By the end of this chapter you should be able to:
- Explain radiative and non-radiative recombination and why an indirect band gap makes silicon a poor emitter and a good detector.
- Convert between band gap and emission wavelength using \(\lambda = hc/E_g\) and identify the material for a required colour.
- Size the series resistor for an LED, compute the drive efficiency and assess the sensitivity to forward-voltage tolerance.
- Define responsivity and quantum efficiency, and compute a photodiode's photocurrent, dark current and noise-equivalent power.
- Distinguish photovoltaic from photoconductive operation and choose between them for a stated application.
- Compute the fill factor and conversion efficiency of a solar cell from its measured \(I_{sc}\), \(V_{oc}\) and maximum-power point.
- Design an optocoupler interface using the worst-case current transfer ratio, including ageing derating.
A p-n junction moves charge across an energy gap. Every chapter so far has been about the charge; this one is about the energy. When an electron in the conduction band recombines with a hole in the valence band it must give up roughly \(E_g\) of energy, and that energy has to go somewhere — into lattice vibrations as heat, or into a photon as light. Run the process backwards and a photon absorbed in the depletion region creates an electron-hole pair that the built-in field sweeps apart, producing a current. One junction, two directions, and an entire family of devices.
The devices divide cleanly. Emitters convert current into light: the light-emitting diode, and by extension the laser diode. Detectors convert light into current: the photodiode, the phototransistor and the photoconductive cell. Energy converters are detectors operated to deliver power rather than to signal: the solar cell is a large photodiode working in a quadrant where it is a source. And couplers put an emitter and a detector in one package with no electrical connection between them, giving isolation that no transformer can match for bandwidth or size. Displays — the seven-segment indicator and the liquid-crystal panel — are emitters and modulators arranged for a human reader.
The chapter derives the wavelength of an LED from its band gap, explains why silicon can detect light efficiently but cannot emit it at all, sizes the series resistor that every LED needs, computes the responsivity and noise floor of a photodiode, works the fill factor and efficiency of a real silicon solar cell from its four measured numbers, and designs a logic isolator around an optocoupler's current transfer ratio including the derating that its ageing demands.
1 Radiative Recombination and the Silicon Problem
An electron in the conduction band is in an excited state, and it will eventually fall into an empty valence state. Chapter 4 treated recombination as a loss mechanism characterised by a lifetime \(\tau\); here the question is what happens to the energy.
In radiative recombination the electron drops directly into the hole and the energy \(E_g\) leaves as a photon. In non-radiative recombination the electron falls in stages through defect or impurity states in the forbidden gap, giving up its energy in small parcels to lattice vibrations — heat. Both processes always occur; the fraction of recombinations that are radiative is the internal quantum efficiency, and it is what separates a light-emitting material from an ordinary one.
Which process dominates is decided not by the size of the gap but by its shape. An electron state is described by its energy and its crystal momentum \(\hbar k\), and both must be conserved in any transition. A photon of visible light carries a large energy and almost no momentum, so a photon-only transition must be vertical on an energy-momentum diagram.
Figure 14.1 contrasts the two cases. In a direct-gap material such as gallium arsenide, the conduction-band minimum lies directly above the valence-band maximum at \(k = 0\). An electron that has relaxed to the bottom of the conduction band therefore sits at the same momentum as the hole it will recombine with, and a photon alone can conserve everything. Radiative recombination is fast, with lifetimes of nanoseconds, and internal quantum efficiencies approach unity.
In an indirect-gap material such as silicon or germanium, the conduction-band minimum is displaced in momentum from the valence-band maximum. Recombination now needs a photon to carry the energy and a phonon to carry the momentum difference, and a process requiring three bodies to meet is far less likely than one requiring two. Radiative lifetimes in silicon are milliseconds, while non-radiative recombination through defect states happens in microseconds, so essentially every recombination in silicon produces heat rather than light.
That is why there is no silicon LED. It is also why silicon is an excellent detector: absorption is the reverse process, and although an indirect material absorbs weakly it does absorb, so a photodiode simply needs to be thick enough — tens of micrometres rather than a fraction of one. The whole of Part 1's argument for silicon as the material of choice for transistors is untouched; silicon is bad at exactly one job, and compound semiconductors do that job instead.
The emitted photon's energy is close to \(E_g\) but not exactly equal to it, because the recombining carriers have thermal energy of their own. The distribution has a full width of roughly \(1.8kT = 46.5\) meV at 300 K, which translates into a wavelength spread of
giving 15 nm at 630 nm and 33 nm at 940 nm. An LED is therefore not monochromatic in the way a laser is, but it is far narrower than a filtered incandescent lamp, which is why LED colours look saturated.
