1. DIODES AND DIODE CIRCUITS

2. What Are Diodes Made Out Of?Si +4
Silicon (Si) and Germanium (Ge) are the two most common single elements that are used to make Diodes. A compound that is commonly used is Gallium Arsenide (GaAs), especially in the case of LEDs because of it’s large bandgap.
Silicon and Germanium are both group 4 elements, meaning they have 4 valence electrons. Their structure allows them to grow in a shape called the diamond lattice.
Gallium is a group 3 element while Arsenide is a group 5 element. When put together as a compound, GaAs creates a zincblend lattice structure.
In both the diamond lattice and zincblend lattice, each atom shares its valence electrons with its four closest neighbors. This sharing of electrons is what ultimately allows diodes to be build. When dopants from groups 3 or 5 (in most cases) are added to Si, Ge or GaAs it changes the properties of the material so we are able to make the P- and N-type materials that become the diode.
The diagram above shows the 2D structure of the Si crystal. The light green lines represent the electronic bonds made when the valence electrons are shared. Each Si atom shares one electron with each of its four closest neighbors so that its valence band will have a full 8 electrons.

3. N-Type Material N-Type Material:
When extra valence electrons are introduced into a material such as silicon an n-type material is produced. The extra valence electrons are introduced by putting impurities or dopants into the silicon. The dopants used to create an n-type material are Group V elements. The most commonly used dopants from Group V are arsenic, antimony and phosphorus.
The 2D diagram to the left shows the extra electron that will be present when a Group V dopant is introduced to a material such as silicon. This extra electron is very mobile. +4 +4 +4 +4 +5 +4 +4 +4 +4

4. P-Type Material P-Type Material:
P-type material is produced when the dopant that is introduced is from Group III. Group III elements have only 3 valence electrons and therefore there is an electron missing. This creates a hole (h+), or a positive charge that can move around in the material. Commonly used Group III dopants are aluminum, boron, and gallium.
The 2D diagram to the left shows the hole that will be present when a Group III dopant is introduced to a material such as silicon. This hole is quite mobile in the same way the extra electron is mobile in a n-type material. +4 +4 +4 +4 +3 +4 +4 +4 +4

5. Metallurgical Junction
The PN Junction
Steady State1
Metallurgical Junction Na Nd P n
Space Charge Region
ionized acceptors
ionized donors
E-Field + + _ _
h+ drift =
h+ diffusion
e- diffusion =
e- drift

6. Metallurgical Junction
The PN Junction
Steady State P n Na Nd
Metallurgical Junction
Space Charge Region
ionized acceptors
ionized donors
E-Field + _
h+ drift
h+ diffusion
e- diffusion
e- drift =
When no external source is connected to the pn junction, diffusion and drift balance each other out for both the holes and electrons
Space Charge Region: Also called the depletion region. This region includes the net positively and negatively charged regions. The space charge region does not have any free carriers. The width of the space charge region is denoted by W in pn junction formula’s.
Metallurgical Junction: The interface where the p- and n-type materials meet.
Na & Nd: Represent the amount of negative and positive doping in number of carriers per centimeter cubed. Usually in the range of 1015 to 1020.

7. Applied Electric Field
The Biased PN Junction
Metal Contact
“Ohmic Contact”
(Rs~0) _ +
Applied Electric Field P n I _ +
Vapplied
The pn junction is considered biased when an external voltage is applied. There are two types of biasing: Forward bias and Reverse bias.
These are described on the next slide.

8. The Biased PN Junction Vapplied > 0 Vapplied < 0 Forward Bias:
In forward bias the depletion region shrinks slightly in width. With this shrinking the energy required for charge carriers to cross the depletion region decreases exponentially. Therefore, as the applied voltage increases, current starts to flow across the junction. The barrier potential of the diode is the voltage at which appreciable current starts to flow through the diode. The barrier potential varies for different materials.
Vapplied > 0
Reverse Bias:
Under reverse bias the depletion region widens. This causes the electric field produced by the ions to cancel out the applied reverse bias voltage. A small leakage current, Is (saturation current) flows under reverse bias conditions. This saturation current is made up of electron-hole pairs being produced in the depletion region. Saturation current is sometimes referred to as scale current because of it’s relationship to junction temperature.
Vapplied < 0

9. Figure 1.10 – The Diode Transconductance Curve2
Properties of Diodes
Figure 1.10 – The Diode Transconductance Curve2 VD ID
(mA)
(nA)
VBR
~V IS
VD = Bias Voltage
ID = Current through Diode. ID is Negative for Reverse Bias and Positive for Forward Bias
IS = Saturation Current
VBR = Breakdown Voltage
V = Barrier Potential Voltage

