Analogue Electronics-Chapter two intro.ppt

analog electronics

1. Chapter 2 Semiconductor DiodesChapter 2 Semiconductor Diodes
Analogue ElectronicsAnalogue Electronics

2. 1.1 Semiconductor structure
1.2 P-type and N-type semiconductorP-type and N-type semiconductor
1.3 The p-n junction
1.4 Diode characteristics
1.5 Diode Models
1.6 Graphical Analysis (Load-line analysis)
1.7 Diode circuits (Diode applications )
Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
ContentContent

3. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Semiconductor structure
The predominant semiconductor material is silicon. Silicon is one of the most
abundant elements on earth, and is always found in compound form in nature
(sand is mainly SiO
2
). It is purified by chemical means so that the concentration
of troublesome impurities is about 1 in a billion. The valence of silicon is 4,
like carbon.
1 silicon atom = 1 nucleus + 4 electrons

4. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Semiconductor structure
It is the valence electrons that participate in chemical bonding when the
atoms form compounds. Silicon crystallizes in a diamond-like structure –
each atom has four neighbours.
Pure silicon will possess the same structure as diamond, but it is nowhere
near as hard as diamond. There are two reasons for this. Firstly, the silicon-
silicon bond is much weaker than the carbon-carbon bond. Secondly, carbon is
a significantly smaller-than-average atom, and there are more bonds per unit of
volume in a diamond than in any other substance.

5. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Semiconductor structure
To simplify things, we can describe the Si crystal in a two-dimensional form.
To simplify things, we can describe the Si crystal in a two-dimensional form. At
very low temperatures, pure silicon behaves as an insulator, since a shared
electron is bound to its locality and there is no source of extra energy to free
itself from its bonds and make itself available for conduction. The extra energy can
be obtained from thermal vibrations of the crystal lattice atoms.

6. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
When a valence electron is freed, two charge carriers are created. The first is
the electron itself. The second one, called a hole, is the charge located in the
area vacated by the electron, left with a net positive charge
(obviously caused by a silicon nucleus). Any one of the other valence electrons
moving nearby can step into the vacated site. This shifts the net positive charge
– the hole – to a new location. Both the free electron and the hole can therefore
move around in the semiconductor crystal. The conductivity of pure silicon is
therefore proportional to the free carrier concentration, and is very small.
Why can semiconductors conduct electricity?

7. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
p-type Semiconductor
To make devices like diodes and transistors, it is necessary to increase the
electron and hole population. This is done by intentionally adding specific
impurities in controlled amounts – a process known as doping.
If Si is doped with an element with only 3 valence electrons, then at the
location of that impurity one of the covalent bonds is missing. One of the other
valence electrons can cross over and complete the missing bond. When this
happens a hole is created at the position vacated by that valence electron.
The impurity atom, having accepted an additional electron, is called an
acceptor and now has a net negative charge. A semiconductor doped with
acceptors is rich in holes, i.e. positive charge carriers, and therefore called
p-type. In this case the holes are called the majority carriers, the electrons are
called the minority carriers.

8. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
n-type Semiconductor
If Si is doped with an element with 5 valence electrons, then four of the valence
electrons will take part in the covalent bonding with the neighbouring Si atoms
while the fifth one will be only weakly attached to the impurity atom location. The
thermal energy of a semiconductor at room temperature is more than
enough to free this electron, making it available for conduction.
The impurity atom, having donated an additional electron, is called a donor.
The semiconductor in this case is called n-type, because it is rich in negative
charge carriers. The electrons are the majority carriers and the holes are the
minority carriers for this type of semiconductor.

9. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
The p-n Junction
What happens when P-type meets N-type?

10. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
The p-n Junction
What happens when P-type meets N-type?
Imagine that we have a gas cylinder full of oxygen. We open the valve to
release the oxygen into the room. What happens? The oxygen and the air
in the room mix – a process known as diffusion. Nature wants things to
spread out in an even fashion. This is what happens with the free, gas-like
particles in the p-n junction.

11. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
The p-n Junction
What happens when P-type meets N-type?
• Holes diffuse from the P-type into the N-type, electrons diffuse from the N-
type
into the P-type, creating a diffusion current.
• Once the holes [electrons] cross into the N-type [P-type] region, they
recombine
with the electrons [holes].
• This recombination “strips” the n-type [P-type] of its electrons near the
boundary, creating an electric field due to the positive and negative bound
charges.
• The region “stripped” of carriers is called the space-charge region, or depletion
region.
• V
0 is the contact potential that exists due to the electric field. Typically, at room

temp, V
0 is 0.5~0.8V.

12. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
The p-n Junction
There are two mechanisms by which mobile carriers move in
semiconductors – resulting in current flow
– Diffusion
• Majority carriers move (diffuse) from a place of higher
concentration to a place of lower concentration
– Drift
• Minority carrier movement is induced by the electric field.
What happens when P-type meets N-type?

13. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
The p-n Junction
Forward bias: apply a positive voltage to the P-type, negative to N-type.
Add more majority carriers to both sides
shrink the depletion region lower V
0
diffusion current increases.
• Decrease the built-in potential, lower
the barrier height.
• Increase the number of carriers able
to
diffuse across the barrier
• Diffusion current increases
• Drift current remains the same. The
drift current is essentially constant, as

it is dependent on temperature.
• Current flows from p to n

14. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
The p-n Junction
Reverse bias: apply a negative voltage to the P-type, positive to N-type.
• Increase the built-in potential, increase the
barrier height.
• Decrease the number of carriers able to
diffuse across the barrier.
• Diffusion current decreases.
• Drift current remains the same
• Almost no current flows. Reverse leakage
current, I
S, is the drift current, flowing from
N to P.

15. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Diodes
A p-n junction forms what is called a diode. Its circuit symbol is:

16. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Diodes v-i Characteristic
The diode’s terminal electrical characteristics can be obtained using the
following circuit:
With the battery as shown, we can vary R and measure V and I to obtain the
forward-bias characteristic. We could also use a curve tracer to obtain the
characteristic. We can reverse the polarity of E to obtain the reverse-bias
characteristic. The total characteristic looks like:

17. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Diodes v-i Characteristic
Typical PN junction diode volt-ampere
characteristic is shown on the left.
– In forward bias, the PN junction has
a “turn on” voltage based on the
“built-in” potential of the PN
junction. turn on voltage is typically
in the range of 0.5V to 0.8V
– In reverse bias, the PN junction
conducts essentially no current until
a critical breakdown voltage is
reached. The breakdown voltage
can range from 1V to 100V.
Breakdown mechanisms include
avalanche and zener tunneling.

18. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Diode Models
The curve describing the diode’s terminal characteristics is non-linear. How
can we use this curve to do circuit analysis? We only know how to analyze
linear circuits. There is therefore a need for a linear circuit model of the
diode.
When we model something, we transform it into something else – usually
something simpler – which is more amenable to analysis and design using
mathematical equations. Modelling mostly involves assumptions and
simplifications, and the only requirement of a model is for it to “work”
reasonably well. By “work” we mean that it agrees with experimental results
to some degree of accuracy.

19. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
The Ideal Diode Model
As a first approximation, we can model the diode as an ideal switch:
The characteristic in this case is approximated by two straight lines – the
vertical representing the “on” state of the diode, and the horizontal
representing the “off” state. To determine which of these states the diode is
in, we have to determine the conditions imposed upon the diode by an
external circuit. This model of the diode is used sometimes where a quick
“feel” for a diode circuit is needed. The above model can be represented
symbolically as:

20. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
The Ideal Diode Model - example
(i) Find the current, I, in the circuit shown below, using the ideal diode
model.
(ii) If the battery is reversed, what does the current become?

21. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
The Constant Voltage Drop Model
A better model is to approximate the forward bias region with a vertical line that
passes through some voltage called e
fd :
This “constant voltage drop” model is better because it more closely
approximates the characteristic in the forward bias region. The “voltage drop”
is a model for the barrier voltage in the p-n junction. The model of the diode
in this case is:

22. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
The Constant Voltage Drop Model - example
(i) Find the current, I, in the circuit shown below, using the constant voltage
drop model of the diode (assume efd = 0.7 V).
(ii) If the battery is reversed, what does the current become?

23. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
The Piece-Wise Linear Model
An even better approximation to the diode characteristic is called a “piece-
wise” linear model. It is made up of pieces, where each piece is a straight
line:
For each section, we use a different diode model (one for the forward bias
region and one for the reverse bias region)
Typical values for the resistances are r
fd
= 5 Ω and r
rd
= 10
9
Ω

24. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
The Piece-Wise Linear Model - example
(i) Find the current, I, in the circuit shown below, using the piece-wise linear
model of the diode (assume efd = 0.7 V, r
fd
= 5 Ω and r
rd
=∞).
(ii) If the battery is reversed, what does the current become?

25. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
The Small Signal Model
Suppose we know the diode voltage and current exactly. Would we still have
a need for a linear diode model? Yes. Suppose the diode has a DC voltage and
current. We may want to examine the behaviour of a circuit when we apply a
signal (a small AC voltage) to it. In this case we are interested in small
excursions of the voltage and current about some “DC operating point” of the
diode. The best model in this instance is the following (the forward bias
region is used as an example, but the method applies anywhere):

26. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
The Small Signal Model
We approximate the curved characteristic by the tangent that passes through
the operating point. It is only valid for small variations in voltage or current.
This is called the small signal approximation. A straight line is a good
approximation to a curve if we don’t venture too far.
The model we get in this case is exactly the same as the piece-wise model
except the values of e
fd
and r
fd
are different for each DC operating point.
Finally, to complete all our models, we can add a capacitance in parallel to
model the forward and reverse capacitance described previously. We will not
in general include the capacitance because it only becomes important at very
high frequencies.

27. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Graphical Analysis (Load-line analysis)
Any linear resistive circuit can be reduced to an equivalent circuit containing
one source and one resistor. When the source is a voltage, the reduction is
obtained using Thévenin's theorem. Consider the following circuit:

28. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Graphical Analysis (Load-line analysis)
The equivalent circuit (as far as the diode is concerned) can be found
using Thévenin's theorem. (Look into the circuit from the diode terminals. What
do you see?) The Thévenin equivalent circuit for the diode is:
Verify that the Thévenin voltage and Thévenin resistance in this case are given by:

29. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Graphical Analysis (Load-line analysis)
When graphed, we call it the load line. It was derived from KVL, and so it is
always valid. The load line gives a relationship between i
D
and v
D
that is
totally determined by the external circuit. The diode’s characteristic gives a
relationship between i
D
and v
D
that is determined purely by the geometry and
physics of the diode. Since both the load line and the characteristic are to be
satisfied, the only place this is possible is the point at which they meet. This
point is called the Q point.

30. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Graphical Analysis (Load-line analysis)
If the source voltage is increased the Thévenin voltage changes to V′ and the
operating point to ′ Q (the DC load line is shifted up).

31. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Graphical Analysis (Load-line analysis) - example










20
15
i D (mA)
1.0
10
0.5 1.5
vD(V)
operating point
Load line

E=1.5V


D
100Ώ I
Q
I
D
=7(mA), V
D
=0.8(V)

32. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Diode Circuits - Limiter

+

+
-
vi
vo

vo

-
+
R
D
When v
i
> V
on
, D on v
o
 v
i
；
v
i
< V
on
， D off  v
o
= 0。

33. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Diode Circuits - Logical circuit (AND Gates)



R
+5V
V2
V1 Vo
D1
D2
V
1
(V) V
2
(V) V
o
(V) Logic output
List all the possible output voltages Vo if V1 and V2
equal to 0 V or 5 V. Define the logic output is 0
when Vo reaches 5 V and the logic output is 1 when
Vo is clamped at 0 V. Assume the diodes are ideal
model.

34. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Diode Circuits - Logical circuit (AND Gates)



R
+5V
V2
V1 Vo
D1
D2
V
1
(V) V
2
(V) V
o
(V) Logic output
0 0 0 0
5 0 0 0
0 5 0 0
5 5 5 1
List all the possible output voltages Vo if V1 and V2
equal to 0 V or 5 V. Define the logic output is 0
when Vo reaches 5 V and the logic output is 1 when
Vo is clamped at 0 V. Assume the diodes are ideal
model.

35. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Diode Circuits - Logical circuit (AND Gates)



R
+5V
V2
V1 Vo
D1
D2
V
1
(V) V
2
(V) V
o
(V) Logic output
List all the possible output voltages Vo if V1 and V2
equal to 0 V or 5 V. Define the logic output is 0
when Vo reaches 5 V and the logic output is 1 when
Vo is clamped at 0.7 V. Assume the diodes are
constant voltage drop model with efd = 0.7 V.

36. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Diode Circuits - Logical circuit (AND Gates)



R
+5V
V2
V1 Vo
D1
D2
V
1
(V) V
2
(V) V
o
(V) Logic output
0 0 0.7 0
5 0 0.7 0
0 5 0.7 0
5 5 5 1
List all the possible output voltages Vo if V1 and V2
equal to 0 V or 5 V. Define the logic output is 0
when Vo reaches 5 V and the logic output is 1 when
Vo is clamped at 0.7 V. Assume the diodes are
constant voltage drop model with efd = 0.7 V.

37. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Diode Circuits - Logical circuit (OR Gates)
Design a circuit using ideal diodes with OR gates?

38. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Diode Circuits - The Clipping Circuit
Assume both diodes are off. KVL then gives:
If the output voltage is less than E1, then diode
D1 cannot be reversed bias, so it will conduct.
This limits or clamps the output voltage to E1:
If the output voltage is more than E2 then
diode D2 cannot be reversed bias, and it turns
on, limiting the output voltage to E2:
Input voltage is sinusoidal. Draw
the output voltage waveform

39. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Diode Circuits - Multivoltage
Determine V and I in the following circuits, when:
a) The diodes are assumed to be ideal.
b) The diodes are modelled with a constant voltage drop model with e
fd
= 0.7 V.

40. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Diode Circuits – Two-diode circuit analysis
Assumption1:. Since the 15-V source appears to
force positive current through D
1
and D
2
, and the -10-
V source is forcing positive current through D
2
,
assume both diodes are on.
The ideal diode model is chosen. Determine the
operation status of the diodes and find the voltage
and current.
Since the voltage at node D is zero due to the short
circuit of ideal diode D
1
,
The Q-points are (-0.5 mA, 0 V) and (2.0 mA, 0 V)
But, I
D1
< 0 is not allowed by the diode, so try again.
 
mA
k
V
I
D
00.2
5
100
2




mAIIII
DDD 500.000.250.1 |
1211 
mA
k
V
I 50.1
10
)015(
1





41. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Diode Circuits - Two-diode circuit analysis
Since the current in D
1
is zero, I
D2
= I
1
,
Assumption2: Since current
in D
2 is always valid, the
second guess is D
1 off and
D
2 on.
0)10(000,5000,1015
21

D
II
Q-Points are D
1 : (0 mA, -1.67 V):off
D
2 : (1.67 mA, 0 V) :on
Now, the results are consistent with the
assumptions.
mA
V
I 67.1
00015
25
1



,
VIV
D
67.17.1615000,1015
11


42. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Diode Circuits - Multivoltage
Determine V and I in the following circuits, when:
a) The diodes are assumed to be ideal.
b) The diodes are modelled with a constant voltage drop model with e
fd
= 0.7 V.

43. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Diode Circuits - Half-Wave Rectifier
What would the waveform
look like if not an ideal
diode?

44. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
•Diode Circuits - Full-Wave Rectifier
To utilize both halves of the input sinusoid use
a center-tapped transformer…

45. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
•Diode Circuits - Bridge Rectifier
Looks like a Wheatstone bridge. Does not require a enter tapped transformer.
Requires 2 additional diodes and voltage drop is double.

46. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Diode Circuits - Peak Rectifier
To smooth out the peaks and obtain a DC voltage, add a capacitor across the
output.

47. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Diode Circuits - Zener Diode
Some diodes are designed to operate in the breakdown region. It is usually a sharper
transition than the forward bias characteristic, and the breakdown voltage is higher
than the forward conduction voltage.
A Zener diode is a diode that exhibits Zener breakdown when it is reverse biased.
Zener breakdown occurs when the electric field in the depletion layer is strong
enough to generate hole-electron pairs, which are accelerated by the field. This
increases the reverse bias current. It gives rise to a sharper transition and steeper
curve than forward biased conduction
Zener symbol
(V
BR)
piecewise linear model
Vz
Rz

48. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Diode Circuits - Zener Diode for voltage regulation
mA
k
V
3
5
)520(






R
VV
I
ZS
S
1
5
5
mA
R
V
I
L
Z
L



k
V
mAIII
LSZ
2

49. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Diode Circuits - Zener Diode for voltage regulation
mA
k
V
3
5
)520(






R
VV
I
ZS
S
0
11











L
Z
S
Z RR
V
R
V
I
For proper regulation, Zener current must be
positive. The Zener diode keeps the voltage
across load resistor R
L
constant. For Zener
breakdown operation, I
Z
> 0.
1
5
5
mA
R
V
I
L
Z
L



k
V
mAIII
LSZ
2
min
1
R
V
V
R
R
Z
S
L















How to ensure I
Z
> 0, in other words, to
keep the load voltage constant?
What is the minimum value for RL?

50. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Diode Circuits - The peak detector

51. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Diode Circuits - The peak detector

52. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Diode Circuits - The Clamp Circuit

53. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Diode Circuits - The Clamp Circuit

54. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Example 2.14 – Determine vo for the network below for the input indicated
(Assume the diode is ideal)
•Note that the frequency is 1000 Hz, resulting in a period of 1 ms and an interval of 0.5 ms between levels.
Applying Kirchhoff’s voltage law around the input loop results in
020V-Vc5V-  25VVc
The capacitor will therefore charge up to 25 V at the time constant determined by τ = RC. In this case,
τ = C = 1 μs. So the capacitor can be assumed to be charging very quickly compared to 0.5 ms.
•The analysis will begin with the period t1 to t2 of the input signal since the diode is in its short-circuit state.
The output is the voltage across the resistor R with direction connection of 5-V battery. Therefore, vo = 5 V.
The circuit will become:

55. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Example 2.14 – Determine vo for the network below for the input indicated
(Assume the diode is ideal)
For the period t2 to t3, the input voltage and the capacitor voltage have the same polarity, forcing
the diode to be “off”. The network will appear as
Applying Kirchhoff’s voltage law around the input loop results in
0v25V-10V-
o 35VV
o
The input voltage source will therefore discharge the capacitor at the time constant determined
by τ = RC. In this case, τ = RC = 0.01s = 10 ms, much larger than 0.5 ms．Therefore, it can
assume that the capacitor hold its voltage during the discharge period.

56. Ch2 - Semiconductor DiodesCh2 - Semiconductor Diodes
Example 2.14 – Determine vo for the network below for the input indicated
(Assume the diode is ideal)