2 LED Materials and Wavelength
An LED is a forward-biased direct-gap junction in which injected minority carriers recombine radiatively. Two design decisions follow: the material sets the colour, and the packaging sets how much of the light escapes.
The colour follows from a single conversion. A photon of energy \(E\) has wavelength \(\lambda = hc/E\), and with \(hc = 1240\) eV·nm the working form is
Worth memorising in both directions: 1.24 eV corresponds to 1000 nm, and the visible range 400 to 700 nm corresponds to 3.10 down to 1.77 eV. Any emitter of visible light therefore needs a gap wider than 1.77 eV, which immediately rules out silicon (1.12 eV) and germanium (0.66 eV) even before the indirect-gap argument.
| Material | \(E_g\) (eV) | \(\lambda\) (nm) | Appearance |
|---|---|---|---|
| Si (indirect) | 1.12 | 1107 | None — heat only |
| GaAs | 1.42 | 873 | Infrared, remote controls |
| AlGaAs | 1.79 | 693 | Deep red |
| GaAsP | 1.90 | 653 | Red |
| GaAsP | 2.03 | 611 | Orange |
| GaAsP:N | 2.10 | 590 | Yellow |
| GaP | 2.26 | 549 | Green |
| InGaN | 2.70 | 459 | Blue |
| InGaN | 3.10 | 400 | Violet / near UV |
The table explains the historical order in which colours appeared. Red arrived in 1962, because GaAsP alloys near 1.9 eV were straightforward. Green followed, using nitrogen-doped GaP, which is indirect but has an isoelectronic trap that makes radiative recombination possible at reduced efficiency. Blue took until the 1990s because it needs a gap above 2.6 eV, and the wide-gap materials that provide it — gallium nitride and its indium alloys — could not be grown with usable crystal quality or doped p-type. Solving that problem produced the white LED, which is not a white emitter at all but a 460 nm InGaN die under a phosphor that converts part of the blue to yellow.
Getting the light out is a separate difficulty. GaAs has a refractive index of 3.6, so the critical angle at a flat semiconductor-air interface is \(\arcsin(1/3.6) = 16.1^\circ\), and only about 2 % of the light generated isotropically inside the die falls within that cone. The rest is totally internally reflected and eventually absorbed. Encapsulating the die in an epoxy dome of index 1.5 widens the critical angle and, because the dome is curved, presents a near-normal surface to most rays; roughening the die surface and adding a reflector cup do the rest. The external quantum efficiency is what remains after all of this, and for a good visible LED it is 20 to 50 %.
An LED's forward drop is not 0.7 V. Conduction requires carriers to be injected across a gap of \(E_g\), so \(V_F\) is roughly \(E_g/q\) plus the ohmic drops: about 1.8 V for a red device, 2.1 V for yellow, 3.2 V for blue and 3.4 V for white. This has a practical consequence that catches people out constantly — a blue or white LED simply cannot be driven from a 3.3 V rail through a resistor, because the available headroom of 0.1 V is smaller than the part-to-part spread in \(V_F\). At 5 V with a 5 \(\Omega\) resistor, a device with \(V_F = 3.2\) V draws 20 mA while one with \(V_F = 3.0\) V draws 60 mA and burns out. Low-headroom LEDs need a current source, not a resistor.
3 Driving an LED
An LED is a diode, so its current rises exponentially with voltage and it must never be driven from a voltage source. The universal solution is a series resistor:
with \(I_F\) chosen from the datasheet — 20 mA is the traditional figure for an indicator, though modern high-efficiency devices are visible at 2 mA. The resistor dissipates \((V_S-V_F)I_F\), which is often more than the LED itself does.
| Supply | LED | \(I_F\) | \(R\) | \(P_R\) | Drive efficiency \(V_F/V_S\) |
|---|---|---|---|---|---|
| 5 V | Red, 2.0 V | 20 mA | 150 \(\Omega\) | 60 mW | 40 % |
| 5 V | White, 3.4 V | 20 mA | 80 \(\Omega\) | 32 mW | 68 % |
| 12 V | Red, 2.0 V | 20 mA | 500 \(\Omega\) | 200 mW | 17 % |
| 5 V | Red, 1.8 V | 10 mA | 320 \(\Omega\) | 32 mW | 36 % |
Two things follow. Resistive drive wastes the headroom, so a 12 V indicator throws away five times the power it uses; where efficiency matters, LEDs are driven from a switching current regulator instead. And the current depends on the difference between two numbers that are both uncertain, so the tolerance on \(V_F\) propagates into the current. With \(V_S = 5\) V and \(R = 150\ \Omega\), a \(V_F\) spread of 1.8 V to 2.2 V gives 21.3 mA to 18.7 mA — a spread of \(\pm 7\ \%\), which is tolerable. The same LED on a 2.5 V supply with a 25 \(\Omega\) resistor would give 28 mA to 12 mA, a spread of \(\pm 40\ \%\), and a visible brightness difference between units.
