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IV. Diodes
We start our study of nonlinear circuit elements. These elements (diodes and transistors) are made of semiconductors. A brief description of how semiconductor devices work is first given to understand their i − v characteristics. You will see a rigorous analysis of semiconductors in the breadth courses.
4.1 Energy Bands in Solids
In every atom, the nucleus (positively charged) is surrounded by a cloud of electrons. Initially scientists envisioned the atomic structure to be similar to the solar system: the nucleus in the center (similar to the sun), electrons revolving on orbits around the nucleus (similar to planets), and the electric attraction force between the nucleus and electrons being similar to the gravitational force between the sun and planets. However, according to Maxwell’s equations, electrons that revolve around a nucleus should emit electromagnetic radiation. As such, the electrons should lose energy gradually and their orbit should decay (in a fraction of a second!). This means that there is a mechanism that would not allow the electrons to lose energy gradually and is one of main reasons that the quantum theory was developed.
According to quantum theory, electrons revolving around an atom can have only discrete levels of energy as is shown below. Electrons are not allowed to gradually lose or gain energy. They can only “jump” from one level to another (by absorbing or emitting a quanta of energy, usually a photon). Furthermore, there are only a finite number of electrons that are allowed in each state (Pauli’s Principle). Therefore, both the energy levels for electrons and the number of electrons at each energy level are specified. Because any system would tend to be in a minimum energy state, electrons in an atom start by “filing” the lowest energy level. Once all allowable slots is filled, electrons start filling the next energy level and so on. Therefore, the electronic structure include “filled” energy levels (all slots are taken by electrons), “empty” energy levels (positions are available but no electron is present), and “partially filled” levels (there are some electrons but there is also room for more electrons).
Isolated Atoms A Solid
When atoms are arranged in a solid, the inter-spacing between atoms can become comparable to the size of electron orbitals of each atom (electrons at each energy level are confined to a region in space called the orbital and the higher energy levels have larger orbital sizes). In this case, the outer orbitals merged into energy bands. Electrons in these bands are not tied to an atom, rather they are free to move around the solid (if space is available per Pauli’s Principle). In addition, instead of discrete energy levels, these “shared” electrons can have continuous values of energy within a “band” of energy. As before, there are range of energies that no electron can occupy. These range of energies are called “Forbidden Gaps.” (see the figure above).
The electric properties of metals, semiconductors, and insulators are can be understood with this picture. As these properties are tied to energy bands and forbidden gaps, we only focus on these regions as are shown in the figure below.
Forbidden Gap
Partially Filled Band
Forbidden Gap
Empty Energy Band
Filled Energy Band Metals
Forbidden Gap
Empty Energy Band
Filled Energy Band Semiconductors
In metals, one of the energy bands is only partially filled. Obvi- ously, all energy bands below (i.e., with lower energies) are com- pletely filled and all energy bands above are completely empty. Electrons in the partially filled band can easily move around the solid with smallest amount of energy as there are lots of spaces available. As such, metals can conduct electricity (and heat) very well.
In a semiconductor, no partially filled band exists. In the filled energy band, there are a lot of electrons, but there are no avail- able slot to move into. In the empty band, there are a lot of slots but no electrons. One would expect that semiconductors to be perfect insulators (for both electricity and heat). However, the size of the forbidden gap in a semiconductor is small. At room temperature, the thermal energy in the material is sufficient to provide sufficient energy to a select number of electrons to move from the filled energy band into the empty energy band.
These electrons that are promoted to the empty energy band now can carry electricity and heat because there are a lot of slots available in this band. In addition, these electrons leave spaces (or “holes”) in the originally filled energy band. As such, electrons in the originally filled energy band can also move around the solid by moving from one hole to another hole and participate in the conduction of electricity. Obviously it is easier to keep track of the small number of holes in the filled energy band as opposed to the large number of electrons in that band (it is much easier to find the attendance in a almost full class room by counting the number of empty seats rather the number of students!). In this picture, when electrons move for example to the left to fill a hole, it would look like the hole is moving to the right direction. So, in describing semiconductors we usually keep track of negatively charged
the above figure) which are generated due to temperature effects. In a n-type semiconductor, the number of free electrons from the dopant is much larger than the number of electrons from electron-hole pairs. As such, a n-type semiconductor is considerably more conductive than the base semiconductor (in this respect, a n-type semiconductor is more like a “resistive” metal than a semiconductor).
