Figure 2.24: Diode I–V characteristics showing breakdown effects
•
Zener diodes are designed and fabricated to provide a specified breakdown
voltage VZo .
•
A Zener diode can be used as a constant-voltage reference in a circuit.
•
The diode breakdown voltage is essentially constant over a wide range of
currents and temperatures.
•
The slope of the I–V characteristics curve in the breakdown is quite large, so
the incremental resistance rz is small. Typically, rz is in the range of a few ohms
or tens of ohms.
•
The circuit symbol of a Zener diode is shown in Figure 2.25.
Figure 2.25: Circuit symbol for Zener diode
41
Example
The shows a constant-voltage reference circuit.
Assume that the Zener diode
breakdown voltage is VZ  5.6V and the Zener resistance is rZ  0 . The current in the
diode is to be limited to 3 mA. Design the value of resistance required to limit the
current in this circuit.
Figure 2.26: Simple circuit containing a Zener diode in which the Zener diode is
biased in the breakdown region
VPS VZ
R
V VZ 10V  5.6V
R  PS

 1.47 k
3mA
I
I
The power dissipated in the Zener diode is:
PZ  I ZVZ  3mA5.6V  16.8mW
The Zener diode must be able to dissipate 16.8 mW of power without being damaged.
42
3
Diode Application
Diodes are used in the design of many electronic circuits. The circuits include:
•
Rectifier circuits
•
Voltage regulation
•
clipper and clamper circuits
•
Multi-diode circuits
•
Photodiode and LED circuits
•
Power Supply circuits
3.1 Power supply circuits
DC power supplies are used to convert alternating current (ac) voltage to dc voltage
for powering electronics circuits such as PCs, televisions, and stereo systems.
A power supply circuit consists of various stages as shown in Figure 3.1.
Figure 3.1: Diagram of an electronic power supply.
Let us look at each part of the power supply circuit.
3.1.1 Diode Rectifier
•
A diode rectifier forms the first stage of a dc power supply.
•
An electrical cord that is plugged into a wall socket and attached to a
television, for example, is connected to a rectifier circuit inside the TV. In
addition, battery chargers for portable electronic devices such as cell phones
and laptop computers contain rectifier circuits.
43
•
Rectification is the process of converting an alternating (ac) voltage into one
that is limited to one polarity.
•
A diode is suited for this application because it conducts in one direction.
•
There are two types of rectifiers: half wave rectifier and full wave rectifier.
3.1.1.1 Half wave rectifier
A half wave rectifier circuit is shown in Figure 3.2. It consists a step-down
transformer, a single diode, and a load connected to the secondary side of the
transformer. The ac power input fed to the primary of the transformer.
Figure 3.2: Half-wave rectifier Circuit
The primary voltage and secondary voltage are related using the transformation ratio
as in Equation (3.1).
v1
N
 1
vs
N2
(3.1)
To fully analyse the diode circuit, the piecewise linear model of the diode is used.
Steps:
1. Determine the input voltage condition such that a diode is conducting (on).
Then find the output signal for this condition.
2. Determine the input voltage condition such that a diode is not conducting (off).
Then find the output signal for this condition.
44
Consider the diode transfer characteristics shown in Figure 3.3.
Figure 3.3: voltage transfer characteristics
•
For vs  0 , the diode is reverse biased, which means that the current is zero and
the output voltage is zero.
•
As long as vs  V the diode will be non-conducting, so the output voltage will
remain zero.
•
When vs  V , the diode becomes forward biased and a current is induced in the
circuit. In this case:
iD 
vs V
R
(3.2)
and
vo  iD R  vs V
(3.3)
•
For vs  V , the slope of the transfer curve is 1.
•
If vs is a sinusoidal signal, the output voltage can be found using the voltage
transfer curve.
•
Consider the secondary voltage signal vs in Figure 3.4. The signal alternates
polarity and has a time average value of zero.
•
The output signal, vo shown in Figure 3.5 is unidirectional and has an average
value that is not zero. The input signal is therefore rectified.
45
•
Since the output voltage appears only during the positive cycle of the output
signal, the circuit is called a half-wave rectifier.
