1 mao-hsu yen [email protected] electrical engineering
TRANSCRIPT
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Expected Context
• Chapter 8 Operational Amplifiers
• Chapter 9 Semiconductors and Diodes
• Chapter 11 Field effect Transistors: Operation, Circuit, Models, and Applications
• Chapter 12 Digital Logic Circuit
Chapter 9Semiconductors
and Diodes
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
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Context
• 9.1 Electrical Conduction in Semiconductor Devices
• 9.2 The pn Jnuction and Semiconductor Diode
• 9.3 Circuit Models or the Semiconductor Diode
• 9.4 Rectifier Circuits• 9.5 DC Power Supplies, Zener Diodes, and
Voltage Regulation
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Electrical Conduction in Semiconductor Devices
• Elemental or intrinsic semiconductors are materials consisting of elements from group IV of the periodic table and having electrical properties falling somewhere between those of conducting and of insulating materials.
• A conducting material is characterized by a large number of conduction band electrons, which have a very weak bond with the basic structure of the material.
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• In a semiconductor one needs to consider the lattice structure of the material.
• Free electrons enable current flow in the semiconductor.
• The number of charge carriers depends on the amount of thermal energy present in the structure. Many semiconductor properties are a function of temperature.
Free electrons and “holes” in the lattice structure
Current flow in a semiconductor
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• An additional phenomenon, called recombination, reduces the number of charge carriers in a semiconductor.
• The number of free electrons available for a given material is called the intrinsic concentration ni. For example, at room temperature, silicon has:
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• To control the number of charge carriers in a semiconductor, the process of doping is usually employed.
• Doping consists of adding impurities to the crystalline structure of the semiconductor. If the dopant is an element from the fifth column of the periodic table, the end result is that wherever an impurity is present, an additional free electron is available for conducting.
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• The elements providing the impurities are called donors in the case of group V elements, since they “donate” an additional free electron to the lattice structure.
• An equivalent situation arises when group III elements are used to dope silicon.
• An additional hole is created by the doping element which is called an acceptor, since it accepts a free electron from the structure and generates a hole in doing so.
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• Semiconductors doped with donor elements conduct current predominantly by means of free electrons and are therefore called n-type semiconductors.
• When an acceptor element is used as the dopant, holes constitute the most common carrier, the resulting semiconductor is said to be a p-type semiconductor.
Doped semiconductor
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• If n is the total number of free electrons and p that of holes, then in an n-type doped semiconductor, we have:
• Free electrons are the majority carriers in an n-type material, while holes are the minority carriers. In a p-type material, the majority and minority carriers are reversed.
14A pn junction
• When a section of p-type material and a section of n-type material are brought in contact to form a pn junction, a number of interesting properties arise. The pn junction forms the basis of the semiconductor diode, a widely used circuit element.
The pn Junction and The Semiconductor Diode
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Drift and diffusion currents in a pn junction
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Forward-and reverse-biased pn junctions
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Equations
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Semiconductor diode i-v characteristic
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Semiconductor diode circuit symbol
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The i-v characteristic of the semiconductor diode
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Large-signal on/off diode model
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Ideal Diode Model
Circuit of Figure 9.12, containing ideal diode
Circuit of Figure 9.13, assuming that the ideal diode
conducts
Circuit of Figure 9.14, assuming that the ideal diode does not conduct
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FOCUS ON METHODOLOGYDETERMING THE CONDUCTION
STATE OF AN IDEAL DIODE 1. Assume a diode conduction state (on or off).2. Substitute ideal circuit model into circuit (short circuit if
“on,” open circuit if “off”).3. Solve for diode current and voltage, using linear circuit
analysis techniques.4. If the solution is consistent with the assumption, then the
initial assumption was correct; if not, the diode conduction state is opposite to that initially assumed. For example, if the diode conduction state is opposite to that initially assumed. For example, if the diode has been assumed to be “off” but the diode voltage computed after replacing the diode with an open circuit is a forward bias, then it must be true that the actual state of the diode is “on.”
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EXAMPLE 9.1 Determining the Conduction State of an Ideal
Diode• Determine whether the ideal diode of figure 9.15 is
conducting.
Figure 9.15 Figure 9.16 Figure 9.17
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EXAMPLE 9.2 Determining the Conduction State of an Ideal Diode
Figure 9.18
Figure 9.19
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CHECK YOUR UNDERSTADING
• Repeat the analysis of example 9.2 assuming that the diode is conducting, and show that this assumption leads to inconsistent results.
• Determine which of the diodes conduct in the circuit shown in the figure, for each of the following voltages. Treat the diodes as ideal.
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CHECK YOUR UNDERSTADING
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Rectification
Half-wave rectifier
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Half-wave rectifier
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Offset diode as an extension of ideal diode model
Offset diode model
Offset Diode Model
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EXAMPLE 9.3 Using the Offset Diode Model in a Half-Wave
Rectifier• Compute and plot the rectified load voltage
νR in the circuit of figure 9.24.