Si Si
Si
Si Si
Si
Si
Si
Si
Hole Electron
Si Si
Si Si
Si
Si
Si
Si
P
Electron
Si Si
Si Si
Si
Si
Si
Si
B
Hole
Pure Semiconductor n-type semiconductor p-type semiconductor
In summary, in a n-type semiconductor there are two charge carriers: “holes” from the base semiconductor (called the “minority” carriers) and electrons from both the n-type dopant and electron-hole pairs (called the “majority” carrier).
Similarly, we can create a p-type semiconductor by adding an element from group IIIB, such as B, to the semiconductor. In this case, the p-type dopant generate holes. We will have two charge carriers: majority carriers are “holes” from the p-type dopant and electron-hole pairs and minority carriers are electrons from the base semiconductor (from electron-hole pairs).
The charge carriers (electrons and holes) move in a semiconductor through two mechanisms: First, charge carriers would move from regions of higher concentration to lower concentration in order to achieve a uniform distribution throughout the semiconductor. This process is called “diffusion” and is characterized by the diffusion coefficient, D. Second, charge carriers move under the influence of an electric field. This motion is called the drift and is characterized by the mobility, μ. The ratio of D and μ is independent of the dopant or the base semiconductor material and is given by Einstein’s Equation
D μ
kT q
T
≈ VT
where k is the Boltzmann’s constant, T is temperature in Kelvin, and q is the electron charge. Parameter VT is called the “volt-equivalent of temperature” and appears in most semiconductor formulas. At room temperature, VT ≈ 0 .026 V.
4.3 The Junction Diode (^) + v −
iD
D
The simplest semiconductor device is a pn- junction diode as is shown in the figure with p- type material on the left and n-type material on the right and metal contacts are attached to both sides. Circuit symbol for the device is also shown with the “inward” arrow depicting the p-type ma- terial side. The convention for the direction of the diode current and the voltage across the diode are also shown.
We see that in the n-type material side, there is a large concentration of free electrons. As such, these electrons would like to diffuse to the p-side where there are few electrons. Similarly there is a large concentration of holes in the p-type material and they would like to diffuse to the n-type ma- terial side. When these electrons and holes meet in the vicinity of the junction, they combine and are neutralized. Since this region is “depleted” of mobile carriers, it is called the “depletion” re- gion (also referred to as the “transition” or “space- charge” region). The width of this region is small, typically ∼ 0. 5 μm.
Since the n-type material has lost electrons, it becomes positively charged. Similarly the p- type material becomes negatively charged. As a result, an electric potential and an electric field appears across the depletion region. This electric field (or the voltage barrier) impedes the flow of electrons from the n-type material and flow of holes from the p-type material. The height of this voltage barriers is related to the forbidden gap of the base semiconductor material (and not the amount of dopant).
i − v Characteristics
Let us attach a voltage source to this junction diode such that the positive terminal of the voltage source is attached to the n-type material. This configuration is called reverse bias. In this case, some of the free electrons from the n-type material will travel to the voltage sources making the n-type material side more positively charged. Similarly, some of the holes from the p-type material would go the voltage source, making the p-type material more negatively charged. As a result, the depletion region becomes wider and the height of the potential barrier increases and no majority carriers can cross the depletion region.
0
1
2
3
4
5
6
7
8
-1 -0.5 0 0.5 1
VD (V)
ID (mA)
0
10
20
30
40
50
60
70
-0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.
VD (V)
ID (nA)
Diode Limitations: In the forward bias region, the power P = iD vD is dissipated in the diode in the form of heat. The diode packaging provides for the conduction of this heat to the outside and diode is cooled by the air. If we increase vD (or iD) diode is heated more. At some point, the generated heat is more than capability of the diode package to conduct it away and diode temperature rises dramatically and diode burns out. As vD changes slowly, this point is usually characterized by the current iD,max, maximum forward current, and is specified by the manufacturer.