Figure 3.4: sinusoidal input voltage
Figure 3.5: Rectified output voltage
•
When the diode is cut off and non-conducting, no voltage drop occurs across
the resistor R; therefore, the entire input signal voltage appears across the
diode as shown in Figure 3.6.
Figure 3.6: Diode voltage
•
Consequently, the diode must be capable of handling the peak current in the
forward direction and sustaining the largest peak inverse voltage (PIV) without
breakdown.
•
For a half-wave rectifier the PIV is equal to the amplitude of vs .
•
We can use a half-wave rectifier circuit to charge a battery.
46
•
Charging current exists whenever the instantaneous ac source voltage is
greater than the battery voltage plus the diode cut-in voltage.
•
The resistance R in the circuit is to limit the current.
•
When the ac source voltage is less than VB .
•
Thus current flows only in the direction to charge the battery.
•
One disadvantage of the half-wave rectifier is that we “waste” the negative
half-cycles. The current is zero during the negative half-cycles, so there is no
energy dissipated, but at the same time, we are not making use of any possible
available energy.
TASK {10 Marks}
Consider the battery charger shown in Figure 3.7. The transformer of the
half-wave rectifier has been replaced by the voltage vs (representing the
secondary voltage of the transformer). Assume VB  12V , R  100 , and
V  0.6V . Also, assume Vs (t )  24sin  t(V )
Figure 3.7: Half-wave rectifier as a battery charger
Conducting mode:
When the diode is conducting, the diode current is given by:
iD (t ) 
vs (t ) VB V 
R
The voltage across the diode is:
47
VD (t )  V 
Non-Conducting Mode
When the diode is not conducting:
The magnitude of the reverse voltage is given as:
vR (t )  vs (t )  VB
and the diode current:
iD (t )  0
The conduction cycle can be described as the period from when the diode
starts conducting to the time when the diode stops conducting over one
cycle of the input signal.
Requirement:
a) Determine the peak diode current
b) Maximum reverse-bias diode voltage
c) The fraction of the cycle over which the diode is conducting.
Copy the script below into MATLAB to plot the curves that define the diode
voltage, diode current, and input voltage. Replace the three dot (…) with the
correct parameters before running the code. Follow through the code to
understand the design process, and use of the fomulas developed in this
section. Identify the conduction cycle (the angles where the diode current
cross the line y=0), and the peak values of diode current and voltage.
Compare the values with those obtained from your calculation above. Submit
the calculations together with the graphs obtained.
V_B=...; %Battery voltage in volts
R=...; %limiting current resistor in ohms
V_gamma=...; %Diode cut-in voltage in Volts
w=2*pi*50; %angular frequency in rad/s
t=linspace(0,0.02,360); %time interval for the plot.
theta=rad2deg(t*w); %the angle theta
48
v_s=24.*sind(theta); %the input signal;
for n=1:360
if (v_s(n)>=V_gamma+V_B) %forward bias condition
i_diode(n)=(v_s(n)-(V_gamma+V_B)).*1000/R;%diode current in mA
%when conduction
v_diode(n)=V_gamma; %diode voltage when conducting is simply
%the cut-in voltage
else %reverse bias condition
v_diode(n)=v_s(n)-V_B; % is the sum of the battery voltage and
%source voltage-the polarity
%of the source changes making the two
%voltages additive in the negative
%direction
i_diode(n)=0;
end
end
figure %plotting the graphs
subplot(2,1,1)
plot(theta,v_s,'b',theta,v_diode,'g');
grid on
% set(gca,'XAxisLocation','origin','YAxisLocation','origin')%uncomment
%if your Matlab version is newer than 2015
title('Source Voltage and Diode Voltage');
xlabel('Conduction angle in Degrees');
ylabel('Voltage in Volts)');
legend('Input signal V_s in Volts','diode voltage v_d in Volts');
subplot(2,1,2)
plot(theta,i_diode,'r');
title('Diode Current');
xlabel('Conduction angle in Degrees');
ylabel('Diode Current in mA)');
legend('diode current i_d in mA');
grid on
%set(gca,'XAxisLocation','origin','YAxisLocation','origin')%uncomment
%this line if your matlab version is newer than 2015
3.1.1.2 Full-Wave Rectification
•
The full-wave rectifier inverts the negative portions of the sine wave so that a
unipolar output signal is generated during both halves of the input sinusoid.