The current's sensitivity to \(V_F\) is \(dI/dV_F = -1/R = -I_F/(V_S-V_F)\), so the fractional current error is the fractional \(V_F\) error multiplied by \(V_F/(V_S-V_F)\). With 3 V of headroom that multiplier is 0.67 and the design is forgiving; with 0.1 V of headroom it is 32 and the design is unbuildable. This single ratio decides whether a resistor will do or whether a current source is required, and it is worth evaluating before choosing the supply rail.
Multiplexed drive. A four-digit seven-segment display has 28 segments, and driving each continuously needs 28 resistors and 28 output pins. Instead the digits are enabled one at a time in rapid rotation, so only eight segment lines and four digit lines are needed. Each digit is lit for a quarter of the time, so to obtain the same apparent brightness as 10 mA of continuous current the peak current must be 40 mA. An LED tolerates this because its rating is thermal and the average dissipation is unchanged; datasheets quote a peak forward current of 100 mA or more at low duty cycle for exactly this purpose. The refresh rate must exceed about 100 Hz per digit to avoid visible flicker, so a four-digit display is scanned at 400 Hz or faster.
Three further practical points. An LED has a reverse breakdown voltage of only about 5 V, so an LED across an AC source needs an ordinary diode in inverse parallel to protect it. Light output falls with temperature at roughly \(-0.5\ \%/^\circ\text{C}\) for red devices, and since \(V_F\) also falls at about \(-2\) mV/°C, a resistor-driven LED gets slightly brighter as its resistor's share of the voltage grows — a mild positive feedback that matters only in power LEDs. And LEDs in parallel must never share one resistor: the device with the lowest \(V_F\) takes most of the current and the others stay dim, so each branch needs its own resistor.
4 Photodiodes: Responsivity, Modes and Noise
Run the junction backwards. A photon with energy greater than \(E_g\) absorbed in or near the depletion region creates an electron-hole pair; the built-in field sweeps the electron towards the n side and the hole towards the p side; the result is a current flowing from cathode to anode inside the device, that is in the reverse direction. The complete characteristic is the ordinary diode equation with that photocurrent subtracted:
The whole family of curves is the dark characteristic translated downwards by \(I_{ph}\), which is proportional to the incident optical power over many decades — typically eight or nine, which is what makes the photodiode a measurement device rather than merely a detector.
Responsivity \(\mathcal{R}\) is the photocurrent per watt of incident light. Each absorbed photon of wavelength \(\lambda\) carries \(hc/\lambda\) joules and yields at most one electron, so
where \(\eta\) is the quantum efficiency. Responsivity therefore rises with wavelength, because longer-wavelength photons are individually weaker and a given power contains more of them. At 900 nm the ideal figure is 0.726 A/W and a good silicon device with \(\eta = 0.8\) gives 0.581 A/W; at 400 nm the ideal is only 0.323 A/W. Silicon's responsivity peaks near 900 nm and falls to zero at 1100 nm, where the photon energy drops below \(E_g\).
The two operating modes are the two shaded quadrants of Figure 14.2, and the choice between them is one of the standard decisions in instrumentation.
- Photoconductive mode reverse biases the diode. The depletion region widens, so the junction capacitance falls and the response is fast; the photocurrent is accurately proportional to illumination over eight decades; but the dark current flows and contributes both an offset and shot noise. Used wherever speed or linearity matters — optical receivers, laser measurement, position sensing.
- Photovoltaic mode operates at zero bias, usually into the virtual earth of a transimpedance amplifier. There is no dark current at all, so the noise floor is set by thermal noise in the feedback resistor; but the junction capacitance is at its largest, so the circuit is slow. Used for low-light photometry and precision radiometry.
