Revision on Basic Transistor Amplifiers

Transcription

1 Electronic Circuits Revision on Basic Transistor Amplifiers Contents Biasing Amplification principles Small-signal model development for BJT

2 Aim of this chapter To show how transistors can be used to amplify a signal. amplifier 2

3 Basic idea Step 1: Set the transistor at a certain DC level biasing Step 2: Inject a small signal to the input and get a bigger output coupling 0.6V amplifier 7V 3

4 Biasing the transistor To set the transistor to a certain DC level = To set V CE and I C R B V BE V CC =10V I B I C V CE Suppose we want the following biasing condition: I C = 10 ma and V CE = 5 V Find R B and Start with V BE 0.7 V. Then, I B = (10 V BE )/ R B = (10 0.7)/ R B I C = βi B = 100 (10 0.7)/ R B = 10 ma So, R B = 94kΩ Transistor: β = 100 Also, V CE = 10 I C Hence, 5 = So, = 0.5kΩ 4

5 β dependent biasing bad biasing V CC =10V Now, let s go to the lab and try using R B = 94kΩ and = 0.5kΩ, and see if we get what we want. R B I C totally wrong! We don t get I C = 10mA and V CE = 5V V BE I B Transistor: β = 100 V CE This is not a good biasing circuit! because it relies on the accuracy of β, but β can be ±50% different from what is given in the databook. 5

6 A slightly better biasing method V CC =10V Again, our objective is to find the resistors such that I C = 10mA and V CE = 5V. First, if I B is small, we can approximately write R B1 R B2 I B V BE I C V CE Suppose we get I C = 10mA. Then = 0.5kΩ. We can start with R B1 = 940Ω and R B2 = 60Ω. Such resistors will make sure I B is much smaller than the current flowing down R B1 and R B2, which is consistent with the assumption. What we need in practice is to fine tune R B1 or R B2 such that V CE is exactly 5V. 6

7 A much better biasing method emitter degeneration V CC =10V Again, our objective is to find the resistors such that I C = 10mA and V CE = 5V. R B1 R B2 I B V E I C V CE R E Set V E = 2V, say. Then, R E = 2V/10mA = 0.2kΩ. Surely, = 0.3kΩ in order to get V CE = 5V. Finally, we have V B = V E 0.6. Therefore, if I B is small compared to I RB1 and I RB2, we have Hence, R B1 = 740Ω and R B1 = 260Ω. NOTE: β is never used in calculation!! 7

8 Stable (good) biasing V CC =10V Summary of biasing with emitter degeneration: Choose V E, I C and V CE. R B1 I C V B R B2 I B V E V CE R E R E Use V BE 0.6 to get V B. Then use to choose R B1 and R B2 such that I B is much smaller the current flowing in R B1 and R B2. 8

9 Terminology The following are the same: Biasing point Quiescent point Operating point (OP) DC point 9

10 Alternative view of biasing V CC V BE I C V CE V CC I C V R V CE I C Load line Slope=1/ operating point V CC V CE 10

11 What controls the operating point? V CC I C V BE I C V CE Load line Slope=1/ operating point a bigger V BE a smaller V CC V CE CONCLUSION: V BE or I B controls the OP also controls the OP 11

12 What happens if V BE dances up and down? V CC V BE I C V CE I C Load line Slope=1/ a bigger V BE = 0.65 a smaller V BE = 0.6 V CC V CE The OP also dances up and down along the load line. V CE also moves up and down. Typically, when V BE moves a little bit, V CE moves a lot! THIS IS CALLED AMPLICATION. 12

13 Derivation of voltage gain Question: what is? V CC Clearly, Ohm s law says that V BE I C V CE Then, what relates I C and V BE? Last lecture: transconductance Hence, 13

14 Common-emitter amplifier The one we have just studied is called COMMON-EMITTER amplifier. V CC SUMMARY: R B1 R B2 v BE v CE I C Small-signal voltage gain = g m That means we can increase the gain by increasing g m and/or. Output waveform is antiphase. 14

15 How do we inject signal into the amplifier? V CC R B1 I C ±20mV? R B2 V BE v CE = V CE v CE or v CE 15

16 Note on symbols = Total signal a v CE = V CE v CE DC point A total signal (large signal) operating point or DC value or quiescent point small signal or ac signal Small signal a or a 16

