Transcript Block D: Semiconductor Electronics

EE2301: Basic Electronic Circuit
Let’s start with diode
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Examples of Diode
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The Basic Property of a Diode
Let’s have a demo
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How does it work?
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Block C Unit 1 Outline
 Semiconductor materials (eg. silicon)
> Intrinsic and extrinsic semiconductors
 How a p-n junction works (basis of diodes)
 Large signal models
> Ideal diode model
> Offset diode model
 Finding the operating point
 Application of diodes in rectification
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Electrical Materials
Insulators
Semiconductor Electronics - Unit 1: Diodes
SemiConductors
Conductors
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Semiconductor Applications
Integrated Circuit
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Semiconductor Applications
TFT (Thin Film Transistor)
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Intrinsic Semiconductor
Si
Si
Si
Si
Covalent Bonds
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Silicon Crystal Lattice
In 3-D, this looks like:
Number atoms per m3: ~ 1028
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Growing Silicon
We can grow very pure silicon
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Conduction
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Currents in Semiconductor
Source: http://hyperphysics.phy-astr.gsu.edu/HBASE/solids/intrin.html
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Carrier Concentration
The number of free electrons available for a
given material is called the intrinsic
concentration ni. For example, at room
temperature, silicon has:
ni = 1.5 x 1016 electrons/m3
1 free electron in about every 1012 atoms
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Doping: n-type
1 Si atom substituted
by 1 P atom
Si
Si
Si
P has 5 valence electrons
(1 electron more)
-
Si
P
Si
1 free electron
created
Electrically neutral
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Doping: p-type
1 Si atom substituted
by 1 B atom
Si
Si
Si
B has 3 valence electrons
(1 electron short)
+
Si
B
Si
1 hole created
Electrically neutral
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p-n Junction
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Diode Physics
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Semiconductor Electronics - Unit 1: Diodes
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Diode Physics
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Website: http://www-g.eng.cam.ac.uk/mmg/teaching/linearcircuits/diode.html
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Biasing and Conventions
vD: Voltage of P (anode) relative to N (cathode)
iD: Current flowing from anode to cathode
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Diode
Diode equation: ID = I0 [exp(eVD/kT) - 1]
Diode begins to conduct a significant amount
of current: Voltage Vγ is typically around 0.7V
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Diode Symbol and Operation
Forward-biased
Current (Large)
Reverse-biased
Current (~Zero)
iD
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Forward Biased:
Reverse Biased:
Diode conducts
Little or no current
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Real diode circuits
+ VD - ID
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VT
RT
To find VL where VT and RT are known,
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VL
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First apply KVL around the loop:
VT = VD + RTID
Then use the diode equation:
ID = I0 [exp(eVD/kT) - 1]
At T = 300K, kT/e = 25mV
We then need to solve these two simultaneous equations, which is not trivial.
One alternative is to use the graphical method to find the value of ID and VD.
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Graphical method
Equation from KVL
1
vT
iD   v D 
RT
RT
Operating point is where
the load line & I-V curve
of the diode intersect
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Diode circuit models
 Simplify analysis of diode circuits which
can be otherwise difficult
 Large-signal models: describe device
behavior in the presence of relatively large
voltages & currents
> Ideal diode model
> Off-set diode mode
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Ideal diode model
vD > 0: Short circuit
In other words, diode is
treated like a switch here
vD < 0: Open circuit
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Ideal diode model
Circuit containing
ideal diode
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Circuit assuming that the
ideal diode conducts
Circuit assuming that
the ideal diode does
not conduct
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Ideal diode example 1
Problems 9.7 and 9.8
Determine whether the diode is conducting or not. Assume diode is ideal
Repeat for Vi = 12V and VB = 15V
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Ideal diode example 1 solution
Option 1:
Assume diode is conducting and find the diode current direction
Outcome 1: If diode current flows from anode to cathode, the assumption is true 
Diode is forward biased
Outcome 2: If diode current flows from cathode to anode, the assumption is false 
Diode is reverse biased
Option 2:
Assume diode is not conducting and find the voltage drop across it
Outcome 1: If voltage drops from cathode to anode, then the assumption is true  Diode
is reverse biased
Outcome 2: If voltage drops from anode to cathode, then the assumption is false  Diode
is forward biased
This slide is meant to be blank
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Ideal diode example 1 solution
Assume diode is conducting
Forward-bias diode current (ie anode to cathode)
= (10 - 12) / (5 + 10) = -2/15 A
Assumption was wrong
Diode is in reverse bias
Assume diode is not-conducting
Reverse-bias voltage (ie cathode referenced to anode)
= 12 - 10 = 2V
Assumption was correct
Diode is in reverse bias
This slide is meant to be blank
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Ideal diode example 1 solution
Assume diode is conducting
Forward-bias diode current (ie anode to cathode)
= (15 - 12) / (5 + 10) = 1/5 A
Assumption was correct
Diode is in forward bias
Assume diode is not-conducting
Reverse-bias voltage (ie cathode referenced to anode)
= 12 - 15 = -3V
Assumption was incorrect
Diode is in forward bias
This slide is meant to be blank
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Ideal diode example 2
Problem 9.14
Find the range of Vin for which D1 is forward-biased. Assume diode is ideal
The diode is ON as long as forward bias voltage
is positive
Now, minimum vin for vD to be positive = 2V
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Offset diode model
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Offset model example
Problem 9.19
The diode in this circuit requires a minimum current of 1 mA to be above the
knee of its characteristic. Use Vγ = 0.7V
What should be the value of R to establish 5 mA in the circuit?
With the above value of R, what is the minimum value of E required to maintain
a current above the knee
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Offset model example solution
ID = (E - VD)/R
When the diode is conducting, VD = Vγ
ID = (5 - 0.7)/R
We can observe that as R increases, ID will decrease
To maintain a minimum current of 5mA,
Rmax = 4.3/5 = 860 Ω
Minimum E required to keep current above the knee (1mA),
Emin = (10-3 * 860) + 0.7 = 1.56V
This slide is meant to be blank
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Rectification: from AC to DC
One common application of diodes is rectification. In rectification, an AC
sinusoidal source is converted to a unidirectional output which is further
filtered and regulated to give a steady DC output.
Supply is AC
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DC required
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Rectifier with regulator diagram
Rectifier
Bi-directional
input
Unidirectional output
Filter
Regulator
Steady DC output
We will look at two types of rectifiers and apply the large signal
models in our analysis:
1) Half wave rectifier
2) Full wave rectifier
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Half-Wave Rectifier
~
VS
RL
~
VS
RL
We can see that the
circuit conducts for
only half a cycle
VL
VS
On the positive cycle
On the negative cycle
Diode is forward biased Diode is reverse biased
Diode conducts
Diode does not conduct
VL will follow VS
VL will remain at zero
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Average voltage in a HW Rectifier
vL
T /2
T

