🌲 The Power Tree & Choosing a Regulator

A board does not have “a” power supply. It has a tree: raw input at the root, splitting into branches, each branch a regulator that hands a clean, fixed voltage to the parts hanging off it. Draw that tree before you place a single part, because the order of the branches and the kind of regulator on each one decides almost everything that follows. This lesson is about reading the tree and picking the right regulator for each rail.

By the end, you can

1. Draw a board's power tree from input protection through bulk regulation to point-of-load and local decoupling
2. Calculate the heat an LDO dissipates from $(V_\text{in} - V_\text{out})\,I$ and its efficiency $V_\text{out}/V_\text{in}$
3. Choose between an LDO and a buck for a given rail using dropout, current, noise, and thermal budget
4. Explain why a switcher is efficient but adds ripple and EMI, and when a post-LDO earns its place

Intuition first

Think of voltage as the height of a water tank and the regulator as the thing that gets water from a high tank down to a lower one.

An LDO (low-dropout linear regulator) is a valve. The high water comes in, the valve throttles it down to the height you want, and all the energy in that drop is simply thrown away as heat. It is dead simple, the stream that comes out is glassy smooth, but a big drop means a lot of wasted energy turned into warmth. Drop a little and the valve is fine. Drop a lot, at high flow, and the valve gets scalding.

A buck (a switching step-down converter) is a bucket on a fast lever. Instead of throttling, it scoops small packets of high water and tips them into the low tank, hundreds of thousands of times a second, using an inductor to smooth the packets into a steady stream. Almost no energy is wasted, so it stays cool even across a huge drop. The catch is the lever is banging up and down at high speed, and that banging shows up as ripple on the output and as electrical noise sprayed into everything nearby.

So the whole tradeoff is one sentence: the LDO is quiet but burns the drop as heat; the buck is efficient but noisy and fussy about layout. Everything else in this lesson is just learning when each one wins.

The power tree, root to leaf

A power tree is the map of how raw input becomes every clean rail on the board. Energy flows in one direction, root to leaf, through four stages.

1. Input protection. The first thing the raw input meets. A fuse opens if the board pulls a dangerous current. A TVS (transient voltage suppressor) clamps the brief high-voltage spikes that come down a cable when something nearby switches. A reverse-polarity FET quietly blocks current if someone plugs the supply in backwards. None of these regulate; they just make sure the rest of the tree never sees something it cannot survive.
2. Bulk regulation. A single big converter (almost always a buck, or a boost if the rail must sit above the input) takes the protected, raw input and brings it to a sensible intermediate rail. This is where the board does its heavy voltage-stepping efficiently, in one place, so the drop is not paid for again downstream.
3. Point-of-load. Right next to each chip that needs a specific voltage sits its own regulator. This can be a small buck for a hungry rail, or a small LDO when the chip wants a quiet supply. “Point-of-load” means physically close to the load, so the clean voltage does not have to travel far and pick up noise on the way.
4. Local decoupling. At the very leaf, small capacitors right at each chip’s power pins. These are the topic of the next lesson; for now, know that they are the last stage of the tree and the reason a regulator three centimetres away still looks steady to a chip switching at megahertz.

Read the tree top to bottom and a design rule falls out: step the voltage down once, in bulk, where you can afford to be efficient, then only do small, cheap, local steps at each load. A board that drops 12 V straight to 1.2 V at every chip with linear regulators would be a space heater. A board that bucks 12 V down to an intermediate rail once, then sips from it, stays cool.

The LDO: simple, quiet, and hot

An LDO is a linear regulator built so it keeps working even when the input is only barely above the output. Inside it there is a pass transistor, an error amplifier, and a stable reference voltage. The amplifier watches the output, compares it to the reference, and continuously adjusts the pass transistor so the output stays put. The pass transistor acts like an automatically adjusted resistor in series with the load.

That word “resistor” is the whole story. Whatever voltage the LDO drops, it drops by burning it across that pass transistor. The heat it makes is the drop times the current flowing through it:

$$P_\text{loss} = (V_\text{in} - V_\text{out})\,I_\text{out}$$

(There is also a tiny term, $V_\text{in} I_Q$, for the quiescent current the LDO needs to run its own internals, but at any real load $I_\text{out}$ dwarfs $I_Q$, so the formula above is what governs the heat.)

