Recall Last Lecture

Biasing of BJT Three types of biasing

Fixed Bias Biasing Circuit Biasing using Collector to Base Feedback Resistor Voltage Divider Biasing Circuit

Applications of BJT As digital logic gates

NOT NOR

CHAPTER 5BASIC BJT AMPLIFIERS

(AC ANALYSIS)

The Bipolar Linear AmplifierThe Bipolar Linear Amplifier

Bipolar transistors have been traditionally used in linear amplifier circuits because of their relatively high gain.

To use the circuit as an amplifier, the transistor needs to be biased with a dc voltage at a quiescent point (Q-point) such that the transistor is biased in the forward-active region.

If a time-varying signal is superimposed on the dc input voltage, the output voltage will change along the transfer curve producing a time-varying output voltage.

If the time-varying output voltage is directly proportional to and larger than the time-varying input voltage, then the circuit is a linear amplifier.

The linear amplifier applies superposition principle Response – sum of responses of the circuit for

each input signals alone So, for linear amplifier,

DC analysis is performed with AC source turns off or set to zero

AC analysis is performed with DC source set to zero

EXAMPLE

iC , iB and iE,

vCE and vBE

Sum of both ac and dc components

Graphical Analysis and ac Equivalent CircuitGraphical Analysis and ac Equivalent Circuit

From the concept of small signal, all the time-varying signals are superimposed on dc values. Then:

and

PERFORMING DC and AC analysis

DC ANALYSIS AC ANALYSIS

Turn off AC SUPPLY = short circuit

Turn off DC SUPPLY = short circuit

DO YOU STILL REMEMBER?

DC equivalent AC equivalent

rd

idIDQ

VDQ = V

Let’s assume that Model 2 is used

DC ANALYSIS

DIODE = MODEL 1 ,2 OR 3

CALCULATE DC CURRENT, ID

CALCULATE rd

AC ANALYSIS

DIODE = RESISTOR, rd

CALCULATE AC CURRENT, id

WHAT ABOUT BJT?

AC equivalent circuit –

Small-Signal Hybrid-Small-Signal Hybrid-ππ Equivalent Equivalent

ib

OR

THE SMALL SIGNAL PARAMETERS

The resistance rπ is called diffusion resistance or B-E input resistance. It is connected between Base and Emitter terminals

The term gm is called a transconductance

ro = VA / ICQrO = small signal transistor output resistance

VA is normally equals to , hence, if that is the case, rO = open circuit

Hence from the equation of the AC parameters, we HAVE to perform DC analysis first in order to calculate them.

EXAMPLE

The transistor parameter are = 125 and VA=200V. A value of gm = 200 mA/V is desired. Determine the collector current, ICQ and then find r and ro

ANSWERS: ICQ = 5.2 mA, r= 0.625 k and ro = 38.5 k

CALCULATION OF GAIN

Voltage Gain, AV = vo / vs

Current Gain, Ai = iout / is

Small-Signal Voltage Gain: Av = Vo / Vs

ib

Common-Emitter Common-Emitter AmplifierAmplifier

Remember that for Common Emitter Amplifier, the output is measured at the collector terminal. the gain is a negative value

Three types of common emitter Emitter grounded With RE

With bypass capacitor CE

STEPSOUTPUT SIDE

1.Get the equivalent resistance at the output side, ROUT

2.Get the vo equation where vo = - gm vbeROUT

INPUT SIDE

3.Calculate Ri

4.Get vbe in terms of vs – eg: using voltage divider.

5.Go back to vo equation and calculate the voltage gain

VCC = 12 V

RC = 6 k93.7 k

6.3 k

0.5 kβ = 100VBE = 0.7VVA = 100 V

Emitter Grounded

Voltage Divider biasing:Change to Thevenin EquivalentRTH = 5.9 kVTH = 0.756 V

Perform DC analysis to obtain the value of IC

BE loop: 5.9IB + 0.7 – 0.756 = 0

IB = 0.00949

IC = βIB = 0.949 mA

Calculate the small-signal parameters

r = 2.74 k , ro = 105.37 k and gm = 36.5 mA/V

VCC = 12 V

RC = 6 k93.7 k

6.3 k

0.5 kβ = 100VBE = 0.7VVA = 100 V

Emitter Grounded

vbe

Follow the steps1. Rout = ro || RC = 5.677 k

2. Equation of vo : vo = - ( ro || RC ) gmvbe= - 36.5 ( 5.677) vbe = -207.21 vbe

3. Calculate Ri RTH||r = 1.87 k

4. vbe in terms of vs use voltage divider: vbe = [ Ri / ( Ri + Rs )] * vs = 0.789 vs

so vs = 1.2674 vbe

so: vs = 1.2674 vbe

5. Go back to equation of vo and calculate the gain

vo / vs = -207.21 vbe / 1. 2674 vbe

vo / vs = - 207.21 / 1.2674

vo / vs = -163.5AV = vo / vs = - 163.5

vbe

TYPE 2: Emitter terminal connected with RE – normally ro = in this type

VCC = 5 V

RC = 5.6 k250 k

75 k

0.5 k

RE = 0.6 k

β = 120VBE = 0.7VVA =

New parameter: input resistance seen from the base, Rib = vb / ib

0.5 k

57.7 k

RC = 6 k

7.46 k

RE = 0.6 k

vb

1. Rout = RC = 6 k

2. Equation of vo : vo = - RC ib= - 720 ib

4. Calculate Ri RTH||Rib = 33.53 k

5. vb in terms of vs use voltage divider: vb = [ Ri / ( Ri + Rs )] * vs = 0.9853 vs

vs = 1.0149 vb

3. Calculate Rib using KVL: ib r + ie RE - vb = 0

but ie = (1+ ) ib = 121 ibso: ib [ 121(0.6) + 7.46 ] = vb Rib = 80.06 k

vb

so: vs = 1.0149 vb

6. Go back to equation of vo

vo = - 720 ib = - 720 [ vb / Rib ] = -720 vb / 80.06 = - 8.993 vb

vo / vs = - 8.993 vb / 1.0149 vb

vo / vs = - 8.86 AV = vo / vs = - 8.86

vb