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ANALOG ELECTRONIC CIRCUITS 1 EKT 204 Frequency Response of BJT Amplifiers (Part 2) 1.

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Presentation on theme: "ANALOG ELECTRONIC CIRCUITS 1 EKT 204 Frequency Response of BJT Amplifiers (Part 2) 1."— Presentation transcript:

1 ANALOG ELECTRONIC CIRCUITS 1 EKT 204 Frequency Response of BJT Amplifiers (Part 2) 1

2 HIGH FREQUENCY The gain falls off at high frequency end due to the internal capacitances of the transistor. Transistors exhibit charge-storage phenomena that limit the speed and frequency of their operation. Small capacitances exist between the base and collector and between the base and emitter. These effect the frequency characteristics of the circuit. C  = C be pF ~ 50 pF C  = C bc pF ~ 5 pF reverse-biased junction capacitance forward-biased junction capacitance 2

3 C ob = C bc C ib = C be  Output capacitance  Input capacitance Basic data sheet for the 2N2222 bipolar transistor 3

4 Miller’s Theorem This theorem simplifies the analysis of feedback amplifiers. The theorem states that if an impedance is connected between the input side and the output side of a voltage amplifier, this impedance can be replaced by two equivalent impedances, i.e. one connected across the input and the other connected across the output terminals. 4

5 Miller Equivalent Circuit I2I2 I1I1 V1V1 V2V2 Impedance Z is connected between the input side and the output side of a voltage amplifier.. 5

6 V1V1 V2V2 Miller Equivalent Circuit (cont).. The impedance Z is being replaced by two equivalent impedances, i.e. one connected across the input (Z M1 ) and the other connected across the output terminals (Z M2 ) 6

7 Miller Capacitance Effect I2I2 I1I1 V1V1 V2V2 C C M = Miller capacitance Miller effect Multiplication effect of Cµ Miller effect Multiplication effect of Cµ 7

8 rr roro CC VV gmVgmV CC - + C  = C be C  = C bc High-frequency hybrid-  model 8

9 High-frequency hybrid-  model with Miller effect rr roro C Mi gmVgmV CC C Mo A : midband gain 9

10 High-frequency in Common- emitter Amplifier vOvO 22 k  V CC = 10V 4.7 k  RSRS C1C1 10  F C2C2 C3C3 600  470  2.2 k  R1R1 RCRC RERE R2R2 vSvS RLRL Given :  = 125, C be = 20 pF, C bc = 2.4 pF, V A = 70V, V BE (on) = 0.7V Determine : 1.Upper cutoff frequencies 2.Dominant upper cutoff frequency Calculation Example 10

11 R 1 ||R 2 RSRS R C ||R L vsvs vovo rr roro CC C Mi C Mo gmVgmV  midband gain  Miller’s equivalent capacitor at the input  Miller’s equivalent capacitor at the output High-frequency hybrid-  model with Miller effect for CE amplifier 11

12  Thevenin’s equivalent resistance at the input  Thevenin’s equivalent resistance at the output  total input capacitance  total output capacitance  upper cutoff frequency introduced by input capacitance  upper cutoff frequency introduced by output capacitance Calculation (Cont..) 12

13 How to determine the dominant frequency The lowest of the two values of upper cutoff frequencies is the dominant frequency. Therefore, the upper cutoff frequency of this amplifier is 13

14 TOTAL AMPLIFIER FREQUENCY RESPONSE f (Hz) f C3 f C1 f C2 f C4 f C5 A (dB) A mid fHfH fLfL ideal actual -3dB 14

15 Total Frequency Response of Common-emitter Amplifier 33 k  V CC = 5V 22 k  RSRS C1C1 1  F C2C2 C3C3 10  F 2  F 2 k  4 k  R1R1 RCRC RERE R2R2 vSvS RLRL 5 k  vOvO Given :  = 120, C be = 2.2 pF, C bc = 1 pF, V A = 100V, V BE (on) = 0.7V Determine : 1.Midband gain 2.Lower and upper cutoff frequencies Calculation Example 15

16 Step 1 - Q-point Values 16

17 Step 2 - Transistor parameters value 17

18 Step 3 - Midband gain 18

19 Step 4 - Lower cutoff frequency ( f L ) Due to C 1 Due to C 2 Due to C 3 SCTC method Lower cutoff frequency 19

20 Step 5 - Upper cutoff frequency (f H ) Miller capacitance Input & output resistances 20

21 Input side Output side Upper cutoff frequency (the smallest value) Step 5 - Upper cutoff frequency (f H ) 21

22 Exercise  Textbook: Donald A. Neamen, ‘MICROELECTRONICS Circuit Analysis & Design’,3rd Edition’, McGraw Hill International Edition, 2007  Exercise


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