Applications Of Semiconductor Devices Amplifiers

Applications of Semiconductor devices – Amplifiers.Electrical and Electronic Principles

© University of Wales Newport 2009 This work is licensed under a Creative Commons Attribution 2.0 License.

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 2002Hambly A , Electronics 2nd Edition, Pearson Education 2000Storey 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


-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

time

input

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

Vs

RcRb

Ic

Ib

Vout

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 ARc = VRc/Irc = (9 - 4.5)/2 mA = 2.25kRb = VRb/Irb = (9 – 0.6)/10 A = 840k


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.


Collector Base BiasThis 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

RcRb

IcIb

Vout

Vbe

NoteThe current flowing through Rc is the sum of Ic and IbAll of the other notes from the previous method apply to this circuit.

Use the same conditions as for the previous example.

Vout = 4.5V (half supply)hfe = 200

So Ib = Ic/hfe = 2 mA/200 = 10 ARc = VRc/Irc = (9 - 4.5)/(2 mA + 10 A) =

2.24kRb = 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 BiasThis 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

Vs

RcRb1

IcIb

Vout

Vbe

Rb2 Re

Ib1

Ib2 Ie

Vb

Notes.The value of the required base voltage must be given.A rule of thumb used to determine Ib1 is that it should be 10 x Ib. This will give us the value of 9 x Ib for Ib2

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 mAVe = Vb – 0.6 = 2.5 – 0.6 = 1.9VVout = Ve + (Vs – Ve)/2 = 1.9 + (9 – 1.9)/2 =

5.45VRc = VRc/Irc = (9 – 5.45)/2 mA = 1.775kRe = VRe/Ire = 1.9/2 mA = 950Rb1 = VRb1/Irb1 = (Vs – 2.5)/(10 x 10 A) =

65kRb2 = VRb2/Irb2 = 2.5/(9 x 10 A) = 27.8k


Once again let us examine what happen when we have a change in hfe due to temperature.

Due to temperaturehfe 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.


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

RcRb1Vout

Rb2 Re Ce

Vin

Ground

The complete amplifier would therefore have the following form.Typically a stage of amplification of the form above would generate a voltage gain of 50 to 150 depending upon the transistor and the load applied to the output.

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:

D

S

G

0V

-ive

+ive

Note we are considering a JFET here but this is also true for a MOSFET.We could set-up the voltages using two individual supplies but in practice we use a single supply and arrange the following voltages:

D

S

G

+ive

0V

++ive

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 V

sRd Vout

Rg Rs

Vin

Notes.Rg has a very large value ~ 1M.-Vgs is determined by the 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.

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

RdVout

Rg Rs

Vin

2.5mA

Gm = 3.19 mA/V

VmA

Vgs 78.019.3

5.2

VVout 39.62

78.01278.0

3125.2

78.0

mA

V

IRs

VRsRs

kmAIRd

VRdRd 24.2

5.2

)39.612(

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.


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© 2009 University of Wales Newport

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