The noise floor is worth computing, because it determines the smallest detectable signal. Dark current \(I_d\) produces shot noise \(i_n = \sqrt{2qI_dB}\). For \(I_d = 1\) nA the spectral density is \(\sqrt{2\times1.602\times10^{-19}\times10^{-9}} = 17.9\) fA/\(\sqrt{\text{Hz}}\). Dividing by the responsivity gives the noise-equivalent power, the optical power that produces a signal equal to the noise:
In a 1 kHz bandwidth that is 0.97 pW — about three million photons per second, which is a very dark room indeed. Note that dark current roughly doubles every 10 °C, so the same device at 75 °C has 32 nA of dark current and a noise floor 5.7 times worse.
A photodiode's output is a current, and the standard interface is a transimpedance amplifier: an op-amp with the diode between its inverting input and ground and a feedback resistor \(R_f\) setting the gain, so that \(V_o = -I_{ph}R_f\). With \(R_f = 1\ \text{M}\Omega\), 1 µA becomes 1 V and the 1 nA dark current becomes a 1 mV offset. Chapter 27 develops the circuit, including the feedback capacitor that stops the diode's capacitance from making it oscillate.
5 Solar Cells, Fill Factor and Efficiency
A solar cell is a photodiode with a very large area, operated in the fourth quadrant of Figure 14.2 where the voltage is positive and the current negative — the quadrant in which the device delivers power rather than absorbing it. Four numbers describe it, and all four are measured under a standard illumination of 1000 W/m\(^2\) at 25 °C.
- Short-circuit current \(I_{sc}\), equal to the photocurrent, proportional to area and to irradiance.
- Open-circuit voltage \(V_{oc} = V_T\ln(1 + I_{ph}/I_S)\), which depends only logarithmically on illumination and is limited by the band gap — about 0.6 to 0.7 V for silicon.
- Maximum power point \((V_{mp}, I_{mp})\), the single point on the curve at which \(VI\) is greatest.
- Fill factor \(FF = V_{mp}I_{mp}/(V_{oc}I_{sc})\), the fraction of the bounding rectangle that the actual operating rectangle fills.
A 156 mm square monocrystalline silicon cell measures \(I_{sc} = 9.13\) A and \(V_{oc} = 0.681\) V under 1000 W/m\(^2\). Its maximum-power point is at \(V_{mp} = 0.571\) V, \(I_{mp} = 8.713\) A. Find the fill factor, the efficiency and the current density, and account for the fill-factor deficit.
Maximum power. \(P_{mp} = 0.571\times8.713 = 4.973\) W.
Fill factor. The bounding rectangle is \(V_{oc}I_{sc} = 0.681\times9.13 = 6.217\) W, so \(FF = 4.973/6.217 = 0.800\).
Area and efficiency. \(A = 0.156^2 = 0.02434\ \text{m}^2 = 243.4\ \text{cm}^2\). Incident power \(= 1000\times0.02434 = 24.34\) W, so \(\eta = 4.973/24.34 = 20.4\ \%\).
Current density. \(J_{sc} = 9.13/243.4 = 37.5\ \text{mA/cm}^2\), a typical figure for a good silicon cell.
Where the missing fill factor went. For an ideal cell with no series resistance, the fill factor is set by \(V_{oc}\) alone through \(FF_0 = (v_{oc} - \ln(v_{oc}+0.72))/(v_{oc}+1)\) with \(v_{oc} = V_{oc}/V_T = 0.681/0.02585 = 26.34\), giving \(FF_0 = 0.843\). The measured 0.800 is 0.043 short, and the shortfall is series resistance in the emitter sheet, the metal fingers and the contacts. Since \(FF \approx FF_0(1 - I_{sc}R_s/V_{oc})\), the implied \(R_s\) is about 3.7 m\(\Omega\) — a few milliohms, which at 9 A is all it takes.
Scaling up. Sixty such cells in series make a module of \(60\times0.571 = 34.3\) V at 8.71 A, that is 298 W, which is exactly the rating on a standard domestic panel.
Three departures from the standard test conditions matter in the field. Irradiance scales \(I_{sc}\) linearly but \(V_{oc}\) only logarithmically, so a cell in half-sunlight loses half its current and only 18 mV of voltage. Temperature is the more serious effect: \(V_{oc}\) falls by about 2.1 mV/°C because \(I_S\) rises with temperature, so a panel at a realistic 65 °C operating temperature has lost \(40\times2.1 = 84\) mV, 12.3 % of its open-circuit voltage and most of that in output power. And because \(I_{sc}\) varies with illumination while the cells are in series, a partly shaded module is limited by its worst cell, which is why bypass diodes are fitted across groups of cells.
