Transistors as Amplifiers
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1 Transistors as Amplifiers
2 The transistor works in the active region (a F ) around the quiescent point QP dc supply (dc voltage sources, dc current sources) asic amplifier with one transistor: S and amplifiers G and amplifiers and amplifiers mitter degeneration amplifier mportant features: voltage (current) gain input resistance output resistance frequency bandwidth 2/19
3 Operation of one transistor amplifier (S, ) dc biasing: QP in the middle of the active region dc supply set the QP: ( O, O ) v i input voltage (to be amplified) v o output voltage (amplified voltage) superposition of the variable signal over the dc regime 3/19
4 Operation of the amplifier (S, ) Who is responsible for the gain? 4/19
5 oltage transfer characteristic v O (v ), inverting amplifier Small signal: operation of the amplifier in the narrow linear region around the QP Maximum swing of the input signal: often determined based on linearity considerations 5/19
6 biasing setting the OP Operation of the transistor as amplifier: the transistor biased as close as possible to the middle of the a F the instantaneous operating point kept in the active region keep small the input variable signal (linear region around OP) OP: stabile and predictibile independent on the transistor parameters 6/19
7 MOSFT biasing 1 st variant G1 β ( ) Th G2 + G2 2 S simple the current in QP,, depends on the transistor parameters, β and Th cannot assure the stability of the quiescent point. 7/19
8 MOSFT biasing 2 nd variant GG G1 G2 + G2 GG S β ( Th ) 2 ( + ) S S unknown: and 2 nd degree equations system. choose the suitable value. 8/19
9 MOSFT biasing 2 nd variant GG S β ( Th ) 2 depends also on the drain current, S,, the circuit withstands to the variation tendency of negative feedback due to S ensure the QP stability for variation of certain parameters increases the complexity of computational relations 9/19
10 Numerical example 1 G1 3MΩ; G2 1MΩ; 3KΩ; S 1KΩ; 20 Th 2; β 0,5mA/ 2. GG? What is the OP? G G1 G2 β ( Th ) 2 GG 5 S S ( + S ) 201,35(3+ 1) 14,6? S? ; in ma 1 6,65mA and 2 1,35mA Q(14,6; 1,35mA) 1 is not suitable; results <0 2 1,35mA 10/19
11 Numerical example 2 MOSFT: Th 2; β 0,25mA/ 2 ; 20? hoose the resistances to obtain 1mA in the OP. GG β ( ) Th Th + β Ssat - Th ,25 4 T- active region S (2; 20). Q : S 9 S 209 S ( + S ) + S 11KΩ 1 11/19
12 Numerical example 2 - cont. MOSFT: Th 2; β 0,25mA/ 2? hoose the resistances to obtain 1mA in the QP, S also sets the gain. For now we can consider S 4 across S : S S KΩ 7KΩ GG + S G 1 300KΩ; G 2 200KΩ 12/19
13 MOSFT biasing 3 rd variant Usual in integrated circuits: biasing with current sources independent of the amplifier transistor parameters S + GG oltage across the current source: GG - + S 13/19
14 JT biasing, usual variant in discrete circuits oppositely to MOSFT, for JT appears: - base current different from zero - through collector and emitter do not flow exactly the same current + β ( β + 1) β + 1 β One can approximate precise calculation: make use of approximate calculation: neglecting compared to the current through the resistive voltage divider in the base of the transistor 14/19
15 Approximate calculation Al ( + ) much smaller than the current flowing through the base divider is very important for setting and stabilization the QP, by the negative feedback mechanism ; ; ; ; 15/19
16 Precise calculation Thevenin theorem:, (β+1) /( β+1) + ( + ) insensitive to β variations: >> > 10 ( β +1) β 1, 2 small values for the independence of OP on β 1, 2 high values for the high input resistance insensitive to temperature variations ( ) >> 0.1 a variation of 0,1 can be neglected vis a vis the /19
17 Numerical example 3 15; 1 10kΩ; 2 4,7kΩ; 1,5kΩ; 1,8kΩ; β 150 Approximate calculation???? 2,73mA 6 10,1 4,1 xact calculation? 2,7mA 17/19
18 Numerical example 4 Usually we choose: 1 3 Al (1/3) KΩ µA β < KΩ 1 2 2? The values of the resistances so that T in a 2mA? 12, β100 erification tion: + /( β+ 1) 2 18KΩ; 1 36KΩ mA /(100+ 1) 18/19
19 JT biasing, differential supply 2 + /(β+ 1) + β +1 + β+1 oltage across the current source: 19/19
Biasing. Biasing: The DC voltages applied to a transistor in order to turn it on so that it can amplify the AC signal.
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