EMT 112 4 ANALOGUE ELECTRONICS BJT Frequency Response

EMT 112 / 4 ANALOGUE ELECTRONICS BJT Frequency Response Common-Emitter Amplifier Jan 20, 2010 1600 – 1700 DKP 2

…. the story so far � Assumption: Coupling capacitors and bypass capacitors act as short circuits to the signal voltages and open circuits to dc voltages. Capacitors do not change instantaneously from a short circuit to an open circuit as the frequency approaches zero. � Assumption: transistors are ideal in that output signal respond instantaneously to input signals. There are internal capacitances in bipolar transistor. Frequency response of amplifier circuits due to circuit capacitors and transistor capacitances.

The Decibel (d. B) A logarithmic measurement of the ration of power or voltage � Power gain is expressed in d. B by the formula: � where Ap is the actual power gain, Pout/Pin � � Voltage gain is expressed by: If Av is greater than 1, the d. B is +ve, and if Av is less than 1, the d. B gain is –ve value & usually called attenuation

Amplifier gain vs. frequency Midband Gain falls of due to the effects of C and C Gain falls of due to the effects of CC and CE

Amplifier gain vs. Frequency � LOW FREQUENCY RANGE gain decreases as the frequency decreases due to coupling and bypass capacitor effects. � HIGH FREQUENCY RANGE gain decreases as the frequency increases due to stray capacitance and transistor capacitance effects. � MIDBAND RANGE gain is almost constant – coupling and bypass capacitors act as short circuits and stray and transistor capacitances act as open circuits.

Definition Frequency response of an amplifier is the graph of its gain versus the frequency. Cutoff frequencies : the frequencies at which the voltage gain equals 0. 707 of its maximum value. Midband : the band of frequencies between 10 f 1 and 0. 1 f 2. The voltage gain is maximum. Bandwidth : the band between upper and lower cutoff frequencies Outside the midband, the voltage gain can be determined by these equations: Below midband Above midband

Gain-bandwidth product : constant value of the product of the voltage gain and the bandwidth. Unity-gain frequency : the frequency at which the amplifier’s gain is 1

LOW FREQUENCY � At low frequency range, the gain falloff due to coupling capacitors and bypass capacitors. � As signal frequency , the XC - no longer behave as short circuits.

Short-circuit time-constant method (SCTC) To determine the lower-cutoff frequency having n coupling and bypass capacitors: Ri. S = resistance at the terminals of the ith capacitor Ci with all the other capacitors replaced by short circuits.

Common-emitter Amplifier VCC = 12 V Given : Q-point values : 1. 73 m. A, 2. 32 V R 1 30 k = 100, VA = 75 V RS 1 k v. S Therefore, r = 1. 45 k , ro = 44. 7 k RC 4. 3 k C 1 C 2 v. O 0. 1 F RL 100 k 2 F R 2 10 k RE 1. 3 k C 3 10 F

Common-emitter Amplifier - Low-frequency ac equivalent circuit C 2 RS vo C 1 RC vs RB RE C 3 RL

Circuit for finding R 1 S RS Rin. CE R 1 S RC RB RL Replacing C 2 and C 3 by short circuits

Circuit for finding R 2 S Rout. CE Replacing C 1 and C 3 by short circuits RC RS RB R 2 S RL

Circuit for finding R 3 S Replacing C 1 and C 2 by short circuits RTH RC||RL RS RB RE Rout. CC R 3 S

Estimation of L

EMT 112 / 4 ANALOGUE ELECTRONICS BJT Frequency Response Common-Base & Common. Collector Amplifier Jan 22, 2010 0800 – 1000 DKP 1

Common-base Amplifier RS 100 v. S 4. 7 F 1 F v. O C 1 RE 43 k -VEE RC 22 k C 2 RL 75 k +VCC Given : Q-point values : 0. 1 m. A, 5 V = 100, VA = 70 V Therefore, gm = 3. 85 m. S, ro = 700 k r = 26

Common-base Amplifier - Low-frequency ac equivalent circuit RS vs vo C 2 C 1 RE RC RL

