Series, Shunt and IC Voltage Regulators

About this problem set

Line and load regulation from bench measurements, discrete series and shunt regulators, op-amp error-amplifier regulators with a sampling divider, short-circuit current limiting, and the 78xx three-terminal regulator in both fixed and raised-output configurations, carried through to the dropout constraint and the heat-sink calculation.

Each problem below is solved in full — the given data is listed first, the governing relation is stated, and the arithmetic is carried through to a boxed answer. Work through the statement yourself before opening the solution. The underlying theory is covered in the Electronic Devices lecture series.

PROBLEM 01

Line and Load Regulation from Bench Data

Problem Statement

A 5 V regulator is put on the bench and two sweeps are recorded.

Sweep	Condition	Measured \(V_O\)
Load	\(V_{IN} = 12~\text{V}\), no load \((I_L = 0)\)	5.030 V
Load	\(V_{IN} = 12~\text{V}\), full load \((I_L = 500~\text{mA})\)	4.955 V
Line	\(V_{IN} = 10~\text{V}\), \(I_L\) held at 500 mA	4.996 V
Line	\(V_{IN} = 15~\text{V}\), \(I_L\) held at 500 mA	5.012 V

Compute the percentage load regulation and the equivalent output resistance.
Compute the line regulation, both in \(\text{mV}/\text{V}\) and as a percentage per volt.
The unregulated input carries 2 V peak-to-peak of ripple. Estimate the output ripple and the ripple rejection in dB.

Solution

No load \(V_{NL} = 5.030~\text{V}\); full load \(V_{FL} = 4.955~\text{V}\) at \(I_L = 500~\text{mA}\)
Line sweep: \(V_{IN}\) from 10 V to 15 V gives \(V_O\) from 4.996 V to 5.012 V
The two sweeps are independent: only one variable is moved at a time.

Part (a) — load regulation

Load regulation is the fractional droop from no load to full load, referred to the full-load value:

\[\begin{aligned} \text{Load reg} &= \frac{V_{NL} - V_{FL}}{V_{FL}} \times 100 = \frac{5.030 - 4.955}{4.955} \times 100 = 1.514~\% \\ R_O &= \frac{\Delta V_O}{\Delta I_L} = \frac{75~\text{mV}}{500~\text{mA}} = 0.150~\Omega \end{aligned}\]

Load regulation \( = 1.51~\% \); Thevenin output resistance \( R_O = 0.15~\Omega \).

Part (b) — line regulation

Line regulation is the output shift per volt of input shift, the load being held fixed:

\[\begin{aligned} \frac{\Delta V_O}{\Delta V_{IN}} &= \frac{5.012 - 4.996}{15 - 10} = \frac{16~\text{mV}}{5~\text{V}} = 3.2~\text{mV}/\text{V} \\ \text{Line reg} &= \frac{\Delta V_O / V_O}{\Delta V_{IN}} \times 100 = \frac{16~\text{mV} / 5.004~\text{V}}{5~\text{V}} \times 100 = 0.0639~\%/\text{V} \end{aligned}\]

Over the whole 10 V to 15 V window the output moves by \(0.320~\%\) in total.

\(3.2~\text{mV}/\text{V}\), i.e. \(0.064~\%/\text{V}\) (\(0.32~\%\) across the full input range).

Part (c) — ripple rejection

Ripple is just a fast line variation, so the same \(\Delta V_O/\Delta V_{IN}\) applies:

\[\begin{aligned} V_{r(\text{out})} &= V_{r(\text{in})} \times \frac{\Delta V_O}{\Delta V_{IN}} = 2~\text{V} \times 0.0032 = 6.4~\text{mV p-p} \\ \text{RR} &= 20\log_{10}\frac{V_{r(\text{in})}}{V_{r(\text{out})}} = 20\log_{10}\frac{2}{0.0064} = 49.9~\text{dB} \end{aligned}\]

Output ripple \(\approx 6.4~\text{mV}\) p-p; ripple rejection \( = 50~\text{dB}\).

