Recall Last Lecture

1. 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

2. Chapter 5basic bjt amplifiers(AC ANALYSIS)

3. 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.

4. 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

5. iC , iB and iE, vCE and vBE EXAMPLE Sum of both ac and dc components

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

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

8. DO YOU STILL REMEMBER?

9. VDQ = V rd IDQ id Let’s assume that Model 2 is used DC equivalent AC equivalent

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

11. What about bjt?

12. ib AC equivalent circuit – Small-Signal Hybrid-π Equivalent OR

13. 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 rO = small signal transistor output resistance VA is normally equals to , hence, if that is the case, rO =   open circuit

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

15. 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

16. CALCULATION OF GAIN Voltage Gain, AV = vo / vs Current Gain, Ai = iout / is

17. ib • Small-Signal Voltage Gain: Av = Vo / Vs

18. Common-Emitter Amplifier

19. 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

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

21. Emitter Grounded VCC = 12 V RC = 6 k 93.7 k β = 100 VBE = 0.7V VA = 100 V 0.5 k 6.3 k Voltage Divider biasing: Change to Thevenin Equivalent RTH = 5.9 k VTH = 0.756 V

22. 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

23. Emitter Grounded VCC = 12 V RC = 6 k 93.7 k β = 100 VBE = 0.7V VA = 100 V 0.5 k 6.3 k

24. Follow the steps vbe • 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. vb in terms of vs use voltage divider: vbe = [ Ri / ( Ri + Rs )] * vs = 0.789 vs

25. vbe so:vb = 0.789 vsreplace in equation from step 1 5. Go back to equation of vo vo = -207.21 vbe vo = - 207.21[0.789 vs] vo = -163.5 vs AV = vo / vs = - 163.5 bring VS over

26. TYPE 2: Emitter terminal connected with RE – normally ro =  in this type New parameter: input resistance seen from the base, Rib = vb / ib VCC = 5 V β = 120 VBE = 0.7V VA =  RC = 5.6 k 250 k 0.5 k 75 k RE = 0.6 k

27. 7.46 k 0.5 k RC = 6 k vb 57.7 k RE = 0.6 k

28. vb • 1. Rout = RC = 6 k 2. Equation of vo : vo = - RC ib= - 720 ib 3. Calculate Rib using KVL: ibr+ ie RE - vb = 0 but ie = (1+ ) ib = 121 ib so: ib[ 121(0.6) + 7.46 ] = vb Rib = 80.06 k 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

29. so:vb = 0.9853vs 6. Go back to equation of vo vo = - 720 ib = - 720 [ vb / Rib ] vo = - 720 [ 0.9853 vs / 80.06] vo = - 8.86vs vb AV = vo / vs = - 8.86 bring VS over

30. TYPE 3: With Emitter Bypass Capacitor, CE • Circuit with Emitter Bypass Capacitor • Theremay be times when the emitter resistor must be large for the purpose of DC design, but degrades the small-signal gain too severely. • An emitter bypass capacitor can be used to effectively create a short circuit path during ac analysis hence avoiding the effect RE

31. vb CE becomes a short circuit path – bypass RE; hence similar to Type 1

32. β = 125 VBE = 0.7V VA = 200 V IC = 0.84 mA VCC = 5 V 20 k RC = 2.3 k 0 k 20 k Bypass capacitor RE = 5k

33. β = 125 VBE = 0.7V VA = 200 V IC = 0.84 mA vbe Short-circuited (bypass) by the capacitor CE 3.87 k RC = 2.3 k 10 k r =3.87 k , ro = 238 k and gm = 32.3 mA/V 238 k

34. Follow the steps vbe • 1. Rout = ro || RC = 2.278 k 2. Equation of vo : vo = - ( ro || RC ) gmvbe= -73.58 vbe 3.87 k 3. Calculate Ri RTH||r = 2.79 k RC = 2.3 k 10 k 238 k 4. vbe in terms of vs vbe = vs since connected in parallel

35. so:vbe = vs 6. Go back to equation of vo vo = -73.58 vs vbe AV = vo / vs = - 73.58 bring VS over 3.87 k RC = 2.3 k 10 k 238 k