•
There are two types of full wave rectifiers: one uses a center-tapped
transformer with a pair of diodes, the other uses four diodes arranged in a
bridge configuration.
•
A full wave rectifier with a center-tap transformer is shown in
49
Figure 3.8: Full-wave rectifier circuit with center-tapped transformer
•
The input to the rectifier consists of a power transformer where an ac input is
fed.
•
Two outputs appear on the secondary side of the power transformer due to the
center-tap, with equal values vs .
•
Notice that the center-tap is grounded (is at zero potential).
•
The primary winding connected to the ac source has N1 windings, and each half
of the secondary winding has N 2 windings.
•
The value of vs is given by Equation (3.4):
vs  v1 
N2
N1
v1
N
 1  turns ratio
vs
N2
•
(3.4)
(3.5)
The input power transformer also provides electrical isolation between the
power line circuit and the electronic circuits to be biased by the rectifier circuit.
This isolation reduces the risk of electrical shock.
•
During the positive half of the input voltage cycle, both output voltages vs are
positive; therefore, diode D1 is forward biased and conducting and D2 is reverse
biased and cut off. The current through D1 and the output resistance produce
a positive output voltage.
50
•
During the negative half cycle, D1 is cut off and D2 is forward biased, or “on,”
and the current through the output resistance again produces a positive output
voltage. If we assume that the forward diode resistance rf of each diode is
small and negligible, we obtain the voltage transfer characteristics, vo versus
vs , shown in Figure 3.9.
Figure 3.9: voltage transfer characteristics
•
For a sinusoidal input voltage, we can determine the output voltage versus
time by using the voltage transfer curve shown in Figure 3.9.
•
When vs  V  , D1 is on and the output voltage is vo  vs V . When vs is
negative, then for vs  v or vs  V , D2 is on and the output voltage is
vo  vs V .
•
The corresponding input and output voltage signals are shown in Figure 3.10.
Figure 3.10: input and output waveforms
•
Since a rectified output voltage occurs during both the positive and negative
cycles of the input signal, this circuit is called a full-wave rectifier.
51
•
Another example of a full-wave rectifier circuit is the bridge rectifier shown in
Figure 3.11.
Figure 3.11: A full-wave bridge rectifier
•
The circuit provides electrical isolation between the input ac power line and the
rectifier output, but does not require a center-tapped secondary winding.
•
It uses four diodes, connected as a bridge.
•
During the positive half of the input voltage cycle, vs is positive, D1 and D2 are
forward biased, D3 and D4 are reverse biased, and the direction of the current
is as shown in Figure 3.11.
•
During the negative half-cycle of the input voltage, vs is negative, and D3 and
D4 are forward biased. The direction of the current, shown in Figure 3.12,
produces the same output voltage polarity as before.
Figure 3.12: current direction for a negative input cycle
•
The sinusoidal voltage vs and the rectified output voltage vo are shown in
Figure 3.13. Because two diodes are in series in the conduction path, the
magnitude of vo is two diode drops less than the magnitude of vs .
52
Figure 3.13: input and output voltage waveforms
•
One difference to be noted in the bridge rectifier circuit in Figure 3.11 and the
rectifier in Figure 3.8 is the ground connection. Whereas the center tap of the
secondary winding of the circuit in Figure 3.8 is at ground potential, the
secondary winding of the bridge circuit is not directly grounded. One side of
the load R is grounded, but the secondary of the transformer is not.
Comparison voltages and the transformer turns ratio in two full-wave rectifier
circuits.
Consider the rectifier circuits shown in Figure 3.8 and Figure 3.11. Assume the
input voltage is from a 120 V (rms), 60 Hz ac source. The desired peak output
voltage vo is 9 V, and the diode cut-in voltage is assumed to be v  0.7V .
solution
For the center-tapped transformer circuit shown in Figure 3.8, a peak voltage of
vo (max)  9V means that the peak value of vs is:
vs  vo (max)  V  9  0.7  9.7V
For a sinusoidal signal, this produces an rms value of:
vs ,rms 
vs
9.7