The maximum power point moves with both irradiance and temperature, so a fixed load resistance is at the optimum only under one condition. Practical installations use a switching converter running a maximum power point tracking algorithm, which perturbs the operating voltage and observes whether the power rises or falls — a hill-climbing search that keeps the cell at the peak of the dashed curve in Figure 14.3 whatever the conditions.
6 Phototransistors, Optocouplers and Displays
A photodiode delivering a microampere often needs amplification, and a phototransistor provides it inside the device. It is an ordinary bipolar transistor with a large, exposed base-collector junction and usually no base connection at all: the photocurrent generated at that junction is the base current, and the transistor multiplies it by \(\beta\).
A photocurrent of 2 µA with \(\beta = 200\) gives 400 µA, so the phototransistor is 200 times more sensitive than the bare photodiode. The price is speed and linearity: the large junction capacitance and the transistor's own charge storage give rise times of microseconds rather than nanoseconds, \(\beta\) varies by three to one between devices and with current, and the dark current is multiplied along with the signal. Where speed or accuracy matters, a photodiode and a separate amplifier win; where a cheap on-off detection is wanted, the phototransistor wins.
An optocoupler puts an infrared LED and a phototransistor in one opaque package, facing each other across an insulating gap. Signal passes; current does not. The isolation is typically 2.5 to 5 kV, limited only by the moulding, and there is no upper limit on the DC level either side, so an optocoupler can carry a logic signal between circuits whose grounds differ by hundreds of volts.
The figure of merit is the current transfer ratio:
A 4N25 specifies CTR \(\ge 20\ \%\); high-gain Darlington couplers reach 1000 %. CTR is the parameter that makes optocoupler design awkward, because it is specified only as a minimum, varies by three to one across a production batch, falls with temperature, and — decisively — degrades as the LED ages, typically by half over the equipment's life. A design that works on the bench and fails after two years has almost always ignored the last of those.
Design an optocoupler interface carrying a logic signal between two 5 V systems. The coupler has CTR \(\ge 20\ \%\) and \(V_F = 1.2\) V. The output must reach a valid logic low, below 0.4 V.
Input side. Choose \(I_F = 10\) mA. \(R_{in} = (5 - 1.2)/0.010 = 380\ \Omega\); the nearest standard value, 390 \(\Omega\), gives \(I_F = 9.74\) mA.
Derate the CTR. The datasheet minimum is 20 %, but ageing and temperature can halve it, so design for 10 %. The worst-case collector current is \(I_C = 0.10\times9.74 = 0.974\) mA.
Output side. To pull the output below 0.4 V, \(R_C \ge (5 - 0.4)/0.974\ \text{mA} = 4.72\ \text{k}\Omega\). Choose 10 k\(\Omega\), which drives the transistor firmly into saturation and gives \(V_{OL} \approx 0.2\) V with margin to spare.
The speed penalty. The output node's rise is set by \(R_C\) and the stray plus input capacitance, about 20 pF: \(\tau = 10\ \text{k}\Omega\times20\ \text{pF} = 200\) ns, so the 10 to 90 % rise time is \(2.2\tau = 440\) ns and the useful data rate is a few hundred kilobits per second. Reducing \(R_C\) to 4.7 k\(\Omega\) halves the rise time to 207 ns but removes the saturation margin. This trade — CTR margin against speed — is the whole of optocoupler design, and it is why fast isolation uses logic-output couplers with an internal amplifier rather than a bare phototransistor.
Figure 14.4 shows the finished interface. Seven-segment displays, by contrast, are seven LEDs (plus a decimal point) arranged in a figure-eight, sharing one terminal. In a common-anode display all anodes join the positive rail and each segment is lit by pulling its cathode low; in a common-cathode display the reverse. Each segment needs its own resistor, for the reason given earlier: sharing one resistor between segments would make the brightness depend on how many segments are lit, so an 8 would be dimmer than a 1. Multi-digit displays are multiplexed as described in the LED section.
Liquid-crystal displays work on an entirely different principle: they emit nothing and instead modulate light that is already there. A twisted-nematic cell holds a liquid crystal between crossed polarisers; with no field the molecular twist rotates the plane of polarisation by 90° and light passes, while an applied field aligns the molecules, the rotation disappears and the cell goes dark. Because the cell is a capacitor rather than a current-carrying junction, an LCD segment consumes microwatts against the tens of milliwatts of an LED segment — which is why every battery-powered instrument uses one. The two costs are that an LCD needs ambient light or a backlight, and that it must be driven with an AC waveform of zero mean, since a sustained DC field causes electrochemical degradation of the cell within hours.
