By the end of this chapter you should be able to:
- Quantify what line and load variation do to an unregulated rail, and identify the worst-case input a regulator must survive.
- Define line regulation, load regulation and ripple rejection, and compute all three in their proper units.
- Analyse the Zener shunt regulator of Chapter 12 and state the three limits that rule it out above a few tens of milliamps.
- Derive the output impedance of an emitter-follower series regulator and explain what it does and does not improve.
- Derive \(V_{out} = V_{ref}(1+R_1/R_2)\) for the feedback regulator and show that the loop divides both regulation figures by \(1+A\beta\).
- Design constant-current and fold-back limiting, and compute the short-circuit dissipation each permits.
- Apply the 78xx, 79xx and LM317 data-sheet parameters, and size a heat sink from the junction-to-ambient thermal chain.
- Derive the buck, boost and buck-boost duty-cycle relations and explain the efficiency difference from a linear regulator.
Chapters 9 and 10 built a d.c. supply out of a transformer, a bridge rectifier and a reservoir capacitor, and Chapter 12 put a Zener diode across the result and called it regulated. It is worth being honest about how little that achieved. The rail produced by a capacitor-input filter is not a constant voltage at all: it sags between rectifier pulses, it sags further as the load draws more, and the whole band rises and falls with the mains. For the supply worked through in this chapter — an 18 V secondary, a bridge and 2200 µF — the rail sits between 21.78 V and 24.06 V at full load on a nominal mains, but the complete range once the mains is allowed its statutory ±10 per cent and the load is allowed to vary from nothing to half an amp is 19.24 V to 26.60 V. A circuit designed for 15 V has no use whatever for a supply like that.
A voltage regulator is the block that stands between the two and holds the output within a few millivolts while its input wanders over seven volts. This chapter builds one four times over: as a Zener shunt, whose limitations are instructive; as a series pass transistor, which fixes the load problem and not the line problem; as a full feedback regulator with a reference, an error amplifier and a sampling divider, which fixes both by a factor equal to the loop gain; and as a switching converter, which fixes the efficiency problem that all three linear circuits share. Along the way it deals with the two things that actually destroy regulators in service — short circuits and heat.
1 What the Regulator Is Given
Start with what the regulator is given. A bridge rectifier fed from an 18 V r.m.s. secondary charges the reservoir capacitor to the peak of the secondary less two diode drops:
Between rectifier pulses the capacitor supplies the load alone and discharges. Taking the discharge as linear, which Chapter 10 justified for a modest ripple, the peak-to-peak ripple is
\(f = 100\) Hz because a bridge rectifies both half-cycles of a 50 Hz supply. The rail therefore sawtooths between 24.06 V and 21.78 V at full load, a ripple of 9.9 per cent of the mean.
Now let the two independent variables move. Line variation scales the whole waveform: the mains is specified to hold ±10 per cent, so \(V_{pk}\) ranges from \(0.9(25.46) - 1.4 = 21.51\) V to \(1.1(25.46) - 1.4 = 26.60\) V. Load variation changes the ripple depth: at no load the capacitor does not discharge at all and the rail sits at the peak, while at 0.5 A it loses 2.27 V between pulses. The two combine at the corners:
| Condition | \(V_{pk}\) | Ripple | Minimum rail |
|---|---|---|---|
| Nominal mains, full load 0.5 A | 24.06 V | 2.27 V | 21.78 V |
| Nominal mains, no load | 24.06 V | 0 | 24.06 V |
| Mains −10 %, full load | 21.51 V | 2.27 V | 19.24 V |
| Mains +10 %, no load | 26.60 V | 0 | 26.60 V |
Figure 30.1 puts all of that on one axis, and three design constraints can be read straight off it.
- The regulator must work at 19.24 V. That is the worst corner, and it is the number that decides whether a 15 V output is possible at all. A 7815 needs about 2 V of headroom, so it needs 17 V and has 2.24 V to spare; an LM317 wants 2.5 V, needs 17.5 V and has 1.74 V to spare. Neither is generous, and a designer who checked only the nominal 24.06 V would have shipped a supply that fails on a bad mains day under full load.
- The regulator must survive 26.60 V. This corner sets the pass transistor’s voltage rating, and — far more painfully — its dissipation, since the heat is \((V_{in}-V_{out})I_L\) and \(V_{in}\) is at its largest exactly when the designer would prefer it small.
- The regulator must reject 2.27 V of 100 Hz ripple. That is a signal, not a d.c. offset, and how much of it appears at the output is a separate specification from either of the two d.c. ones.
2 Line Regulation, Load Regulation and Ripple Rejection
Three numbers describe a regulator’s performance, and they are quoted in a confusing variety of units. All three are worth writing out carefully once.