10. Properties of Diodes ID = IS(eVD/VT – 1) The Shockley Equation
The transconductance curve on the previous slide is characterized by the following equation:
ID = IS(eVD/VT – 1)
As described in the last slide, ID is the current through the diode, IS is the saturation current and VD is the applied biasing voltage.
VT is the thermal equivalent voltage and is approximately 26 mV at room temperature. The equation to find VT at various temperatures is:
VT = kT q
k = 1.38 x J/K T = temperature in Kelvin q = 1.6 x C
 is the emission coefficient for the diode. It is determined by the way the diode is constructed. It somewhat varies with diode current. For a silicon diode  is around 2 for low currents and goes down to about 1 at higher currents

11. Diode Circuit Models The Ideal Diode Model
The diode is designed to allow current to flow in only one direction. The perfect diode would be a perfect conductor in one direction (forward bias) and a perfect insulator in the other direction (reverse bias). In many situations, using the ideal diode approximation is acceptable.
Example: Assume the diode in the circuit below is ideal. Determine the value of ID if a) VA = 5 volts (forward bias) and b) VA = -5 volts (reverse bias)
a) With VA > 0 the diode is in forward bias and is acting like a perfect conductor so:
ID = VA/RS = 5 V / 50  = 100 mA
b) With VA < 0 the diode is in reverse bias and is acting like a perfect insulator, therefore no current can flow and ID = 0.
RS = 50  ID + VA _

12. The Ideal Diode with Barrier Potential
Diode Circuit Models
The Ideal Diode with Barrier Potential
This model is more accurate than the simple ideal diode model because it includes the approximate barrier potential voltage. Remember the barrier potential voltage is the voltage at which appreciable current starts to flow. + V
Example: To be more accurate than just using the ideal diode model include the barrier potential. Assume V = 0.3 volts (typical for a germanium diode) Determine the value of ID if VA = 5 volts (forward bias).
RS = 50 
With VA > 0 the diode is in forward bias and is acting like a perfect conductor so write a KVL equation to find ID:
0 = VA – IDRS - V
ID = VA - V = 4.7 V = 94 mA
RS  ID + VA _ + V

13. Diode Circuit Models
The Ideal Diode with Barrier Potential and Linear Forward Resistance
This model is the most accurate of the three. It includes a linear forward resistance that is calculated from the slope of the linear portion of the transconductance curve. However, this is usually not necessary since the RF (forward resistance) value is pretty constant. For low-power germanium and silicon diodes the RF value is usually in the 2 to 5 ohms range, while higher power diodes have a RF value closer to 1 ohm. ID + V RF
Linear Portion of transconductance curve
RF = VD
ID
ID VD
VD

14. The Ideal Diode with Barrier Potential and Linear Forward Resistance
Diode Circuit Models
The Ideal Diode with Barrier Potential and Linear Forward Resistance
Example: Assume the diode is a low-power diode with a forward resistance value of 5 ohms. The barrier potential voltage is still: V = 0.3 volts (typical for a germanium diode) Determine the value of ID if VA = 5 volts.
RS = 50 
Once again, write a KVL equation for the circuit:
0 = VA – IDRS - V - IDRF
ID = VA - V = 5 – = mA
RS + RF ID + VA _ + V RF

15. Diode Circuit Models ID
Values of ID for the Three Different Diode Circuit Models
Ideal Diode Model
Ideal Diode Model with Barrier Potential Voltage
Ideal Diode Model with Barrier Potential and Linear Forward Resistance ID
100 mA
94 mA
85.5 mA
These are the values found in the examples on previous slides where the applied voltage was 5 volts, the barrier potential was 0.3 volts and the linear forward resistance value was assumed to be 5 ohms.

16. The Q Point
The operating point or Q point of the diode is the quiescent or no-signal condition. The Q point is obtained graphically and is really only needed when the applied voltage is very close to the diode’s barrier potential voltage. The example 3 below that is continued on the next slide, shows how the Q point is determined using the transconductance curve and the load line.
First the load line is found by substituting in different values of V into the equation for ID using the ideal diode with barrier potential model for the diode. With RS at 1000 ohms the value of RF wouldn’t have much impact on the results.
ID = VA – V  RS
Using V  values of 0 volts and 1.4 volts we obtain ID values of 6 mA and 4.6 mA respectively. Next we will draw the line connecting these two points on the graph with the transconductance curve. This line is the load line.
RS = 1000  ID + VA
= 6V _ + V