Part 4 now leaves the single junction behind. The phototransistor in this section was the first three-terminal device to appear in this course, and Chapter 16 begins the proper treatment of the bipolar transistor, in which the control is electrical rather than optical but the underlying physics of injection and transport is exactly what Chapters 4 and 5 established.
7 Summary and Key Results
| Quantity | Expression | Value |
|---|---|---|
| Emission wavelength | \(\lambda = 1240/E_g\) with \(E_g\) in eV | 873 nm for GaAs; 653 nm for 1.90 eV GaAsP; 459 nm for 2.70 eV InGaN |
| Visible-light requirement | \(E_g > 1.77\) eV | Rules out Si (1.12 eV) and Ge (0.66 eV) |
| Emission linewidth | \(\Delta\lambda \approx \lambda^2\Delta E/hc\), \(\Delta E \approx 1.8kT\) | 15 nm at 630 nm; 33 nm at 940 nm |
| Light-extraction cone | \(\theta_c = \arcsin(1/n)\) | 16.1° for GaAs; about 2 % escapes a flat surface |
| LED series resistor | \(R = (V_S - V_F)/I_F\) | 150 \(\Omega\) for 2.0 V at 20 mA from 5 V |
| Drive efficiency | \(V_F/V_S\) | 40 % from 5 V; 17 % from 12 V |
| Current sensitivity | \(\delta I/I = (\delta V_F/V_F)\,V_F/(V_S-V_F)\) | \(\pm 7\ \%\) at 3 V headroom; unusable at 0.1 V |
| Multiplexed peak current | \(I_{pk} = I_{av}/\text{duty}\) | 40 mA peak for 10 mA average at 1/4 duty |
| Responsivity | \(\mathcal{R} = \eta\lambda/1240\) A/W | 0.726 A/W ideal at 900 nm; 0.581 A/W at \(\eta\) = 0.8 |
| Shot noise | \(i_n = \sqrt{2qI_dB}\) | 17.9 fA/\(\sqrt{\text{Hz}}\) for \(I_d\) = 1 nA |
| Noise-equivalent power | \(\text{NEP} = i_n/\mathcal{R}\) | \(3.08\times10^{-14}\) W/\(\sqrt{\text{Hz}}\); 0.97 pW in 1 kHz |
| Fill factor | \(FF = V_{mp}I_{mp}/(V_{oc}I_{sc})\) | 4.973/6.217 = 0.800 (ideal 0.843) |
| Cell efficiency | \(\eta = P_{mp}/(GA)\) | 4.973/24.34 = 20.4 % on 243.4 cm\(^2\) |
| Temperature effect | \(dV_{oc}/dT \approx -2.1\) mV/°C | −84 mV at 65 °C, 12.3 % of \(V_{oc}\) |
| Current transfer ratio | \(\text{CTR} = I_C/I_F\) | 20 % minimum; derate to 10 % for ageing |
| Optocoupler load resistor | \(R_C \ge (V_S - V_{OL})/I_{C(\min)}\) | 4.72 k\(\Omega\) minimum; 10 k\(\Omega\) chosen, 440 ns rise |
8 Common Mistakes
The forward drop of any diode is set by the energy carriers must acquire to cross the junction, and in an LED that energy is \(E_g\), which is by definition large enough to make visible light. A red LED drops 1.8 V, a blue one 3.2 V, a white one 3.4 V. Using 0.7 V in \(R = (V_S-V_F)/I_F\) gives a resistor that is far too large — for a white LED on 5 V it gives 215 \(\Omega\) instead of 80 \(\Omega\), and a barely visible display. The related error is worse: attempting to drive a 3.2 V LED from a 3.3 V rail leaves 0.1 V of headroom, so a \(\pm 0.2\) V spread in \(V_F\) swings the current by a factor of three or turns the LED off altogether.
CTR is specified as a minimum, not a typical, because it varies by three to one across a batch, falls with temperature and degrades by roughly half over the life of the LED. A design that just works at the datasheet minimum of 20 % will fail after a year or two in service. The correct procedure is to halve the specified minimum, design at that figure, and choose \(R_C\) generously so that the output transistor saturates with margin. The cost is speed, since a large \(R_C\) slows the output edge, and that trade is the reason data isolation above about 1 Mbit/s uses couplers with an integrated amplifier rather than a bare phototransistor.