Circuit for finding R 1 S RS R 1 S Replacing C 2 by short circuit RE RC || RL Rin. CB

Circuit for finding R 2 S Rout. CB Replacing C 1 by short circuit RS || RE RC R 2 S RL

Estimation of L

Common-collector Amplifier +VCC RS 1 k v. S C 1 C 2 0. 1 F RB 100 k 100 F RE 3 k Given : Q-point values : 1 m. A, 5 V = 100, VA = 70 V v. O RL 47 k -VEE Therefore, r = 2. 6 k , ro =70 k

Common-collector Amplifier - Low-frequency ac equivalent circuit RS C 1 C 2 vs vo RB RE RL

Circuit for finding R 1 S Rin. CC Replacing C 2 by short circuit

Circuit for finding R 2 S Replacing C 1 by short circuit Rout. CC R 2 S RTH = RS || RB

Estimation of L

Example Given : Q-point values : 1. 6 m. A, 4. 86 V VCC = 10 V = 100, VA = 70 V Therefore, R 1 62 k r = 1. 62 k , ro = 43. 75 k , gm = 61. 54 m. S RS v. S Determine the total lowfrequency response of the amplifier. RC 2. 2 k C 1 C 2 0. 1 F RL 10 k 600 0. 1 F R 2 22 k v. O RE 1. 0 k C 3 10 F

Low frequency due to C 1 and C 2 C 3 Low frequency due to C 1 Low frequency due to C 2

Low frequency due to C 3

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 = Cbe ------ 2 p. F ~ 50 p. F C = Cbc ------ 0. 1 p. F ~ 5 p. F reversebiased junction capacitance forwardbiased junction capacitance

Basic data sheet for the 2 N 2222 bipolar transistor Cob = Cbc Output capacitance Cib = Cbe Input capacitance

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.

Miller Equivalent Circuit Impedance Z is connected between the input side and the output side of a voltage amplifier. . I 1 V 1 I 2 V 2

Miller Equivalent Circuit (cont) V 1 . . The impedance Z is being replaced by two equivalent impedances, i. e. one connected across the input (ZM 1) and the other connected across the output terminals (ZM 2) V 2

Miller Capacitance Effect C I 1 I 2 V 1 V 2 -A CM = Miller capacitance Miller effect Multiplication effect of Cµ V 1 C M 2 V 2

High-frequency hybrid- model C + r V C gm. V ro - C = Cbe C = Cbc

High-frequency hybrid- model with Miller effect r C CMi gm. V A : midband gain ro CMo

High-frequency in Commonemitter Amplifier Calculation Example VCC = 10 V Given : = 125, Cbe = 20 p. F, Cbc = 2. 4 p. F, VA = 70 V, VBE(on) = 0. 7 V R 1 22 k RS Determine : 1. Upper cutoff frequencies RC 2. 2 k C 1 C 2 10 F RL 2. 2 k 600 10 F v. S 2. Dominant upper cutoff frequency R 2 4. 7 k v. O RE 470 C 3 10 F

High-frequency hybrid- model with Miller effect for CE amplifier RS vo vs R 1||R 2 C CMi r gm. V ro CMo RC||RL midband gain Miller’s equivalent capacitor at the input Miller’s equivalent capacitor at the output

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

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

TOTAL AMPLIFIER FREQUENCY RESPONSE A (d. B) ideal Amid actual -3 d. B f. C 1 f. C 2 f. C 3 f. L f. C 4 f. C 5 f. H f (Hz)

Total Frequency Response of Common-emitter Amplifier Calculation Example VCC = 5 V Given : = 120, Cbe = 2. 2 p. F, Cbc = 1 p. F, VA = 100 V, VBE(on) = 0. 7 V R 1 33 k RS 2 k Determine : 1. Midband gain 2. Lower and upper cutoff frequencies v. S RC 4 k C 1 C 2 2 F RL 5 k 1 F R 2 22 k v. O RE 4 k C 3 10 F

Step 1 - Q-point Values

Step 2 - Transistor parameters value

Step 3 - Midband gain

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

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

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