Because the output power delivered is $V_\text{out} I_\text{out}$ and the input power drawn is $V_\text{in} I_\text{out}$, the efficiency of an LDO is just the ratio of the two voltages:

$$\eta = \frac{V_\text{out}}{V_\text{in}}$$

This is a brutal, simple truth. A 5 V to 3.3 V LDO is $3.3/5 = 66\%$ efficient at best. A 5 V to 1.8 V LDO is $1.8/5 = 36\%$ efficient. The bigger the drop, the worse it gets, and the wasted fraction all becomes heat you have to get off the board.

So why use one at all? Because the output is clean. There is no switch banging away inside, so there is no switching ripple and very little noise added to the rail. An LDO will also reject a lot of the junk riding on its input, a spec called PSRR (power supply rejection ratio). That makes it the natural choice for sensitive analog: a sensor’s reference rail, an oscillator, an ADC. When the drop is small and the rail must be quiet, the LDO wins on simplicity and noise.

The buck: efficient, but it switches

A buck converter does not throttle. It chops. A transistor switches fully on and fully off at a high frequency (typically 100 kHz to a few MHz), and an inductor stores energy during the on-time and releases it during the off-time. Because the transistor is either all the way on (almost no voltage across it) or all the way off (almost no current through it), it spends very little time in the dissipative middle. That is the trick: the ideal switching elements waste almost nothing, so a real buck routinely runs above 90% efficient, even across a big voltage drop.

The output voltage is set by the duty cycle $D$, the fraction of each cycle the switch is on. For an ideal buck in continuous conduction:

$$V_\text{out} = D\,V_\text{in} \qquad\Rightarrow\qquad D = \frac{V_\text{out}}{V_\text{in}}$$

So to make 3 V from 12 V you run a 25% duty cycle; the feedback loop nudges $D$ up and down to hold the output steady as the input or load wanders. Because efficiency stays high regardless of the ratio, the buck does not care that it is dropping 12 V to 1.2 V the way an LDO would. The heat it makes is small and roughly fixed, not proportional to the drop.

Nothing is free. The switching that makes the buck efficient also makes it noisy. Every cycle the output rises a little during the on-state and falls during the off-state, leaving a sawtooth ripple at the switching frequency riding on top of the DC. Worse, the sharp current edges radiate EMI (electromagnetic interference) into nearby traces and antennas, so a buck must be filtered and, crucially, laid out carefully. The loops where current slams on and off must be kept tiny, or the board sprays noise. A buck is a layout problem as much as a circuit.

See it / Try it

Here is the choice made numerical. Set an input voltage, an output voltage, and a load current, and watch what an LDO and a buck each do with the same job. The LDO’s heat is $(V_\text{in} - V_\text{out})\,I$ and its efficiency is $V_\text{out}/V_\text{in}$; the buck’s heat is whatever the chosen efficiency $\eta$ leaves over. Start at the worked case below: 12 V in, 1.2 V out, 3 A.

Watch the LDO heat number. At 12 V to 1.2 V and 3 A it reads over 32 W, which no small package can shed. Now widen the gap or push the current and it gets worse linearly. Pull the output close to the input, say 3.6 V from 5 V, and the LDO heat collapses to something a tiny part handles easily, while its efficiency climbs toward the buck’s. That crossover is the whole decision: small drop, low current, want quiet, use an LDO. Big drop or big current, use a buck. The calculator just shows you where you are on that line.

?An LDO regulates 5 V down to 1.8 V at 0.5 A. About how much power does it dissipate as heat, and what is its efficiency?

• A 1.6 W and 36%: heat is (5 − 1.8) × 0.5, efficiency is 1.8 / 5
• B 2.5 W and 64%: heat is 5 × 0.5, efficiency is the drop over the input
• C 1.6 W and 90%: a linear regulator is efficient as long as the current is low
• D 1.6 W and 64%: efficiency is the drop divided by the input

?A robot-hand board has a 12 V input. The brushless motor driver wants 5 V at 4 A; a tiny IMU wants a very clean 1.8 V at 20 mA. Which regulators fit best?