Fill factor measures the squareness of the I-V curve — how much of the \(V_{oc}I_{sc}\) rectangle the cell actually delivers — and it says nothing about how much sunlight was captured. Efficiency is \(P_{mp}\) divided by the incident optical power, and it involves the area and the irradiance. The cell in Worked Example 14.1 has \(FF = 0.800\) and \(\eta = 20.4\ \%\): the two numbers describe different things and neither can be derived from the other without \(I_{sc}\), \(V_{oc}\), the area and the irradiance. A cell can have an excellent fill factor and poor efficiency if it reflects most of the light, or a poor fill factor and respectable efficiency if it has high series resistance but good absorption.
9 Chapter Review
1. Explain why silicon is used for photodiodes but never for LEDs, and find the longest wavelength a silicon photodiode can detect.
Both emission and detection involve a transition across the band gap, and both must conserve energy and crystal momentum. A visible or infrared photon carries a great deal of energy and almost no momentum, so a photon-only transition must be vertical on an energy-momentum diagram. Silicon is an indirect-gap material: its conduction-band minimum lies at a different crystal momentum from its valence-band maximum, so a recombining electron and hole cannot conserve momentum by emitting a photon alone and must also involve a phonon. A three-body process is far less likely than a two-body one, so the radiative lifetime in silicon is of the order of milliseconds while non-radiative recombination through defect states takes microseconds. Essentially every recombination in silicon therefore produces heat, and the internal quantum efficiency for light emission is of order \(10^{-4}\). Detection is the same transition run backwards, and here the weakness is only a matter of degree: silicon absorbs indirectly and therefore weakly, but a photodiode can simply be made thick — tens of micrometres of depletion region rather than a fraction of one — and every absorbed photon still yields a carrier pair that the field sweeps out. The long-wavelength limit is set by the photon energy falling below the gap: \(\lambda_{\max} = 1240/1.12 = 1107\) nm. Beyond that, silicon is transparent, which is exactly why fibre-optic systems working at 1310 and 1550 nm use germanium or InGaAs detectors instead.
2. A green LED (\(V_F = 2.1\) V) is to be run at 15 mA from a 5 V supply. Find the resistor and its dissipation, then find the current if \(V_F\) is actually 2.3 V, and comment on whether a resistor is adequate.
The resistor takes the surplus voltage: \(R = (V_S - V_F)/I_F = (5 - 2.1)/0.015 = 193.3\ \Omega\), so specify the nearest standard value, 180 \(\Omega\) or 200 \(\Omega\). Taking 180 \(\Omega\), the actual current is \((5-2.1)/180 = 16.1\) mA. The resistor dissipates \((V_S-V_F)I_F = 2.9\times0.0161 = 46.7\) mW, so a 0.125 W part is ample; the LED itself dissipates \(2.1\times0.0161 = 33.8\) mW, so 58 % of the power drawn goes into the resistor as heat. If \(V_F\) is really 2.3 V rather than 2.1 V, the current becomes \((5-2.3)/180 = 15.0\) mA, a fall of 6.8 % for a 9.5 % change in \(V_F\). The sensitivity is \(V_F/(V_S-V_F) = 2.1/2.9 = 0.72\), so fractional errors in \(V_F\) are attenuated rather than amplified, and a resistor is entirely adequate here. The answer would be different on a 2.5 V supply, where the headroom is only 0.4 V, the sensitivity is 5.25, and the same 0.2 V spread would change the current by 50 %. The rule that emerges is to keep \(V_S - V_F\) above about 1 V if a resistor is to be used, and to fit a current source otherwise.
3. A silicon photodiode with \(\eta = 0.75\) and an active area of 5 mm\(^2\) is illuminated at 850 nm with an irradiance of 2 W/m\(^2\). Its dark current is 2 nA. Find the responsivity, the photocurrent, the output of a 1 M\(\Omega\) transimpedance amplifier, and the signal-to-noise ratio in a 10 kHz bandwidth.
