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SMALL SIGNAL BJT AMPLIFIER SMALL SIGNAL OPERATION.

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Presentation on theme: "SMALL SIGNAL BJT AMPLIFIER SMALL SIGNAL OPERATION."— Presentation transcript:

1 SMALL SIGNAL BJT AMPLIFIER SMALL SIGNAL OPERATION

2 INTRODUCTION ☺One of the primary uses of a transistor is to amplify ac signals. ☺This could be an audio signal or perhaps some high frequency radio signal. ☺It has to be able to do this without distorting the original input. ☺Most transistors amplifiers are designed to operate in the linear region ☺The common-emitter amplifier has high voltage and current gain

3 Definition of Small Signal ac input signal voltages and currents are in the order ±10% of Q-point voltages and currents. eg –If dc current is 10mA, the ac current (peak to peak) is less than 0.1 mA.

4 HOW BJT AMPLIFIES? When a weak ac signal is given to the base, a small ac base current start flowing. Due to BJT, a much larger ac current flow through R C Therefore a large voltage appear across the collector circuit.

5 R 1, R 2, R E – form the biasing circuit & stabilization. C 1 (Coupling capacitor) – couple the signal to the base. C 2 (Bypass capacitor) - use in parallel with RE to provide low reactance path to amplified ac signal C 3 (Coupling capacitor)- couple the signal to the next stage

6 Circuit Diagram Common-Emitter Biasing(CE)Common-Emitter Biasing(CE) –input = V BE & I B –output= V CE & I C +_ ICICICIC V CE IBIBIBIB VCCVCCVCCVCC b= Common-emitter current gain = = Common-base current gain = The relationships between the two parameters are:

7 Example Given the common-emitter circuit below with I B = 25  A, V CC = 15V and  = 100. Find the ideal collector current.Given the common-emitter circuit below with I B = 25  A, V CC = 15V and  = 100. Find the ideal collector current.  = 100 = I C /I B  = 100 = I C /I B I C = 100 * I B = 100 * (25x10 -6 A) = 2.5 mA

8 Common-Emitter V CE Collector-Current Curves ICICICIC Active Region IBIBIBIB Saturation Region Cutoff Region I B = 0 Region of Operation Description ActiveSmall base current controls a large collector current SaturationV CE(sat) ~ 0.2V, V CE increases with I C CutoffAchieved by reducing I B to 0, Ideally, I C will also equal 0.

9 DC Analysis of Transistor Circuits Common-Emitter Circuit Common-emitter circuit with an npn transistor Assume B-E junction: forward biased  V drop is the cut-in / turn-on V [V BE (on)] I C represented as a dependent I source (function of I B ) Neglect reverse-biased junction leakage current & Early effect

10 dc equivalent circuit, with piecewise linear parameters

11 Base-emitter Junction Characteristics And The Input Load Line

12 Common- Emitter Transistor Characteristics And The Collector-emitter Load Line

13 Common-emitter circuit Load Line

14 Kirchoff’s voltage law equation (around B-E loop): Both load line & quiescent I B change as either or both V BB & R B change Kirchoff’s voltage law equation (around C-E loop): I C & V CE relationship represents DC load line

15 Amplifier Operation The boundary between cutoff and saturation is called the linear region. A transistor which operates in the linear region is called a linear amplifier. Only the ac component reaches the load because of the capacitive coupling and that the output is 180º out of input phase.

16 Example Assume Q-point value of i B =10  A, ac signal peak value is 5  A and β=100. So at any instant, i C =100i B iBiB i C =100i B V CE =V CC -i C R C Output voltage 15  A1.5mA4 VPositive peak value 10  A1.0mA6 VSignal current zero 5A5A0.5mA8 VNegative peak value

17 dc equivalent circuit of transistor amplifier Only dc condition are to be considered No signal is applied (All ac source reduce to zero) dc cannot flow through a capacitor (open circuit) To calculate the dc currents & voltages

18 DC Equivalent for the BJT Amplifier All capacitors in the original amplifier circuit are replaced by open circuits, disconnecting v I, R I, and R 3 from the circuit and leaving R E intact. The the transistor Q will be replaced by its DC model. DC Equivalent Circuit

19 ac equivalent circuit of transistor amplifier Only ac condition are to be considered dc voltage not important, considered zero or ground. (V CC =0) the capacitors used to couple or bypass the ac signal. (short circuit)

20 AC Equivalent for the BJT Amplifier The coupling and bypass capacitors are replaced by short circuits. The DC voltage supplies are replaced with short circuits, which in this case connect to ground.

21 (continued) By combining parallel resistors into equivalent R B and R, the equivalent AC circuit above is constructed. Here, the transistor will be replaced by its equivalent small-signal AC model (to be developed).

22 RCRC RBRB vsvs vOvO v ce v be icic ibib + + - - AC equivalent circuit of C-E with npn transistor Graphical Analysis & AC Equivalent Circuit Equations –Input loop –Output loop 0.026 V

23 Small-signal hybrid-  equivalent circuit Using transconductance (g m ) parameter g m =I CQ /V T r  =V T /I CQ v be = i b r π r π = diffusion resistance / base-emitter input resistance 1/r π = slope of i B – V BE curve

24 Using common-emitter current gain (β) parameter Small-signal hybrid-  equivalent circuit

25 Small-signal equivalent circuit rr VsVs RBRB RCRC VoVo IcIc IbIb g m V be V be V ce + + -- Small-signal hybrid-  equivalent circuit Output signal voltage Input signal voltage

26 RCRC RBRB vsvs vOvO V BB VCCVCC Given :  = 100, V CC = 12V V BE = 0.7V, R C = 6k , R B = 50k , and V BB = 1.2V Calculate the small-signal voltage gain. Example

27 Solutions 1. 2. 3. 4. 5. 6.

28 Hybrid-  Model and Early Effect transconductance parameter current gain parameter r o =V A /I CQ r o = small-signal transistor output resistance V A = early voltage

29 Early Voltage (V A ) Hybrid-  Model and Early Effect Early Voltage (pg 109)

30 Example vsvs RSRS R1R1 R2R2 RCRC C vOvO V CC Given :  = 100, V CC = 12V V BE(on) = 0.7V, R S = 0.5k , R C = 6k , R 1 = 93.7k , R 2 = 6.3k  and V A = 100V. Calculate the small-signal voltage gain.

31 R 1 \\ R 2 VsVs RSRS RCRC rOrO rr gmVgmV VoVo RiRi RoRo Solution

32 Exercises Neamen book: Ch. 4 1.Question 4.1 (pg. 177) 2.Question 4.3 (pg. 179)


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