Optoelectronic and Special-Purpose Diodes

Six worked problems on special-purpose diodes: sizing an LED ballast resistor, seven-segment display current and power budgets, photodiode responsivity and quantum efficiency, optocoupler current transfer ratio driving a transistor to saturation, varactor capacitance and the LC tuning range it produces, and a loss comparison between Schottky and silicon rectifiers at 100 kHz.

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.

Design. A red LED with a forward drop of \( 2.1~\text{V} \) is to be run at \( 25~\text{mA} \) from a \( 12~\text{V} \) supply through a single series resistor.

1. Compute the exact resistance required and pick the nearest standard (E24) value.

2. Find the actual LED current with that standard resistor, and the power rating the resistor needs.

3. What fraction of the power drawn from the supply actually reaches the LED?

• \( V_S = 12~\text{V} \), \( V_F = 2.1~\text{V} \), target \( I_F = 25~\text{mA} \).

• The LED is modelled as a constant \( 2.1~\text{V} \) drop — its dynamic resistance is negligible beside the series resistor.

Step 1 — KVL around the loop

The E24 values straddling this are 390 Ω and 430 Ω. Take 390 Ω, the nearer of the two: it gives 25.38 mA, only 1.5% above the target. Had 25 mA been an absolute ceiling, 430 Ω would be used instead, giving 23.02 mA.

(a) \( R = 396.0~\Omega \) exactly; use the standard \( 390~\Omega \).

Step 2 — Actual current and resistor rating

(b) \( I_F = 25.38~\text{mA} \) (1.5% above target); \( P_R = 0.251~\text{W} \), so a \( \tfrac12~\text{W} \) resistor is needed — a \( \tfrac14~\text{W} \) part would run right at its limit.

Step 3 — Where the power goes

(c) Only \( 17.5~\% \) of the supplied power reaches the LED; the rest is heat in the resistor.

That ratio is just \( V_F/V_S \). Driving a 2 V LED from a 12 V rail through a resistor is inherently wasteful — which is why multi-LED strings are wired in series, or driven by a switching constant-current source.

A common-anode seven-segment display is driven from \( +5~\text{V} \). Each segment LED has \( V_F = 1.7~\text{V} \) and its own \( 330~\Omega \) series resistor to the driver.

1. Find the current in one segment, and the total supply current and power when the digit 8 (all seven segments) is displayed.

2. Split that power between the LEDs and the resistors.

3. A four-digit display shows 8.8.8.8. (seven segments plus the decimal point on every digit). Find the static supply current, and then the peak segment current and resistor value needed if the four digits are instead multiplexed at 25% duty cycle for the same average brightness.

• \( V_{CC} = 5~\text{V} \), \( V_F = 1.7~\text{V} \) per segment, \( R = 330~\Omega \) per segment.

• Driver saturation voltages neglected; \( V_F \) assumed constant with current.

Part (a) — Segment current and total

(a) \( I_{seg} = 10.00~\text{mA} \); displaying "8" draws \( 70.0~\text{mA} \) and \( 350~\text{mW} \).

Part (b) — Split of the power

(b) \( 119.0~\text{mW} \) in the LEDs, \( 231.0~\text{mW} \) burnt in the resistors — the ballast wastes more than the LEDs use, because \( 3.3~\text{V} > 1.7~\text{V} \).

Part (c) — Four digits, static versus multiplexed

With the decimal point lit, each digit uses 8 LEDs:

Multiplexing scans one digit at a time, so each digit is lit for \( D = 25\% \) of the frame. Average brightness tracks average current, so the peak must be scaled up by \( 1/D \):

The nearest standard value is \( 82~\Omega \), giving \( I_{seg,pk} = 40.24~\text{mA} \) and a peak digit-driver current of \( 8 \times 40.24 = 281.7~\text{mA} \).

(c) Static: \( 320~\text{mA} \). Multiplexed: \( R = 82~\Omega \), \( I_{seg,pk} = 40.24~\text{mA} \), digit-driver peak \( \approx 281.7~\text{mA} \), average supply current still \( \approx 320~\text{mA} \).

Multiplexing does not save power for a given brightness — the average current is the same. What it saves is wiring: seven segment lines plus four digit lines instead of thirty-two independent drives.

A silicon PIN photodiode has a responsivity of \( 0.55~\text{A/W} \) at \( 850~\text{nm} \) and a dark current of \( 2~\text{nA} \). It is reverse-biased from a \( 5~\text{V} \) rail through a load resistor \( R_L \), and the output is taken across \( R_L \). The incident optical power is \( 20~\mu\text{W} \).