Line regulation measures how much the output moves when the input moves, with the load held constant. It appears in three forms:
For a regulator whose output moves 6 mV as the input sweeps the 7.36 V from 19.24 to 26.60 V while delivering 15 V: 0.815 mV/V, or 0.0054 %/V, or 0.040 per cent over the full range. All three describe the same measurement; a data sheet will pick one and a question will ask for another.
Load regulation measures how much the output moves when the load changes from nothing to full, with the input held constant. It is a disguised statement of output impedance, since \(\Delta V_{out}/\Delta I_{out}\) is an impedance in ohms:
A regulator whose output falls from 15.020 V at no load to 14.970 V at 0.5 A has a load regulation of 0.334 per cent, or 100 mV/A, or an output impedance of 0.100 Ω.
Ripple rejection is the attenuation the regulator gives to an a.c. component riding on its input, quoted in decibels:
An LM317 is specified at about 65 dB, a factor of 1778, so the 2.273 V of ripple on our rail becomes 1.278 mV at the output. A 78xx typically manages 78 dB, a factor of 7943, giving 0.286 mV. Both figures assume the regulator is not in dropout on the ripple troughs; a regulator whose input dips below the dropout voltage for part of each cycle passes the ripple almost undiminished, which is the commonest cause of a "faulty" regulator that is in fact merely starved.
Line regulation is measured at d.c., ripple rejection at 100 Hz, and load regulation at neither. A regulator has plenty of loop gain at d.c. and progressively less as frequency rises, so its ripple rejection is always worse than its line regulation would suggest, and both collapse above a few kilohertz. That is why every three-terminal regulator data sheet insists on a 0.1 µF capacitor close to the input pin and a 1 µF on the output: above the frequency at which the loop still works, only the capacitors are regulating.
3 The Zener Shunt and the Series Pass Transistor
Chapter 12 built the simplest possible regulator: a series resistor \(R_S\) from the unregulated rail and a Zener diode across the load. It is worth revisiting with the numbers of this chapter, because its failure modes are the reason every other circuit in this chapter exists.
The design rule is that the Zener must stay in breakdown at the worst corner, which is minimum input and maximum load, when the least current is available and the load is taking most of it. Taking \(V_Z = 15\) V, a knee current \(I_{ZK} = 5\) mA and a maximum load of 30 mA:
so take \(R_S = 120\ \Omega\). Now check the opposite corner, maximum input and no load, when the entire current goes through the Zener:
A 2 W Zener is needed to supply a load that never takes more than \(15 \times 0.030 = 0.45\) W, and \(R_S\) must dissipate up to \((26.60-15)^2/120 = 1.12\) W as well. Roughly six watts of components are being installed to deliver less than half a watt.
The regulation figures follow from the Zener’s dynamic resistance \(r_z\), which for a 15 V device is about 9 Ω. For line variation the circuit is a divider between \(R_S\) and \(r_z\); for load variation the Zener sees \(r_z\) in parallel with \(R_S\):
Over the 7.36 V input range that gives \(\Delta V_{out} = 0.514\) V, a line regulation of 3.43 per cent; over 30 mA of load it gives 0.251 V, a load regulation of 1.67 per cent. Both are an order of magnitude worse than any modern requirement, and the efficiency is 66 per cent at full load falling to zero at no load, because the current drawn from the rail does not depend on the load at all.
The fatal objection appears when the load grows. Asked to deliver 0.5 A instead of 30 mA, the same design needs \(R_S = (19.24-15)/0.505 = 8.4\ \Omega\), and at maximum input with no load the Zener would have to absorb \((26.60-15)/8.4 = 1.38\) A, dissipating 20.7 W. No Zener diode is made for that duty at any sensible price, and the whole 20.7 W would be dissipated even with the load disconnected.
The series pass transistor. The cure is to stop the Zener carrying the load current and let it carry only a base current. Connect an emitter follower with its base at the Zener and its collector on the raw rail, and take the output from the emitter: \(V_{out} = V_Z - V_{BE}\), so a 15.7 V Zener gives 15.0 V. Now the Zener chain need only supply
instead of 500 mA, a reduction by the factor \((1+\beta)\), and the load current comes through the transistor from the rail. The reflection rule of Chapter 19 gives the output impedance directly: the base circuit resistance divided by \((1+\beta)\), in series with \(r_e = V_T/I_E = 26/500 = 0.052\ \Omega\):
The load regulation is now \(0.5 \times 0.1381 = 69\) mV, or 0.46 per cent — a sixty-fold improvement on the shunt circuit and at sixteen times the current. But the line regulation is unchanged at 3.43 per cent, because the base still follows the Zener and the Zener still follows the rail through \(r_z/(R_S+r_z)\).
That last sentence is the whole motivation for the next section. The follower has fixed one of the two problems and left the other exactly where it was, and it has done nothing at all about the fact that \(V_{out}\) is set by a Zener voltage minus a temperature-dependent \(V_{BE}\) and cannot be adjusted.