Figure 9.24 Figure 9.25
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EXAMPLE 9.3 Using the Offset Diode Model in a Half-Wave
Rectifier
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CHECK YOUR UNDERSTANDING
• Compute the DC value of the rectified waveform for the circuit of figure 9.20 for νi = 52 cos ωt V.
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EXAMPLE 9.4 Using the Offset Diode Model Problem
• Using the offset diode model to determine the value of ν1 for which diode D1 first conducts in the circuit of figure 9.27.
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CHECK YOUR UNDERSTANDING
• Determine which of the diode conduct in the circuit in the circuit shown below. Each diode has an offset voltage of 0.6 V.
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Small-Signal Diode Models
Diode circuit for illustration of load-line analysis
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Graphical solution of equations 9.13 and 9.14
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FOCUS ON METHODOLOGYDETERMING THE OPERATING
POINT OF A DIODE• Reduce the circuit to a Thévenin or Norton equivalent
circuit with the diode as the load.
• Write the load-line equation.
• Solve numerically two simultaneous equations in two unknowns (the load-line equations and the diode equation) for the diode current and voltage.
or• Solve graphically by finding the intersection of the
diode curve (e.g., from a data sheet) with the load-line curve. The intersection of the two curves is the diode operating point.
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EXAMPLE 9.5 Using Load-Line Analysis and Diode Curves to
Determine the Operating Point of a Diode
• Determine the operating point of the 1N914 diode in the circuit of Figure 9.31, and compute the total power output of the 12-V battery.
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CHECK YOUR UNDERSTANDING
• Use load –line analysis to determine the operating point (Q point) of the diode in the circuit shown in figure. The diode has the characteristic curve of figure 9.32.
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Piecewise Linear Diode Model
Piecewise linear diode model
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EXAMPLE 9.6 Computing the Incremental (Small-Signal)
Resistance of a Diode• Determine the incremental resistance of a
diode, using the diode equation.
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CHECK YOUR UNDERSTANDING
• Computer the incremental resistance of the diode of Example 9.6 if the current through the diode is 250mA.
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EXAMPLE 9.7 Using the Piecewise Linear Diode Model
• Determine the load voltage in the rectifier of figure9.36, using a piecewise linear approximation.
Figure 9.36 Figure 9.37
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CHECK YOUR UNDERSTANDING
• Consider a half-wave rectifier similar to that of Figure 9.20, with νi =18 cost V, and a 4-Ώ load resistor. Sketch the output waveform if the piecewise linear diode modle is used to represent the diode, with Vγ= 0.6V and γD=1Ώ.what is the peak value of the rectifier output waveform?
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The Full-Wave Rectifier
Full-wave rectifier
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Full-wave rectifier current and voltage waveforms (RL = 1 Ω)
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Operation of bridge rectifier
The Bridge Rectifier
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(a) Unrectified source voltage; (b) rectified load voltage (ideal diodes); (c) rectified load voltage
(ideal and offset diodes)
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Bridge rectifier with filter circuit
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EXAMPLE 9.8 Half-Wave Rectifiers
• A half-wave rectifier, similar to that in figure 9.25, is used to provide a DC supply to a 50-Ώ load. If the AC source voltage is 20V (rms), find the peak and average current in the load.Auusme an ideal diode.
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CHECK YOUR UNDERSTANDING
• What is the peak current if an offset diode model is used with offset voltage equal to 0.6V ?
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EXAMPLE 9.9 Bridge Rectifier
• A bridge rectifier, similar to that in figure9.41, is used to provide a 50-V, 5-A DC supply. What is the resistance of the load that will draw exactly 5A? What is the required rms source voltage to achieve the desired DC voltage? Assume an ideal diode.
Figure 9.41
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CHECK YOUR UNDERSTANDING
• Show that the DC output voltage of the full-wave rectifier of figure 9.39 is 2Nνspeak /π. Compute the peak voltage output of the bridge rectifier of figure9.40, assuming diode with 0.6-V offset voltage and a 110-V rms AC supply.
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DC power supply
DC Power Supplies, Zener Diode, and Voltage Regulation
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EXAMPLE 9.10 Determining the Power Rating of a Zener Diode
• We wish to design a regulator similar to the one depicted in figure9.49(a). Determine the minimum acceptable power rating of the Zener diode.
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CHECK YOUR UNDERSTANDING
• How would the power rating change if the load were reduced to 100Ώ?
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EXAMPLE 9.11 Calculation of Allowed Load Resistances for a
Given Zener Regulator • Calculate the allowable range of load
resistances for the Zener regulator of figure 9.50 such that the diode power rating is not exceeded.
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CHECK YOUR UNDERSTANDING
• What should the power rating of the Zener diode be to withstand operation with open-circuit load?
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EXAMPLE 9.12 Effect of Nonzero Zener Resistance in a Regulator
• Calculate the amplitude of the ripple present in the output voltage of the regulator of figure9.51. The unregulated supply voltage is depicted in figure 9.52.
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CHECK YOUR UNDERSTANDING
• Compute the actual DC load voltage and the percent of ripple reaching the load (relative to the initial 100-mV ripple) for the circuit of example 9.12 if γz = 1Ώ
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Homework Problem