Responsivity is \(\mathcal{R} = \eta\lambda/1240 = 0.75\times850/1240 = 0.514\) A/W. The optical power falling on the diode is the irradiance times the area: \(P = 2\ \text{W/m}^2\times5\times10^{-6}\ \text{m}^2 = 10\) µW. The photocurrent is therefore \(I_{ph} = 0.514\times10\times10^{-6} = 5.14\) µA. Through a 1 M\(\Omega\) feedback resistor the transimpedance amplifier gives \(V_o = 5.14\ \mu\text{A}\times1\ \text{M}\Omega = 5.14\) V, with a further 2 mV of offset from the dark current. For the noise, the dominant term at this dark current is shot noise on the total current, \(I_{ph} + I_d = 5.142\) µA: \(i_n = \sqrt{2qIB} = \sqrt{2\times1.602\times10^{-19}\times5.142\times10^{-6}\times10^4} = \sqrt{1.647\times10^{-20}} = 1.284\times10^{-10}\) A, that is 128 pA RMS. The signal-to-noise ratio is \(5.14\ \mu\text{A}/128\ \text{pA} = 4.0\times10^{4}\), or 92 dB. Note that with this much light the shot noise of the signal itself dominates the dark current entirely, so cooling the diode would gain nothing; at a thousand times less light the dark-current term would dominate and cooling would help. Note also that the thermal noise of the 1 M\(\Omega\) feedback resistor, \(\sqrt{4kTB/R} = 12.9\) pA in the same bandwidth, is ten times smaller than the shot noise, so the choice of \(R_f\) is not limiting here.
4. A solar cell measures \(I_{sc} = 5.0\) A, \(V_{oc} = 0.62\) V, and delivers maximum power at 0.50 V and 4.55 A. Its area is 156 cm\(^2\). Find the fill factor and efficiency at 1000 W/m\(^2\), and state what limits each.
Maximum power is \(P_{mp} = 0.50\times4.55 = 2.275\) W. The bounding rectangle is \(V_{oc}I_{sc} = 0.62\times5.0 = 3.10\) W, so the fill factor is \(FF = 2.275/3.10 = 0.734\). The area is \(156\ \text{cm}^2 = 0.0156\ \text{m}^2\), so the incident power is \(1000\times0.0156 = 15.6\) W and the efficiency is \(\eta = 2.275/15.6 = 14.6\ \%\). What limits each is quite different. The fill factor is limited from above by \(V_{oc}\) itself through the shape of the diode exponential: with \(v_{oc} = 0.62/0.02585 = 23.98\), the ideal value is \(FF_0 = (23.98 - \ln(24.70))/(24.98) = 0.831\). The measured 0.734 falls short by 0.097, and the deficit is resistive — series resistance in the fingers, busbars and contacts, plus any shunt leakage across the junction. The efficiency is limited by three quite separate things: the fraction of the solar spectrum with photon energy above 1.12 eV that can be absorbed at all (photons below the gap pass straight through, and photons well above it waste their excess as heat, which together cap a single-junction silicon cell at about 33 %); the fraction of carriers collected before they recombine; and finally the fill factor. This cell's 14.6 % against the 20.4 % of the chapter's example is mostly explained by the poorer fill factor and the lower \(V_{oc}\), both of which point to a cell with higher series resistance and more recombination — probably multicrystalline rather than monocrystalline silicon.
5. Design an optocoupler interface to carry a 4 kHz square wave from a 12 V control circuit to a 5 V logic input, using a coupler with CTR \(\ge 50\ \%\) and \(V_F = 1.3\) V. Include ageing derating and check the speed.
Input side. Choose \(I_F = 8\) mA, comfortably within any small coupler's rating. \(R_{in} = (12 - 1.3)/0.008 = 1338\ \Omega\); use 1.3 k\(\Omega\), which gives \(I_F = 8.23\) mA and dissipates \((12-1.3)\times0.00823 = 88\) mW, so a 0.25 W resistor is right. Derating. The 50 % minimum CTR must be halved for ageing and temperature, so design at 25 %. The worst-case collector current is \(I_C = 0.25\times8.23 = 2.06\) mA. Output side. To pull below 0.4 V from a 5 V rail, \(R_C \ge (5-0.4)/0.00206 = 2.23\ \text{k}\Omega\). Choosing 4.7 k\(\Omega\) doubles the saturation margin and gives \(V_{OL}\approx0.2\) V. Speed check. The limiting edge is the rising one, set by \(R_C\) charging the output capacitance of perhaps 20 pF: \(\tau = 4.7\ \text{k}\Omega\times20\ \text{pF} = 94\) ns, so the 10 to 90 % rise time is \(2.2\tau = 207\) ns. A 4 kHz square wave has a period of 250 µs, so the edge occupies 0.08 % of it — entirely adequate, with three orders of magnitude of margin. The phototransistor's own storage time, typically 2 to 5 µs, is in fact the real limit and still leaves a factor of fifty. Final check. The isolation voltage must exceed the worst-case difference between the two grounds plus any transient, and a 2.5 kV coupler is standard. Note that if the data rate were 1 Mbit/s rather than 4 kHz, the phototransistor's storage time alone would make this design unworkable and a logic-output coupler would be needed.