17 Solution: Add the same biasing DC level Exactly the same biasing V BE R B1 V CC I C But, it is impossible to find a voltage source which is equal to the exact biasing voltage across B-E. ±20mV R B2 v BE v CE V BE could actually be V, which is determined by the network RB1, RB2 and the transistor characteristic!! How to apply the exact V BE? 17

18 The wonderful voltage source: capacitor V CC The capacitor voltage is exactly equal to V BE because DC current must be zero V C 0A R B1 R B2 V BE I C V CE 18

19 Solution insert coupling capacitor This is called a coupling capacitor DC voltage equal to exactly the same biasing V BE ±20mV R B1 R B2 V CC v BE v CE I C 19

20 Complete common emitter amplifier V CC R B1 I C R B2 v BE v CE coupling capacitors (large enough so that they become short-circuit at signal frequencies) 20

21 Can we simplify the analysis? We are mainly interested in the ac signals. The DC bias does not matter! Can we create a simple circuit just to look at ac signals? common emitter amplifier 21

22 Small-signal model Two basic questions: common emitter v?? in amplifier 1 2 What is the loading (resistance) seen here? What is the Thévenin or Norton equivalent circuit seen here? 22

23 Small-signal model of BJT: objectives To find: r in R o G m or r in R o A m r in R o r in R o G m A m Norton form Thévenin form 23

24 Derivation of the small-signal model Input side: i B v BE For small-signal, r π r π β/g m where g m is the BJT s transconductance 24

25 Derivation of the small-signal model Output side: VCC I C v CE = V CC I C For small-signal, g m v BE v CE v CE where g m is the BJT s transconductance 25

26 Derivation of the small-signal model Output side: V CC Including BJT s Early effect g m v BE r o v CE I C v CE where r o is the Early resistor of the BJT. Recall: r o = V A /I C, where V A is typically about 100V. A very rough approx. is r o =. 26

27 Initial small-signal model for BJT MUST REMEMBER Small-signal BJT parameters: v BE B r π g m v BE C r o v CE E BJT model 27

28 Initial small-signal model for FET Similar to BJT, but input resistance is. v GS G g m v GS S D r o v DS Small-signal FET parameters: FET model All amplifier configurations using BJT can be likewise constructed using FET. 28

29 Example: common-emitter amplifier Assume the coupling caps are large enough to be considered as short-circuit at signal frequency V CC B V CC is ac 0V. C R B1 R B2 v BE R B1 R B1 r π E g m v BE r o E 29

30 Complete model for common-emitter amplifier Complete model: R B1 R B1 v BE r π r o g m v BE Total input resistance R in = R B1 R B1 r π Simplified model: Total output resistance R o = r o R B1 R B1 r π v BE g m v BE r o Voltage gain 30

31 Alternative model for common-emitter amplifier Output in Thévenin form: r o Total input resistance R B1 R B1 r π v BE R in = R B1 R B1 r π Total output resistance g m ( r o ) v BE R o = r o Voltage gain 31

32 More about common-emitter amplifier Because the output resistance is quite large (equal to r o ), the common-emitter amplifier is a POOR voltage driver. That means, it is not a good idea to use such an amplifier for loads which are smaller than. This makes it not suitable to deliver current to load. 1kΩ, for example R B1 R B1 r π v BE r o practically no output!! 10Ω g m ( r o ) v BE 32

33 Bad idea wrong use of common-emitter amplifier 10V Transconductance g m = I C /(25mV) = 5/25 = 0.2 A/V R B1 5mA 1kΩ Expected gain = g m = (0.2)(1k) = 200 or 46dB But the output circuit is: R B2 v BE speaker 10Ω 200 1kΩ 10Ω The effective gain drops to 33

34 Proper use of common-emitter amplifier 10V The load must be much larger than. R B1 5mA 1kΩ Now the output circuit is: R B2 v BE nearly open circuit 10MΩ 200 1kΩ 10MΩ The effective gain is 34

35 How can we use the amplifier in practice? 10V How to connect the output to load? R B1 5mA 1kΩ? R B2 v BE speaker 10Ω 35

