Transcript Applications of Semiconductor devices.

Applications of
Semiconductor
devices – Amplifiers.
Electrical and Electronic
Principles
The following presentation is a part of the level 4 module -- Electrical and Electronic Principles. This resources is a part
of the 2009/2010 Engineering (foundation degree, BEng and HN) courses from University of Wales Newport (course
codes H101, H691, H620, HH37 and 001H). This resource is a part of the core modules for the full time 1 st year
undergraduate programme.
The BEng & Foundation Degrees and HNC/D in Engineering are designed to meet the needs of employers by placing
the emphasis on the theoretical, practical and vocational aspects of engineering within the workplace and beyond.
Engineering is becoming more high profile, and therefore more in demand as a skill set, in today’s high-tech world. This
course has been designed to provide you with knowledge, skills and practical experience encountered in everyday
engineering environments.
Contents
 Bipolar Junction Transistors (BJT).
 Supply Base Bias
 Collector Base Bias
 Four Resistor Bias
 Field Effect Transistors.
 Credits
In addition to the resource below, there are supporting documents which should be used in combination with this
resource. Please see:
Green D C, Higher Electrical Principles, Longman 1998
Hughes E , Electrical & Electronic, Pearson Education 2002
Hambly A , Electronics 2nd Edition, Pearson Education 2000
Storey N, A Systems Approach, Addison-Wesley, 1998
Applications of Semiconductor
Devices- Amplifiers
Bipolar Junction Transistors (BJT).
In order to use a transistor as an amplifying device we
must ensure that the DC voltages around the transistor
are at an acceptable level. This process is called biasing.
Typical input signals will be AC i.e. they will go both
positive and negative. The transistor will only work with
one polarity npn – positive and pnp – negative inputs. We
therefore need to ensure that as the input swings positive
and negative that the input to the transistor remains
positive (npn). This is done by raising the starting point on
the base to a mid point on the characteristic. A positive
input will simple increase this value and a negative value
reduce it. See below
Applications of Semiconductor
Devices- Amplifiers
0.7
0.6
0.5
input
0.4
0.3
0.2
0.1
0
-0.1
time
The starting point will be 0.6V which is the point
identified from the characteristic.
The other thing we will need to do is to convert the
output current into an output voltage. This is done by
passing the current through a resistor.
There are three methods used to bias a transistor.
Supply Base Bias.
This uses a resistor from the supply to the base to
provide the bias for the base and use a resistor
connected to the collector to supply our output. See
below
Ic
Rb
Vs
Rc
Vout
Ib
Vbe
Notes.
 With a voltage on the base the transistor will be
conducting and therefore a current will be flowing
through the transistor. This is called the quiescent
current and will be given for a particular application.
 The supply Vs will be under your control or will be
given.
 The current gain of the transistor Ic/Ib (hfe) is one
of the characteristics of the transistor and will be
given.
 The value of Vout will normally be set at half the
supply. This will allow the output to swing positively
and negatively by the same amount.
 Vbe is taken to be 0.6 for normal operation.
Applications of Semiconductor
Devices- Amplifiers
Example
A 2N2222 is used to produce an amplifier. The bias
method adopted is supply base. If the supply is 9v and
the quiescent current must equal 2 mA determine the
value for the two resistors Rc and Rb.
Vout = 4.5V
(half supply)
hfe = 216 use 200 (biasing is not an exact science) (see
previous notes)
So
Ib = Ic/hfe = 2 mA/200 = 10 A
Rc = VRc/Irc = (9 - 4.5)/2 mA = 2.25k
Rb = VRb/Irb = (9 – 0.6)/10 A = 840k
Applications of Semiconductor
Devices- Amplifiers
This method is simple but can be affected by drift in the
transistor due to temperature or aging.
Due to temperature
hfe  Ic  VRc  Vout 
This drift can cause unacceptable changes in the bias
conditions.
Note the base is “sitting” at 0.6V and the output is
“sitting” at 4.5V. These voltages will affect anything
connected to the input or output and therefore capacitors
are used to block the DC.
Note if electrolytic capacitors are used the positive side
is connected to the transistor side. As below:
Vs
If non-electrolytic
capacitors are used
then it doesn’t matter
which way round the
capacitors are.
Applications of Semiconductor
Devices- Amplifiers
Collector Base Bias
This uses a resistor from the collector to the base to
provide the bias for the base and again uses a resistor
connected to the collector to supply our output. See
below
Vs
Rc
Rb
Ib
Vbe
Ic
Vout
Note
The current flowing