 6.86V
2
2
The turns ratio of the primary to each secondary winding must then be:
N1
120

 17.5
N 2 6.86
53
For the bridge circuit shown in Figure 3.11, a peak voltage of vo (max)  9V means
that the peak value of vs is:
vs  vo (max)  2V  9  2(0.7)  10.4V
For a sinusoidal signal, this produces an rms value of:
vs ,rms 
vs
2

10.4
 7.35V
2
The turns-ratio should then be:
N1
120

 16.3
N 2 7.35
For the center-tapped rectifier, the peak inverse voltage (PIV) of a diode is:
PIV  vR (max)  2vs (max) V  2(9.7)  0.7  18.7V
For the bridge rectifier, the peak inverse voltage of a diode is:
PIV  vR (max)  vs (max) V  10.4  0.7  9.7V
Observation
•
Only half as many turns are required for the secondary winding in the bridge
rectifier. This is true because only half of the secondary winding of the centertapped transformer is utilized at any one time.
•
For the bridge circuit, the peak inverse voltage that any diode must sustain
without breakdown is only half that of the center-tapped transformer circuit.
Note: There are times when a negative dc voltage is also required. We can produce
negative rectification by reversing the direction of the diodes in either full-wave
rectifier circuit as shown in Figure 3.14.
The corresponding input and output waveforms are shown in Figure 3.15
54
Figure 3.14: Full-wave bridge rectifier circuit to produce negative output voltages.
Figure 3.15: Input and output waveforms.
3.1.2 Filter
The output voltage from the diode rectifier, though unipolar, contains ripples. These
ripples can be reduced by incorporating a filter circuit.
A filter is a capacitor connected in parallel to the load as shown in Figure 3.16.
Figure 3.16: Simple filter circuit
The filtered output of a half-wave rectifier is shown in Figure 3.17.
The filtered output of a half-wave rectifier is shown in Figure 3.18.
55
Figure 3.17: steady-state input and output voltages
Figure 3.18: Output voltage of a full-wave rectifier with an RC filter showing the
ripple voltage
•
The capacitor charges to its peak voltage value when the input signal is at its
peak value.
•
As the input decreases, the diode becomes reverse biased and the capacitor
discharges through the output resistance R.
•
Determining the ripple voltage is necessary for the design of a circuit with an
acceptable amount of ripple.
•
For a half-wave rectifier, the ripple voltage is given as:
vr 
•
(3.6)
For a full wave rectifier, the ripple voltage is given as:
vr 
•
VM
2 fRC
VM
fRC
(3.7)
From the formula for ripple voltage, the larger the capacitor, the smaller the
ripple voltage. The ripple voltage is also related to the load resistance.
56
Rectifier Design
Consider the equivalent circuit of a full wave rectifier during capacitor charging
time shown in the Figure 3.19. The corresponding output voltage and diode
current are also shown in Figure 3.20.
Figure 3.19: Equivalent circuit of a full-wave rectifier during capacitor charging
cycle
Figure 3.20: Output of a full-wave rectifier with an RC filter: diode conduction time
and diode current
As you can see in Figure 3.20 the diode in a filtered full-wave rectifier only conducts
for a short time t .
Let us assume that the diodes are ideal and V  0 .
iD  iC  iR  C
dvo vo

dt
R
(3.8)
During the diode conduction time near t = 0, we can write:
v0  VM cos  t
57
(3.9)
For small ripple voltages, the diode conduction time is small, so we can approximate
the output voltage as (using Taylor Series expansion of coswt):
 1

v0  VM cos  t  VM 1 ( t) 2 
 2

(3.10)
The charging current through the capacitor is:
iC  C
•
dv0
 CVM
dt
 1
  (2)( t)()   CVM t
 2 
(3.11)
The diode conduction occurs during the time t  t  0 , so that the capacitor
current is positive and is a linear function of time. At t=0, the capacitor current
is iC  0.
•
At t  t , the capacitor charging current is at a peak value and is given by:
iC  iC , peak  CVM [( t)]   CVM t
•
We can write that the voltage VL is given by (from equation (3.10)):


1
VL  VM cos[( t)]  VM 1 ( t) 2 


2
•
(3.12)
(3.13)
Solving for  t , we find:
t 
2Vr
VM
(3.14)
where Vr  VM VL
•
From equation (3.6), we can write:
fC 
VM
2 RVr
(3.15)
or
2fC  C 
•
VM
RVr
Substituting Equations (3.15) and (3.16) into Equation (3.12), we have
58
(3.16)
 V   2Vr
iC , peak   M VM 
 RVr   VM



