Semiconductor Diodes and Applications

Semiconductor Diodes

• Basic vehicle electronics use diodes, transistor circuits, FETs, microcontrollers, and digital signal processors.
• Diodes allow unidirectional current flow, acting as switches and rectifying AC in alternators.
• Photodiodes adjust interior lighting, and diode lasers weld body structures.

Semiconductor Diodes

• A two-terminal device allowing current in one direction only.
• It is a p-n junction, but the focus is on diode circuits rather than solid-state physics.
• Schematically represented by an arrowhead and a vertical bar.
• Allows current flow in the direction of the arrow, preventing flow in the opposite direction.
• Forward Biased: Arrowhead is positive relative to the vertical bar.
• Reverse Biased: Arrowhead is negative relative to the vertical bar, preventing current flow.
• Acts like a closed switch when forward biased and an open switch when reverse biased.
• Forward Resistance: Resistance under forward bias.
• Reverse Resistance: Resistance under reverse bias.
• Ideal Diode: Zero forward resistance and infinite reverse resistance.
• Practical diodes have negligible forward resistance compared to reverse resistance.
• Germanium diodes have a forward to reverse resistance ratio of approximately 1:40,000.
• Silicon diodes have a ratio of approximately 1:1,000,000.
• Forward resistance is low (below $25 \Omega$).

Diode Forward Characteristics

• Determined from the forward characteristic, a plot of current vs. voltage under forward bias.
• The diode is a nonlinear element.
• $I_F$: Forward current.
• $V_F$: Voltage across the diode under forward bias.
• P: An operating point on the forward characteristic curve.

DC Forward Resistance

• $R_f$: DC forward resistance at point P, defined as the ratio of DC voltage to current, i.e., $\frac{OA}{OB}$.

AC Forward Resistance

• Determined by perturbing the operating voltage equally on either side of P to get perturbed voltages (Oa and Oc).
• Corresponding currents are Od and Of.
• $r_f$: AC forward resistance at P, defined as the ratio of the change in voltage to the change in current, i.e., $\frac{ac}{df}$.

Reverse Resistance

• Defined similarly to forward resistance when the diode is in reverse bias.
• Ideal diode has infinite reverse resistance.
• Practical diodes have small reverse bias current (in $\mu A$).
• Reverse resistance is very large compared to forward resistance.

Diode Models

• Equivalent circuit models simulate diode behavior for solving diode-containing networks.

Ideal Diode Model

• Behaves as a short circuit under forward bias and an open circuit under reverse bias.
• Barrier potential is zero, forward resistance is zero, and reverse resistance is infinitely large.

Simplified Diode Model

• Assumes a barrier potential at the p-n junction but retains other assumptions of the ideal diode model.
• Forward biased only after overcoming the barrier potential $V_o$.
• $V_o$ is about 0.7 V for silicon diodes and 0.3 V for germanium diodes.

Practical Diode Model

• Incorporates the forward resistance $r_f$ into the simplified diode model.

Diode Specifications

• Maximum forward current is specified to avoid thermal failure, as semiconductor diodes are heat-sensitive.
• Silicon diodes can withstand up to $200^\circ C$, while germanium diodes can withstand up to $80^\circ C$.
• Reverse (leakage) current flows in reverse bias; it's low until the peak inverse voltage is reached, then sharply increases.
  • Below $1 \mu A$ in silicon diodes and about $100 \mu A$ in germanium diodes.
• Peak inverse voltage varies from 10 V to 10 kV, depending on diode type.

Examples

Example 1

• Circuit with a silicon diode (barrier potential 0.7 V) to determine if it is forward biased and to find currents or voltage across it.
• If forward biased, $I_2 = \frac{0.7V}{1 k\Omega} = 0.7 mA$.
• Using Kirchhoff’s law in the outer loop, $I_1 = \frac{2V - 0.7V}{4 k\Omega} = 0.325 mA$.
• $I<em>D = I</em>1 - I_2 = -0.375 mA$. Diode is not forward biased (open circuit).
• Voltage across the diode is $\frac{2V \times 1 k\Omega}{1 k\Omega + 4 k\Omega} = 0.4V$, less than the barrier potential.
• $I_1 = \frac{2V}{1 k\Omega + 4 k\Omega} = 0.4 mA$.

Example 2

• Determine the current through the resistor, assuming the forward resistance of a silicon diode is $1 \Omega$.
• Diodes $D<em>1$ and $D</em>3$ are forward biased; $D<em>2$ and $D</em>4$ are reverse biased.
• Current through the resistor is $\frac{10 - 0.7 - 0.7}{1 + 48 + 1} = 0.172 A = 172 mA$.

Example 3

• Silicon diodes are identical and must be limited to $30 mA$ to prevent thermal failure.
• With two diodes in parallel, the diode current is $\frac{1}{2} \times \frac{10 - 0.7}{200} = 23.25 mA$, so the diodes are safe.
• With a single diode, the current is $\frac{10 - 0.7}{200} = 46.5 mA$, and the diode will fail.