36 Emitter follower R B1 R B2 10V I C biased to 10mA V CE biased to 5V R E Biasing conditions: Base voltage 5.6V Emitter voltage 5V Collector current 10mA R E = 500Ω R B1 :R B2 44:56 Say, R B1 = 440kΩ R B2 = 560kΩ V E = V B 0.6 Thus, for small signal, V E = V B or = Gain = / = 1 36

37 Small-signal model of emitter follower 10V B C R B1 R B2 R E R B1 R B2 r π E R E g m v BE r o E 37

38 Small-signal model of emitter follower i B B C Input resistance is r r π g m v BE R B1 R B2 E E r o R E which is quite large (good)!! 38

39 Small-signal model of emitter follower B C Output resistance is r π g m v BE r o i m E E R E r out v m which is quite small (good)!! 39

40 Small-signal model of emitter follower Thevenin form: very large R E (1/g m ) very small r π (1β)R E 1 Large input resistance Small output resistance Voltage gain = 1 Draw no current from previous stage Good for any load 40

41 A better emitter follower 10V Input resistance is very LARGE because R E =. R B1 Output resistance is 1/g m. R B2 I E 1/g m Gain = 1. This circuit is also called CLASS A output stage. Details to be studied in second year EC2. 41

42 Common-emitter amplifier with emitter follower as buffer 10V R B1 emitter follower (unit gain) R B2 v BE common-emitter amplifier (high gain) I E 1 g m speaker 10Ω 42

43 FET amplifiers (similar to BJT amplifiers) 10V R G1 source follower (unit gain) R G2 v GS common-source amplifier (high gain = g m ) I S 1 g m speaker 10Ω 43

44 Further thoughts Will the biasing resistors affect the gain? Seems not, because R bias Gain = g m which does not depend on R bias. However, a realistic voltage source has finite internal resistance. This will affect the gain. 44

45 Input source with finite resistance The input has a voltage divider network. R bias R s Therefore, the gain decreases to assuming r o very large. R s R bias v BE r π g m v BE r o 45

46 Example 10V 94kΩ 5mA 1kΩ By how much does the gain drop? 50Ω 50Ω 600Ω R bias v BE 94k 600 = 596Ω r π g m = 5mA/25mV = 0.2A/V r π = β/g m = 100/0.2 = 500Ω Voltage divider attenuation = Hence, the gain is reduced to 0.845(g m ) =

47 Further thoughts Recall that the best biasing scheme should be β independent. 10V 84kΩ 16kΩ 5mA 1kΩ 200Ω One good scheme is emitter degeneration, i.e., using R E to fix biasing current directly. Here, since V B is about 1.6V, as fixed by the base resistor divider, V E is about 1V. Therefore, I C V E /R E = 5mA (no β needed!) Question: Will this biasing scheme affect the gain? 47

48 Common-emitter amplifier with emitter degeneration V CC R B1 Exercise: Find the small-signal gain of this amplifier. Answer: R B2 R E The gain is MUCH smaller. We have a good biasing, but a poor gain! Can we improve the gain? 48

49 Common-emitter amplifier with emitter by-pass V CC Add C E such that the effective emitter resistance becomes zero at signal frequency. R B1 R B2 R E C E So, this circuit has good biasing, and the gain is still very high! Gain = g m which is unaffected by R E because effectively R E is shorted at signal frequency. C E is called bypass capacitor. 49

50 Summary Basic BJT model (small-signal ac model): B C v BE r π g m v BE r o E E 50

51 Summary Basic FET model (small-signal ac model): Similar to the BJT model, but with infinite input resistance. Therefore, the FET can be used in the same way as amplifiers. G S v GS g m v GS r o D S 51

52 Summary Common-emitter (CE) amplifier small-signal ac model: R bias R bias v BE r π g m v BE r o Gain = g m Input resistance = R bias r π (quite large desirable) Output resistance = r o (large undesirable) 52

53 Summary Emitter follower (EF) small-signal ac model: R bias R E v BE r π g m v BE R bias R E Gain = 1 Input resistance = R bias [r π (1β)R E ] Output resistance = R E (1/g m ) (quite large desirable) (small desirable) 53