1. Find the photocurrent and the output voltage with \( R_L = 100~\text{k}\Omega \).

2. Find the quantum efficiency of the detector.

3. Design: choose \( R_L \) for a full-scale output of \( 2.5~\text{V} \) at \( 20~\mu\text{W} \), pick a standard value and check that the diode remains reverse biased.

• \( \mathcal{R} = 0.55~\text{A/W} \), \( P_{opt} = 20~\mu\text{W} \), \( \lambda = 850~\text{nm} \), \( I_D = 2~\text{nA} \), \( V_{bias} = 5~\text{V} \).

• The reverse-biased photodiode behaves as a current source: its current is set by the light, not by the voltage across it.

Part (a) — Photocurrent and output

The dark current contributes only \( (2~\text{nA})(100~\text{k}\Omega) = 0.2~\text{mV} \), a fraction \( 1/5500 \) of the signal.

(a) \( I_{ph} = 11.0~\mu\text{A} \), \( V_o = 1.10~\text{V} \).

Part (b) — Quantum efficiency

Responsivity counts amps per watt; quantum efficiency counts electrons per photon:

(b) \( \eta = 80.2~\% \) — about four photons in five produce a collected carrier pair.

Part (c) — Choosing \( R_L \)

Take the standard \( 220~\text{k}\Omega \):

(c) \( R_L = 220~\text{k}\Omega \) gives \( V_o = 2.42~\text{V} \) at full scale, leaving \( 2.58~\text{V} \) of reverse bias across the diode — still comfortably reverse biased.

Pushing \( R_L \) higher would give more volts per microwatt but eats into the reverse bias and, with the diode capacitance, slows the detector: the bandwidth falls as \( 1/(2\pi R_L C_d) \). That trade-off is why fast receivers use a transimpedance amplifier instead of a plain load resistor.

An optocoupler isolates a 5 V logic output from a 12 V circuit. Its input LED has \( V_F = 1.2~\text{V} \) and is driven through a series resistor \( R_1 \) from the logic gate (assume the gate pulls all the way to 0 V and 5 V). The output phototransistor has a collector resistor \( R_C = 4.7~\text{k}\Omega \) to \( +12~\text{V} \), an emitter at ground, a guaranteed minimum current transfer ratio \( \text{CTR} = 50\% \), and \( V_{CE(sat)} = 0.3~\text{V} \).

1. Choose \( R_1 \) for a nominal LED current of 10 mA and find the actual current with the nearest standard value.

2. Find the available collector current and show that the output transistor saturates.

3. State the output logic levels and find the smallest CTR for which the design still works.

• Input: \( V_{logic} = 5~\text{V} \), \( V_F = 1.2~\text{V} \), target \( I_F = 10~\text{mA} \).

• Output: \( V_{CC} = 12~\text{V} \), \( R_C = 4.7~\text{k}\Omega \), \( \text{CTR}_{\min} = 50\% \), \( V_{CE(sat)} = 0.3~\text{V} \).

• \( \text{CTR} \equiv I_C/I_F \) at the specified operating point.

Part (a) — Input resistor

(a) \( R_1 = 390~\Omega \), giving \( I_F = 9.744~\text{mA} \).

Part (b) — Output current and the saturation test

What the phototransistor can deliver:

What the collector circuit needs to pull the output down to \( V_{CE(sat)} \):

Since \( 4.872~\text{mA} > 2.489~\text{mA} \), the transistor is driven into saturation with an overdrive factor

(b) \( I_{C,\text{avail}} = 4.872~\text{mA} \) against \( 2.489~\text{mA} \) required: saturated, ODF \( = 1.96 \).

Part (c) — Logic levels and CTR margin

• LED on \( \Rightarrow \) transistor saturated \( \Rightarrow V_o = V_{CE(sat)} = 0.3~\text{V} \) (logic LOW).

• LED off \( \Rightarrow \) only the dark current flows \( \Rightarrow V_o \approx V_{CC} = 12~\text{V} \) (logic HIGH).

(c) Output swings \( 0.3~\text{V} \leftrightarrow 12~\text{V} \); the link keeps working down to \( \text{CTR} = 25.5~\% \), so there is roughly a 1.96-fold margin on the 50% specification.

That margin matters: optocoupler CTR falls with age and temperature, often by half over the life of the part, so designing with an ODF of about 2 is normal practice.