4 The Feedback Series Regulator
The feedback regulator is the same emitter follower with an amplifier inserted between the reference and the base, and a divider that samples the output rather than assuming it. Figure 30.2 shows the four blocks that every linear regulator, discrete or integrated, contains.
- A reference, here a Zener fed from the rail through \(R_{ref}\), establishing a fixed voltage that the output is compared against.
- A sampling divider \(R_1\), \(R_2\) across the output, returning a fixed fraction of it.
- An error amplifier of gain \(A\) that subtracts the sample from the reference.
- A series pass element whose conduction the error amplifier controls.
The output-voltage expression comes from the virtual-short argument of Chapter 26. If the loop has enough gain, the error amplifier drives the pass transistor until its two inputs are equal, so the sampled fraction of the output equals the reference:
With a 5.1 V reference, \(R_2 = 10\ \text{k}\Omega\) and \(R_1 = 19.6\ \text{k}\Omega\), \(V_{out} = 5.1(1+1.96) = 15.10\) V. The exact \(R_1\) for 15.00 V would be 19.41 kΩ; in practice \(R_1\) is a fixed resistor plus a trimmer.
The regulating action, step by step. Suppose the load increases and the output starts to fall by \(\delta\). The divider passes \(0.3378\delta\) of that fall to the inverting input; the error amplifier, with gain 1000, therefore raises its output by \(337.8\delta\); that raises the pass transistor’s base, which conducts harder, which raises the emitter. The output is restored to within \(\delta/(1+A\beta)\) of where it started. Exactly the same argument applies to a rise on the input rail. The improvement is the same factor in both cases, and it is the ordinary negative-feedback result of Chapter 26:
Take the follower of Section 3, with open-loop \(Z_{out} = 0.1381\ \Omega\) and open-loop line regulation 3.43 per cent, and wrap it in the loop of Figure 30.2 with an error amplifier of gain \(A = 1000\).
The sampled fraction is \(\beta = 10/(19.6+10) = 0.3378\), so the loop gain is \(A\beta = 337.8\) and the improvement factor is \(1 + A\beta = 338.8\).
a load regulation of 0.0014 per cent. The line regulation falls from 3.43 per cent to \(3.43/338.8 = 0.0101\) per cent, that is 1.52 mV over the whole 7.36 V input range.
These figures are better than any real regulator achieves, and the reason is worth understanding: the feedback loop corrects everything downstream of the reference but nothing inside it. The reference itself still moves with the rail through \(r_z/(R_{ref}+r_z)\), and it moves with temperature, and the divider resistors have their own tempcos. Real discrete regulators land near 0.01 to 0.1 per cent, and the way to do better is to feed the reference from the regulated output rather than from the raw rail — a bootstrap that removes the reference’s line sensitivity entirely, at the cost of needing a starting circuit, since neither the reference nor the output can come up without the other.
5 Current Limiting, Fold-Back and the Shunt Regulator
A regulator with a 0.4 mΩ output impedance will, if its output is shorted to ground, attempt to deliver whatever current that implies. The pass transistor then has the entire input voltage across it and destroys itself in milliseconds. Current limiting is therefore not an optional refinement but part of the regulator.
Constant-current limiting. Put a small resistor \(R_{SC}\) in series with the output and connect a second transistor \(Q_2\) with its base-emitter junction across it. While the load current is small the drop across \(R_{SC}\) is below \(Q_2\)’s turn-on voltage and \(Q_2\) is off. When \(I_LR_{SC}\) reaches about 0.7 V, \(Q_2\) conducts and steals base drive from the pass transistor, preventing any further increase:
For a 0.6 A limit, \(R_{SC} = 0.7/0.6 = 1.167\ \Omega\); the nearest preferred value 1.2 Ω gives \(I_{\lim} = 0.583\) A. \(R_{SC}\) drops \(0.5 \times 1.2 = 0.6\) V at the rated 0.5 A, which is 0.6 V of headroom permanently given away, and dissipates 0.30 W.
The circuit is now safe against a short, but only just. At the limit with the output shorted the pass transistor drops the whole input and carries the limit current:
which is nearly three times the 5.9 W it dissipates in normal service. The heat sink must therefore be sized for a fault condition that may last indefinitely, or the regulator must be made to reduce its current when the output collapses.
