ANALOG ELECTRONIC CIRCUITS 1

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

2. Small capacitances exist between the base and collector and between the base and emitter. These effect the frequency characteristics of the circuit. 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. reverse-biased junction capacitance C = Cbe------2 pF ~ 50 pF forward-biased junction capacitance C= Cbc------0.1 pF ~ 5 pF

3. Basic data sheet for the 2N2222 bipolar transistor  Output capacitance Cob = Cbc Cib = Cbe  Input capacitance

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.

5. I1 I2 V1 V2 Miller Equivalent Circuit Impedance Z is connected between the input side and the output side of a voltage amplifier..

6. V1 V2 Miller Equivalent Circuit (cont) .. The impedance Z is being replaced by two equivalent impedances, i.e. one connected across the input (ZM1) and the other connected across the output terminals (ZM2)

7. C I1 I2 V1 V2 Miller Capacitance Effect CM = Miller capacitance Miller effect Multiplication effect of Cµ

8. C + r V C gmV ro - C = Cbe C= Cbc High-frequency hybrid- model

9. r C CMi ro CMo gmV High-frequency hybrid- model with Miller effect A : midband gain

10. VCC = 10V R1 RC C2 22 k 2.2 k RS C1 10 F RL 600  10 F 2.2 k vS R2 RE C3 4.7 k 10 F 470  High-frequency in Common-emitter Amplifier Calculation Example Given :  = 125, Cbe = 20 pF, Cbc = 2.4 pF, VA = 70V, VBE(on) = 0.7V vO • Determine : • Upper cutoff frequencies • Dominant upper cutoff frequency

11. RS vo vs R1||R2 C CMi r ro RC||RL CMo gmV High-frequency hybrid- model with Miller effect for CE amplifier  midband gain  Miller’s equivalent capacitor at the input  Miller’s equivalent capacitor at the output

12. Calculation (Cont..)  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

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

14. A (dB) ideal Amid -3dB actual f (Hz) fC4 fC5 fC1 fC2 fC3 fL fH TOTAL AMPLIFIER FREQUENCY RESPONSE

15. VCC = 5V R1 RC C2 33 k 4 k RS C1 2 F RL 2 k 1 F 5 k vS R2 RE C3 22 k 10 F 4 k Total Frequency Response of Common-emitter Amplifier Calculation Example Given :  = 120, Cbe = 2.2 pF, Cbc = 1 pF, VA = 100V, VBE(on) = 0.7V vO • Determine : • Midband gain • Lower and upper cutoff frequencies

16. Step 1 - Q-point Values

17. Step 2 - Transistor parameters value

18. Step 3 - Midband gain

19. Step 4 - Lower cutoff frequency (fL) Due to C1 Due to C2 Due to C3 SCTC method Lower cutoff frequency

20. Step 5 - Upper cutoff frequency (fH) Miller capacitance Input & output resistances

21. Step 5 - Upper cutoff frequency (fH) Input side Output side Upper cutoff frequency (the smallest value)

22. Exercise • Textbook: Donald A. Neamen, ‘MICROELECTRONICS Circuit Analysis & Design’,3rd Edition’, McGraw Hill International Edition, 2007 • Exercise 7.11