Note that load regulation and output resistance are the same measurement expressed two ways — a 0.15 \(\Omega\) source is what produces a 75 mV droop at 500 mA. Both figures improve in direct proportion to the loop gain of the regulator's error amplifier.

PROBLEM 02

Emitter-Follower Series Regulator Currents

Problem Statement

A discrete series regulator uses an npn pass transistor as an emitter follower. The unregulated input is \(V_{IN} = 15~\text{V}\) nominal, swinging between 14 V and 16 V. A 470 \(\Omega\) resistor runs from the input rail to the base, and an 8.2 V Zener holds the base to ground. The emitter feeds a 100 \(\Omega\) load. Take \(\beta = 50\), \(V_{BE} = 0.7~\text{V}\) and \(I_{ZK} = 2~\text{mA}\).

Find the regulated output voltage and the load current.
Find the resistor current, the base current and the Zener current at nominal input, and the worst-case Zener dissipation.
Find the worst-case power dissipated in the pass transistor.
Find the lowest input voltage at which regulation is still held.

Solution

\(V_{IN} = 15~\text{V}\) (14–16 V), \(V_Z = 8.2~\text{V}\), \(R = 470~\Omega\)
\(R_L = 100~\Omega\), \(\beta = 50\), \(V_{BE} = 0.7~\text{V}\)
Ideal Zener (zero \(r_Z\)) above the knee; pass transistor in the active region throughout.

Part (a) — output voltage

The Zener fixes the base; the emitter follows it one junction drop below:

\[\begin{aligned} V_O &= V_Z - V_{BE} = 8.2 - 0.7 = 7.5~\text{V} \\ I_L &= \frac{V_O}{R_L} = \frac{7.5~\text{V}}{100~\Omega} = 75~\text{mA} \end{aligned}\]

\(V_O = 7.5~\text{V}\), \(I_L = 75~\text{mA}\)

Part (b) — resistor, base and Zener currents

The resistor carries the sum of the Zener current and the base current. Since the pass transistor is a follower, \(I_B = I_E/(\beta+1)\):

\[\begin{aligned} I_R &= \frac{V_{IN} - V_Z}{R} = \frac{15 - 8.2}{470~\Omega} = 14.468~\text{mA} \\ I_B &= \frac{I_L}{\beta + 1} = \frac{75~\text{mA}}{51} = 1.471~\text{mA} \\ I_Z &= I_R - I_B = 14.468 - 1.471 = 12.997~\text{mA} \end{aligned}\]

The Zener is worst off at maximum input, where \(I_R\) is largest:

\[\begin{aligned} I_{R(\max)} &= \frac{16 - 8.2}{470} = 16.596~\text{mA}, \qquad I_{Z(\max)} = 15.125~\text{mA} \\ P_{Z(\max)} &= V_Z I_{Z(\max)} = (8.2)(15.13~\text{mA}) = 124~\text{mW} \end{aligned}\]

\(I_R = 14.47~\text{mA}\), \(I_B = 1.47~\text{mA}\), \(I_Z = 13.00~\text{mA}\); \(P_{Z(\max)} = 124~\text{mW}\) — a 400 mW device is ample.

Part (c) — pass-transistor dissipation

The pass device carries the full load current and drops the difference between input and output:

\[\begin{aligned} P_Q &= (V_{IN} - V_O)\,I_E \approx (V_{IN} - V_O)\,I_L \\ &= (15 - 7.5)(75~\text{mA}) = 0.5625~\text{W} \quad \text{(nominal)} \\ P_{Q(\max)} &= (16 - 7.5)(75~\text{mA}) = 0.6375~\text{W} \quad \text{(high line)} \end{aligned}\]

\(P_Q = 0.562~\text{W}\) nominal, \(0.637~\text{W}\) worst case — a TO-220 part without a heat sink will do.

