Why switch — Voltage Regulation

Along the way

1
Part 2 was the dropper. Series R (or a FET in triode used as a resistor) hits one voltage at one current and dumps the rest as heat. That is the linear tax.
2
Linear loss is the difference times current. \(P_{\mathrm{lin}}=(V_{\mathrm{in}}-V_{\mathrm{out}})I_{\mathrm{load}}\). Efficiency is just \(V_{\mathrm{out}}/V_{\mathrm{in}}\) for an ideal LDO. 12 V → 3.3 V is ~27% even if the silicon is perfect.
3
Switching spends Part 4 packets. Store energy in L (and C), dump it to the load, repeat. Ideal switches and inductors do not eat the \((V_{\mathrm{in}}-V_{\mathrm{out}})I\) term. Real converters still have \(R_{\mathrm{DS(on)}}\), gate charge, magnetics — later Part 18.
4
Duty is the ratio knob. For a buck, \(D=V_{\mathrm{out}}/V_{\mathrm{in}}\) in CCM. You are not “turning the FET into a resistor”; you are choosing what fraction of the period VIN is connected.
5
Topology is which job and which port is noisy. Next pages: buck, boost, inverting buck-boost, SEPIC, Ćuk, Zeta, flyback, forward — each as when / operation / conducted ports. Noise is page 3, not the pitch.

Linear still wins for tiny \(\Delta V\), tiny current, or when switching ripple is unacceptable (post-reg after a switcher). This page is why we bother with the rest of the family.

What we ignored: switching loss (Part 5d), min on-time, magnetics, EMI as a system. Control of the loop is Control Loops, not this series.

Linear heat vs a switcher

Same VIN, VOUT, load · left bar = linear dissipation · right = switcher at the efficiency knob

\(V_{\mathrm{IN}}\) (V) \(V_{\mathrm{OUT}}\) (V) \(I_{\mathrm{load}}\) (A) Switcher efficiency (%)

If VOUT ≥ VIN the linear bar is a fictional step-up (an LDO cannot boost). The switcher bar still assumes some topology can do the job — pick it on the next pages.

Power · load (teal) vs heat (rose = linear, gold = switcher)

P_load—

P_lin heat—

η linear—

P_sw heat—

Buck D—

\[ P_{\mathrm{load}} = V_{\mathrm{out}} I \]

→ —

\[ P_{\mathrm{lin}} = (V_{\mathrm{in}}-V_{\mathrm{out}})I \]

→ —

\[ D_{\mathrm{buck}} = V_{\mathrm{out}}/V_{\mathrm{in}} \]

→ —

Family map · next seven clusters

#	Cluster	Job	Quiet port
12	Buck	always down	output
13	Boost	always up	input
14	Inverting buck-boost	negative rail	neither
15	SEPIC	up or down	input
16	Ćuk	negative, quiet cables	both
17	Zeta	up or down	output
18	Flyback	isolated	neither

Try it: 12 V → 3.3 V @ 2 A — linear heat is huge, switcher heat is small. Walk VOUT up toward VIN: linear becomes reasonable (that is when an LDO still wins). Drop efficiency to 80% — the switcher bar grows, but it is still not \((V_{\mathrm{in}}-V_{\mathrm{out}})I\).

Analog Intuition

Why switch — packets instead of a dropper

Linear heat vs a switcher

Family map · next seven clusters