L10 BJT Applications and Biasing

8/10/2019 L10 BJT Applications and Biasing

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Basic TransistorApplications


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Switch

A bipolar circuit called an inverter, in which the transistor in the

circuit is switched between cutoff and saturation.

The load, for example, could be a

motor, a light-emitting diode or

some other electrical device.

If vI< VBE(on), then iB= iC= 0 and

the transistor is cut off.

Since

iC= 0, the voltage dropacross the load is zero, so the

output voltage is vO= VCC.

Since the currents in the transistor are zero, power dissipation in

the transistor is zero that would turn off the LED if the load is LED.


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If vI

= VCC

and if the ratio of IC

to IB

less than ,

then the transistor is usually driven into

saturation, means that:

In this case, the current ICwill flow that would turn on the load.

Power Dissipated by the transistor, P = IB( VBE) + IC( VCE)


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Digital LogicNOT GATE

In the simple inverter circuit, if the input is approximately zero volts,

the transistor is in cutoff and the output is high and equal to VCC.

If the input is high and equal to VCC, the transistor is driven into

saturation, and the output is low and equal to VCE(sat).


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Digital LogicNOR Gate

If the two inputs are zero,

both transistors Q1and Q2

are in cutoff, and VO= 5 V.

When V1= 5 V and V2= 0,transistor Q1can be driven

into saturation, and Q2

remains in cutoff. With Q1

in saturation, the output

voltage VO= VCE(sat).

If V1= 0 and V2= 5 V, then Q1is in cutoff,

and Q2can be driven in saturation, and

VO= VCE(sat).


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If both inputs are high,

meaning V1= V2= 5 V,

then both transistors can

be driven into saturation,

and VO= VCE(sat).

In a positive logic system,

meaning that the larger

voltage is a logic 1 and the

lower voltage is a logic 0,

the circuit performs the NOR logicfunction.

The circuit is then a two-input bipolar

NOR logic circuit.


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Amplifier: The BJT inverter circuit can also be used to amplify a time-varying

input signal

(a) A bipolar transistor inverter circuit, (b) the voltage transfer

characteristics


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Bipolar TransistorBiasing


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Biasing refers to the DC voltages applied tothe transistor for it to turn on and operate in

the forward active region, so that it can

amplify the input AC signal

Bipolar Transistor Biasing


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Biasing CircuitsFixed Bias Biasing

Circuit

The circuit is one of the simplest transistor circuits is known as

fixed-bias biasing circuit.

There is a single dc power supply, and the quiescent base current

is established through the resistor RB.

The coupling capacitor C1

acts as an open circuit to dc,

isolating the signal source

from the base current.

Typical values of C1are in

the rage of 1 to 10 F,

although the actual value

depends on the frequency

range of interest.


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Determine the following:

(a) IBQand ICQ

(b) VCEQ

(c) VBand VC

(d) VBC

Example


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Adding a resistor to the emitter circuit stabilizes the bias circuit.

Emitter-Stabilized Biasing Circuit


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Applying Kirchoffs voltage law:

Knowing:

Combining these two formulas:

Grouping terms and solving for IB:EB

BECCB

R)1(R

VVI

0EEBEBBCC

RIVRIV

BE II )1(

0)1( EBBEBBCC RIVRIV

B-E Loop


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Applying Kirchoffsvoltage law:

Finding VE:

Finding VC:

or

Finding VB:

or

0CCCCCEEE

VRIVRI

EEE RIV

ECEC VVV

CCCCC RIVV

BBCCB RIVV

EBEB VVV

C-E Loop


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Adding REto the Emitter improves the stability of a

transistor.

Stabi l i ty refers to a bias circuit in which the currentsand voltages will remain fairly constant for a wide

range of temperatures and transistor Betas ().

The temperature surrounding the transistor circuit is

not always constant; the Beta () of a transistor is nota fixed value.

Improved Bias Stability


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This is a very stable bias circuit.

The currents and voltages are almost independent of variations in .

Voltage Divider Biasing Circuit


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Redrawing the input side of the network bychanging it into Thevenin Equivalent

RTh: the voltage source is replacedby a short-circuit equivalent

Analysis


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VTh: open-circuit Thevenin voltageis determined.

Inserting the Thevenin

equivalent circuit

Analysis

VTH

Use voltage divider

VTH


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The Thevenin equivalent circuit

Analysis


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BJT Biasing in Amplifier

Example

Find VCE ,IE,IC and IBgiven=100, VCC=10V, R1= 56 kW, R2= 12.2 kW,

RC= 2 kW andRE= 0.4 kW

VTH= R2/(R1+ R2 )VCC

VTH = 12.2k/(56k+12.2k).(10)

VTH = 1.79V

RTH= R1// R2

= 10 kW


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BJT Biasing in Amplifier

CircuitsV

TH= R

THI

B+ V

BE+ R

EI

E1.79= 10k IB + 0.7+ 0.4k (+1)IB

IB = 21.62mA

IC= I

B= 100(21.62m)=2.16mA

IE= IC+ IB= 2.18mA

VCC= RCIC + VCE + REIE

10 = 2k(2.16m)+VCE +0.4(2.18m)

VCE= 4.8 Vactive region


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Biasing using Collector to Base Feedback

Resistor

Find RBand RCsuch that IE = 1mA , VCE= 2.3 V, VCC= 10 V

and =100.


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Biasing using Collector to Base Feedback

Resistor

VCE= IBRB+ VBE 2.3 = (IE/ (+1))RB+ 0.7

RB= 161.6 kW

VCC= IERC+ VCE 10 = 1m RC+ 2.3

RC= 7.7 kW

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L10 BJT Applications and Biasing