Derive input load line IB versus VBE Formulas

Derive input load line, IB versus VBE Formulas KVL at BE Loop DC Analysis npn KVL at CE Loop pnp Bipolar Junction Transistor Cutoff Mode of operation CHAPTER 4 Derive output load line, IC versus VCE Active Saturation What happened to each junctions?

Fixed Biasing Circuit Collector to Base Feedback Resistor Voltage Divider Biasing Circuit Bipolar Junction Transistor (Biasing) CHAPTER 4 CONTINUE DC Analysis Voltage transfer characteristic, VO versus VI

CHAPTER 5 BASIC BJT AMPLIFIERS (AC ANALYSIS)

The Bipolar Linear Amplifier Ø Bipolar transistors have been traditionally used in linear amplifier circuits because of their relatively high gain. Ø 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. Ø If a time-varying signal is superimposed on the dc input voltage, the output voltage will change along the transfer curve producing a timevarying output voltage. Ø 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.

l The linear amplifier applies superposition principle l l Response – sum of responses of the circuit for each input signals alone So, for linear amplifier, l l DC analysis is performed with AC source turns off or set to zero AC analysis is performed with DC source set to zero

EXAMPLE l i. C , i. B and i. E, l v. CE and v. BE Sum of both ac and dc components

Graphical Analysis and ac Equivalent Circuit Ø From the concept of small signal, all the time-varying signals are superimposed on dc values. Then:

PERFORMING DC and AC analysis DC ANALYSIS Turn off AC SUPPLY = short circuit AC ANALYSIS Turn off DC SUPPLY = short circuit

DO YOU STILL REMEMBER?

VDQ = V + - IDQ DC equivalent rd id AC equivalent

DC ANALYSIS DIODE = MODEL 1 , 2 OR 3 CALCULATE DC CURRENT, ID AC ANALYSIS CALCULATE rd DIODE = RESISTOR, rd CALCULATE AC CURRENT, id

WHAT ABOUT BJT?

AC equivalent circuit – Small-Signal Hybrid-π Equivalent 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 / ICQ r. O = small signal transistor output resistance VA is normally equals to , hence, if that is the case, r. O = open circuit Hence from the equation of the AC parameters, we HAVE to perform DC analysis first in order to calculate them.

EXAMPLE l The transistor parameter are = 125 and VA=200 V. A value of gm = 200 m. A/V is desired. Determine the collector current, ICQ and then find r and ro ANSWERS: ICQ = 5. 2 m. A, r = 0. 625 k and ro = 38. 5 k

CALCULATION OF GAIN Voltage Gain = vo / vs Current Gain = io / is

Common-Emitter Amplifier

l Remember that for Common Emitter Amplifier, l l l the output is measured at the collector terminal. the gain is a negative value Three types of common emitter l l l Emitter grounded With RE With bypass capacitor CE

STEPS TO OBTAIN VOLTAGE AMPLIFIER COMPONENTS OUTPUT SIDE 1. Get the equivalent resistance at the output side, RO 2. Get the vo equation where vo = - gm vbe RO INPUT SIDE 3. Calculate Ri 4. Get vbe in terms of vi 5. Get the open circuit voltage gain

Emitter Grounded 93. 7 k 0. 5 k 6. 3 k Voltage Divider biasing: Change to Thevenin Equivalent RTH = 5. 9 k VTH = 0. 756 V VCC = 12 V RC = 6 k β = 100 VBE = 0. 7 V VA = 100 V

l Perform DC analysis to obtain the value of IC BE loop: l 5. 9 IB + 0. 7 – 0. 756 = 0 IB = 0. 00949 IC = βIB = 0. 949 m. A Calculate the small-signal parameters r = 2. 74 k , ro = 105. 37 k and gm = 36. 5 m. A/V

Emitter Grounded β = 100 VBE = 0. 7 V VA = 100 V off - becomes short circuit off becomes short circuit CC becomes short circuit during AC vo v. S RTH RC

1. Ro = ro || RC 2. Equation of vo : vo = - ( ro || RC ) gmvbe 3. Calculate Ri = RTH||r 4. vbe = vi 5. Av vi = vo = - ( ro || RC ) gmvbe open circuit voltage Av vi = - ( ro || RC ) gmvbe because vi = vbe Av = - ( ro || RC ) gm open circuit voltage gain

RS v. S Ri vo To find new voltage gain, vo/vs with input signal voltage source, vs 6. vi in terms of vs use voltage divider: vi = [ Ri / ( Ri + Rs )] * vs = RTH||r vs (RTH||r ) + RS 7. vo = Avvi because there is no load resistor vs RTH||r vo = - ( ro || RC ) gm (RTH||r ) + RS vo/vs = - ( ro || RC ) gm RTH||r (RTH||r ) + RS vs

v. S RS = 0. 5 k + vi RTH 5. 9 k 2. 74 k vbe - v. O gmvbe 105. 37 k RC = 6 k Follow the steps 1. Ro = 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 = vi

Equation of vo : vo = - ( ro || RC ) gmvbe= - 36. 5 ( 5. 677) vbe = -207. 21 vbe = vi 5. Av vi = vo open circuit voltage Avvi = -207. 21 vbe = -207. 21 vi Av = -207. 21 open circuit voltage gain

RS = 0. 5 kΩ v. S 5. 677 k Ri = 1. 87 k vo To find new voltage gain, vo/vs with input signal voltage source, vs 6. vi in terms of vs use voltage divider: vi = [ Ri / ( Ri + Rs )] * vs = 0. 789 vs 7. vo = Avvi because there is no load resistor vo = -207. 21 (0. 789 vs) vo/vs = -163. 49

Example β = 139 VBE = 0. 668 V VA =

Voltage Divider biasing: Change to Thevenin Equivalent RTH = 4 k β = 139 VBE = 0. 668 V VA = VTH = 0. 7 V l Perform DC analysis to obtain the value of IC BE loop: l 4 IB + 0. 668 – 0. 7 = 0 IB = 0. 008 IC = βIB = 1. 112 m. A Calculate the small-signal parameters r = 3. 25 k , ro = and gm = 42. 77 m. A/V

V 1 0. 5 k + vi 4 k vbe 3. 25 k gmvbe - RC 0. 3 k Follow the steps 1. Ro = RC = 0. 3 k 2. Equation of vo : vo = - (RC ) gmvbe= - 0. 3 ( 42. 77) vbe = -12. 831 vbe 3. Calculate Ri RTH||r = 4 || 3. 25 = 1. 793 k 4. vbe = vi

Equation of vo : vo = - (RC ) gmvbe= - 0. 3 ( 42. 77) vbe = -12. 831 vbe = vi 5. Avvi = vo open circuit voltage Avvi = - 12. 831 vbe = -12. 831 vi Av = -12. 831 open circuit voltage gain

0. 5 k v 1 0. 3 k 1. 793 k RL = 100 k To find new voltage gain, vo/v 1 now with signal voltage, v 1 and RL 6. vi in terms of vs use voltage divider: vi = [ Ri / ( Ri + Rs )] * v 1 = 0. 782 v 1 7. vo = [ RL / ( RL + Ro )] * Avvi this is because we have load resistor RL vo = 0. 997 (-12. 831 ) (0. 782 v 1) vo/v 1 = -10