An abrupt-junction varactor obeys

with \( C(0) = 100~\text{pF} \), \( V_{bi} = 0.7~\text{V} \) and \( m = 0.5 \). It tunes a parallel resonant circuit with \( L = 1~\mu\text{H} \); the reverse bias is applied through an RF choke so the inductor sees only the diode capacitance.

1. Find \( C_T \) at \( V_R = 2~\text{V} \) and at \( V_R = 10~\text{V} \).

2. Find the resonant frequency at each extreme and the tuning ratio.

3. What reverse bias tunes the circuit to exactly \( 27~\text{MHz} \)?

• \( C(0) = 100~\text{pF} \), \( V_{bi} = 0.7~\text{V} \), \( m = 0.5 \) (abrupt junction), \( L = 1~\mu\text{H} \).

• Stray and holder capacitance neglected; the diode is always reverse biased so no conduction occurs.

Part (a) — Capacitance at the two bias points

(a) \( C_T = 50.92~\text{pF} \) at 2 V and \( 25.58~\text{pF} \) at 10 V — a capacitance ratio of \( 1.991 \).

Part (b) — Resonant frequencies

The frequency ratio is the square root of the capacitance ratio:

(b) The circuit tunes from \( 22.30~\text{MHz} \) to \( 31.47~\text{MHz} \), a ratio of \( 1.411 : 1 \).

Part (c) — Bias for 27 MHz

Work backwards: first the capacitance, then the voltage.

(c) \( V_R = 5.10~\text{V} \) puts the circuit on 27 MHz.

Because \( m = \tfrac12 \), frequency scales as \( (1+V_R/V_{bi})^{1/4} \) — very slow. A hyperabrupt varactor with \( m \approx 1 \) or 2 is used where a wide band must be swept with a modest tuning voltage.

The output rectifier of a \( 5~\text{V} \), \( 2~\text{A} \) switching supply running at \( 100~\text{kHz} \) is to be either a silicon PN diode (\( V_F = 0.7~\text{V} \), reverse-recovery time \( t_{rr} = 500~\text{ns} \)) or a Schottky diode (\( V_F = 0.3~\text{V} \), negligible \( t_{rr} \), junction capacitance \( C_j = 300~\text{pF} \)). The diode carries the full \( 2~\text{A} \) average output current and blocks \( 20~\text{V} \) when off; during recovery the reverse current peaks at \( 2~\text{A} \).

1. Compare the conduction losses and the rectifier efficiency of the two devices, ignoring switching.

2. Estimate the reverse-recovery loss of the silicon diode and the capacitive loss of the Schottky.

3. Compare total loss and overall efficiency, and state the penalty that comes with the Schottky.

• \( V_o = 5~\text{V} \), \( I_o = 2~\text{A} \Rightarrow P_o = 10~\text{W} \); \( f_{sw} = 100~\text{kHz} \), \( V_R = 20~\text{V} \).

• Si: \( V_F = 0.7~\text{V} \), \( t_{rr} = 500~\text{ns} \), \( I_{RM} = 2~\text{A} \). Schottky: \( V_F = 0.3~\text{V} \), \( C_j = 300~\text{pF} \).

• Diode \( V_F \) taken as constant over the conduction interval; the recovery current is modelled as a triangle of height \( I_{RM} \) and base \( t_{rr} \) at full reverse voltage (a deliberately pessimistic upper bound).

Part (a) — Conduction loss

(a) The Schottky saves \( 1.4 - 0.6 = 0.8~\text{W} \) of conduction loss, raising the efficiency from \( 87.7~\% \) to \( 94.3~\% \).

Part (b) — Switching loss

The silicon diode must sweep out its stored minority charge before it can block. Approximating the recovery current as a triangle of height \( I_{RM} \) and base \( t_{rr} \), all of it flowing while the diode sustains \( V_R \):

A Schottky is a majority-carrier device: there is no stored charge and hence no recovery current. Only its junction capacitance must be charged each cycle:

(b) \( P_{rr} = 1.00~\text{W} \) for the silicon diode against \( 6.0~\text{mW} \) of capacitive loss for the Schottky.

Part (c) — Totals

(c) Total diode loss falls from \( 2.40~\text{W} \) to \( 0.606~\text{W} \), a factor of \( 4.0 \); overall efficiency rises from \( 80.6~\% \) to \( 94.3~\% \).

The penalties are real, though: a Schottky's reverse leakage is orders of magnitude larger than a PN diode's and roughly doubles every 10 °C, and its reverse breakdown rarely exceeds a few tens of volts. That is exactly why Schottkys own the low-voltage, high-frequency rectifier slot and nothing above it.