1
   V peak sin(t )dt   0dt
T0
T /2


2


1

  V peak sin( )d   0dt
2  0

 st
1 half of
V peak
2Vrms
period




2nd half
of period
NB: This is equal to the DC term of the Fourier series
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Full-Wave Rectifier
VS
~
 Also known as BRIDGE rectifier
 Comprises 2 sets of diode pairs
 Each pair conducts in turn on each half-cycle
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Full-Wave Rectifier
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Full-Wave Rectifier
Repeats for every half a period:
Integrate through half a period
vL
2

T

1
T /2
v
peak
sin(t )dt
0

v


peak
sin( )d
Half a period
0

2v peak


2 2vrms

NB: This is equal to the DC term of the Fourier series
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T/2 – time
π – phase
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Full-Wave Rectifier (offset)
VD-on (only one diode is on)
2VD-on (two diodes are on)
With ideal diodes
With offset diodes
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Ripple filter
Anti-ripple filter is used to smoothen out the rectifier output
Charging
Discharging
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Ripple filter
Approximation: abrupt change in the voltage
From transient analysis: VMexp(-t/RC)
VL
Ripple voltage Vr = VM - VL min
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Ripple filter example
Problem 9.40
Find the turns ratio of the transformer and the value of C given that:
IL = 60mA, VL = 5V, Vr = 5%, Vline = 170cos(ωt) V, ω = 377rad/s
Diodes are fabricated from silicon, Vγ = 0.7V
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Ripple filter example solution
a) TURNS RATIO: To find the turns ratio, we need to find VS1 and VS2
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Vm  VL  Vr  5  0.125  5.125 V
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VL  min  VL  Vr  5  0.125  4.875 V
2
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Ripple filter example solution
But VM is not equal to VS1 due to voltage drop across diodes
So we now apply KVL on the secondary coil side:
VS1 - VD - VM = 0
VS1 = 5.825 V (VD = 0.7V)
Turns ratio, n = Vline / VS1 ~ 29
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Ripple filter example solution
b) Value of C: Need to find the RC time constant associated with
the ripple
RL = VL/IL = 83.3 Ω
We know it decays by VMexp(-t/RC), we now just need to know how long this lasts (t2)
VL-min = - VSOcos(ωt2) - VD-on
vso is negative at this point
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Ripple filter example solution
t2 = (1/ω) cos-1{-(VL-min + VD-on)/VSO} = 7.533 ms
Decaying exponential:
VL-min = VM exp(-t2/RLC)
 t2 
 VL min 
  1.8m F
C     ln
 RL 
 VM 
VL-min = - VSOcos(ωt2) - VD-on
2nd half of the sinusoid
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Examples of Diode
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Electrical Materials
Insulators
Electrons are bound to the nucleus and are therefore
not free to move
With no free electrons, conduction cannot occur
Conductors
Sea of free electrons not bound to the atoms
Ample availability of free electrons allows for
electrical conduction
Semiconductors
Electrons are bound to the nucleus but vacancies are
created due to thermal excitation
Electrical conduction occurs through positive (called
holes) and negative (electrons) charge carriers
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Conduction in Semiconductors
Silicon is the dominant semiconductor material used in the
electronics industry. In a cubic meter of silicon, there are
roughly 1028 atoms. Among these 1028, there will be about
1.5×1016 vacancies at room temperature. This is known as
the intrinsic carrier concentration: n = 1.5×1016 electrons/m3.
This corresponds to 1 free electron for every 1012 atoms.
There will be same number of electrons as holes in intrinsic
silicon since it is overall electrically neutral.
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Extrinsic semiconductors
A semiconductor material that has been subjected to the doping process is
called an extrinsic material. Both n-type and p-type materials are formed by
adding a predetermined number of impurity atoms to a silicon base.
An n-type material is created by introducing impurity elements that have five
valence electrons. In an n-type material, the electron is called the majority carrier.
An p-type material is created by introducing impurity elements that have three
valence electrons. In a p-type material, the hole is the majority carrier.
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p-n Junction
The pn junction forms the basis of the semiconductor diode
Within the depletion region, no free carriers exist since the holes and electrons at
the interface between the p-type and n-type recombine.
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Response of the depletion region
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----+ --------+
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+++++
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Forward biased:
Reverse biased:
Voltage on the p-type side is higher than the
n-type side
Voltage on the p-type side is lower than the ntype side
Depletion width reduces, lowering barrier for
majority carriers to move across the depletion
region
Depletion width increases, increasing the
barrier for majority carriers to move across
the depletion region
Large conduction current
Very small leakage current
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Analogy from tides
Depletion region
Forward Biased
Reverse Biased
Depletion region
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