Part (d) — dropout

Regulation fails when the Zener falls below its knee current \(I_{ZK} = 2~\text{mA}\). The resistor must still deliver \(I_{ZK} + I_B\):

\[\begin{aligned} V_{IN(\min)} &= V_Z + R\,(I_{ZK} + I_B) \\ &= 8.2 + (470)(2 + 1.471)~\text{mA} = 9.831~\text{V} \end{aligned}\]

Regulation is held down to \(V_{IN} = 9.83~\text{V}\).

Efficiency here is only \(V_O/V_{IN} = 50.0~\%\): a linear series regulator burns the input–output difference as heat, and the price of a low dropout figure is a large \(R\), which in turn starves the base at high load. That trade-off is exactly what the error-amplifier version removes.

PROBLEM 03

Series Regulator with Error Amplifier

Problem Statement

A series regulator uses an op-amp error amplifier. A 5.1 V Zener reference drives the non-inverting input; the inverting input is fed from a sampling divider across the output, with \(R_1\) from the output node to the divider tap and \(R_2\) from the tap to ground. The op-amp output drives the base of an npn pass transistor whose collector sits on the unregulated rail \(V_{IN} = 18~\text{V}\). Initially \(R_1 = 22~\text{k}\Omega\) and \(R_2 = 18~\text{k}\Omega\); the load draws 250 mA and the pass transistor has \(\beta = 60\).

Derive \(V_O = V_{REF}(1 + R_1/R_2)\) and evaluate it.
Find the pass-transistor dissipation and the current the op-amp must source into the base. Comment.
Design. Keeping \(R_2 = 10~\text{k}\Omega\), choose \(R_1\) from E24 values for \(V_O = 15.0~\text{V}\), and quote the error.

Solution

\(V_{REF} = 5.1~\text{V}\), \(V_{IN} = 18~\text{V}\), \(I_L = 250~\text{mA}\), \(\beta = 60\)
Ideal op-amp: infinite gain, no input current, negative feedback closed through the pass transistor.
The divider current is much larger than the op-amp bias current.

Part (a) — the sampling relation

With negative feedback closed, the op-amp forces its two inputs to the same potential, so the divider tap must sit at \(V_{REF}\):

\[\begin{aligned} V_O\,\frac{R_2}{R_1 + R_2} &= V_{REF} \\ \Rightarrow\; V_O &= V_{REF}\left(1 + \frac{R_1}{R_2}\right) \\ &= 5.1\left(1 + \frac{22}{18}\right) = 5.1 \times 2.2222 = 11.333~\text{V} \end{aligned}\]

\(V_O = 11.33~\text{V}\)

Part (b) — dissipation and base drive

\[\begin{aligned} P_Q &= (V_{IN} - V_O) I_L = (18 - 11.333)(250~\text{mA}) = 1.6667~\text{W} \\ I_B &= \frac{I_L}{\beta+1} = \frac{250~\text{mA}}{61} = 4.098~\text{mA} \end{aligned}\]

A general-purpose op-amp is typically guaranteed to source only about 5 mA, so 4.10 mA leaves almost no margin. The fix is a Darlington pass pair: with an added \(\beta = 60\) driver the required base current falls to \(I_L/(\beta+1)^2 = 67~\mu\text{A}\).

\(P_Q = 1.67~\text{W}\), \(I_B = 4.10~\text{mA}\) — marginal for a single pass transistor; use a Darlington.