Fold-back limiting does exactly that. A divider \(R_3\), \(R_4\) is added so that \(Q_2\)’s base is driven not by the drop across \(R_{SC}\) alone but by that drop plus a fraction of the output voltage. Writing \(K = R_4/(R_3+R_4)\), \(Q_2\) turns on when
The second term vanishes when the output collapses, so the two limits are
Aim for a knee at 0.6 A and a short-circuit current of 0.3 A. Solving the two equations simultaneously with \(V_{out} = 15\) V gives \(R_3/R_4 = 0.0490\) and \(R_{SC} = 2.45\ \Omega\). Take \(R_4 = 10\ \text{k}\Omega\), \(R_3 = 470\ \Omega\) and \(R_{SC} = 2.4\ \Omega\), so \(K = 10\,000/10\,470 = 0.9551\). Then
The short-circuit dissipation falls from 15.5 W to \((26.60)(0.305) = 8.1\) W, a factor of 1.91, and the heat sink can be sized accordingly. The price is the larger \(R_{SC}\): 2.4 Ω drops 1.2 V at the rated 0.5 A instead of 0.6 V, which comes straight out of the already tight headroom of Section 1. Fold-back always costs more headroom than simple limiting, because only the fraction \(K\) of the sense voltage reaches \(Q_2\).
One warning. A fold-back characteristic can latch into its low-current state when driving a load that draws a large current at low voltage, such as a lamp filament or a capacitive input stage. The load demands more than \(I_{SC}\) at start-up, the regulator folds back, the output never rises, and the demand never falls. Loads of that kind need plain constant-current limiting, or a fold-back knee placed well above the inrush.
The shunt regulator. The topology of Section 3, in which the regulating element is across the load rather than in series with it, does not disappear entirely. Its virtues are that it is inherently short-circuit proof — a shorted output simply removes the shunt current, and \(R_S\) limits everything — and that it can sink current as well as source it, which a series regulator cannot. Its vice is that it draws full current from the supply whether or not the load wants any. It survives as the reference element inside other circuits (the TL431 is a shunt regulator in a three-pin package) and in low-power circuits with a nearly constant load, and nowhere else.
6 Three-Terminal Regulators and Thermal Design
Everything in Sections 4 and 5 is available in a three-pin package for well under a pound, and no sensible designer builds it from discrete parts. The parameters that matter are on the first page of the data sheet.
- The 78xx series gives a fixed positive output, the last two digits being the voltage: 7805, 7812, 7815 and so on, in TO-220 at 1 A or TO-92 at 100 mA. The 79xx series is its negative counterpart, and its pinout is different — a fact that has cost many students a working circuit.
- Dropout voltage is the minimum \(V_{in} - V_{out}\) at which regulation is maintained, about 2 V for a standard 78xx and 2.0 to 2.5 V for an LM317. Below it the pass transistor saturates, the loop opens, and the output simply follows the input, ripple and all. Low-dropout (LDO) regulators use a PNP or p-channel pass element and manage 0.1 to 0.5 V, at the cost of needing a specific output capacitor with a specific equivalent series resistance to stay stable.
- Quiescent current is what the device draws for itself and passes to ground rather than to the load, typically 5 mA for a 78xx. It contributes \(V_{in}I_Q\) to the dissipation and it never reaches the load, so it dominates the efficiency of a lightly loaded regulator.
The LM317 is adjustable. It holds a fixed 1.25 V between its output and adjust pins, so a divider from output to ground with its tap on the adjust pin sets the output. The adjust pin also draws a small current \(I_{adj}\) of about 50 µA which flows through the lower resistor, so
\(R_1\) is the resistor between the output and adjust pins, and is fixed at 240 Ω by convention because \(1.25/240 = 5.21\) mA is the minimum load the device needs to regulate at all. \(R_2\) runs from the adjust pin to ground.
Setting the voltage. Ignoring \(I_{adj}\), 15 V needs \(R_2 = 240(15/1.25 - 1) = 2640\ \Omega\). The nearest E96 value is 2.61 kΩ, for which
The \(I_{adj}\) term contributes 0.131 V, or 0.87 per cent — small but not negligible, and it is the reason \(R_2\) should be kept modest.
Dissipation. The worst case is maximum input at full load:
At the nominal 24.06 V rail it is 4.65 W, and the efficiency is \(7.5/(24.06 \times 0.505) = 61.7\) per cent.
The thermal chain. Heat flows from the silicon to the air through three resistances in series, and thermal resistance behaves exactly like electrical resistance with power in place of current and temperature in place of voltage:
\(\theta_{JC}\) is junction-to-case, fixed by the package (5 °C/W for a TO-220); \(\theta_{CS}\) is case-to-sink, set by the mounting (1 °C/W with thermal compound, 2 to 3 °C/W with an insulating mica washer); \(\theta_{SA}\) is sink-to-ambient, which is what the designer buys.
Rearranging for the sink, with \(T_{J(\max)} = 125\) °C and an ambient of 45 °C:
Designing instead to \(T_J = 110\) °C for margin gives \(\theta_{SA} \le 4.95\) °C/W. Choosing a 4 °C/W extruded sink makes \(\theta_{JA} = 10\) °C/W, a rise of \(5.93 \times 10 = 59.3\) °C and a junction at 104 °C — comfortable. Without a heat sink at all, a TO-220 in free air has \(\theta_{JA} \approx 50\) °C/W, which predicts a junction temperature of 342 °C. In practice the internal thermal shutdown cuts in long before that and the regulator oscillates on and off, which is exactly what an unheatsunk regulator does on the bench.