17. The Q Point ID (mA) VD (Volts)
The transconductance curve below is for a Silicon diode. The Q point in this example is located at 0.7 V and 5.3 mA.
ID (mA) 12 10 8
Q Point: The intersection of the load line and the transconductance curve. 6
5.3
4.6 4 2
VD (Volts)
0.2
0.4
0.6
0.8
1.0
1.2
1.4
0.7

18. Dynamic Resistance rF = VT ID vF = vac rF rF + RS VD = V + vF
The dynamic resistance of the diode is mathematically determined as the inverse of the slope of the transconductance curve. Therefore, the equation for dynamic resistance is:
rF = VT ID
The dynamic resistance is used in determining the voltage drop across the diode in the situation where a voltage source is supplying a sinusoidal signal with a dc offset.
The ac component of the diode voltage is found using the following equation:
vF = vac rF
rF + RS
The voltage drop through the diode is a combination of the ac and dc components and is equal to:
VD = V + vF

19. Dynamic Resistance
Example: Use the same circuit used for the Q point example but change the voltage source so it is an ac source with a dc offset. The source voltage is now, vin = 6 + sin(wt) Volts. It is a silicon diode so the barrier potential voltage is still 0.7 volts.
The DC component of the circuit is the same as the previous example and therefore ID = 6V – 0.7 V = 5.2 mA
1000 
rF = VT = 1 * 26 mV = 4.9 
ID mA
 = 1 is a good approximation if the dc current is greater than 1 mA as it is in this example.
RS = 1000  ID +
vin + V
vF = vac rF = sin(wt) V  = sin(wt) mV
rF + RS  
Therefore, VD = sin (wt) mV (the voltage drop across the diode)

20. Types of Diodes and Their Uses
PN Junction Diodes:
Are used to allow current to flow in one direction while blocking current flow in the opposite direction. The pn junction diode is the typical diode that has been used in the previous circuits. A K P n
Schematic Symbol for a PN Junction Diode
Representative Structure for a PN Junction Diode
Zener Diodes:
Are specifically designed to operate under reverse breakdown conditions. These diodes have a very accurate and specific reverse breakdown voltage. A K
Schematic Symbol for a Zener Diode

21. Types of Diodes and Their Uses
Schottky Diodes:
These diodes are designed to have a very fast switching time which makes them a great diode for digital circuit applications. They are very common in computers because of their ability to be switched on and off so quickly. A K
Schematic Symbol for a Schottky Diode
Shockley Diodes:
The Shockley diode is a four-layer diode while other diodes are normally made with only two layers. These types of diodes are generally used to control the average power delivered to a load. A K
Schematic Symbol for a four-layer Shockley Diode

22. Types of Diodes and Their Uses
Light-Emitting Diodes:
Light-emitting diodes are designed with a very large bandgap so movement of carriers across their depletion region emits photons of light energy. Lower bandgap LEDs (Light-Emitting Diodes) emit infrared radiation, while LEDs with higher bandgap energy emit visible light. Many stop lights are now starting to use LEDs because they are extremely bright and last longer than regular bulbs for a relatively low cost.
The arrows in the LED representation indicate emitted light. A K
Schematic Symbol for a Light-Emitting Diode

23. Types of Diodes and Their Uses Schematic Symbols for Photodiodes
While LEDs emit light, Photodiodes are sensitive to received light. They are constructed so their pn junction can be exposed to the outside through a clear window or lens.
In Photoconductive mode the saturation current increases in proportion to the intensity of the received light. This type of diode is used in CD players.
In Photovoltaic mode, when the pn junction is exposed to a certain wavelength of light, the diode generates voltage and can be used as an energy source. This type of diode is used in the production of solar power. A K  A K
Schematic Symbols for Photodiodes

24. Sources
Dailey, Denton. Electronic Devices and Circuits, Discrete and Integrated. Prentice Hall, New Jersey: (pp 2-37, )
2 Figure The diode transconductance curve, pg. 7
Figure Determination of the average forward resistance of a diode, pg 11
3 Example from pages 13-14
Liou, J.J. and Yuan, J.S. Semiconductor Device Physics and Simulation. Plenum Press, New York:
Neamen, Donald. Semiconductor Physics & Devices. Basic Principles. McGraw-Hill, Boston: (pp 1-15, )
1 Figure The space charge region, the electric field, and the forces acting on the charged carriers, pg 213.

