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

The Decibel (dB) A logarithmic measurement of the ration of power or voltage Power gain is expressed in dB by the formula: where a p is the actual power gain, Pout/Pin Voltage gain is expressed by: If a v is greater than 1, the dB is +ve, and if a v is less than 1, the dB gain is –ve value & usually called attenuation 2

Amplifier gain vs frequency Midband range Gain falls of due to the effects of C C and C E Gain falls of due to the effects of stray capacitance and transistor capacitance effects 3

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 of its maximum value. Midband : the band of frequencies between 10f L and 0.1f H where the voltage gain is maximum. The region where coupling & bypass capacitors act as short circuits and the stray capacitance and transistor capacitance effects act as open circuits. 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 4

Lower & Upper Critical frequency  Critical frequency a.k.a the cutoff frequency  The frequency at which output power drops by 3 dB. [in real number, 0.5 of it’s midrange value.  An output voltage drop of 3dB represents about a drop from the midrange value in real number.  Power is often measured in units of dBm. This is decibels with reference to 1mW of power. [0 dBm = 1mW], where; 5

Gain & frequencies 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 6

LOW FREQUENCY At low frequency range, the gain falloff due to coupling capacitors and bypass capacitors. As signal frequency , the reactance of the coupling capacitor, X C  - no longer behave as short circuits. 7

Short-circuit time-constant method (SCTC) To determine the lower-cutoff frequency having n coupling and bypass capacitors: R iS = resistance at the terminals of the ith capacitor C i with all the other capacitors replaced by short circuits. 8

Common-emitter Amplifier 30 k  V CC = 12V 10 k  RSRS C1C1 2  F C2C2 C3C3 10  F 0.1  F 1 k  1.3 k  4.3 k  R1R1 RCRC RERE R2R2 vSvS vOvO RLRL 100 k  Given : Q-point values : 1.73 mA, 2.32 V  = 100, V A = 75 V Therefore, r  = 1.45 k , r o =44.7 k  9

Common-emitter Amplifier - Low-frequency ac equivalent circuit RSRS RBRB RCRC RLRL RERE C1C1 C2C2 C3C3 vsvs vovo In the above circuit, there are 3 capacitors (coupling plus bypass capacitors). Hence we need to find 3 resistances at the terminals of the 3 capacitors in order to find the lower cut-off frequency of the amplifier circuit. 10

Circuit for finding R 1S RSRS RBRB R 1S RCRC RLRL R inCE Replacing C 2 and C 3 by short circuits 11

Circuit for finding R 2S RLRL RSRS RBRB R 2S RCRC R outCE Replacing C 1 and C 3 by short circuits 12

Circuit for finding R 3S Replacing C 1 and C 2 by short circuits RSRS RBRB R 3S RERE R C ||R L R outCC R TH 13

Estimation of  L 14

Common-base Amplifier Given : Q-point values : 0.1 mA, 5 V  = 100, V A = 70 V Therefore, g m = 3.85 mS, r o = 700 k  r  = 26  RERE RSRS RCRC RLRL C1C1 C2C2 vOvO vSvS 22 k  100  43 k  75 k  1  F4.7  F +V CC -V EE 15

Common-base Amplifier - Low-frequency ac equivalent circuit RERE RSRS RCRC RLRL C1C1 C2C2 vovo vsvs 16

Circuit for finding R 1S Replacing C 2 by short circuit R C || R L R 1S RERE RSRS R inCB 17

Circuit for finding R 2S Replacing C 1 by short circuit R S || R E R 2S RCRC RLRL R outCB 18

Estimation of  L 19

Common-collector Amplifier RBRB RSRS RERE RLRL C1C1 C2C2 vSvS vOvO -V EE +V CC 100 k  1 k  3 k  47 k  100  F 0.1  F Given : Q-point values : 1 mA, 5 V  = 100, V A = 70 V Therefore, r  = 2.6 k , r o =70 k  20

Common-collector Amplifier - Low-frequency ac equivalent circuit RBRB RSRS RERE RLRL C2C2 C1C1 vovo vsvs 21

Circuit for finding R 1S Replacing C 2 by short circuit RBRB RSRS R E || R L R 1S R inCC 22

Circuit for finding R 2S Replacing C 1 by short circuit RERE RLRL R TH = R S || R B R 2S R outCC 23

Estimation of  L 24

Example vSvS 62 k  V CC = 10V 22 k  RSRS C1C1 0.1  F C2C2 C3C3 10  F 0.1  F 600  1.0 k  2.2 k  R1R1 RCRC RERE R2R2 vOvO RLRL 10 k  Given : Q-point values : 1.6 mA, 4.86 V  = 100, V A = 70 V Therefore, r  = 1.62 k , r o = k , g m = mS Determine the total low- frequency response of the amplifier. 25

Low frequency due to C 1 and C 2 C 3 Low frequency due to C 1 Low frequency due to C 2 26

Low frequency due to C 3 27