• A A buck for the motor rail and a small LDO for the IMU rail
• B An LDO for the motor rail and a buck for the IMU rail
• C One LDO from 12 V feeding both rails to keep it simple
• D One buck from 12 V feeding both rails so nothing runs hot

Lab: budget the tree before you place a part

On the bench, the power tree is a spreadsheet you fill in before you route anything. List every rail the board needs: its voltage, its worst-case current, and whether it is “quiet” (analog, sensor, clock) or “rough” (motor, logic, LEDs). For each rail compute the candidate LDO heat as $(V_\text{in} - V_\text{out})\,I$ from whatever node would feed it. Any rail where that number climbs past roughly a watt in a small package is a buck. Anything quiet and low-current is a candidate LDO. Then look for the bulk step: one buck from the raw input to an intermediate rail (often 3.3 V or 5 V) that every small regulator can sip from, so no rail pays for the full 12 V drop linearly. Finish with a one-line verdict per rail (“1.2 V/3 A sensor core, fed from 3.3 V: buck, then 1.2 V LDO post-reg for the ADC reference”). That table is the power tree, and it is far cheaper to get right on paper than on a four-layer board.

↘ Why a switcher beats a linear supply on efficiency, and what 'low-dropout' really buys you

A linear regulator, including an LDO, regulates by dissipating the excess. Its pass transistor sits in its active region, behaving like a controlled resistor in series with the load, and the difference between input and output voltage is wasted as heat across that transistor. That is why the best a linear supply can ever do is

$$\eta_\text{linear} = \frac{V_\text{out}}{V_\text{in}}$$

A switched-mode supply takes a completely different path. Its pass transistor is never allowed to loiter in the dissipative middle: it is driven fully on (low resistance, little voltage across it) or fully off (no current through it), and it crosses the lossy region only briefly during transitions. Ideal switches paired with ideal inductors and capacitors would convert power at 100% efficiency, moving energy by storing and releasing it rather than throttling it. Real parts have conduction losses ($I^2R$ in the switch, inductor, and capacitor ESR) and switching losses (the energy spent charging and discharging the transistor gate, paid every cycle, so it rises with switching frequency), which is why a real buck lands in the high 80s to high 90s of percent rather than at 100, and why pushing the switching frequency higher shrinks the inductor but costs efficiency.

Now the “low-dropout” part. An ordinary linear regulator built as an emitter follower needs a couple of volts of headroom to keep its pass transistor in regulation, so it cannot make 3.3 V from a 3.6 V battery. An LDO changes the topology to an open-drain or open-collector pass element that can be driven into saturation, so the input can sit only a few hundred millivolts above the output and the regulator still holds. “Low dropout” does not mean low loss; the efficiency is still $V_\text{out}/V_\text{in}$. It means the regulator keeps working when $V_\text{in}$ is close to $V_\text{out}$, which is exactly the regime where a linear regulator is most efficient and runs coolest. That is the LDO’s real niche: a small, quiet final step where the drop is already small.

Putting the two together explains the favourite trick on a sensitive rail. Use a buck for the efficient bulk drop, accept its ripple, then feed a small LDO from the buck’s output to scrub the switching noise off the rail before it reaches the analog part. The combination runs at roughly the buck’s efficiency (because the LDO only drops a small amount) yet delivers the LDO’s clean output. Worked the other way, our headline example shows why you cannot skip the buck: a 12 V to 1.2 V LDO at 3 A would dissipate $(12 - 1.2)\times 3 = 32.4$ W, an impossible amount for a point-of-load part, so the bulk step must be a switcher.

A note on the authoritative numbers versus the source material: textbook efficiency figures vary (older references quote linear supplies at 30–40% and switchers at 70–90% for mains-derived designs, while modern point-of-load bucks reach the high 90s). The relationships here, $P_\text{loss} = (V_\text{in} - V_\text{out})I$ and $\eta_\text{linear} = V_\text{out}/V_\text{in}$, are exact for the ideal linear case and are what governs your rail decisions regardless of which datasheet you read.

Grounded in Wikipedia: “Low-dropout regulator”, “Buck converter”, “Switched-mode power supply” (CC BY-SA).

Key takeaways

• ▸The power tree is input protection → bulk regulation → point-of-load → local decoupling; draw it before placing parts.
• ▸An LDO burns the drop as heat: $P_\text{loss} = (V_\text{in} - V_\text{out})\,I$ and $\eta = V_\text{out}/V_\text{in}$. Simple and quiet, but hot on a big drop.
• ▸A buck is efficient (often over 90%) across any drop because it switches instead of throttling, but it adds ripple and EMI and is layout-sensitive.
• ▸Choose by dropout, current, noise, and thermal budget: small drop / low current / quiet → LDO; big drop or big current → buck.
• ▸12 V to 1.2 V at 3 A in an LDO is 32.4 W, impractical. Use a buck, optionally with a small post-LDO to clean a sensitive rail.
• ▸Robot-hand: the motor rail wants a buck; the sensor rail may want a clean LDO.