25. Diodes and diode circuits with examples and applications
Reading Assignment
Diodes and diode circuits with examples and applications

26. SERIES ELEMENTS
KCL tells us that all of the elements in a single branch carry the same current.
We say these elements are in series.
Current entering node = Current leaving node
i1 = i2

27. Thus, equivalent resistance of resistors in series is the sum
Circuit with several resistors in series: Find “equivalent resistance” R 2 1
VSS I 3 4  +
KCL tells us same current flows through every resistor
KVL tells us
Clearly,
Thus, equivalent resistance of resistors in series is the sum

28. VOLTAGE DIVIDER
Circuit with several resistors in series
We know I  + V1 R 1
Thus, R 2  +
VSS  + V3 R 3 R 4
and
etc…

29. WHEN IS VOLTAGE DIVIDER FORMULA CORRECT?2 1 V SS I 3 4 - + I R 2 1 3 4 - + V2 + 2 V - + - V I SS 3 R 5 SS 4 3 2 1 V R × + = SS 4 3 2 1 V R × + ≠
Correct if nothing else
connected to nodes
because R
removes condition of 5
resistors in series I 3

30. MEASURING CURRENT
To measure current in a circuit, insert DMM (in current mode) into circuit, in series with measured element.
But ammeters change the circuit. Ammeters are characterized by their “ammeter input resistance,” Rin. Ideally this should be very low. Typical value 1W.
Ideal Ammeter
Real Ammeter ?
Rin

31. Example: V = 1 V: R1= R2 = 500 W, Rin = 1W
MEASURING CURRENT
Potential measurement error due to non-zero input resistance: I I
meas
ammeter R R 1 1 R in _ + _ + V V R R 2 2
undisturbed circuit
with ammeter
Example: V = 1 V: R1= R2 = 500 W, Rin = 1W

32. PARALLEL ELEMENTS Va = Vb
KVL tells us that any set of elements which are connected at both ends carry the same voltage.
We say these elements are in parallel.
KVL clockwise,
start at top:
Vb – Va = 0
Va = Vb

33. Resistors in parallel can be made into one equivalent resistorx
KCL tells us
Iss = I1 + I2 I I 1 2
ISS R R 1 2
The two resistors are in parallel; they have the same voltage
VX = I1 R1 = I2 R2
ground
Iss = VX / R1 + VX / R VX = Iss R1 R2 / (R1+R2)
Generally, Req = (R1-1 + R2-1 + R3-1 + …)-1

34. For resistors in series:
IMPORTANT FACTS
For resistors in series:
Current through Req is equal to the current through each of the original resistors (all have same current)
Voltage over Req is the sum of the voltages over the original resistors
For resistors in parallel:
Current through Req is equal to the sum of the currents through each of the original resistors
Voltage over Req is equal to the voltage over the original resistors (all have same voltage)

35. CURRENT DIVIDER
There is a simple equation for the way current splits between two parallel resistors: x
Remember
VX = I1 R1 = ISS Req I I 1 2
ISS R R 1 2
ground

36. REAL VOLTMETERS
How is voltage measured? Digital multimeter (DMM) in parallel with measured element.
Connecting a real voltmeter across two nodes changes the circuit. The voltmeter may be modeled by an ideal voltmeter (open circuit) in parallel with a resistance: “voltmeter input resistance,” Rin. Typical value: 10 MW
Ideal Voltmeter
Real Voltmeter
Rin

37. Computation of voltage (uses ideal voltmeter)
REAL VOLTMETERS
Computation of voltage (uses ideal voltmeter)
Measurement of voltage (with loading by real voltmeter) R R 1 1 +  - + + - +
VSS
VSS R V R 2 2 2 R in - -
Example:
But if
a 1% error

38. CAPACITORS IN PARALLEL+ + C2
i(t) C1
| (
| (
| ( V
i(t)
Ceq
V(t)  
Equivalent capacitance defined by

39. CAPACITORS IN SERIES | ( | ( C1 V1 i(t) C2 V2 +  Veq + 
Equivalent to
Ceq
i(t)

40. DIGITAL CIRCUIT: DRAM + 
initially uncharged -

41. FINDING VOLTAGES IN A CIRCUIT
So far, we have been applying KVL and KCL “haphazardly” to find voltages and currents in a circuit.
Good for developing intuition, finding things quickly…
…but what if the circuit is complicated? What if you get stuck?
Systematic way to find all node voltages in a circuit:
Nodal Analysis

42. NOTATION: NODE VOLTAGES
The voltage drop from node X to a reference node (ground) is called the node voltage Vx.
Example: a b + + + Vb Va _ _ _
ground

