ee2301: basic electronic circuit let’s start with diode ee2301: block c unit 11
TRANSCRIPT
EE2301: Basic Electronic Circuit
The Basic Property of a Diode
Let’s have a demo
EE2301: Block C Unit 1 3
EE2301: Basic Electronic Circuit
EE2301: Block C Unit 1 5
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
EE2301: Basic Electronic Circuit
Semiconductor Electronics - Unit 1: Diodes 6
Electrical Materials
Insulators ConductorsSemi-
Conductors
EE2301: Basic Electronic Circuit
Semiconductor Electronics - Unit 1: Diodes 7
Semiconductor Applications
Integrated Circuit
EE2301: Basic Electronic Circuit
Semiconductor Electronics - Unit 1: Diodes 8
Semiconductor Applications
TFT (Thin Film Transistor)
EE2301: Basic Electronic Circuit
Semiconductor Electronics - Unit 1: Diodes 9
Intrinsic Semiconductor
Si Si
Si Si
Covalent Bonds
EE2301: Basic Electronic Circuit
Semiconductor Electronics - Unit 1: Diodes 10
Silicon Crystal Lattice
In 3-D, this looks like:
Number atoms per m3: ~ 1028
EE2301: Basic Electronic Circuit
Semiconductor Electronics - Unit 1: Diodes 11
Growing Silicon
We can grow very pure silicon
EE2301: Basic Electronic Circuit
Semiconductor Electronics - Unit 1: Diodes 13
Currents in Semiconductor
Source: http://hyperphysics.phy-astr.gsu.edu/HBASE/solids/intrin.html
EE2301: Basic Electronic Circuit
Semiconductor Electronics - Unit 1: Diodes 14
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
EE2301: Basic Electronic Circuit
Semiconductor Electronics - Unit 1: Diodes 15
Doping: n-type
1 Si atom substituted by 1 P atom
P has 5 valence electrons (1 electron more)
1 free electron created
Si Si
Si P
Si
Si
-
Electrically neutral
EE2301: Basic Electronic Circuit
Semiconductor Electronics - Unit 1: Diodes 16
Doping: p-type
Si Si
Si B
Si
Si
1 Si atom substituted by 1 B atom
+B has 3 valence electrons (1 electron short)
1 hole created
Electrically neutral
EE2301: Basic Electronic Circuit
Semiconductor Electronics - Unit 1: Diodes 18
Diode Physics
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EE2301: Basic Electronic Circuit
Semiconductor Electronics - Unit 1: Diodes 19
Diode Physics
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Website: http://www-g.eng.cam.ac.uk/mmg/teaching/linearcircuits/diode.html
EE2301: Basic Electronic Circuit
EE2301: Block C Unit 1 20
Biasing and Conventions
vD: Voltage of P (anode) relative to N (cathode)
iD: Current flowing from anode to cathode
EE2301: Basic Electronic Circuit
EE2301: Block C Unit 1 21
Diode
Diode begins to conduct a significant amount of current: Voltage Vγ is typically around 0.7V
Diode equation: ID = I0 [exp(eVD/kT) - 1]
EE2301: Basic Electronic Circuit
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Diode Symbol and Operation
Forward-biased Current (Large)
Reverse-biased Current (~Zero)
+ -
Forward Biased:
Diode conducts
- +
Reverse Biased:
Little or no current
iD
EE2301: Basic Electronic Circuit
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Real diode circuits
+-
+ VL
-
+ VD - ID
VT RT
To find VL where VT and RT are known,
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.
EE2301: Basic Electronic Circuit
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Graphical method
T
TD
TD R
vv
Ri
1
Operating point is where the load line & I-V curve of the diode intersect
Equation from KVL
EE2301: Basic Electronic Circuit
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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
EE2301: Basic Electronic Circuit
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Ideal diode model
In other words, diode is treated like a switch herevD > 0: Short circuit
vD < 0: Open circuit
EE2301: Basic Electronic Circuit
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Ideal diode model
Circuit containing ideal diode
Circuit assuming that the ideal diode conducts
Circuit assuming that the ideal diode does
not conduct
EE2301: Basic Electronic Circuit
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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
EE2301: Basic Electronic Circuit
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Ideal diode example 1 solution
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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
EE2301: Basic Electronic Circuit
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Ideal diode example 1 solution
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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
EE2301: Basic Electronic Circuit
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Ideal diode example 1 solution
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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
EE2301: Basic Electronic Circuit
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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
EE2301: Basic Electronic Circuit
EE2301: Block C Unit 1 34
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
EE2301: Basic Electronic Circuit
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Offset model example solution
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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
EE2301: Basic Electronic Circuit
EE2301: Block C Unit 1 36
Rectification: from AC to DC
Supply is AC DC required
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.