Heat generation is not a problem in the revers bias as iD ' 0 (and, thus, P ' 0). However, if we increase the reverse bias voltage, at some voltage, a large current can flow through the diode. This voltage is called the reverse breakdown voltage or the Zener voltage.
This large reverse current is produced through two processes. First, in the reverse bias, minority carriers enter the depletion region. These minority carriers are accelerated by the voltage across the depletion region. If the reverse bias voltage is high enough, these minority carrier can accelerate to a sufficiently high energy, impact an atom, and disrupt a covalent band, thereby generating new electron-hole pairs. The new electron-hole pairs can accelerate, impact other atoms and generate new electron-hair pairs. In this manner, the number of minority carriers increases exponentially (an avalanches process), leading to a large reverse current. This is called the avalanche breakdown. Second, when the strength of the electric field across the junction becomes too large, electrons can be pulled out of the covalent bonds directly, generating a large number of electron-hole pairs and a large reverse current. This is known as the Zener effect or Zener breakdown.
Regular diodes are usually destroyed when operated in the Zener or reverse breakdown region. These diode should be operated such that vD > −vZ. A special type of diodes, Zener diodes, are manufactured specifically to operate in the Zener region. We will discuss these diodes later. In Zener diodes, heat generation sets a maximum for the allowable diode current in the Zener region.
4.4 Solving Diode Circuits
iD Rest of vD Circuit −
iD vD −
V +
R
T
T
−
With diode i−v characteristic in hand, we now attempt to solve diode circuits. Consider any linear circuit with a diode. The box labeled “the rest of the circuit” in the figure can be replaced by its Thevenin equivalent, giving the simple diode circuit below.
Because the three elements are in series, current iD flows through all elements. Writing KVL around the loop we have:
VT = iDRT + vD
This is an equation with two unknowns (iD and vD) as VT and RT are known. The second needed equation (to get two equations in the two unknown) is the diode characteristic equation:
iD = Is
( evD^ /ηVT^ − 1
)
The above two equations in two unknown cannot be solved analytically as the diode i − v equation is non-linear. PSpice solves these equations numerically. As analytical solutions can provide insight in the circuit behavior and may be also needed for circuit design, we develop two methods to solve diode circuits without numerical analysis.
4.4.1 Load Line
The i − v characteristic of the diode is plotted in the figure. The iD and vD values of any point on the curve satisfies the diode equation above. Also plotted is the line representing VT = iDRT + vD. This line is called the “load line.” The iD and vD values of any point on the load line satisfies the first equation above. Since the solution to the above two equations in two unknowns is a pair of iD and vD that satisfies both equations, such a point should be both on the iv plot of the diode and on the load line, i.e., at the intersection of the two. This point is called the Q-point or the Quiescent point (or the operating point) of the diode.
The load line technique is not accu- rate in finding numerical values of iD and vD. However, it is a power- ful tool to get qualitative measures of the circuit behavior e.g., ensur- ing that the diode operating point is safety away from the maximum forward current.
0
1
2
3
4
5
6
7
8
-0.5 0 0.5 1 1.5 2 2.
VD
ID
Q
Load Line
VQ
IQ
VT
VT/RT
iD vD −
V +
RS
S −
Let’s use this method for the Si diode circuit shown with vs = 5 V and Rs = 1 kΩ.
Step 1: KCL tells that current iD flow in all elements and by KVL:
vs = iDRs + vD → 5 = 10^3 iD + vD
Step 2: Assume diode is OFF, iD = 0 and vD < vγ. Substituting for iD = 0 in the circuit equation, we get:
5 = 10^3 × 0 + vD → vD = 5 V
Step 3: Since vD = 5 > 0 .7 = vγ , diode vD does NOT satisfy range of validity and the assumed diode state is incorrect.
Step 2: Assume diode is ON, vD = vγ and iD > 0. Substituting for vD = vγ = 0.7 V in the circuit equation, we get:
5 = 10^3 × iD + 0. 7 → iD = 4.3 mA
Step 3: Although we know that since diode was not OFF, it should be ON, it is a good practice to check in case we might have made a math mistake. For this case, iD > 0 and satisfies the range of validity. So diode is ON and vD = 0.7 V and iD = 4.3 mA.