43. FORMAL CIRCUIT ANALYSIS USING KCL: NODAL ANALYSIS
(Memorize these steps and apply them rigorously!)
Choose a reference node (ground, node 0)
(look for the one with the most connections!)
2 Define unknown node voltages (those not fixed by voltage sources)
Write KCL at each unknown node, expressing current in terms of the node voltages (using the I-V relationships of branch elements*)
Solve the set of equations (N equations for N unknown node voltages)
* with floating voltages we will use a modified Step 3

44. EXAMPLE OF NODE ANALYSIS
node voltage set  Va V b
 reference node R 4 V1 2 + - IS 3 R1
What if we used different ref node?
Choose a reference node.
Define the node voltages (except reference node and the one set by the voltage source).
Apply KCL at the nodes with unknown voltage.
Solve for Va and Vb in terms of circuit parameters.

45. EXAMPLE OF NODE ANALYSISV 2 1 R 4 5 3 I Va

46. NODAL ANALYSIS WITH “FLOATING” VOLTAGE SOURCES
A “floating” voltage source is a voltage source for which neither side is connected to the reference node. VLL in the circuit below is an example. R 4 2 I V a b + - LL 1
Problem: We cannot write KCL at node a or b because there is no way to express the current through the voltage source in terms of
Solution: Define a “supernode” – that chunk of the circuit containing nodes a and b. Express KCL at this supernode.

47. FLOATING VOLTAGE SOURCES (cont.)
Use a Gaussian surface to enclose the floating voltage source; write KCL for that surface.
supernode V V a LL V b - + I R R I 1 2 4 2
Two unknowns: Va and Vb.
Get one equation from KCL at supernode:
Get a second equation from the property of the voltage source:

48. Can we choose ground to avoid a floating voltage source?
ANOTHER EXAMPLE Va Vb | + + 
Can we choose ground to avoid a floating voltage source?
Not in this circuit.

49. DEPENDENT VOLTAGE AND CURRENT SOURCES
A linear dependent source is a voltage or current source that depends linearly on some other circuit current or voltage.
Example: You watch a certain voltmeter V1 and manually adjust a voltage source Vs to be 2 times this value. This constitutes a voltage-dependent voltage source. V1 + -
Circuit A
2V1 + -
Circuit B
This is just a manual example, but we can create such dependent sources electronically.
We will create a new symbol for dependent sources.

50. DEPENDENT VOLTAGE AND CURRENT SOURCES
We can have voltage or current sources depending on voltages or currents elsewhere in the circuit.
Here, the voltage V provided by the dependent source (right) is proportional to the voltage drop over Element X. The dependent source does not need to be attached to the Element X in any way. + - V X
Element X
A diamond-shaped symbol is used for dependent sources, just as a reminder that it’s a dependent source.
Circuit analysis is performed just as with independent sources.

51. THE 4 BASIC LINEAR DEPENDENT SOURCES
Constant of proportionality
Parameter being sensed
Output
Voltage-controlled voltage source … V = Av Vcd
Current-controlled voltage source … V = Rm Ic
Current-controlled current source … I = Ai Ic
Gm Vcd
Voltage-controlled current source … I = Gm Vcd
Ai Ic + _ + _
Av Vcd
Rm Ic

52. EXAMPLE OF THE USE OF DEPENDENT SOURCE IN THE MODEL FOR AN AMPLIFIER
V0 depends only on input (V+  V-) +  A V+ V V0
Differential Amplifier
AMPLIFIER SYMBOL +  V0
A(V+-V-)
V+-V- Ri
Circuit Model in linear region
AMPLIFIER MODEL
This model when used correctly mimics the behavior of an amplifier but omits the complication of the many transistors and other components.
We will learn more about the amplifier next week.

53. NODAL ANALYSIS WITH DEPENDENT SOURCESV R V R5 V 1 a 3 b c R 2 +
VSS
ISS R R  + 4 6 A V  v c

54. NODAL ANALYSIS WITH DEPENDENT SOURCESV R V R a 1 b 3 R I2
ISS I R + m 2 2 R 4 -

55. ANOTHER EXAMPLE Va Vb | + +  I2

56. INTUITION
A dependent source is like a “krazy resistor”, except its voltage (current) depends on a different current (voltage).
If the controlling voltage or current is zero, the dependent voltage source will have zero voltage (and the dependent current source will have zero current).
There must be independent sources in the circuit for nonzero voltages or currents to happen!
Math explanation: Node voltage analysis leads to
The A matrix is nonsingular for real circuits, and is made of resistor values and dependent source parameters. The b vector is made of independent voltage & current source values. If b is zero and A is nonsingular, the solution must be zero!
Ax = b