let’s start with diode

EE2301: Basic Electronic Circuit

Let’s start with diode

EE2301: Block C Unit 1 1


Examples of Diode



The Basic Property of a Diode

Let’s have a demo



How does it work?




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


Semiconductor Electronics - Unit 1: Diodes 6

Electrical Materials

Insulators ConductorsSemi-

Conductors



Semiconductor Applications

Integrated Circuit



Semiconductor Applications

TFT (Thin Film Transistor)



Intrinsic Semiconductor

Si Si

Si Si

Covalent Bonds



Silicon Crystal Lattice

In 3-D, this looks like:

Number atoms per m3: ~ 1028



Growing Silicon

We can grow very pure silicon



Conduction



Currents in Semiconductor

Source: http://hyperphysics.phy-astr.gsu.edu/HBASE/solids/intrin.html



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



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



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



p-n Junction



Diode Physics

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Diode Physics

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Website: http://www-g.eng.cam.ac.uk/mmg/teaching/linearcircuits/diode.html



Biasing and Conventions

vD: Voltage of P (anode) relative to N (cathode)

iD: Current flowing from anode to cathode



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]



Diode Symbol and Operation

Forward-biased Current (Large)

Reverse-biased Current (~Zero)

+ -

Forward Biased:

Diode conducts

- +

Reverse Biased:

Little or no current

iD



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.



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



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



Ideal diode model

In other words, diode is treated like a switch herevD > 0: Short circuit

vD < 0: Open circuit



Ideal diode model

Circuit containing ideal diode

Circuit assuming that the ideal diode conducts

Circuit assuming that the ideal diode does

not conduct



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



Ideal diode example 1 solution

This slide is meant to be blank

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





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





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



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



Offset diode model



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



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



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.



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



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



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



Full-Wave Rectifier

Also known as BRIDGE rectifier Comprises 2 sets of diode pairs Each pair conducts in turn on each half-cycle

VS ~



Full-Wave Rectifier



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



Full-Wave Rectifier (offset)

VD-on (only one diode is on)

2VD-on (two diodes are on)

With offset diodes

With ideal diodes



Ripple filter

Charging

Discharging

Anti-ripple filter is used to smoothen out the rectifier output



Ripple filter

Approximation: abrupt change in the voltage

From transient analysis: VMexp(-t/RC)

VL

Ripple voltage Vr = VM - VL min



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



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




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




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




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


Examples of Diode




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



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.


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.



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: 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



Analogy from tides

Depletion region

Depletion region

Forward Biased

Reverse Biased

let’s start with diode

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