Diode circuit Model: The above method can become cumbersome if we have a compli- cated circuit and/or several diodes as it is not strightforward to use advcned circuit solution methods (node-voltage, mesh current). Another approach would be to use circuit models (instead of equation) for the diode as is shown below:
iD vγ + v − iD
D
Diode ON: iD > 0
Diode OFF: vD < vγ ,
If you use the above circuit models, weneed to modify our recipe for solving diode circuits accordingly:
Recipe for solving diode circuits with diode circuit models:
- Draw a circuit for each state of the diode (If more than one diode, you should draw circuits to all possible combinations of states of diodes).
- Solve each diode circuit and find iD and vD.
- Check the range of validity inequality with the values of iD and vD that was found. If iD and vD values satisfy the range of validity, the assumption was correct. Otherwise, the assumed diode state is incorrect. Go to step 2 above and start with another circuit.
As an example, let’s solve the diode circuit of the previous page with this method:
vD
iD
−
V +
RS
S −
Diode OFF: from the circuit by KVL, vD = vs = 5 V. Since vD > vγ = 0.7 V, our assumption is not correct.
iD vγ −
V +
RS
S −
Diode ON: from the circuit by Ohm’s Law, iD = (vs − + vγ )/R = 4.3 mA. Since iD > 0, our assumption is correct.
Therefore, the diode is ON with vD = 0.7 V and iD = 4.3 mA.
Parametric solution of diode circuits
It is very useful if we can derive the circuit solution parametrically, i.e., with values of various circuit elements as parameters. This approach would allows us to solve the circuit ONCE. For example, for the voltage divider of Page 20, we found Vo = Vs(R 2 ‖ RL)/(Rs +R 1 +R 2 ‖ RL). So, for example if we encounter a voltage divider circuit with RS = 1 kΩ, R 1 = R 2 = 10 kΩ, VS = 10 V, we do not need the circuit again. We just plug in the values in the formula above. (Note that up to now, we have solved all of the circuit parametrically with circuit elements as parameters).
We would like to derive solution to diode circuit in the same manner, e.g., in the diode circuit above we would like to find iD and vD in terms of vs and Rs. The problem is that diode can be either of its two states which in principle would depend on the values of circuit parameters (vs and Rs in the example above). As a result, since the diode has two states, we will find TWO solutions to the circuit, each being valid over a range of parameters as described in the recipe below:
4.5 Diode Logic Gates
You have seen binary mathematics and logic gates in ECE25. We will explore some electronic logic gates in this course. Binary mathematics is built upon two states: 0, and 1. We need to relate the binary states to currents or voltages as these are the parameters that we can manipulate in electronic circuits. Similar to our discussion of analog circuits, it is advantageous (from power point of view) to relate these the binary states to voltages. As such, we “choose” two voltages to represent the binary states: VL for state 0 or Low state and VH for state 1 or High state (for example, 0 V to represent state 0 and 5 V to represent state 1). These voltages are quite arbitrary and can be chosen to have any value. We have to be careful as it is extremely difficult, if not impossible, to design an electronic circuit to give exactly a voltage like 5 V (what if the input voltage was 4.99 V?). So, we need to define a range of voltages (instead of one value) to represent high and low states. We will discuss logic gates more thoroughly in the transistor section. Here, we consider a simplr diode logic gate.
o
R
CC
A
1
1
A
2
2
1
2
D
i
D
i
i
v
V
v
v
Diode AND Gate: To study the behavior of the gate we will consider the state of the circuit with VCC = 5 V and RA = 1 kΩ for different values of v 1 and v 2 (either 0 or 5 V corresponding to low and high states). We also assume that in the output any voltage < 1 V would be considered a low state and any volagte > 4 V is considered a high state. We note that by KCL, iA = i 1 + i 2 (assuming that there is no current drawn from the circuit).
Case 1, v 1 = v 2 = 0: Since the 5-V supply (Vcc) will tend to forward bias both D 1 and D 2 , let’s assume that both diodes are forward biased. Thus, vD 1 = vD 2 = vγ = 0.7 V and i 1 > 0, i 2 > 0. In this case:
vo = v 1 + vD 1 = v 2 + vD 2 = 0.7 V
iA = VCC − vo RA
= 4.3 mA
Current iA will be divided between two diodes by KCL, each carrying one half of iA (because of symmetry). Thus, i 1 = i 2 = 2.1 mA. Since diode currents are positive, our assumption of both diode being forward biased is justified and, therefore, vo = 0.7 V.