7 Switching Regulators and the Efficiency Argument
The 4.65 W of heat is not a defect of the design; it is arithmetic. The pass transistor is a variable resistor in series with the load, it carries the full load current, and it drops \(V_{in}-V_{out}\). The only escape is to stop operating the transistor in its linear region and operate it as a switch instead. A switch that is fully on has current through it and no voltage across it; a switch that is fully off has voltage across it and no current through it; either way the product is nearly zero. An inductor and a capacitor then average the chopped waveform back into d.c.
All three topologies of Figure 30.4 are analysed by the same argument. In the steady state the average voltage across an inductor over one cycle must be zero, because otherwise its current would ramp indefinitely. Write that condition for each circuit with the switch on for a fraction \(D\) of the period and off for \((1-D)\).
Buck. On: the inductor sees \(V_{in}-V_{out}\). Off: the diode clamps the left-hand end to ground and it sees \(-V_{out}\). Volt-second balance gives \((V_{in}-V_{out})D = V_{out}(1-D)\), hence
Boost. On: the inductor sees \(V_{in}\) and stores energy. Off: it sees \(V_{in}-V_{out}\), which must be negative. Balance gives \(V_{in}D = (V_{out}-V_{in})(1-D)\), hence
Buck-boost. On: the inductor sees \(V_{in}\). Off: it sees \(V_{out}\), which is negative. Balance gives \(V_{in}D = -V_{out}(1-D)\), hence
For the running example, a buck taking 24.06 V to 15 V needs \(D = 15/24.06 = 62.4\) per cent. A boost from 12 V to 24 V needs \(D = 1 - 12/24 = 50.0\) per cent. An inverting buck-boost from 12 V to \(-15\) V needs \(D = 15/27 = 55.6\) per cent. Note that none of these depends on the load: the duty cycle sets the ratio, and the control loop adjusts it to hold the ratio as the input moves.
Deliver 15 V at 0.5 A from the 24.06 V rail. The linear regulator of Worked Example 30.3 draws 0.505 A from the rail, delivers 7.5 W and dissipates 4.65 W, for an efficiency of 61.7 per cent.
A buck converter at \(D = 0.624\) and 100 kHz has a quite different loss budget. Taking a MOSFET of \(R_{DS(on)} = 50\) mΩ, a Schottky of \(V_F = 0.45\) V, an inductor of 0.1 Ω series resistance, a 50 ns transition and 2 mA of controller current:
| Loss | Expression | Value |
|---|---|---|
| Switch conduction | \(D I^2 R_{DS}\) | 0.008 W |
| Diode conduction | \((1-D)V_FI\) | 0.085 W |
| Switching | \(V_{in}I\,t_{sw}f_{sw}\) | 0.060 W |
| Inductor copper | \(I^2R_L\) | 0.025 W |
| Controller | \(V_{in}I_Q\) | 0.048 W |
| Total | 0.226 W |
The efficiency is \(7.5/7.726 = 97.1\) per cent, the input current is 0.321 A instead of 0.505 A, and the heat is 0.226 W instead of 4.65 W — a factor of twenty. The heat sink disappears entirely. The 220 µH inductor carries a ripple current of \(\Delta I_L = (V_{in}-V_{out})D/(Lf_{sw}) = 0.257\) A peak to peak, and with a 220 µF output capacitor the resulting output ripple is \(\Delta I_L/(8f_{sw}C) = 1.5\) mV.
That last figure is optimistic, and honesty requires the qualification: it assumes an ideal capacitor. Real electrolytics have equivalent series resistance, and the ripple is nearer \(\Delta I_L \times \text{ESR}\) — with 100 mΩ of ESR that is 26 mV, twenty times the ideal figure and still riding on a 100 kHz switching residue that a linear regulator would never produce.
| Linear (series feedback) | Switching (buck) | |
|---|---|---|
| Efficiency, 24 V to 15 V at 0.5 A | 61.7 % | ≈ 97 % |
| Efficiency, 12 V to 3.3 V at 2 A | 27.4 % | ≈ 85 % |
| Heat to remove | 4.65 W, heat sink essential | 0.23 W, usually none |
| Can it step up or invert? | No — step-down only | Yes, all three |
| Output noise | microvolts, no switching residue | tens of mV at \(f_{sw}\) plus EMI |
| Transient response | fast — loop bandwidth is high | slower, limited by \(L\), \(C\) and \(f_{sw}\) |
| Parts count and layout | three pins and two capacitors | inductor, diode, controller, careful layout |
| Where it wins | small drops, low noise, analogue and RF rails | large drops, high current, battery equipment |
The two are not rivals so much as a sequence. Almost every modern instrument uses a switching converter to get the bulk voltage conversion done at 90-odd per cent efficiency, then a small low-dropout linear regulator on each sensitive rail to strip off the switching residue. The linear regulator is asked for a drop of a few hundred millivolts instead of nine volts, so its own efficiency is high and its heat negligible, and it contributes the one thing the switcher cannot: a quiet output.
