassingly simple: in steady state the average voltage across an inductor must be zero, so the switch node's average — which is VDDH for a fraction D of the time and zero otherwise — has to equal V_o.

Two things the averaged model in a textbook will not show you, and this one will. The startup transient: the L and C form a resonant tank with almost nothing damping it, so switching on into a discharged output produces an inductor current far larger than anything in steady state. And the settling time: that tank rings for a long time, so an average taken too early is not an average of anything.

THINGS WORTH TRYING

Drag the duty cycle slowly across its range. Every value you stop at gets simulated and dropped onto the two bottom-right panels as a red point, so the curve draws itself as you explore. Compare where the points land against VDDH·D, and against VDDH·D − R_s·I_o — the second line is the one that fits, and the gap between them is the switch resistance.
Untick start from steady state and set t_end to its minimum, around 10 µs. This is what the notebook used to do, and the efficiency readout goes negative while the state readout says the averages are unreliable. That is not a bug: the tank Q here is around a thousand, the inductor current is still swinging when the second half of the run begins, and averaging over a ring is not averaging over anything. Wind t_end up and watch the numbers become sensible.
Still cold-starting, compare the peak inductor current on the first panel against the load current. Switching on into a discharged output draws hundreds of milliamps where the converter only ever needs one — the notebook's warning that "these high currents can kill our device". Then raise V_o initial towards the target and watch the overshoot go away. That is what soft-start is for.
Look at ripple I_L against I_L / I_o. With a 1 µH inductor at 10 MHz the ripple current is tens of times the DC load current, so the inductor current swings hard negative every cycle. The converter works anyway because the capacitor absorbs it — but the switches carry all of it, which is why the efficiency is what it is.
Raise R_s. V_o sags below VDDH·D by R_s·I_o and the efficiency falls, because that is where the power is going. Then lower the R load to draw more current and watch the sag grow in proportion.
Change L and C and watch only the ripple change, not the average. Ripple current goes as 1/L and ripple voltage as 1/C, but the DC output is set purely by the duty cycle. This is why picking the inductor is a ripple-and-size question rather than a voltage question.

THE PYTHON

for i in range(1,N):
    ts = t[i]

    # Model switch
    if(ts % T < dtc*T):
        pmos = 1
    else:
        pmos = 0

    dt = (t[i]-t[i-1])

    # Current voltage across the inductor
    vx[i] = pmos*(VDDH) - Rs*ix[i-1] - vo[i-1]

    # Current in inductor, trapezoidal approximation of area
    ix[i] = ix[i-1] + 1/L * (vx[i] + vx[i-1])/2*dt

    io[i] = 1/R * vo[i-1]
    vo[i] = vo[i-1] + 1/C * ((ix[i] + ix[i-1])/2 - (io[i] + io[i-1])/2)*dt

Source: jupyter/buck.ipynb. The integration is the notebook's, verbatim. Building this page turned up a real problem in that notebook, since fixed: it averaged over the second half of a 10 µs run while R·C is 1 ms, so the tank was still ringing and it printed a negative efficiency. Its settled mode now starts at the operating point and runs long enough to mean something. You can still reproduce the old behaviour here by unticking the warm start and shortening t_end. For the same power stage under a completely different control scheme, see the PFM buck.