The ideal diode
Lecture #13 Diodes 1. Topics covered (a) Introduction (b) The ideal diode (c) The method of assumed states (d) The half wave rectifier (e) The full wave rectifier (f) The diode limiter (g) Diodes with capacitors rectifier filtering 2. Introduction (a) At this point in the course we introduce a new class of circuit elements-nonlinear elements (b) The main ones that we shall consider are diodes and various types of transistors. (c) We can combine these elements, along with R’s, L’s, and C’s to produce a common macro-element known as the operational amplifier (d) The combination of diodes, transistors, and operational amplifiers constitute the basic building blocks of electronics (e) There are an astounding number of applications using these devices. (f) The remainder of the term is concerned with understanding (1) the electrical properties of the basic building blocks and (2) a reasonably wide range of electronics applications. (g) This lecture thus represents the transition from the “circuits” portion of the course to the “electronics” portion. 3. The ideal diode (a) The ideal diode is the simplest nonlinear element that we will consider 1
(b) It is a single port element; that is, it has just two terminals (c) The ideal diode is a device that allows current to flow in one direction (i.e. the forward direction). When this current flows, the voltage drop across the ideal diode is zero. In the forward direction the diode acts like a short circuit. (d) The ideal diode does not allow any current to flow in the opposite direction (i.e. the reverse direction). When a voltage is applied across the diode with a polarity chosen to drive reverse current, the diode acts like an open circuit. (e) The symbol for the diode and the equivalent circuits for the two modes of operation are shown below I
-V+
Figure 13.1 (f) The I − V characteristics for an ideal diode are illustrated below I
V
Figure 13.2 (g) In practice there are a number ways to construct devices that accurately approximate the behavior of the ideal diode 2
(h) Practical diodes can be constructed with vacuum tubes or semiconductors and can be designed for both high and low power operation. (i) We shall be primarily concerned with semiconductor diodes. (j) These devices are often fabricated with silicon, or other elements in Column IV of the periodic table. (k) Silicon itself is a rather poor conductor of electricity. By doping silicon with small amounts of elements in either Column III (e.g. boron) or Column V (e.g. phosphorous) the conductivity greatly increases. (l) The change in conductivity is associated with the freedom of electrons to move through the material. The electrons in silicon are rather tightly bound because of the crystal lattice structure. Adding phosphorous adds another electron to the crystal structure. This “extra” electron is not required to maintain the crystal structure and thus has considerably more freedom to move from site to site in the material. Materials doped with elements in Column V are known as “n type semiconductors” the “n” designating a freely moving negative charge, i.e. an electron. (m) Doping silicon with elements in Column III produces a crystal with a deficit of one electron. Electrons in the material can easily flow to fill this deficit, leaving behind new deficits at the places where they started. Note that a deficit of electrons looks like a net positive charge and is called a “hole”. The motion of these holes looks like an effective flow of positive charge. Thus, materials doped with elements in Column III are known as “p type semiconductors” the “p” designating a freely moving net positive charge, i.e. a hole. (n) A diode is constructed by fabricating a p − n junction through material processing. A simple schematic diagram is shown below
3
p-type n-type I -V+
Figure 13.3 (o) The operation of the diode can be qualitatively explained in an admittedly oversimplified manner as follows. When a positive voltage is applied across the diode, as shown below, the n region is attracted to the positive potential and the p region to the negative potential. The resulting “overlap” makes it very easy for freely moving n region electrons to migrate into the p region. Similarly, holes in the p region can migrate into the n region. This freedom of charged particle motion implies that it is easy for current to flow across the boundary of the junction in the forward, conducting direction. This looks like a short circuit. I p
n
overlap
{ p R
n
V
Figure 13.4 (p) When the polarity of the voltage is switched, the p and n regions are attracted in the opposite direction leaving a gap of free electrons and holes at the interface. This absence of free charge carriers makes it impossible for current to flow across the interface. Thus, since no current can flow, the device behaves like an open circuit in the reverse direction.