EE2301: Basic Electronic Circuit
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Rectifier with regulator diagram
Rectifier
Bi-directional input Steady DC output
Filter Regulator
Unidirectional output
We will look at two types of rectifiers and apply the large signal models in our analysis:1) Half wave rectifier2) Full wave rectifier
EE2301: Basic Electronic Circuit
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Half-Wave Rectifier
VS~ RL
On the positive cycle
Diode is forward biased
Diode conducts
VL will follow VS
VS~ RL
On the negative cycle
Diode is reverse biased
Diode does not conduct
VL will remain at zero
VSVL
We can see that the circuit conducts for only half a cycle
EE2301: Basic Electronic Circuit
EE2301: Block C Unit 1 39
Average voltage in a HW Rectifier
rmspeak
peak
T
T
T
peakL
VV
dtdV
dtdttVT
v
2
0)sin(2
1
0)sin(1
2
0
2/
2/
0
1st half of period
2nd half of period
NB: This is equal to the DC term of the Fourier series
EE2301: Basic Electronic Circuit
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Full-Wave Rectifier
Also known as BRIDGE rectifier Comprises 2 sets of diode pairs Each pair conducts in turn on each half-cycle
VS ~
EE2301: Basic Electronic Circuit
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Full-Wave Rectifier
rmspeak
peak
T
peakL
vv
dv
dttvT
v
222
)sin(1
)sin(2
0
2/
0
Half a period
T/2 – time
π – phase
Repeats for every half a period:
Integrate through half a period
NB: This is equal to the DC term of the Fourier series
EE2301: Basic Electronic Circuit
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Full-Wave Rectifier (offset)
VD-on (only one diode is on)
2VD-on (two diodes are on)
With offset diodes
With ideal diodes
EE2301: Basic Electronic Circuit
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Ripple filter
Charging
Discharging
Anti-ripple filter is used to smoothen out the rectifier output
EE2301: Basic Electronic Circuit
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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
EE2301: Basic Electronic Circuit
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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
EE2301: Basic Electronic Circuit
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Ripple filter example solution
a) TURNS RATIO: To find the turns ratio, we need to find VS1 and VS2
V125.5125.052
1 rLm VVV
V875.4125.052
1min rLL VVV
EE2301: Basic Electronic Circuit
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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
EE2301: Basic Electronic Circuit
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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
EE2301: Basic Electronic Circuit
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Ripple filter example solution
VL-min = - VSOcos(ωt2) - VD-on
2nd half of the sinusoid
t2 = (1/ω) cos-1{-(VL-min + VD-on)/VSO} = 7.533 ms
Decaying exponential: VL-min = VM exp(-t2/RLC)
mFV
VRt
CM
L
L
8.1ln min2
EE2301: Basic Electronic Circuit
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Electrical MaterialsInsulators
Electrons are bound to the nucleus and are therefore not free to move
With no free electrons, conduction cannot occur
ConductorsSea of free electrons not bound to the atomsAmple 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
EE2301: Basic Electronic Circuit
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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 about1.5×1016 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.
EE2301: Basic Electronic Circuit
Extrinsic semiconductors
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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.
EE2301: Basic Electronic Circuit
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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.
EE2301: Basic Electronic Circuit
EE2301: Block C Unit 1
Response of the depletion region
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Forward biased:
Voltage on the p-type side is higher than the n-type side
Depletion width reduces, lowering barrier for majority carriers to move across the depletion region
Large conduction current
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Reverse biased:
Voltage on the p-type side is lower than the n-type side
Depletion width increases, increasing the barrier for majority carriers to move across the depletion region
Very small leakage current