Part (c) — design of the divider for 15.0 V

Invert the sampling relation:

\[\begin{aligned} \frac{R_1}{R_2} &= \frac{V_O}{V_{REF}} - 1 = \frac{15.0}{5.1} - 1 = 1.94118 \\ R_1 &= 19.4118~\text{k}\Omega \;\Rightarrow\; 18~\text{k}\Omega + 1.5~\text{k}\Omega = 19.5~\text{k}\Omega\ \text{(E24 in series)} \end{aligned}\]

Check the value actually obtained:

\[\begin{aligned} V_O &= 5.1\left(1 + \frac{19.5}{10}\right) = 15.0450~\text{V} \quad (\text{error} = +0.30~\%) \\ I_{\text{div}} &= \frac{V_O}{R_1 + R_2} = \frac{15.045}{29.5~\text{k}\Omega} = 510~\mu\text{A} \end{aligned}\]

\(R_1 = 19.5~\text{k}\Omega\) (18 k\(\Omega\) + 1.5 k\(\Omega\)), \(R_2 = 10~\text{k}\Omega\); \(V_O = 15.045~\text{V}\), error \(+0.30~\%\). Pass dissipation rises to \(0.74~\text{W}\).

The divider current of 510 \(\mu\)A is three orders of magnitude above a typical bipolar op-amp input bias current, so loading of the tap is negligible; making the divider much higher in resistance to save power would start to introduce an offset error.

PROBLEM 04

Short-Circuit Current Limiting Design

Problem Statement

A series regulator delivers \(V_O = 12~\text{V}\) from \(V_{IN} = 20~\text{V}\). Constant-current limiting is added: a sense resistor \(R_{SC}\) sits in series with the output between the pass-transistor emitter and the load, and the base–emitter junction of a limiting transistor is connected directly across \(R_{SC}\). When that junction reaches \(V_{BE} = 0.7~\text{V}\) the limiter steals base drive from the pass device.

With \(R_{SC} = 1~\Omega\), find the limiting current, and comment on the normal-load condition of 500 mA.
Find the pass-transistor dissipation with the output shorted, and compare it with the normal-load figure.
Design. Re-specify \(R_{SC}\) for a limit of 500 mA with a rated load of 400 mA. Choose an E24 value and give its power rating.
State the short-circuit dissipation with the redesigned sense resistor and what you would do about it.

Solution

\(V_{IN} = 20~\text{V}\), \(V_O = 12~\text{V}\), \(V_{BE(\text{on})} = 0.7~\text{V}\) for the limiting transistor
The limiter conducts hard once \(I_L R_{SC}\) reaches \(V_{BE}\); below that it is off.
Constant-current (brick-wall) limiting, not foldback.

Part (a) — limit set by the sense resistor

\[I_{\lim} = \frac{V_{BE}}{R_{SC}} = \frac{0.7~\text{V}}{1~\Omega} = 0.700~\text{A}\]

At the normal load of 500 mA the sense drop is \((500~\text{mA})(1~\Omega) = 0.5~\text{V}\). That is already within 0.2 V of the turn-on threshold, so the limiter has begun to conduct weakly and 0.5 V of headroom has been thrown away in the sense resistor. The limit sits only 1.40 times above the rated load — too close.

\(I_{\lim} = 700~\text{mA}\); at 500 mA the 0.5 V sense drop already encroaches on the limit.

Part (b) — dissipation under short circuit

With the output shorted, \(V_O = 0\) and the current is held at \(I_{\lim}\). The pass transistor then drops everything except the sense-resistor volt drop:

\[\begin{aligned} P_{Q(\text{short})} &= (V_{IN} - I_{\lim}R_{SC})\,I_{\lim} = (20 - 0.7)(0.7~\text{A}) = 13.51~\text{W} \\ P_{Q(\text{normal})} &= (V_{IN} - V_O - I_L R_{SC})\,I_L = (20 - 12 - 0.5)(0.5~\text{A}) = 3.75~\text{W} \end{aligned}\]

\(P_Q\) jumps from 3.75 W to 13.5 W — a factor of 3.6 — when the output is shorted.