4
p
n
gap
{
-
+ p R
n
V
Figure 13.5 (q) The I − V characteristics of a semiconductor diode can be obtained using the atomic theory of energy band structure in solids. This analysis shows that the relation between I and V in a diode corresponds to that of a nonlinear resistor and can be written as I = Is (eV /VT − 1) VT = kT /e = 26mV @ 27o C Is = 10−13 A (saturation current)
(1)
(r) The I − V characteristic is sketched below using two different scales. 1013I
I
3 2 1
1
0.03
V
0.8
V
Figure 13.6 (s) Observe that except for very, very small voltages and currents, the I − V characteristic closely approximates that of the ideal diode. The main difference is that the diode does not become conducting at V = 0+ . Instead it requires a positive voltage of about V ∼ 0.6 volts (corresponding to I ∼ 1mA). (t) This leads to a slightly more accurate approximation to model a real diode consisting of an ideal diode in series with a small battery: i.e. the constant voltage model. The equivalent circuit and I − V characteristics are shown below 5
I
0.6
V
Figure 13.7 (u) For our applications in the course we shall use the ideal diode model to maximize simplicity. More accurate models, including the nonlinear exponential model are incorporated in the numerical tools for circuit analysis. 4. The method of assumed states (a) The presence of even a single nonlinear element in a circuit greatly complicates the analysis. (b) Superposition and linearity no longer apply. KCL and KVL do apply but the resulting nodal equations are difficult to solve because the voltages and currents appear nonlinearly. (c) A reasonably general method has been devised to analyze diode circuits when the diodes are treated as ideal diodes. (d) The method is based on the observation that an ideal diode enters the analysis either as a short circuit or an open circuit. (e) In either of these two positions, the resulting circuit is linear and can be treated by standard techniques. The trick, as we shall see, is to determine when the diode is open and when it is closed. (f) The essence of the method is to consider all possible options for the diodes, analyze the resulting circuits, determine when the assumptions of the diode state are consistent with the actual currents and voltages, and then overlay the self-consistent portions of the solutions for each appropriate region. (g) Specifically, the steps in the method of assumed states are as follows:
6
• Assume each diode is either on (i.e. short circuit) or off (i.e. open circuit). There are 2n possibilities for n diodes so the method quickly becomes cumbersome for more than a few diodes. • Perform network analysis for each combination • Determine the range of applicability for each combination by examining the off-state diode voltages (each must be negative) and the on-state diode currents (each must be positive). When these conditions are simultaneously satisfied for every diode in the circuit, the diode state is self-consistent and the resulting analysis valid. If any condition is not satisfied, the corresponding diode state is not self-consistent and the analysis is not valid. • Splice the various self-consistent analyses together over their respective ranges of validity. 5. The half wave rectifier (a) A rectifier is a circuit that converts an AC voltage into a DC voltage. (b) A half wave rectifier uses 1 diode and potentially wastes large amounts of the AC voltage. (c) A full wave rectifier uses 4 diodes and makes more efficient use of the AC signal. (d) Adding a capacitor greatly improves the performance of the circuit. It also substantially complicates the analysis. (e) Analysis of resistive rectifier circuits is often sufficiently simple that one can determine the state of each diode at any instant of time by simple intuition. (f) However, in the examples that follow, we shall use the method of assumed states in order to show how it actually works and to give us some practice for more complicated examples. (g) Consider the half wave rectifier circuit illustrated below
7
I + V i sin(ωt) -
R
+ V -
Figure 13.8 (h) With one diode there are two possible states to consider. Either the diode is on or it is off. (i) Assume first that the diode is on. The circuit reduces to I
+ V -
+ V i sin(ωt) -
R
+ V -
Figure 13.9 (j) The voltage across the resistor is simply V = Vi sin(ωt) and the circuit current is given by I = V /R. These are illustrated below. These signals must be tested for applicability. Recall that the test for a diode being “on” is that the current flowing through it must be positive. The curve in the regions of validity is shaded darker in the diagram
8
V
t
I
t
Figure 13.10 (k) Now assume the diode is off. The resulting circuit diagram is given by + V0 V i sin(ωt) R
+ V -
Figure 13.11 (l) For this state the current through the circuit is zero: I = 0. Therefore the voltage across the resistor is also zero: V = 0. To test for applicability of a diode in the “off” position, we must examine when the voltage across it is negative. The voltages across the resistor and the diode are shown below, with the darker curves indicating validity of the solution
9
V
t
V0
t
Figure 13.12 (m) The complete output voltage is obtained by overlaying the separate solutions in their regions of validity. V
t
Figure 13.13 (n) This is a half wave rectified voltage. (o) Since the voltage is everywhere positive, it clearly has a non-zero average DC component. It also has a lot of higher harmonics present since the signal is far from a constant. (p) Note also that the voltage is zero over half the period. This wastes a large portion of the input signal. (q) The conclusion is that a single diode does rectify an AC voltage but the output is far from a pure DC voltage and it is not very efficient in using the input signal. 10