Rectification

• Diodes convert alternating current to direct current.

Half Wave Rectifier

• AC voltage applied across a diode in series with a load produces voltage across the load only during positive half cycles when the diode is forward biased.
• When input A is negative, the diode is reverse biased, resulting in zero output current.
• When A is positive, the diode conducts, creating an output voltage across the load.
• The output is unidirectional and pulsating with the same frequency as the input.
• Known as a half-wave rectifier because the output is available for only half of the input cycle.
• For a half-wave rectified output current with peak value $I_o$ and period T:
  • RMS value: $I<em>{rms} = \frac{I</em>o}{2}$.
  • Average value: $I<em>{ave} = \frac{I</em>o}{\pi}$.
• Proof:
  • $I<em>{rms} = \sqrt{\frac{1}{T} \int</em>{0}^{T/2} {I<em>o sin(\frac{2\pi}{T}t)}^2 dt} = \frac{I</em>o}{2}$
  • $I<em>{ave} = \frac{1}{T} \int</em>{0}^{T/2} I<em>o sin(\frac{2\pi}{T}t) dt = \frac{I</em>o}{\pi}$

Rectification Efficiency

• The ratio of DC output power to AC input power.
• Output DC power (across resistive load $R<em>L$) = $I</em>{ave}^2 R<em>L = (\frac{I</em>o}{\pi})^2 R<em>L = \frac{I</em>o^2 R<em>L}{\pi^2}$, where $I</em>o$ is the peak load current.
• Input AC power (across resistive load $R<em>L$ and diode AC forward resistance $r</em>f$) = $I<em>{rms}^2(R</em>L + r<em>f) = (\frac{I</em>o}{2})^2 (R<em>L + r</em>f) = \frac{I<em>o^2 (R</em>L + r_f)}{4}$.
• Rectification efficiency $= \frac{\frac{I<em>o^2 R</em>L}{\pi^2}}{\frac{I<em>o^2 (R</em>L + r<em>f)}{4}} = \frac{0.405}{1 + \frac{r</em>f}{R_L}}$.

Ripple Factor

• The ratio of the RMS value of the AC component in the output to the DC component.
• If RMS value of output is $I<em>{rms}$ and mean value is $I</em>{ave}$, then $I<em>{dc} = I</em>{ave}$ and $I<em>{ac} = \sqrt{I</em>{rms}^2 - I_{dc}^2}$.
• For a half-wave rectifier:
  • $I<em>{dc} = I</em>{ave} = \frac{I_o}{\pi}$.
  • $I<em>{rms} = \frac{I</em>o}{2}$.
  • $I<em>{ac} = \sqrt{(\frac{I</em>o}{2})^2 - (\frac{I_o}{\pi})^2}$.
• Ripple factor $= \frac{I<em>{ac}}{I</em>{dc}} = \frac{\sqrt{(\frac{I<em>o}{2})^2 - (\frac{I</em>o}{\pi})^2}}{\frac{I_o}{\pi}} = \sqrt{(\frac{\pi}{2})^2 - 1} = 1.21$.

Full Wave Rectifier

• Two diodes provide rectified voltage over the full input cycle.
• One diode operates during the positive cycle, and the other operates during the negative cycle.
• The load current flows in the same direction during both half cycles.
• The secondary of the transformer is connected to the p-side of the diodes and has a center tap connected through the load to the n-side of the diodes.
• The input to each diode is half the secondary voltage, and the input waveforms are completely out of phase.
• Each diode is active for only half the input cycle.
• The output waveform is unidirectional but pulsating, with twice the frequency of the input.
• For a full-wave rectified output current with peak value $I_o$:
  • RMS value: $I<em>{rms} = \frac{I</em>o}{\sqrt{2}}$.
  • Average value: $I<em>{ave} = \frac{2I</em>o}{\pi}$.
• A capacitor is used to filter the output (smoothing), charged during the rising part of the half cycle and discharged through the load as the output voltage falls.
• When the time constant is large, the rise and fall will be slow, and the output will be nearly constant and close to the peak of the rectified voltage.

Full Wave Rectifier Bridge Circuit

• Rectifies the full input voltage, not just half as with a center tap transformer.
• Employs four diodes arranged in a bridge configuration.
• The AC input voltage $V<em>{in}$ is applied between points P and Q, and the DC output $V</em>{out}$ is tapped across the load resistor R between points X and Y.
• When P is positive, diodes 3 and 4 are forward biased, completing the circuit P-A-D-Z-Y-X-B-C-Q, while diodes 1 and 2 are reverse biased.
• When P is negative, diodes 1 and 2 are forward biased, completing the circuit Q-C-D-Z-Y-X-B-A-P, while diodes 3 and 4 are reverse biased.
• Load current flows in the same direction during both half cycles.

