Transistors

1. Transistors • They are unidirectional current carrying devices with capability to control the current flowing through them • The switch current can be controlled by either current or voltage • Bipolar Junction Transistors (BJT) control current by current • Field Effect Transistors (FET) control current by voltage • They can be used either as switches or as amplifiers

2. NPN Bipolar Junction Transistor • One N-P (Base Collector) diode one P-N (Base Emitter) diode

3. PNP Bipolar Junction Transistor • One P-N (Base Collector) diode one N-P (Base Emitter) diode

4. NPN BJT Current flow

5. BJT  and  • From the previous figureiE = iB + iC • Define  = iC / iE • Define  = iC / iB • Then  = iC / (iE –iC) =  /(1- ) • TheniC =  iE; iB = (1-) iE • Typically   100 for small signal BJTs (BJTs that • handle low power) operating in active region (region • where BJTs work as amplifiers)

6. BJT in Active Region Common Emitter(CE) Connection • Called CE because emitter is common to both VBB and VCC

7. BJT in Active Region (2) • Base Emitter junction is forward biased • Base Collector junction is reverse biased • For a particular iB, iCis independent of RCC • transistor is acting as current controlled current source (iC is controlled by iB, and iC=  iB) • Since the base emitter junction is forward biased, from Shockley • equation

8. Early Effect and Early Voltage • As reverse-bias across collector-base junction increases, width of the collector-base depletion layer increases and width of the base decreases (base-width modulation). • In a practical BJT, output characteristics have a positive slope in forward-active region; collector current is not independent of vCE. • Early effect: When output characteristics are extrapolated back to point of zero iC, curves intersect (approximately) at a common point vCE = -VA which lies between 15 V and 150 V. (VA is named the Early voltage) • Simplified equations (including Early effect): Chap 5 - 8

9. BJT in Active Region (3) • Normally the above equation is never used to calculateiC, iB • Since for all small signal transistors vBE  0.7. It is only useful for deriving the small signal characteristics of the BJT. • For example, for the CE connection, iB can be simply calculated as, or by drawing load line on the base –emitter side

10. iB 100 A 0 5V vBE Deriving BJT Operating points in Active Region –An Example In the CE Transistor circuit shown earlier VBB= 5V, RBB= 107.5 k, RCC = 1 k, VCC = 10V. Find IB,IC,VCE,and the transistor power dissipation using the characteristics as shown below By Applying KVL to the base emitter circuit By using this equation along with the iB / vBE characteristics of the base emitter junction, IB = 40 A

11. iC 100 A 10 mA 80 A 60 A 40 A 20 A 0 20V vCE Deriving BJT Operating points in Active Region –An Example (2) By Applying KVL to the collector emitter circuit By using this equation along with the iC / vCE characteristics of the base collector junction,iC = 4 mA, VCE = 6V Transistor power dissipation = VCEIC = 24 mW We can also solve the problem without using the characteristics if and VBE values are known

12. BJT in Cutoff Region • Under this condition iB= 0 • As a result iC becomes negligibly small • Both base-emitter as well base-collector junctions may be reverse • biased • Under this condition the BJT can be treated as an off switch

13. BJT in Saturation Region • Under this condition iC / iB   in active region • Both base emitter as well as base collector junctions are forward • biased • VCE 0.2 V • Under this condition the BJT can be treated as an on switch

14. BJT in Saturation Region (2) • A BJT can enter saturation in the following ways (refer to the CE circuit) • For a particular value ofiB,if we keep on increasing RCC • For a particular value ofRCC,if we keep on increasing iB • For a particular value ofiB,if we replace the transistor with one with higher 

15. iC 100 A 10 mA 80 A 60 A 40 A 20 A 0 20V vCE BJT in Saturation Region – Example 1 In the CE Transistor circuit shown earlier VBB= 5V, RBB= 107.5 k, RCC = 10 k, VCC = 10V. Find IB,IC,VCE,and the transistor power dissipation using the characteristics as shown below Here even though IB is still 40 A; from the output characteristics, IC can be found to be only about 1mA and VCE 0.2V( VBC  0.5V or base collector junction is forward biased (how?))  = IC / IB = 1mA/40 A = 25 100

16. BJT in Saturation Region – Example 2 In the CE Transistor circuit shown earlier VBB= 5V, RBB= 43 k, RCC = 1 k, VCC = 10V. Find IB,IC,VCE,and the transistor power dissipation using the characteristics as shown below Here IB is 100 A from the input characteristics; IC can be found to be only about 9.5 mA from the output characteristics and VCE 0.5V( VBC  0.2V or base collector junction is forward biased (how?))  = IC / IB = 9.5 mA/100 A = 95  100 Transistor power dissipation = VCEIC  4.7 mW Note: In this case the BJT is not in very hard saturation

17. iC iB 100 A 100 A 80 A 60 A 40 A 20 A 0 0 20V vCE 5V vBE Output Characteristics BJT in Saturation Region – Example 2 (2) 10 mA Input Characteristics