6. The full wave rectifier (a) The full wave rectifier fills up the wasted signal of the half wave rectifier (b) To do this the circuit requires 4 diodes. (c) Assume the load is a pure resistor. The full wave rectifier circuit is given by
3
1
R
V i sin(ωt)
4
2
Figure 13.14 (d) Since there are 4 diodes there are 24 = 16 possible states to consider. (e) In order to save time we shall simply state the solution. Of the 16 possible states, only 2 are self-consistent. (f) The first is as follows. When the input voltage is positive diodes #1 and #4 are closed and diodes #2 and #3 are open. The circuit and current flow are illustrated below I + Vin>0 -
+ V -
+ Vin -
R
+ V -
Figure 13.15 (g) Note that the voltage across the resistor is given by V = +Vi sin(ωt) when Vi sin(ωt) > 0. 11
(h) The second self-consistent state occurs when the input voltage is negative. In this case diodes #2 and #3 are closed and diodes #1 and #4 are open. The circuit and current flow are illustrated below
+ Vin 1. (o) The above solution remains valid as long as IS > 0. We note from the solutions above that IS crosses zero when sin(ωt) + ωτ cos(ωt) = 0
(6)
(p) In the limit of ωτ >> 1 this transition occurs just after ωt ≈ π/2. At this point the voltage across the capacitor is approximately Vout ≈ Vi . See the diagram below.
20
Vout
Vi
π/2 IS
ωt
ωCVi
ωt
Figure 13.30 (q) Once this transition occurs, the assumption of the diode being closed is no longer valid. The circuit switches to the second state where the diode is open as illustrated below
+ Vin -
R
C Vout
Figure 13.31 (r) In this regime we have a simple RC circuit where the capacitor has an initial voltage across it VC ≈ Vi . The voltage decays exponentially in accordance with the familiar relation Vout = Vi e−t/τ ≈ Vi (1 − t/τ )
(7)
(s) This solution remains valid until the voltage across the diode becomes positive: Vin − Vout > 0. The voltage signals are shown below. The point labeled with a black dot represents the transition point 21
Vout
Vout
Vin π
2π
ωt
Figure 13.32 (t) We can estimate the amplitude of the signal at the transition point as follows. The transition takes place when t Vi (1 − ) ≈ Vi sin(ωt) τ
(8)
(u) When ωτ >> 1 the transition point occurs just before ωt = 5π/2. Thus, if we expand ωt = 5π/2 − δ with δ
(b) It is a single port element; that is, it has just two terminals (c) The ideal diode is a device that allows current to flow in one direction (i.e. the forward direction). When this current flows, the voltage drop across the ideal diode is zero. In the forward direction the diode acts like a short circuit. (d) The ideal diode does not allow any current to flow in the opposite direction (i.e. the reverse direction). When a voltage is applied across the diode with a polarity chosen to drive reverse current, the diode acts like an open circuit. (e) The symbol for the diode and the equivalent circuits for the two modes of operation are shown below I
-V+
Figure 13.1 (f) The I − V characteristics for an ideal diode are illustrated below I
V
Figure 13.2 (g) In practice there are a number ways to construct devices that accurately approximate the behavior of the ideal diode 2
(h) Practical diodes can be constructed with vacuum tubes or semiconductors and can be designed for both high and low power operation. (i) We shall be primarily concerned with semiconductor diodes. (j) These devices are often fabricated with silicon, or other elements in Column IV of the periodic table. (k) Silicon itself is a rather poor conductor of electricity. By doping silicon with small amounts of elements in either Column III (e.g. boron) or Column V (e.g. phosphorous) the conductivity greatly increases. (l) The change in conductivity is associated with the freedom of electrons to move through the material. The electrons in silicon are rather tightly bound because of the crystal lattice structure. Adding phosphorous adds another electron to the crystal structure. This “extra” electron is not required to maintain the crystal structure and thus has considerably more freedom to move from site to site in the material. Materials doped with elements in Column V are known as “n type semiconductors” the “n” designating a freely moving negative charge, i.e. an electron. (m) Doping silicon with elements in Column III produces a crystal with a deficit of one electron. Electrons in the material can easily flow to fill this deficit, leaving behind new deficits at the places where they started. Note that a deficit of electrons looks like a net positive charge and is called a “hole”. The motion of these holes looks like an effective flow of positive charge. Thus, materials doped with elements in Column III are known as “p type semiconductors” the “p” designating a freely moving net positive charge, i.e. a hole. (n) A diode is constructed by fabricating a p − n junction through material processing. A simple schematic diagram is shown below
3
p-type n-type I -V+
Figure 13.3 (o) The operation of the diode can be qualitatively explained in an admittedly oversimplified manner as follows. When a positive voltage is applied across the diode, as shown below, the n region is attracted to the positive potential and the p region to the negative potential. The resulting “overlap” makes it very easy for freely moving n region electrons to migrate into the p region. Similarly, holes in the p region can migrate into the n region. This freedom of charged particle motion implies that it is easy for current to flow across the boundary of the junction in the forward, conducting direction. This looks like a short circuit. I p
n
overlap
{ p R
n
V
Figure 13.4 (p) When the polarity of the voltage is switched, the p and n regions are attracted in the opposite direction leaving a gap of free electrons and holes at the interface. This absence of free charge carriers makes it impossible for current to flow across the interface. Thus, since no current can flow, the device behaves like an open circuit in the reverse direction.