(3.17)
or
iC , peak  
•
VM
R
2VM
Vr
(3.18)
Since the charging current through the capacitor is triangular, we have that the
average capacitor current during the diode charging time is:
iC , peak 
•
 VM
2 R
2VM
Vr
(3.19)
During the capacitor charging time, there is still a current through the load.
This current is also being supplied through the diode. Neglecting the ripple
voltage, the load current is approximately:
iL 
•
VM
R
(3.20)
Therefore, the peak diode current during the diode conduction time for a fullwave rectifier is approximately:
iD , peak 
VM
R

1   2VM

Vr




(3.21)
and the average diode current during the diode conduction time is:
iD ,avg 
VM
R

1  

2

2VM
Vr



(3.22)
The average diode current over the entire input signal period is:
iD ,avg 
VM
R

1  

2

2VM
Vr
 t

 T
For the full-wave rectifier, we have 1/2T = f , so from equation (3.14)
59
(3.23)
t 
2Vr
1

VM
2f
1

2Vr
VM
(3.24)
Then:
2Vr
1
2f 

VM
t
1

T
2f
2Vr
VM
(3.25)
Then the average current through the diode during the entire cycle for a fullwave rectifier is:
iD ,avg 
1

2Vr VM 

1 
VM R 
2
2VM
Vr



(3.26)
Design Example:
A full-wave rectifier is to be designed to produce a peak output voltage of 12V,
deliver 120 mA to the load, and produce an output with a ripple of not more
than 5 percent. An input line voltage of 240V (rms), 50 Hz is available.
The effective load resistance is:
R
VO
12

 100
0.12
IL
Assuming a diode cut-in voltage of 0.7 V, the peak value of vs is:
vs max   vO max   2V  12(07)  13.4 V
For a sinusoidal signal, this produces an rms voltage value of:
vs ,rms 
13.4
 9.48V
2
The transformer turns ratio is then:
N1
240

 25.3
N2
9.48
For a 5 percent ripple, the ripple voltage is:
Vr  (0.05) VM  (0.05)(12)  0.6 V
60
The required filter capacitor is found to be:
C
VM
12

 2000F
2 fRVr
250100 0.6
The peak diode current:
iD , peak 
VM
R

1   2VM

Vr

 12 

1   2(12)   2.50 A
 

 100 
0.6 
and the average diode current over the entire signal period
iD ,avg 
1

2Vr VM 

1 
2
VM R 
2VM
Vr
 1
 
 
2(0.6) (12) 

1 
12 100 
2
2(12) 

0.6 
 132mA
Finally, the peak inverse voltage that each diode must sustain is:
PIV  vR (max)  vs (max) V  13.4  0.7  12.7V
3.1.3 Voltage regulator
•
Zener Diode circuits are used in power supplies to provide voltage regulation.
•
A Zener Voltage regulation circuit is shown in Figure 3.21.
Figure 3.21: A Zener diode voltage regulator circuit
•
For this circuit, the output voltage should remain constant, even when the
output load resistance varies over a fairly wide range, and when the input
voltage varies over a specific range.
•
The variation in VPS may be the ripple voltage from a rectifier circuit.
61
•
We determine, initially, the proper input resistance Ri . The resistance Ri limits
the current through the Zener diode and drops the “excess” voltage between
VPS and VZ .
•
We can write:
Ri 
•
VPS VZ
V VZ
 PS
I1
IZ  IL
(3.27)
Solving this equation for the diode current, I Z , we get:
IZ 
VPS VZ
 IL
Ri
(3.28)
where I L  VZ / RL , and the variables are the input voltage source VPS and the
load current I L .
•
For proper operation of this circuit, the diode must remain in the breakdown
region and the power dissipation in the diode must not exceed its rated value.
In other words:
1. The current in the diode is a minimum, I Z (min), when the load current is a
maximum, I L (max), and the source voltage is a minimum, VPS (min).
2. The current in the diode is a maximum, I Z (min), when the load current is a
minimum, I L (min), and the source voltage is a maximum, VPS (max).
•
Inserting these two specifications into Equation (3.27), we obtain
Ri 
VPS (min) VZ
I Z (min)  I L (max)
(3.29)
Ri 
VPS (max) VZ
I Z (max)  I L (min)
(3.30)
and
•
Equating these two expressions, we then obtain
62
Ri  VPS (min) VZ  I Z (max)  I L (min) 
 VPS (max) VZ  I Z (min)  I L (max) 
•
(3.31)
Reasonably, we can assume that we know the range of input voltage, the range
of output load current, and the Zener voltage. Equation (3.31) then contains
two unknowns, I Z (min) and I Z (max) .
•
Further, as a minimum requirement, we can set the minimum Zener current to
be one-tenth the maximum Zener current, or I Z (min)  0.1I Z (max) . (More
stringent design requirements may require the minimum Zener current to be
20 to 30 percent of the maximum value.)
•
We can then solve for I Z (max) , using Equation (3.32), as follows:
I Z (max) 
•
I L (max) VPS (max) VZ  I L (min) VPS (min) VZ 
VPS (min)  0.9VZ  0.1VPS (max)
(3.32)
Using the maximum current thus obtained from Equation (3.32), we can determine
the maximum required power rating of the Zener diode.
Example
The voltage regulator is to power a car radio at VL = 9 V from an automobile battery
whose voltage may vary between 11 and 13.6 V. The current in the radio will vary
between 0 (off ) to 100 mA (full volume).
Solution:
We use equation (3.32):
PZ (max)  I Z (max)  VZ  3009  2.7W
The maximum power dissipated in the Zener diode is then:
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I Z (max) 
100 13.6  9 0
 300mA
11 0.9 9  0.1(13.6)
The value of the current-limiting resistor Ri ,
Ri 
13.6  9
 15.3
0.3  0
The maximum power dissipated in this resistor is:
2
PRi 
VPS (max) VZ 
Ri
2

