1. Small Signal – Low Frequency Transistor amplifier Circuits UNIT-IV

2. contents  Transistor as an Amplifier  Analysis of Transistor Amplifier Circuits using h – parameters  Miller’s theorem and it’s dual  Simplified CE and CC hybrid model  The CE amplifier with emitter resistance  Darlington pair  Analysis of Single Stage Amplifiers

3. Hybrid Parameter Model Linear Two port Device ViViViVi IiIiIiIi IoIoIoIo VoVoVoVo

4. h-Parameters h 11 = h i = Input Resistance h 12 = h r = Reverse Transfer Voltage Ratio h 21 = h f = Forward Transfer Current Ratio h 22 = h o = Output Admittance

5. Use of h – parameters to describe a transistor have the following advantages.  h – parameters are real numbers up to radio frequencies  They are easy to measure  They can be determined from the transistor static characteristics curves.  They are convenient to use in circuit analysis and design.  Easily convert able from one configuration to other  Readily supplied by manufactories

6. Analysis of CE amplifier using hybrid model h re v ce - VSVS

7.  For finding h parameters quantities V be and i c are taken independent variables  For finding voltage gain, current gain, input impedance, output impedance h parameter equations are v be = h ie i b + h re v ce -------------------------------- (1) i c = h fe i b + h oe v ce---------------------------------- (2)  Putting v ce =0 in above equations we will get h ie and h fe, i b =0 in above equations we will get h re and h oe  current gain

8.  Apply current divider rule to the output circuit  To find input resistance Applying KVL to input circuit V s = h ie i b + h re v ce

9. V s = i b h ie + h re i L R L (v ce = i L R L ) V s = i b h ie + h re A i i b R L ( i L =A i i b ) Substituting in equation (2) R i = h ie + h re A i R L h re v ce v ce 1/h oe h fe i b RiRi i b h ie + V S - R L E C -

10.  To find voltage gain Av = since

11. R L + v ce - 1/h oe h fe i b h re v ce i b h ie RO1RO1 RORO  To find output resistance  Applying KC L to the output circuit. i C = h fe i b + i 1 i C = h fe i b + v ce h oe  Applying KVL to input circuit - ( h ie i b + h re v ce ) =0 substituting for i b in equation substituting in equation (Since R L is in parallel with the voltage source, total output resistance is the parallel combination of R L and R O )  To find output resistance with R L R O 1 = R O ||R L

12. PROBLEM 1.A common emitter amplifier has the following h- parameters. h ie =1KΩ, h re = 10 -4, h fe =100, hoe = 12µmho Find current gain, Voltage gain, R i, R o, power gain. Take R L = 2KΩ. Also find output power take v S = 500 mV ( rms).

14.  MILLER’S THEOREM Miller’s theorem states that when a resistance ( or capacitance ) is connected across input and output terminals, the same can be replaced by two independent resistances ( or capacitances ) connected one across the input terminals and the other across output terminals. These are called Miller equivalent resistances ( or capacitances). R VSVS - R L + v S 1 - RS RS Amplifier A VOVO + i PROOF: Applying Ohm’s law Substituting in (1)

15. ...  comparing equations (1) and (2), the same current i will flow through a resistance when it is between input terminal and ground. Therefore, the KCL at the input terminal is not affected by replacing the resisitance R by is called Miller equivalent resistance on the input side.  To find Miller equivalent resistance on the output side. R - R L + v S 1 - RS RS Amplifier A VOVO + i is the Miller equivalent resistance on the output side  The same current I will flow through the resistance if R is replaced by which is connectedacross output terminal and ground.

16.  Replacing R by Miller equivalent resistances, the circuit is as follows. i R MI + - R MO R L + v S 1 - RS RS Amplifier A i + - VoVo + -  Dual of Miller’s Theorem  Consider a general amplifier as shown below. In this amplifier, the resistance R is common to the input and output circuits do not have a common resistance.  The purpose of the following analysis is to remove the inter dependence of input and output circuits. So that either input circuits or output circuits can be solved independently Applying KVL to input circuit But i 2 = -i L

17. - + i1i1 ViVi - + i2i2 v O 1 + R L - + R RSRS Amplifier vO vO iLiL VSVS - From the above equation, is the resistance which when connected in series with the input circuit instead of R will not affect the input circuit.  Repeating the above analysis for the output circuit. From the equation, the resistance  when connected in the output circuit alone does not alter the KVL equation.

18.  Thus, without affecting the electrical nature of the circuit, the circuit can be redrawn a i1i1 R L R MO = R( Ai – 1) Ai R MI = R( 1-A i ) RSRS Amplifier i L VOVO + -  Approximate h-parameter equivalent circuit.  h re is the reverse voltage gain or reciprocal of voltage gain. For an amplifier, voltage gain is high therefore h re is very low for approximation take h re Vce is zero or voltage source is replaced by short circuit  In h parameter equivalent is internal resistance of the current source h fe i b for ideal condition it is infinity so it is open circuit

19.  Darlington pair Allow for much greater gain in a circuit β = β 1 * β 2