8 Summary and Key Results
| Quantity | Expression | Value |
|---|---|---|
| Unregulated rail | \(V_{pk} = \sqrt2V_s - 1.4\), \(V_r = I_L/fC\) | 24.06 V, 2.273 V ripple; full range 19.24 to 26.60 V |
| Line regulation | \(\Delta V_{out}/\Delta V_{in}\) | 6 mV over 7.36 V = 0.815 mV/V = 0.040 % of output |
| Load regulation | \((V_{NL}-V_{FL})/V_{FL}\times100 = Z_{out}\) | 50 mV over 0.5 A = 100 mV/A = 0.334 % |
| Ripple rejection | \(20\log_{10}(v_{r,in}/v_{r,out})\) | 65 dB (×1778) → 1.278 mV from 2.273 V |
| Zener shunt | \(R_S = (V_{in\min}-V_Z)/(I_{L\max}+I_{ZK})\) | 120 Ω; \(P_Z = 1.45\) W for a 0.45 W load |
| Zener regulation | \(r_z/(R_S+r_z)\) and \(r_z\|R_S\) | line 3.43 %, load 1.67 % (\(r_z = 9\ \Omega\)) |
| Emitter-follower regulator | \(Z_o = (r_z\|R_S)/(1+\beta) + r_e\) | 0.1381 Ω → load reg 0.46 %; line reg unchanged |
| Feedback regulator | \(V_{out} = V_{ref}(1+R_1/R_2)\) | 5.1 V, 19.6 kΩ, 10 kΩ → 15.10 V |
| Loop improvement | divide both by \(1+A\beta\) | \(\beta = 0.3378\), \(A = 1000\) → 338.8; \(Z_o = 0.41\) mΩ |
| Constant-current limit | \(I_{\lim} = 0.7/R_{SC}\) | 1.2 Ω → 0.583 A; short-circuit \(P_D = 15.5\) W |
| Fold-back limit | \(I_{SC} = 0.7/KR_{SC}\), \(I_{knee} = I_{SC}+V_{out}R_3/R_4R_{SC}\) | 470 Ω, 10 kΩ, 2.4 Ω → 0.305 A and 0.599 A; \(P_D\) 8.1 W |
| LM317 | \(V_{out} = 1.25(1+R_2/R_1) + I_{adj}R_2\) | 240 Ω, 2.61 kΩ → 14.974 V |
| Dissipation | \(P_D = (V_{in}-V_{out})I_L + V_{in}I_Q\) | 5.93 W worst case, 4.65 W nominal, 61.7 % efficient |
| Thermal chain | \(T_J = T_A + P_D(\theta_{JC}+\theta_{CS}+\theta_{SA})\) | \(\theta_{SA}\le 7.48\) °C/W for \(T_J = 125\) °C; 4 °C/W gives \(T_J = 104\) °C |
| Buck / boost / buck-boost | \(DV_{in}\) / \(V_{in}/(1-D)\) / \(-V_{in}D/(1-D)\) | \(D = 62.4\) %, 50.0 %, 55.6 % for the three worked cases |
| Switching efficiency | \(P_{out}/(P_{out}+\sum\text{losses})\) | 97.1 % against 61.7 %; 0.226 W of heat against 4.65 W |
9 Common Mistakes
The rail in this chapter is 24.06 V nominally and a 7815 with 9 V of headroom looks luxurious. The design must instead survive both extremes at once. At mains −10 per cent under full load the rail bottoms out at 19.24 V, leaving only 2.24 V of headroom above the 7815’s 2 V dropout — and the moment it drops below that, the regulator stops regulating and passes the input, ripple included, straight to the output. At mains +10 per cent with no load the rail is 26.60 V, which is what sets the pass transistor’s voltage rating and its short-circuit dissipation. Both corners must be checked, and the ripple trough, not the mean, is the number that matters for dropout.
The regulator dissipates 5.93 W at its worst normal operating point, which a 7.5 °C/W sink handles. With the output shorted and simple 1.2 Ω current limiting, the pass transistor drops the entire 26.60 V at 0.583 A, that is 15.5 W — two and a half times as much, indefinitely, until someone notices. A sink sized for 5.93 W would let the junction reach \(45 + 15.5(13.5) = 254\) °C. Either size the sink for the fault, add fold-back so the fault dissipation is only 8.1 W, or rely on the integrated regulator’s own thermal shutdown — but decide which, rather than discovering the answer in service.