4
p
n
gap
{
-
+ p R
n
V
Figure 13.5 (q) The I − V characteristics of a semiconductor diode can be obtained using the atomic theory of energy band structure in solids. This analysis shows that the relation between I and V in a diode corresponds to that of a nonlinear resistor and can be written as I = Is (eV /VT − 1) VT = kT /e = 26mV @ 27o C Is = 10−13 A (saturation current)
(1)
(r) The I − V characteristic is sketched below using two different scales. 1013I
I
3 2 1
1
0.03
V
0.8
V
Figure 13.6 (s) Observe that except for very, very small voltages and currents, the I − V characteristic closely approximates that of the ideal diode. The main difference is that the diode does not become conducting at V = 0+ . Instead it requires a positive voltage of about V ∼ 0.6 volts (corresponding to I ∼ 1mA). (t) This leads to a slightly more accurate approximation to model a real diode consisting of an ideal diode in series with a small battery: i.e. the constant voltage model. The equivalent circuit and I − V characteristics are shown below 5
I
0.6
V
Figure 13.7 (u) For our applications in the course we shall use the ideal diode model to maximize simplicity. More accurate models, including the nonlinear exponential model are incorporated in the numerical tools for circuit analysis. 4. The method of assumed states (a) The presence of even a single nonlinear element in a circuit greatly complicates the analysis. (b) Superposition and linearity no longer apply. KCL and KVL do apply but the resulting nodal equations are difficult to solve because the voltages and currents appear nonlinearly. (c) A reasonably general method has been devised to analyze diode circuits when the diodes are treated as ideal diodes. (d) The method is based on the observation that an ideal diode enters the analysis either as a short circuit or an open circuit. (e) In either of these two positions, the resulting circuit is linear and can be treated by standard techniques. The trick, as we shall see, is to determine when the diode is open and when it is closed. (f) The essence of the method is to consider all possible options for the diodes, analyze the resulting circuits, determine when the assumptions of the diode state are consistent with the actual currents and voltages, and then overlay the self-consistent portions of the solutions for each appropriate region. (g) Specifically, the steps in the method of assumed states are as follows:
6
• Assume each diode is either on (i.e. short circuit) or off (i.e. open circuit). There are 2n possibilities for n diodes so the method quickly becomes cumbersome for more than a few diodes. • Perform network analysis for each combination • Determine the range of applicability for each combination by examining the off-state diode voltages (each must be negative) and the on-state diode currents (each must be positive). When these conditions are simultaneously satisfied for every diode in the circuit, the diode state is self-consistent and the resulting analysis valid. If any condition is not satisfied, the corresponding diode state is not self-consistent and the analysis is not valid. • Splice the various self-consistent analyses together over their respective ranges of validity. 5. The half wave rectifier (a) A rectifier is a circuit that converts an AC voltage into a DC voltage. (b) A half wave rectifier uses 1 diode and potentially wastes large amounts of the AC voltage. (c) A full wave rectifier uses 4 diodes and makes more efficient use of the AC signal. (d) Adding a capacitor greatly improves the performance of the circuit. It also substantially complicates the analysis. (e) Analysis of resistive rectifier circuits is often sufficiently simple that one can determine the state of each diode at any instant of time by simple intuition. (f) However, in the examples that follow, we shall use the method of assumed states in order to show how it actually works and to give us some practice for more complicated examples. (g) Consider the half wave rectifier circuit illustrated below
