Why a Resistor Divider Fails

Part 2 · Voltage Regulation

Key point: a series resistor can hit 3.3 V at one current. A load step moves the settled voltage by \(IR\), and the resistor already burns more power than the load.

\(R = (12-3.3)/0.1 = 87~\Omega\). Window is 3.3 V ±5%, so about ±2 mA.

1
One current looks fine.
ΔI = 0. The rail sits at 3.3 V, inside the band. At that same point the load gets 330 mW and the resistor burns 870 mW.
2
Two milliamps leaves the window.
102 mA settles at 3.13 V. 98 mA settles at 3.47 V. The window is 3.14 V to 3.46 V, so both plateaus are out.
3
The capacitor only slows the miss.
±10 mA. 110 mA settles at 2.43 V, 90 mA at 4.17 V. τ is about 870 µs, so the edge is slow, and the plateau is still wrong.
4
Flat and in band still wastes the power.
Back to 100 mA. The trace is legal. \(P_R\) is still 870 mW against 330 mW in the load. That is the dropper, not the step.

Stiff 12 V, series \(R\), 10 µF, and a current load. Settled voltage is \(12 - IR\). Basso, Switch-Mode Power Supplies (2008), on converting power with a resistor. The .asc matches this ΔI.

Series dropper

12 V through 87 Ω · 10 µF · 100 mA, then up, back, and down

Load step ΔI (mA around 100)

12 V source, 87 ohm series resistor, 10 µF, and a 100 mA load

\(V_{\mathrm{OUT}}\) · blue in the ±5% band · red outside

\(I_{\mathrm{LOAD}}\) · 100 mA, then up, back, and down

Status—

\(R_{\mathrm{series}}\)—

Settled plateaus—

\(\tau = RC\)—

\(P_{\mathrm{load}}\) @ 100 mA—

\(P_R\) @ 100 mA—

\(R=(V_{\mathrm{IN}}-V_{\mathrm{nom}})/I_0\)

→ —

\(V_{\mathrm{ss}}=V_{\mathrm{IN}}-IR\)

→ —

\(P_{\mathrm{load}}=V_{\mathrm{nom}} I_0\), \(P_R=(V_{\mathrm{IN}}-V_{\mathrm{nom}})I_0\)

→ —