The efficiency figures are so lopsided — 97 per cent against 62 — that it is tempting to use a switcher everywhere. Two things stop it. The output carries tens of millivolts of residue at the switching frequency and its harmonics, which is fatal to an ADC reference, an oscillator supply or a low-noise amplifier rail, and it radiates: a badly laid out buck converter is an efficient transmitter, because the loop carrying the chopped current is a loop antenna. And the transient response is limited by \(L\), \(C\) and the switching frequency, so a switcher recovers from a load step far more slowly than a linear regulator does. The standard answer is to use both: a switcher for the bulk conversion, then a small LDO with a few hundred millivolts of drop on each quiet rail.
10 Chapter Review
1. Design a Zener shunt regulator for 9.1 V from an input of 12 to 18 V, supplying 0 to 40 mA, with \(I_{ZK} = 5\) mA and \(r_z = 7\ \Omega\). Find \(R_S\), the Zener power rating, and the line and load regulation.
The worst corner for keeping the Zener alive is minimum input with maximum load, so \(R_{S(\max)} = (12-9.1)/(0.040+0.005) = 2.9/0.045 = 64.4\ \Omega\); take the preferred value 62 Ω. The opposite corner is maximum input with no load, when the whole current goes through the Zener: \(I_{Z(\max)} = (18-9.1)/62 = 143.5\) mA and \(P_Z = (0.1435)(9.1) = 1.31\) W, so a 2 W Zener is required. \(R_S\) itself dissipates \((18-9.1)^2/62 = 1.28\) W and also needs a 2 W rating. Checking the first corner, \(I_Z = (12-9.1)/62 - 0.040 = 46.8 - 40 = 6.8\) mA, safely above the 5 mA knee. For line regulation the circuit is a divider: \(\Delta V_{out}/\Delta V_{in} = r_z/(R_S+r_z) = 7/69 = 0.1014\), so a 6 V input change moves the output by 0.609 V, a line regulation of \(100(0.609)/9.1 = 6.69\) per cent. For load regulation the load sees \(r_z\|R_S = (7)(62)/69 = 6.29\ \Omega\), so 40 mA moves the output by 0.252 V, a load regulation of 2.76 per cent. The efficiency at full load is \(100(9.1)(0.040)/[(12)(0.0468)] = 64.9\) per cent, and at no load it is zero. Every one of these figures is poor, which is the argument for Section 4.
2. A feedback series regulator uses a 2.5 V band-gap reference and a divider with \(R_2 = 10\ \text{k}\Omega\). Choose \(R_1\) for a 12 V output, and find the load regulation over 1 A for error-amplifier gains of 200 and 2000, given an open-loop output impedance of 0.15 Ω.
From \(V_{out} = V_{ref}(1+R_1/R_2)\), \(R_1/R_2 = 12/2.5 - 1 = 3.80\), so \(R_1 = 38\ \text{k}\Omega\) and \(V_{out} = 2.5(1+3.8) = 12.000\) V exactly. The sampled fraction is \(\beta = R_2/(R_1+R_2) = 10/48 = 0.2083\). With \(A = 200\), the loop gain is \(A\beta = 41.67\) and the improvement factor \(1+A\beta = 42.67\), so \(Z_{out} = 0.15/42.67 = 3.52\) mΩ and the output falls by 3.52 mV over 1 A, a load regulation of 0.029 per cent. With \(A = 2000\), \(A\beta = 416.7\), \(1+A\beta = 417.7\), \(Z_{out} = 0.359\) mΩ and the drop is 0.359 mV, a load regulation of 0.0030 per cent. A tenfold increase in error-amplifier gain has bought very nearly a tenfold improvement, which is the general property of negative feedback and the reason integrated regulators use error amplifiers with gains in the thousands. Note also what \(\beta\) does: raising the output voltage lowers \(\beta\) and therefore the loop gain, so the same regulator has poorer regulation at 12 V than at 5 V. That is visible in the LM317 data sheet, whose load regulation is quoted as a percentage precisely because it scales with the output voltage.
3. A 12 V regulator running from a 20 V rail is to limit at about 1 A. Design simple current limiting, compute the short-circuit dissipation, then redesign it as fold-back for a 0.35 A short-circuit current and compare.