So, when v 1 and v 2 are low, D 1 and D 2 are ON and vo is low.
Case 2, v 1 = 0, v 2 = 5 V: Again, we note that the 5-V supply (Vcc) will tend to forward bias D 1. Assume D 1 is ON: vD 1 = vγ = 0.7 V and i 1 > 0. Then:
vo = v 1 + vD 1 = 0.7 V
vo = v 2 + vD 2 → vD 2 = − 4 .3 V < vγ
and D 2 will be OFF (i 2 = 0). Then:
iA = VCC − vo RA
= 4.3 mA
i 1 = iA − i 2 = 4. 3 − 0 = 4.3 mA
Since i 1 > 0, our assumption of D 1 being forward biased is justified and, therefore, vo = 0.7 V.
So, when v 1 is low and v 2 is high, D 1 is ON and D 2 is OFF and vo is low.
Case 3, v 1 = 5 V, v 2 = 0 V: Because of the symmetry in the circuit, this is exactly the same as case 2 with roles of D 1 and D 2 reversed.
So, when v 1 is high and v 2 is low, D 1 is OFF and D 2 is ON and vo is low.
Case 4, v 1 = v 2 = 5 V: Examining the circuit, it appears that the 5-V supply (Vcc) will NOT be able to forward bias D 1 and D 2. Assume D 1 and D 2 are OFF: i 1 = i 2 = 0, vD 1 < vγ and vD 2 < vγ. Then:
iA = i 1 + i 2 = 0 vo = VCC − i 1 RA = 5 − 0 = 5 V vD 1 = vo − v 1 = 5 − 5 = 0 < vγ and vD 2 = vo − v 2 = 5 − 5 = 0 < vγ
Thus, our assumption of both diodes being OFF are justified.
So, when v 1 and v 2 are high, D 1 and D 2 are OFF and vo is high.
Overall, the output of this circuit is high only if both inputs are high (Case 4) and the output is low in all other cases (Cases 1 to 3). Thus, this is an AND gate. This analysis can be easily extended to cases with three or more diode inputs.
This is actually not a good gate as for input we used low states of 0 V and the output low state was 0.7 V. We need to make sure that the input low state voltage is similar to the output low state voltage so that we can put these gates back to back. (You can easily show that if we had assumed low states of 0.7 V for input, the output low state would have been 1.4 V.) This gate is not usually used by itself but as part of diode-transistor logic gates that we will discuss in the BJT section.
4.7 Peak Detector
Another popular diode circuit is the peak detector which is similar to the half-wave rectifier circuit with the addition of a capacitor. Using diode circuit models, we find: vD iL
−
VS +
−
V^ C^ RL VL −
C
V − VS γ iL
−
+ −
C V
VC RL L −
iL
−
+ −
VS V^ RL VL
C −
C
Diode is ON Diode is OFF
Note that in the diode ON case, the two voltage sources in series are combined into one. So, when diode is ON, we have a parallel RC circuit and capacitor will be charged up (middle circuit). Current iD is positive as long as the capacitor is charging up, i.e., vs − vγ > vC or vs > vC + vγ. When vs < vC + vγ , diode will be OFF and we have an RC circuit (circuit to the right) and the capacitor discharges in the resistor with a time constant τ = RC.
The output voltage wave form is shown if the value of the capacitor is chosen such that τ = RC T (T is the period of the input AC voltage). To understand the shape of output waveform, let’s assume that voltage across the capacitor was zero at t = 0.
As the input voltage increases in the first quarter wave period (vγ < vs < Vs), the capacitor charges up as vs > vC + vγ and diode is ON. In the second quarter of the wave period, the input voltage vs is decreasing. In this case, diode is OFF and the capacitor slowly discharges
in RL. This continues until the next period when vs becomes again larger than vC + vγ and capacitor charges for a short period of time.
Exercise: What would have if τ = RC T where T is the period of the input AC voltage?
This circuit has two applications. First, by adding a large capacitor (10s to 100s of μF) to our rectifier circuit, we get a relatively smooth DC output. Similarly such a capacitor can be added to a full-wave rectifier circuit. (All AC to DC converters have such a large capacitor).