Rectification Efficiency

• The ratio of DC output power to AC input power.
• Output DC power (across resistive load $R<em>L$) = $I</em>{ave}^2 R<em>L = (\frac{2I</em>o}{\pi})^2 R<em>L = \frac{4I</em>o^2 R<em>L}{\pi^2}$, where $I</em>o$ is peak load current (DC output).
• Input AC power (across resistive load $R<em>L$ and diode AC forward resistance $r</em>f$) = $I<em>{rms}^2(R</em>L + r<em>f) = (\frac{I</em>o}{\sqrt{2}})^2(R<em>L + r</em>f) = \frac{I<em>o^2(R</em>L + r<em>f)}{2}$, where $I</em>o$ is peak load current (DC output).
• Rectification efficiency $= \frac{\frac{4I<em>o^2 R</em>L}{\pi^2}}{\frac{I<em>o^2 (R</em>L + r<em>f)}{2}} = \frac{0.811}{1 + \frac{r</em>f}{R_L}}$.

Ripple Factor

• The ratio of the RMS value of the AC component in the output to the DC component.
• If the RMS value of the output is $I<em>{rms}$ and the mean value of the output is $I</em>{ave}$, then $I<em>{dc} = I</em>{ave}$ and $I<em>{ac} = \sqrt{I</em>{rms}^2 - I_{dc}^2}$.
• For a full-wave rectifier:
  • $I<em>{dc} = I</em>{ave} = \frac{2I_o}{\pi}$.
  • $I<em>{rms} = \frac{I</em>o}{\sqrt{2}}$.
  • $I<em>{ac} = \sqrt{(\frac{I</em>o}{\sqrt{2}})^2 - (\frac{2I_o}{\pi})^2}$.
• Ripple factor $= \frac{I<em>{ac}}{I</em>{dc}} = \frac{\sqrt{(\frac{I<em>o}{\sqrt{2}})^2 - (\frac{2I</em>o}{\pi})^2}}{\frac{2I_o}{\pi}} = \sqrt{\frac{\pi^2}{8} - 1} = 0.48$.

Example 4

• Given $V_{in} = 60 V$ (RMS), 50 Hz AC input, and assuming ideal diodes:
• (i) Determining the DC output voltage $V_{out}$ across resistive load $R = 200 \Omega$
  • Peak input voltage = $60\sqrt{2} = 84.85 V$.
  • Peak current $I_o = \frac{84.85}{200} = 0.424 A$.
  • $I<em>{dc} = \frac{2I</em>o}{\pi} = \frac{2 \times 0.424}{\pi} = 0.27 A$.
  • $V<em>{out} = I</em>{dc} \times R = 0.27 \times 200 = 54 V$.
• (ii) Peak inverse voltage on each diode = 84.85 V.
• (iii) Output frequency is 100 Hz.

Zener Diodes and Voltage Regulation

• Reverse current increases significantly beyond the peak inverse voltage (breakdown voltage).
• Heavily doped p-n junctions have a thin depletion layer and lower breakdown voltage.
• Lighter doping leads to a thicker depletion layer and higher breakdown voltage.
• Zener diode: Heavily doped diode with a sharp breakdown voltage, operating in reverse bias.
• Zener voltage: $V_Z$, the breakdown voltage of the Zener diode.
• Under forward bias, it behaves like a regular diode.
• In reverse bias, the Zener diode's characteristic has a knee with a sharp increase in current as it enters the breakdown region.
• Maximum and minimum Zener current are specified by the manufacturer.
• Turns on when the reverse bias exceeds the breakdown voltage and can be modeled as a battery of EMF $V_Z$.
• Provides a constant voltage $V_Z$ to the load across which it is connected, acting as a voltage stabilizer.
• A series resistance is provided to absorb output voltage fluctuations.

Example 5

• A Zener diode is designed to maintain $V_o = 30 V$ across a $2 k\Omega$ resistive load.
• The current rating of the Zener diode is $25 mA$, and the series resistance is $200 \Omega$.
• To turn on, the voltage across the load must exceed 30 V:
  • $\frac{V<em>i \times 2,000}{2,200} \geq 30 \implies V</em>i \geq 33 V$
• When the Zener current is at its maximum (25 mA), and $V_o = 30 V$:
  • Load current: $\frac{30}{2} = 15 mA$.
  • Current through the series resistor: 40 mA, with a voltage drop of $40 \times 0.2 = 8 V$.
  • $V_i = 38 V$.
• Voltage regulation occurs between $33 V \leq V_i \leq 38 V$.

Special Purpose Diodes

• Include optoelectronic devices such as photodiodes, Light Emitting Diodes (LEDs), and photovoltaic devices (solar cells).
• Photodiodes: Used as photodetectors.
• LEDs: Convert electrical energy into light, available in various colors (blue, green, yellow, orange, red), made of semiconductors (GaAs, GaP, etc.), and used in burglar alarms, remote controls, etc.
  • Long life, low power consumption, low operating voltage, and fast on-off switching capability.
• Photovoltaic devices (solar cells): Convert light into electricity, working like photodiodes but without external bias and with a larger junction area for more solar radiation incidence.
  • Examples of semiconductors used: Se, GaAs, CdTe, etc.