For simple limiting, \(R_{SC} = 0.7/1.0 = 0.7\ \Omega\); the preferred 0.68 Ω gives \(I_{\lim} = 0.7/0.68 = 1.03\) A and drops 0.68 V at the rated current. Into a dead short the pass transistor carries 1.03 A with the full 20 V across it, so \(P_D = (20)(1.03) = 20.6\) W — a duty that needs a substantial sink and a device well inside its safe operating area. For fold-back, use \(I_{SC} = 0.7/(KR_{SC})\) and \(I_{knee} = I_{SC} + V_{out}R_3/(R_4R_{SC})\) with \(K = R_4/(R_3+R_4)\). Aiming for \(I_{SC} = 0.35\) A and \(I_{knee} = 1.05\) A with \(V_{out} = 12\) V, the two equations give \(R_3/R_4 = 0.132\) and \(R_{SC} = 2.26\ \Omega\). Taking \(R_4 = 10\ \text{k}\Omega\), \(R_3 = 560\ \Omega\) and \(R_{SC} = 2.2\ \Omega\) gives \(K = 10\,000/10\,560 = 0.947\), \(I_{SC} = 0.7/[(0.947)(2.2)] = 0.336\) A and \(I_{knee} = 0.336 + (12)(560)/[(10\,000)(2.2)] = 0.642\) A. That knee is below the wanted 1 A, so \(R_3\) must be raised — 1.0 kΩ gives \(I_{SC} = 0.350\) A with \(I_{knee} = 0.896\) A, and 1.2 kΩ gives 0.356 A and 1.011 A. The short-circuit dissipation is \((20)(0.336) = 6.7\) W against 20.6 W, a saving of about three times, but \(R_{SC}\) now drops 2.2 V at 1 A instead of 0.68 V. That 1.5 V of extra headroom is the real cost of fold-back, and on a 20 V rail feeding a 12 V output it is affordable; on a rail with 2 V of margin it would not be.
4. An LM317 supplies 9 V at 0.8 A from a 15 V rail in a 40 °C ambient. Choose the divider, find the dissipation and efficiency, and specify the heat sink for a junction temperature of 110 °C.
With the customary \(R_1 = 240\ \Omega\), \(R_2 = 240(9/1.25 - 1) = 240(6.2) = 1488\ \Omega\); the preferred 1.5 kΩ gives \(V_{out} = 1.25(1+1500/240) + (50\times10^{-6})(1500) = 9.0625 + 0.075 = 9.14\) V, which is 1.5 per cent high — use 1.47 kΩ (E96) for 8.98 V, or a trimmer. The dissipation is \(P_D = (15-9)(0.8) + (15)(0.005) = 4.80 + 0.075 = 4.875\) W, and the efficiency is \(100(9)(0.8)/[(15)(0.805)] = 59.6\) per cent — 4.88 W of heat to deliver 7.2 W of output. For the heat sink, \(\theta_{JA(\max)} = (110-40)/4.875 = 14.36\ ^\circ\)C/W, and subtracting \(\theta_{JC} = 5\) and \(\theta_{CS} = 1\ ^\circ\)C/W leaves \(\theta_{SA} \le 8.36\ ^\circ\)C/W. A common 6 °C/W clip-on sink gives \(\theta_{JA} = 12\ ^\circ\)C/W and \(T_J = 40 + 4.875(12) = 98.5\ ^\circ\)C, with 26 °C of margin to the 125 °C limit. Without any sink, \(\theta_{JA} \approx 50\ ^\circ\)C/W predicts \(T_J = 40 + 244 = 284\ ^\circ\)C, so the device would shut down thermally within seconds. If the mica washer is used for isolation, \(\theta_{CS}\) rises to about 2.5 °C/W and the sink must be 6.9 °C/W or better instead.
5. Compare a linear regulator and a buck converter for 3.3 V at 2 A from a 12 V rail, in a 40 °C ambient. Include the heat-sink implication.
The linear regulator passes the full 2 A and drops 8.7 V, so \(P_D = (12-3.3)(2) + (12)(0.005) = 17.4 + 0.06 = 17.46\) W to deliver \(3.3 \times 2 = 6.6\) W, an efficiency of \(100(6.6)/[(12)(2.005)] = 27.4\) per cent. The thermal requirement is brutal: \(\theta_{JA(\max)} = (125-40)/17.46 = 4.87\ ^\circ\)C/W, and subtracting \(\theta_{JC}+\theta_{CS} = 6\ ^\circ\)C/W leaves a negative allowance for the sink. No heat sink of any size will do it; the job needs forced air, a lower-\(\theta_{JC}\) package, or two regulators sharing the current. The buck converter runs at \(D = 3.3/12 = 27.5\) per cent. With a 30 mΩ MOSFET, a 0.4 V Schottky, a 50 mΩ inductor, 50 ns transitions at 300 kHz and 3 mA of controller current, the losses are 0.033 W of switch conduction, 0.580 W in the diode, 0.360 W of switching, 0.200 W in the inductor and 0.036 W in the controller: 1.21 W in total, for an efficiency of \(100(6.6)/7.81 = 84.5\) per cent and an input current of 0.651 A instead of 2.005 A. The heat has fallen by a factor of 14 and no heat sink is needed. Two things are worth noticing. The diode is now the largest single loss, which is why a synchronous buck replaces it with a second MOSFET and reaches above 90 per cent. And the efficiency is markedly lower than the 97 per cent of Worked Example 30.4 because the conversion ratio is much larger and the fixed diode drop is a bigger fraction of a 3.3 V output — a switching regulator is not equally good at every ratio.