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

Chapter 5 basic bjt amplifiers (AC ANALYSIS)

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

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

DO YOU STILL REMEMBER?

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

CALCULATE DC CURRENT, ID CALCULATE AC CURRENT, id DC ANALYSIS AC ANALYSIS DIODE = MODEL 1 ,2 OR 3 CALCULATE rd DIODE = RESISTOR, rd CALCULATE DC CURRENT, ID 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 = small signal transistor output resistance VA is normally equals to , hence, if that is the case, rO =   open circuit ro = VA / ICQ 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

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

Common-Emitter Amplifier

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

STEPS OUTPUT SIDE Get the equivalent resistance at the output side, ROUT Get the vo equation where vo = - gm vbeROUT INPUT SIDE Calculate Ri Get vbe in terms of vs – eg: using voltage divider. Go back to vo equation and calculate the voltage gain

Emitter Grounded β = 100 VBE = 0.7V VA = 100 V VCC = 12 V RC = 6 k 93.7 k 6.3 k 0.5 k β = 100 VBE = 0.7V VA = 100 V Voltage Divider biasing: Change to Thevenin Equivalent RTH = 5.9 k VTH = 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

Emitter Grounded β = 100 VBE = 0.7V VA = 100 V vo RC RTH vS off - becomes short circuit off - becomes short circuit CC becomes short circuit during AC vS RTH RC vo

Follow the steps 1. Rout = ro || RC = 5.677 k vbe gmvbe RTH RC = 6 k RS = 0.5 k vS vO 5.9 k 2.74 k 105.37 k Follow the steps 1. 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

5. Go back to equation of vo and calculate the gain vbe gmvbe RTH RC = 6 k RS = 0.5 k vS vO 5.9 k 2.74 k 105.37 k 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 = -163.5 AV = vo / vs = - 163.5

Current Gain Output side: io = vo / RC = vo / 6 vbe gmvbe RTH RC = 6 k RS = 0.5 k vS vO 5.9 k 2.74 k 105.37 k is io Output side: io = vo / RC = vo / 6 Input side: is = vs / (RS + Ri ) = vS / 2.37 RS Ri is vS Current gain = iout / is = vo (2.37) = -163.5 * 0.395 = - 64.6 vs (6)

RECALL CHAPTER 1 RS vS Ri RS vS Rload

Example β = 139 VBE = 0.668 V VA = 

Voltage Divider biasing: Change to Thevenin Equivalent RTH = 4 k VTH = 0.7 V β = 139 VBE = 0.668 V VA =  Perform DC analysis to obtain the value of IC BE loop: 4 IB + 0.668 – 0.7 = 0 IB = 0.008 IC = βIB = 1.112 mA Calculate the small-signal parameters r = 3.25 k , ro =  and gm = 42.77 mA/V

Follow the steps 1. Rout = RC || RL = 0.3 || 100 = 0.3 k vbe 0.3 k 100 k 0.5 k 4 k 3.25 k V1 gmvbe RC RL Follow the steps 1. Rout = RC || RL = 0.3 || 100 = 0.3 k 2. Equation of vo : vo = - (RC || RL ) gmvbe= - 0.3 ( 42.77) vbe = -12.831 vbe 3. Calculate Ri  RTH||r = 4 || 3.25 = 1.793 k 4. vbe in terms of vs  use voltage divider: vbe = [ Ri / ( Ri + Rs )] * v1 = 0.782 v1 so v1 = 1.279 vbe

5. Go back to equation of vo and calculate the gain vbe 0.3 k 100 k 0.5 k 4 k 3.25 k V1 gmvbe RL RC so: v1 = 1.279 vbe 5. Go back to equation of vo and calculate the gain vo / v1 = -12.831 vbe / 1.279 vbe vo / v1 = - 12.831 / 1.279 vo / v1 = -10 AV = vo / v1 = - 10

Output side: io = vo / 100 = vo / 100 RS = 0.5 k v1 100k Ri = 1.793 k Output side: io = vo / 100 = vo / 100 Input side: ii = v1 / (RS + Ri ) = v1 / 2.293 Current gain = io / ii = vo (2.293) = -10 * 0.02293 = - 0.2293 v1 (100)