Part (c) — design of \(R_{SC}\)

\[\begin{aligned} R_{SC} &= \frac{V_{BE}}{I_{\lim}} = \frac{0.7~\text{V}}{0.5~\text{A}} = 1.4~\Omega \;\Rightarrow\; R_{SC} = 1.5~\Omega\ \text{(E24)} \\ I_{\lim} &= \frac{0.7}{1.5} = 0.4667~\text{A} = 467~\text{mA} \end{aligned}\]

Check the headroom at the 400 mA rated load and size the resistor:

\[\begin{aligned} V_{\text{sense}} &= (0.4~\text{A})(1.5~\Omega) = 0.60~\text{V} \;<\; V_{BE} \quad \checkmark \\ P_{R_{SC}} &= I^2 R_{SC} = (0.4667)^2(1.5) = 0.327~\text{W} \quad \text{(worst case, at the limit)} \end{aligned}\]

\(R_{SC} = 1.5~\Omega\), giving \(I_{\lim} = 467~\text{mA}\); dissipation 0.327 W (0.240 W at rated load), so a 1 W wirewound part.

Part (d) — the residual problem

\[P_{Q(\text{short})} = (20 - 0.7)(0.4667) = 9.01~\text{W}\]

Still 9.0 W into a short — use foldback limiting, or heat-sink the pass device for continuous short-circuit operation.

Constant-current limiting has a perverse property: the pass transistor dissipates most when the output is shorted, because the current is held up while the voltage across the device rises to the full input. Foldback limiting reduces the current as \(V_O\) collapses and removes this worst case entirely.

PROBLEM 05

Shunt Regulator Currents and Efficiency

Problem Statement

In a shunt regulator the control element sits across the load rather than in series with it. An unregulated rail \(V_{IN} = 15~\text{V}\) (range 13–17 V) feeds a series resistor \(R_S = 220~\Omega\) to the output node. The control element is an npn transistor whose collector is tied to the output node and whose emitter returns to ground through a 8.3 V Zener; its base is fed from the output node through a small resistor \(R_B\), so that \(V_B \approx V_O\). The load is \(R_L = 1~\text{k}\Omega\) and \(V_{BE} = 0.7~\text{V}\).

Find \(V_O\), the series-resistor current, the load current and the shunt-transistor current at nominal input.
Find the worst-case dissipation in the shunt transistor, in the Zener and in \(R_S\), and say when it occurs.
Find the largest load current the regulator can hold if the shunt branch must retain at least 2 mA.
Find the efficiency at nominal input and full load.

Solution

\(V_{IN} = 15~\text{V}\) (13–17 V), \(R_S = 220~\Omega\), \(R_L = 1~\text{k}\Omega\)
\(V_Z = 8.3~\text{V}\), \(V_{BE} = 0.7~\text{V}\); Zener stiff, transistor in the active region
Base current negligible, so \(V_B = V_O\) and the whole shunt current flows collector-to-emitter and on through the Zener.
The current through \(R_S\) splits between the load and the shunt branch.

Part (a) — output voltage and the three currents

The emitter sits at \(V_Z\) and the base, tied to the output, sits one junction drop above it. That is the whole regulating mechanism: if \(V_O\) tries to rise, \(V_{BE}\) rises, the shunt current increases sharply, the drop across \(R_S\) grows and \(V_O\) is pulled back:

\[\begin{aligned} V_O &= V_Z + V_{BE} = 8.3 + 0.7 = 9.0~\text{V} \\ I_S &= \frac{V_{IN} - V_O}{R_S} = \frac{15 - 9.0}{220~\Omega} = 27.27~\text{mA} \\ I_L &= \frac{V_O}{R_L} = \frac{9.0}{1~\text{k}\Omega} = 9.0~\text{mA} \\ I_{sh} &= I_S - I_L = 27.27 - 9.0 = 18.27~\text{mA} \end{aligned}\]

\(V_O = 9.0~\text{V}\), \(I_S = 27.3~\text{mA}\), \(I_L = 9.0~\text{mA}\), \(I_{sh} = 18.3~\text{mA}\)