through Rc is the sum
of Ic and Ib
All of the other notes
from the previous
method apply to this
circuit.
Use the same conditions as for the previous example.
Vout = 4.5V
hfe = 200
So
(half supply)
Ib = Ic/hfe = 2 mA/200 = 10 A
Rc = VRc/Irc = (9 - 4.5)/(2 mA + 10 A) = 2.24k
Rb = VRb/Irb = (4.5 – 0.6)/10 A = 390k
Let us consider the same potential problem with thermal
drift.
Due to temperature
hfe  Ic  VRc  Vout  VRb  Ib  Ic 
The feedback from the output to the input reduces the
drift effect as, if Ic increases the feedback tends to
reduce it again.
This method is preferred over the supply base method.
Four Resistor Bias
This uses four resistors – two to set the base voltage,
one to lift the emitter voltage so that it sits 0.6 below
the base and one on the collector to convert the output
current into an output voltage. See below
Notes.
Vs
The value of the required
base voltage must be
Rb1
Rc
given.
Vout
Ib1
Ic
A rule of thumb used to
Vb
Ib
determine Ib1 is that it
should be 10 x Ib. This will
Ib2
Vbe
Ie
give us the value of 9 x Ib
for Ib2
Rb2
Re
Vout will be set at the mid
point between the emitter
voltage and the supply.
Carry out the same example but with the additional
information that Vb should equal 2.5V.
hfe = 200
So
Ib = Ic/hfe = 2 mA/200 = 10 A
Ie = Ic + Ib = 2.01 mA use 2 mA
Ve = Vb – 0.6 = 2.5 – 0.6 = 1.9V
Vout = Ve + (Vs – Ve)/2 = 1.9 + (9 – 1.9)/2 = 5.45V
Rc = VRc/Irc = (9 – 5.45)/2 mA = 1.775k
Re = VRe/Ire = 1.9/2 mA = 950
Rb1 = VRb1/Irb1 = (Vs – 2.5)/(10 x 10 A) = 65k
Rb2 = VRb2/Irb2 = 2.5/(9 x 10 A) = 27.8k
Applications of Semiconductor
Devices- Amplifiers
Once again let us examine what happen when we have a
change in hfe due to temperature.
Due to temperature
hfe  Ic  Ie  VRe  Vbe  Ib  Ic 
The feedback from the output to the input again reduces
the drift effect as, if Ic increases the feedback tends to
reduce it again.
Applications of Semiconductor
Devices- Amplifiers
Problems with this form of biasing when using the
transistor as an amplifier.
This bias method is the best as it reduces dramatically
the effect of drift. This though causes a problem when
amplifying an input. Consider an input applied to the
amplifier. Consider an increase in input:
Vin  Vbe  Ib  Ic  Ie  VRe  Vbe 
When we apply the input to the base the emitter also rises
which reduces the effect of the input (actual input to the
transistor is Vbe)
e.g. If the input rises by 100 mV then it is quite possible
for the emitter voltage to rise by 90 mV which means that
the transistor only sees 10 mV.
This problem can be removed by placing a capacitor across
the emitter resistor. This does not effect the bias as the
capacitor for DC acts like an open circuit but for AC it
acts like a short circuit. This means that the AC signal Ie
does not alter the emitter voltage but is simply bypassed
down to ground.
Vs
The complete amplifier
would therefore have
Rb1
Rc
the following form.
Vout
Typically a stage of
amplification of the form Vin
above would generate a
voltage gain of 50 to 150
depending upon the
Rb2
Re
Ce
transistor and the load
applied to the output.
Ground
Field Effect Transistors.
Unlike the BJT transistors considered so far, the FET
requires both positive and negative voltages applied to its
connections for it to operate correctly. See below:
Note we are
+ive D
++ive D
considering a JFET
here but this is also
true for a MOSFET.
We could set-up the
G
G
voltages using two
-ive
0V
individual supplies
but in practice we
0V S
+ive S
use a single supply
and arrange the
following voltages:
With this arrangement
the drain is positive with respect to the source
and the gate is negative with respect to the source
This is arranged by using the quiescent current to lift the
voltage on the source whilst holding the gate at 0V
see below
Notes.
Rg has a very large value ~ 1M.
-Vgs is determined by the
Vout voltage across Rs which equals
Rs times the quiescent current.
Vout will be set at the mid point
between Vsource and Vs.
The value of gm will given and
will give you the relationship
between Vgs and Id.
Vs
Rd
Vin
Rg
Rs
Example.
A 2N5484 is used to construct an amplifier. The supply
voltage to be used is 12 v and the quiescent current is to
equal 2.5 mA. Determine the values of the components.
12V
Rd
2.5mA
Vin
Vout
Gm = 3.19 mA/V
2.5mA
Vgs 
 0.78V
3.19
12  0.78
Vout  0.78 
 6.39V
2
VRs 0.78V
Rs 

 312 
IRs 2.5mA
Rg
Rs
Rd 
VRd (12  6.39)

 2.24k
IRd
2.5mA
Rg will have the value of 1M
The input and output capacitors are in place for the same
reasons as before though the input capacitor does not
need to block a voltage as the gate is at 0V.
A capacitor is required across Rs, once again, for the
same reason as given for the transistor.
The MOS devices will operate in the same way as the
FET.
Applications of Semiconductor
Devices- Amplifiers