13.6  9
15.3
 1.4W
We find:
I z min  
11 9
 0.10  30.7 mA
15.3
Task
Design a DC power supply with the following specifications:
•
Load Current is to vary between 25 and 50mA.
•
Output voltage is to remain in the range 12  vO  12.2V
The circuit configuration to be designed is shown in Figure 3.22. A diode bridge circuit
with an RC filter will be used and a Zener diode will be in parallel with the output load.
Figure 3.22: DC power supply circuit for design application
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An ac input voltage with an rms value in the range 110  VL  120V and at 60 Hz is
available. A Zener diode with a Zener voltage of VZO  12V and a Zener resistance of
2 that can operate over a current range of 10  I Z  100mA is available.
Also, a transformer with an 8:1 turns ratio is available.
Solution:
With an 8:1 transformer turns ratio, the peak value of vS is in the range
19.4  vs  21.2V . Assuming diode turn-on voltages of V  0.7V , the peak value of VO1
is in the range 18  vO1  19.8V .
For vO1 (max) and minimum load current, let I Z  90mA . Then:
vO  VZO  I Z rZ  12  0.090(2)  12.18V
The input current is:
I i  I Z  I L  90  25  115mA
The input resistance Ri must then be:
Ri 
vo1  vo 19.8 12.18

 66.3
115mA
Ii
The minimum Zener current occurs for I L (max) and vO1 (min). The voltage vO1 (min)
occurs for vS (min) and must also take into account the ripple voltage. Let
I Z (min)  20 mA . Then the output voltage is:
vO  VZO  I Z rZ  12  0.020(2)  12.04V
The output voltage is within the specified range of output voltage.
We now find:
I i  I Z  I L  20  50  70mA
and
65
vO1 (min)  vo  I i Ri  12.04  (0.070)(66.3)  16.68 V
The minimum ripple voltage of the filter is then
vr (min)  vs (min) 1.4  vo1 (min)  19.4 1.4 16.68  1.32V
Let R  500 . The effective resistance to ground from vo1 is Ri ,eff where Ri ,eff is the
effective resistance to ground through Ri and the other circuit elements. We can
approximate:
Ri ,eff 
vs (avg ) 1.4 20.3 1.4

 164
I i (max)
0.115
Then Ri || Ri ,eff  500 ||164  123.5 . The required filter capacitance is found from
C
VM
19.8

 1012F
2 fRtVr 2 60123.51.32
Requirement: Simulate this design in MATLAB Simulink [10 marks]
4
The Bipolar Junction Transistor
•
The bipolar junction transistor (BJT) has three separately doped regions and
contains two pn junctions.
•
The basic transistor principle is that the voltage between two terminals controls
the current through the third terminal.
•
Current in the transistor is due to the flow of both electrons and holes, hence
the name bipolar.
4.1 Transistor Structures
The npn bipolar transistor contains a thin p-region between two n-regions (Figure
4.1).
66