Part (b) — worst-case dissipation

\(I_S\) is largest at high line, and the shunt branch must absorb all of it when the load is removed. So the worst case is high line, no load. The transistor drops \(V_O - V_Z = V_{BE}\), and the Zener drops \(V_Z\):

\[\begin{aligned} I_{S(\max)} &= \frac{17 - 9.0}{220} = 36.36~\text{mA} \\ P_{Q(\max)} &= (V_O - V_Z) I_{S(\max)} = (0.7)(36.36~\text{mA}) = 25.5~\text{mW} \\ P_{Z(\max)} &= V_Z I_{S(\max)} = (8.3)(36.36~\text{mA}) = 302~\text{mW} \\ P_{R_S(\max)} &= \frac{(V_{IN(\max)} - V_O)^2}{R_S} = \frac{(8.0)^2}{220} = 291~\text{mW} \end{aligned}\]

Worst case is high line with no load: \(P_Q = 25~\text{mW}\), \(P_Z = 302~\text{mW}\) (use a 500 mW Zener), \(P_{R_S} = 291~\text{mW}\) (use a 0.5 W series resistor).

Part (c) — maximum load current

The tightest condition is low line, where \(I_S\) is smallest:

\[\begin{aligned} I_{S(\min)} &= \frac{13 - 9.0}{220} = 18.18~\text{mA} \\ I_{L(\max)} &= I_{S(\min)} - I_{sh(\min)} = 18.18 - 2 = 16.18~\text{mA} \end{aligned}\]

\(I_{L(\max)} = 16.2~\text{mA}\) — the actual 9.0 mA load is comfortably inside it.

Part (d) — efficiency

\[\begin{aligned} P_{\text{in}} &= V_{IN} I_S = (15)(27.27~\text{mA}) = 409.1~\text{mW} \\ P_{\text{out}} &= V_O I_L = (9.0)(9.0~\text{mA}) = 81.0~\text{mW} \\ \eta &= \frac{P_{\text{out}}}{P_{\text{in}}} \times 100 = 19.80~\% \end{aligned}\]

\(\eta = 19.8~\%\)

The shunt regulator draws a nearly constant current from the source whatever the load does, so its efficiency is poor and it is at its hottest when doing the least work. Against that, it is inherently short-circuit proof — a shorted output simply turns the shunt element off and \(R_S\) limits the current to \(V_{IN}/R_S = 68.2~\text{mA}\).

PROBLEM 06

Adjustable 7812 Regulator with Heat Sink

Problem Statement

A 7812 three-terminal regulator holds \(V_{REG} = 12~\text{V}\) between its output and ground pins, draws a quiescent current \(I_Q = 4.2~\text{mA}\) out of the ground pin, and needs at least \(2~\text{V}\) of dropout between input and output. Its unregulated supply delivers \(18~\text{V}\) DC with \(3~\text{V}\) peak-to-peak of ripple, and the load draws 500 mA.

Used as a fixed 12 V regulator, is the dropout requirement met over the whole ripple cycle?
Design. The output is raised by lifting the ground pin on a divider: \(R_1 = 240~\Omega\) from the output pin to the ground pin, \(R_2\) from the ground pin to ground, giving \(V_O = V_{REG}(1 + R_2/R_1) + I_Q R_2\). Choose \(R_2\) from E24 values for \(V_O = 18~\text{V}\).
State the minimum DC input the raised-output version needs.
With the supply raised to \(24~\text{V}\), find the device dissipation and the heat-sink thermal resistance required for \(T_{J(\max)} = 125^\circ\)C in a \(50^\circ\)C ambient. Take \(\theta_{JC} = 5^\circ\)C/W and \(\theta_{CS} = 1^\circ\)C/W.

Solution

\(V_{REG} = 12~\text{V}\), \(I_Q = 4.2~\text{mA}\), dropout \(= 2~\text{V}\)
Unregulated rail: \(18~\text{V}\) DC with \(3~\text{V}\) p-p ripple; \(I_L = 500~\text{mA}\)
\(I_Q\) is taken as constant; the regulator holds \(V_{REG}\) between the output and ground pins whatever the ground pin sits at.