Second, the above circuit can be used as a “peak-detector” circuit. For example, consider the signal transmitted from an AM station. Such a signal includes a carrier wave (radio station frequency). The amplitude of the carrier is modulated according the sound signal. An example of such a signal is shown below (assuming that the sound signal is a triangular wave). If we apply this modulated voltage to our “peak-detector” circuit above and choose the value of capacitor such that τ = RC Trf where Trf is the period of radio-frequency carrier wave but τ = RC Tso where Tso is the period of the sound wave, the output of the circuit would be an approximation of the the initial sound wave as is shown below. The circuit is called the peak-detector because the output voltage is the envelope of the peak amplitudes of the input signal.
4.8 Zener Diodes
Zener diodes are specially manufactured to operate in the Zener region. These diode are made by means of heavily doped regions near the metal contacts to the semiconductor. The high density of charge carriers provides the means for a substantial reverse breakdown current to be sustained. These diodes are useful in applications where one would like to hold some load voltage constant, for example, in voltage regulators.
The circuit symbol, the i − v characteristics of Zener diodes and a piece-wise linear model are shown below:
Case 1: Diode is in the Zener region. In this case, vD = −vZ and iD < 0. Substituting for vD = −vZ in the third equation above, we find vL = vZ and is independent of iL. So, if the diode is in the Zener region, this circuit would behave like an independent voltage source. To find the range of iL for which the diode is in the Zener region, we calculate iD from equations above:
i = vs + vD R
vs − vZ R iD = iL − i < 0 → iL < i = vs − vZ R
Therefore, as long as iL is smaller than the value iL,max = (vs − vZ )/R, the diode would remain in the Zener region and the circuit would act as an independent voltage source.
Case 2: Diode is in the reverse bias region. In this case, iD = 0 and −vZ < vD < vγ. Substituting for iD = 0 in the above circuit equations, we get:
KCL: i = −iD + iL = iL KVL: vs = Ri − vD → −vD = vs − RiL KVL: vL = −vD = vs − RiL
Thus, in this region the output voltage does not remain constant, rather it decreases with iL. The diode is in the the reverse bias region as long as vD > −vZ or iL > iL,max = (vs − vZ )/R.
Case 3: Diode is in the forward bias region. Since vs > 0, the diode CANNOT be in the forward bias region.
vL
iL,max v /RS
Diode in Zener Region
Diode in Reverse Bias
vZ
i L
The i − v characteristics of this circuit is shown in the figure. The circuit resembles an independent voltage source with a strength of vZ as long as the diode remains in the Zener region (iL < (vs−vZ )/R. In practice, because vZ changes slightly when the diode current changes, the output voltage of circuit (vZ ) changes slightly. Much better power supply circuits can be build by addition of a transistor or OpAmps to the circuit (e.g. Experiment 4 of Lab
4.9 Other Wave-form Shaping Circuits
A wide-variety of waveform shaping circuits are used in electronics. These circuits are used to transform one waveform into another. Two examples of such circuits made with diodes are below.
4.9.1 Clipper Circuit
Diodes can be used to “clip” off a portion of the input signal. Consider the circuit below:
vi = RiD + vD + vDC vo = vD + vDC = vi − RiD Diode ON: vD = vγ and iD > 0
vo = vγ + vDC and iD =
vi − vγ − vDC R
0 → vi > vγ + vDC Diode OFF: iD = 0 and vD < vγ vo = vi − 0 × R = vi and vD = vi − RiD − vDC < vγ → vi < vγ + vDC
So, when vi < vγ + vDC , the diode is OFF and vo = vi: the output signal is exactly the same as the input signal. However, when vi exceeds this value of vγ + vDC , the diode is ON and vo = vγ + vDC. As a result, the input signal is “clipped” when it exceeds vγ + vDC. The shape of the output signal for a sinusoidal input signal is shown below.
vi vo^ v J^ + vDC
V −+ V DC Vo
R +
−
i −
Similarly, the circuit below will clip the input at −vγ − vDC
vi
- v J - vDC^ vo
−^ Vo S + DC
V V
R +
−
−