18. BJT in Saturation Region – Example 3 In the CE Transistor circuit shown earlier VBB= 5V, VBE = 0.7V RBB= 107.5 k, RCC = 1 k, VCC = 10V,  = 400. Find IB,IC,VCE, and the transistor power dissipation using the characteristics as shown below By Applying KVL to the base emitter circuit Then IC= IB= 400*40 A = 16000 A and VCE = VCC-RCC* IC =10- 0.016*1000 = -6V(?) But VCE cannot become negative (since current can flow only from collector to emitter). Hence the transistor is in saturation

19. BJT in Saturation Region – Example 3(2) Hence VCE 0.2V IC = (10 –0.2) /1 = 9.8 mA Hence the operating = 9.8 mA / 40 A = 245

20. BJT Operating Regions at a Glance (1)

21. BJT Operating Regions at a Glance (2)

22. BJT Large-signal (DC) model

23. BJT ‘Q’ Point (Bias Point) • Q point means Quiescent or Operating point • Very important for amplifiers because wrong ‘Q’ point selection increases amplifier distortion • Need to have a stable ‘Q’ point, meaning the the operating point should not be sensitive to variation to temperature or BJT, which can vary widely

24. Four Resistor bias Circuit for Stable ‘Q’ Point  By far best circuit for providing stable bias point

25. Analysis of 4 Resistor Bias Circuit 

26. Analysis of 4 Resistor Bias Circuit (2) Applying KVL to the base-emitter circuit of the Thevenized Equivalent form VB - IB RB -VBE - IE RE = 0(1) Since IE = IB + IC = IB + IB= (1+ )IB (2) Replacing IE by (1+ )IB in (1), we get (3) If we design(1+ )RE  RB (say (1+ )RE  100RB) Then (4)

27. Analysis of 4 Resistor Bias Circuit (3) And (for large ) (5) Hence IC and IE become independent of ! Thus we can setup a Q-point independentof which tends to vary widely even within transistors of identical part number (For example,  of 2N2222A, a NPN BJT can vary between 75 and 325 for IC = 1 mA and VCE = 10V)

28. 4 Resistor Bias Circuit -Example A 2N2222A is connected as shown with R1 = 6.8 k,R2 = 1 k,RC = 3.3 k, RE = 1 k and VCC = 30V. Assume VBE = 0.7V. Compute VCC and IC for  = i)100 and ii) 300

29. 4 Resistor Bias Circuit –Example (1) i)  = 100 ICQ = IB = 3.09 mA IEQ = (1+)IB = 3.12 mA VCEQ = VCC-ICRC-IERE = 30-3.09*3.3-3.12*1=16.68V

30. 4 Resistor Bias Circuit –Example (2) i)  = 300 ICQ = 300IB = 3.13 mA IEQ = (1+)IB = 3.14 mA VCEQ = VCC-ICRC-IERE = 30-3.13*3.3-3.14*1=16.53V

31.  = 300 ICQ  = 100 VCEQ 16.68 V 16.53 V 0.9 % 3.09 mA 3.13 mA 1.29 % % Change 4 Resistor Bias Circuit –Example (3) The above table shows that even with wide variation of  the bias points are very stable.

32. Four-Resistor Bias Network for BJT F. A. region correct - Q-point is (201 mA, 4.32 V)

33. Four-Resistor Bias Network for BJT (cont.) • All calculated currents > 0, VBC = VBE - VCE = 0.7 - 4.32 = - 3.62 V • Hence, base-collector junction is reverse-biased, and assumption of forward-active region operation is correct. • Load-line for the circuit is: The two points needed to plot the load line are (0, 12 V) and (314 mA, 0). Resulting load line is plotted on common-emitter output characteristics. IB = 2.7 mA, intersection of corresponding characteristic with load line gives Q-point.

34. Four-Resistor Bias Network for BJT: Design Objectives • We know that • This implies that IB << I2, so that I1 = I2. So base current doesn’t disturb voltage divider action. Thus, Q-point is independent of base current as well as current gain. • Also, VEQ is designed to be large enough that small variations in the assumed value of VBE won’t affect IE. • Current in base voltage divider network is limited by choosing I2 ≤ IC/5. This ensures that power dissipation in bias resistors is < 17 % of total quiescent power consumed by circuit and I2 >> IB for b > 50.

35. Four-Resistor Bias Network for BJT: Design Guidelines • Choose Thévenin equivalent base voltage • Select R1 to set I1 = 9IB. • Select R2 to set I2 = 10IB. • RE is determined by VEQ and desired IC. • RC is determined by desired VCE.

36. Four-Resistor Bias Network for BJT: Example • Problem: Design 4-resistor bias circuit with given parameters. • Given data: IC = 750 mA,bF = 100, VCC = 15 V, VCE = 5 V • Assumptions: Forward-active operation region, VBE = 0.7 V • Analysis: Divide (VCC - VCE) equally betweenREandRC. Thus, VE = 5 V • and VC = 10 V

37. Two-Resistor Bias Network for BJT: Example • Problem: Find Q-point for pnp transistor in 2-resistor bias circuit with • given parameters. • Given data: bF = 50, VCC = 9 V • Assumptions: Forward-active operation region, VEB = 0.7 V • Analysis: Forward-active region operation is correct Q-point is : (6.01 mA, 2.88 V)