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


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 RC Therefore a large voltage appear across the collector circuit.

5 R1, R2, RE – form the biasing circuit & stabilization.
C1(Coupling capacitor) – couple the signal to the base. C2(Bypass capacitor) - use in parallel with RE to provide low reactance path to amplified ac signal C3(Coupling capacitor)- couple the signal to the next stage

6 Circuit Diagram IC VCE IB VCC Common-Emitter Biasing(CE)
input = VBE & IB output = VCE & IC + _ IC VCE IB VCC = Common-emitter current gain = = Common-base current gain = The relationships between the two parameters are:

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

8 Collector-Current Curves
Common-Emitter Collector-Current Curves IC Active Region IB Saturation Region Cutoff Region IB = 0 Region of Operation Description Active Small base current controls a large collector current Saturation VCE(sat) ~ 0.2V, VCE increases with IC Cutoff Achieved by reducing IB to 0, Ideally, IC will also equal 0. VCE

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 [VBE (on)] IC represented as a dependent I source (function of IB) 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 IB change as either or both VBB & RB change Kirchoff’s voltage law equation (around C-E loop): IC & VCE 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. Fig 6-2 amplifier circuit

16 Example Assume Q-point value of iB=10A, ac signal peak value is 5A and β=100. So at any instant, iC =100iB iB iC=100iB VCE=VCC-iCRC Output voltage 15A 1.5mA 4 V Positive peak value 10A 1.0mA 6 V Signal current zero 5A 0.5mA 8 V Negative 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
DC Equivalent Circuit All capacitors in the original amplifier circuit are replaced by open circuits, disconnecting vI, RI, and R3 from the circuit and leaving RE intact. The the transistor Q will be replaced by its DC model.

19 ac equivalent circuit of transistor amplifier
Only ac condition are to be considered dc voltage not important, considered zero or ground. (VCC=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 RB 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 Graphical Analysis & AC Equivalent Circuit
Equations Input loop Output loop RC RB vs vO vce vbe ic ib + - 0.026 V AC equivalent circuit of C-E with npn transistor

23 Small-signal hybrid- equivalent circuit
vbe = ibrπ = diffusion resistance / base-emitter input resistance 1/rπ = slope of iB – VBE curve gm=ICQ/VT r=VT/ICQ Using transconductance (gm) parameter

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

25 Small-signal equivalent circuit
Small-signal hybrid- equivalent circuit Small-signal equivalent circuit r Vs RB RC Vo Ic Ib gmVbe Vbe Vce + - Output signal voltage Input signal voltage

26 Example Given :  = 100, VCC = 12V VBE = 0.7V, RC = 6k,
RB = 50k, and VBB = 1.2V Calculate the small-signal voltage gain. RC RB vs vO VBB VCC

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

28 Hybrid- Model and Early Effect
transconductance parameter ro=VA/ICQ current gain parameter ro = small-signal transistor output resistance VA = early voltage

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

30 Example vs Given :  = 100, VCC = 12V VBE(on) = 0.7V, RS = 0.5k,
RC = 6k, R1 = 93.7k, R2 = 6.3k and VA = 100V. Calculate the small-signal voltage gain. vs RS R1 R2 RC CC vO VCC

31 Solution R1 \\ R2 Vs RS RC rO r gmV Vo Ri Ro

32 Exercises Neamen book: Ch. 4 Question 4.1 (pg. 177)


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