Practice 1 warm-up

A 3.3 V LDO is fed from 5 V and supplies 200 mA to a microcontroller. How much power does it dissipate as heat, and what is its efficiency?

↘ Show worked solution

Heat is the dropped voltage times the current: $P_\text{loss} = (V_\text{in} - V_\text{out})\,I = (5 - 3.3)\,\text{V} \times 0.2\,\text{A} = 1.7\,\text{V} \times 0.2\,\text{A} = 0.34\,\text{W}.$ Efficiency is the voltage ratio: $\eta = \frac{V_\text{out}}{V_\text{in}} = \frac{3.3}{5} = 0.66 = 66\%.$ A third of a watt is comfortable for a small SOT-23 LDO, and 66% is acceptable for a small drop. This is exactly the rail where an LDO is the right call.

Practice 2 core

The robot hand’s motor driver needs 5 V at 4 A, and a junior engineer proposes an LDO straight from the 12 V input “to keep the board simple.” Calculate the heat, then say what you would do instead and why.

↘ Show worked solution

Heat from a linear drop: $P_\text{loss} = (12 - 5)\,\text{V} \times 4\,\text{A} = 7\,\text{V} \times 4\,\text{A} = 28\,\text{W}.$ Twenty-eight watts in a point-of-load package is impossible to cool; the part would go into thermal shutdown or fail. Efficiency would be only $5/12 = 42\%$, so well over half the input power is wasted as heat.

Instead, use a buck for the 5 V/4 A rail. At, say, 92% efficiency the buck’s own heat is roughly the output power times the wasted fraction: output power is $5 \times 4 = 20\,\text{W}$, input power is $20 / 0.92 \approx 21.7\,\text{W}$, so the buck dissipates about $1.7\,\text{W}$, sixteen times less heat than the LDO, and a level a small switcher sheds easily. The buck adds ripple and needs a careful hot-loop layout, but for a big drop at high current it is the only sane choice.

Practice 3 stretch

A precision strain-gauge amplifier on the robot hand needs a very clean 1.2 V at 15 mA, and your board already has an efficient 3.3 V rail (from the bulk buck). You are choosing between (a) a small buck 3.3 V → 1.2 V, (b) an LDO 3.3 V → 1.2 V, and (c) a buck 3.3 V → 1.5 V followed by an LDO 1.5 V → 1.2 V. Reason through heat, noise, and which you would ship.

↘ Show worked solution

First the heat of the LDO options, since current is tiny:

• Option (b), LDO 3.3 V → 1.2 V at 15 mA: $P_\text{loss} = (3.3 - 1.2)\times 0.015 = 2.1 \times 0.015 \approx 0.032\,\text{W}$, about 32 mW. Negligible heat. But efficiency is only $1.2/3.3 = 36\%$, which at 15 mA simply does not matter.
• Option (c)‘s final LDO 1.5 V → 1.2 V at 15 mA: $P_\text{loss} = (1.5 - 1.2)\times 0.015 = 0.3 \times 0.015 \approx 0.0045\,\text{W}$, about 4.5 mW. Even less heat, and a smaller drop means it scrubs the buck’s ripple while wasting almost nothing.

Now noise. The amplifier is precision analog, so switching ripple on its rail directly corrupts the measurement. Option (a), a bare buck, leaves ripple on exactly the rail that must be quiet, so reject it. Option (b), a plain LDO, is clean and its heat is trivial, so it is a perfectly good answer for 15 mA. Option (c), buck-then-LDO, is the textbook clean rail: the buck does an efficient pre-step and the LDO scrubs the ripple, giving the cleanest output at the cost of one more part.

What to ship: at only 15 mA the efficiency difference between (b) and (c) is meaningless (tens of milliwatts either way), so the simplest clean option, (b) the LDO, is the right call. Reserve the buck-plus-post-LDO (c) for when the load current is large enough that a bare LDO’s heat would matter and the rail must stay quiet. The decision was driven by noise first, then by whether the LDO’s heat is tolerable, which, at 15 mA, it always is.