Part (a) — dropout on the fixed connection

The input must stay above \(V_O + V_{\text{dropout}}\) at the trough of the ripple, not on average:

\[\begin{aligned} V_{IN(\min)} &= V_{DC} - \tfrac{1}{2}V_r = 18 - 1.5 = 16.5~\text{V} \\ V_{IN}\text{ required} &= V_O + V_{\text{dropout}} = 12 + 2 = 14~\text{V} \end{aligned}\]

\(16.5~\text{V} > 14~\text{V}\) — the dropout requirement is met, with 2.5 V to spare.

Part (b) — design of \(R_2\) for 18 V

The regulator forces \(V_{REG}\) across \(R_1\), so \(R_1\) carries \(V_{REG}/R_1\); \(R_2\) carries that current plus \(I_Q\). Solving the given relation for \(R_2\):

\[\begin{aligned} V_O &= V_{REG} + \left(\frac{V_{REG}}{R_1} + I_Q\right) R_2 \\ R_2 &= \frac{V_O - V_{REG}}{V_{REG}/R_1 + I_Q} = \frac{18 - 12}{12/240 + 0.0042} = \frac{6}{0.0542~\text{A}} = 110.70~\Omega \end{aligned}\]

The nearest E24 value is 110 \(\Omega\). Back-substitute:

\[\begin{aligned} V_O &= 12\left(1 + \frac{110}{240}\right) + (4.2~\text{mA})(110~\Omega) \\ &= 17.500 + 0.462 = 17.962~\text{V} \end{aligned}\]

\(R_2 = 110~\Omega\) gives \(V_O = 17.96~\text{V}\), an error of \(-0.21~\%\) on the 18 V target.

The \(I_Q R_2\) term contributes 0.46 V here — 2.6% of the output. It cannot be ignored, and because \(I_Q\) drifts with temperature and load it is the main source of error in this configuration.

Part (c) — dropout on the raised output

Dropout is measured between the input pin and the output pin, which now sits at \(V_O\):

\[\begin{aligned} V_{IN(\min)} &= V_O + V_{\text{dropout}} = 17.962 + 2 = 19.96~\text{V} \\ V_{DC} &\ge V_{IN(\min)} + \tfrac{1}{2}V_r = 19.96 + 1.5 = 21.46~\text{V} \end{aligned}\]

The rail must supply at least \(21.5~\text{V}\) DC — the original 18 V supply is no longer adequate.

Part (d) — dissipation and heat sink

\[\begin{aligned} P_D &= (V_{IN} - V_O)I_L + V_{IN} I_Q \\ &= (24 - 17.962)(0.5~\text{A}) + (24)(4.2~\text{mA}) = 3.0190 + 0.1008 = 3.120~\text{W} \end{aligned}\]

The thermal path is junction \(\to\) case \(\to\) sink \(\to\) ambient in series:

\[\begin{aligned} T_J &= T_A + P_D(\theta_{JC} + \theta_{CS} + \theta_{SA}) \\ \theta_{SA} &\le \frac{T_{J(\max)} - T_A}{P_D} - \theta_{JC} - \theta_{CS} \\ &= \frac{125 - 50}{3.120} - 5 - 1 = 24.04 - 6 = 18.04~^\circ\text{C/W} \end{aligned}\]

Without a sink a TO-220 part has \(\theta_{JA} \approx 65^\circ\)C/W, which would give \(T_J = 50 + 65 \times 3.120 = 253^\circ\)C — far beyond the rating, so the internal thermal shutdown would fold the output back.

\(P_D = 3.12~\text{W}\); a heat sink of \(\theta_{SA} \le 18.0~^\circ\text{C/W}\) is required.

Notice how much the 24 V rail costs: every extra volt of input–output difference at 500 mA adds 500 mW of heat. Sizing the transformer so that the rail sits just above the dropout limit at the ripple trough is the single most effective thermal decision in a linear supply.