7
I + V i sin(ωt) -
R
+ V -
Figure 13.8 (h) With one diode there are two possible states to consider. Either the diode is on or it is off. (i) Assume first that the diode is on. The circuit reduces to I
+ V -
+ V i sin(ωt) -
R
+ V -
Figure 13.9 (j) The voltage across the resistor is simply V = Vi sin(ωt) and the circuit current is given by I = V /R. These are illustrated below. These signals must be tested for applicability. Recall that the test for a diode being “on” is that the current flowing through it must be positive. The curve in the regions of validity is shaded darker in the diagram
8
V
t
I
t
Figure 13.10 (k) Now assume the diode is off. The resulting circuit diagram is given by + V0 V i sin(ωt) R
+ V -
Figure 13.11 (l) For this state the current through the circuit is zero: I = 0. Therefore the voltage across the resistor is also zero: V = 0. To test for applicability of a diode in the “off” position, we must examine when the voltage across it is negative. The voltages across the resistor and the diode are shown below, with the darker curves indicating validity of the solution
9
V
t
V0
t
Figure 13.12 (m) The complete output voltage is obtained by overlaying the separate solutions in their regions of validity. V
t
Figure 13.13 (n) This is a half wave rectified voltage. (o) Since the voltage is everywhere positive, it clearly has a non-zero average DC component. It also has a lot of higher harmonics present since the signal is far from a constant. (p) Note also that the voltage is zero over half the period. This wastes a large portion of the input signal. (q) The conclusion is that a single diode does rectify an AC voltage but the output is far from a pure DC voltage and it is not very efficient in using the input signal. 10
6. The full wave rectifier (a) The full wave rectifier fills up the wasted signal of the half wave rectifier (b) To do this the circuit requires 4 diodes. (c) Assume the load is a pure resistor. The full wave rectifier circuit is given by
3
1
R
V i sin(ωt)
4
2
Figure 13.14 (d) Since there are 4 diodes there are 24 = 16 possible states to consider. (e) In order to save time we shall simply state the solution. Of the 16 possible states, only 2 are self-consistent. (f) The first is as follows. When the input voltage is positive diodes #1 and #4 are closed and diodes #2 and #3 are open. The circuit and current flow are illustrated below I + Vin>0 -
+ V -
+ Vin -
R
+ V -
Figure 13.15 (g) Note that the voltage across the resistor is given by V = +Vi sin(ωt) when Vi sin(ωt) > 0. 11
(h) The second self-consistent state occurs when the input voltage is negative. In this case diodes #2 and #3 are closed and diodes #1 and #4 are open. The circuit and current flow are illustrated below
+ Vin 1. (o) The above solution remains valid as long as IS > 0. We note from the solutions above that IS crosses zero when sin(ωt) + ωτ cos(ωt) = 0
(6)
(p) In the limit of ωτ >> 1 this transition occurs just after ωt ≈ π/2. At this point the voltage across the capacitor is approximately Vout ≈ Vi . See the diagram below.
20
Vout
Vi
π/2 IS
ωt
ωCVi
ωt
Figure 13.30 (q) Once this transition occurs, the assumption of the diode being closed is no longer valid. The circuit switches to the second state where the diode is open as illustrated below
+ Vin -
R
C Vout
Figure 13.31 (r) In this regime we have a simple RC circuit where the capacitor has an initial voltage across it VC ≈ Vi . The voltage decays exponentially in accordance with the familiar relation Vout = Vi e−t/τ ≈ Vi (1 − t/τ )
(7)
(s) This solution remains valid until the voltage across the diode becomes positive: Vin − Vout > 0. The voltage signals are shown below. The point labeled with a black dot represents the transition point 21
Vout
Vout
Vin π
2π
ωt
Figure 13.32 (t) We can estimate the amplitude of the signal at the transition point as follows. The transition takes place when t Vi (1 − ) ≈ Vi sin(ωt) τ
(8)
(u) When ωτ >> 1 the transition point occurs just before ωt = 5π/2. Thus, if we expand ωt = 5π/2 − δ with δ
