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MULTISTAGE AMPLIFIERS Multistage Amplifiers Two or more amplifiers can be connected to increase the gain of an ac signal. The overall gain can be calculated by simply multiplying each gain together. A’v = Av1Av2Av3 …… Introduction Many applications cannot be handle with single- transistor amplifiers in order to meet the specification of a given amplification factor, input resistance and output resistance As a solution – transistor amplifier circuits can be connected in series or cascaded amplifiers This can be done either to increase the overall small- signal voltage gain or provide an overall voltage gain greater than 1 with a very low output resistance Multistage Amplifiers Multi-stage amplifiers are amplifier circuits cascaded to increased gain. We can express gain in decibels(dB). Two or more amplifiers can be connected to increase the gain of an ac signal. The overall gain can be calculated by simply multiplying each gain together. A’v = Av1Av2Av3 …… Example Find Vout Vin = 5 mV Av1 = 5 Av2 = 10 Av3 = 6 Avn = 8 Multistage Amplifier Cutoff Frequencies and Bandwidth When amplifiers having equal cutoff frequencies are cascaded, the cutoff frequencies and bandwidth of the multistage circuit are found using f C 2 (T ) f C 2 21/ n 1 f C1(T ) f C1 21 / n 1 BW f C 2 (T ) f C1(T ) 6 Introduction (cont.) Multistage amplifier configuration: RL Rc1 RB R1 Rc2 Vo Q2 Q2 R2 Vi Q1 Vi Q1 R3 Cascade /RC coupling Vo T Cascode Q2 R1 R2 R1 Vo Vi Q1 Q2 Vi Transformer coupling Q1 Darlington/Direct coupling i) Cascade Connection -The most widely used method -Coupling a signal from one stage to the another stage and block dc voltage from one stage to the another stage -The signal developed across the collector resistor of each stage is coupled into the base of the next stage -The overall gain = product of the individual gain **refer page 219 i) Cascade Connection (cont.) small signal gain is: by determine the voltage gain at stages 1 & stage 2 AV AV 1 AV 2 therefore Vo Ri AV g m1 g m 2 ( RC1 // r 2 )( RC 2 // RL )( ) VS Ri RS - the gain in dB AV ( dB) 20 log( AV ) input resistance Ris R1 // R2 // r 1 Output resistance - assume VS 0 ,so Therefore R0 RC 2 V 1 V 2 0 also gm1V 1 gmV 2 0 Exercise 1: Draw the ac equivalent circuit and calculate the voltage gain, input resistance and output resistance for the cascade BJT amplifier in above Figure. Let the parameters are: VCC 20V , Q1 Q 2 200,VBE (ON ) 0.7V , r0 R1 R3 15k, R2 R4 4.7k, RC1 RC 2 2.2k, RE1 RE 2 1k Solution 1 (cont.): Ri Vi C1 B1 R1 // R2 V 1 r 1 g m1V 1 E1 C2 RC1 RB1 B2 RB 2 V 2 RC 2 r 2 R3 // R4 g m 2V 2 E2 Ac equivalent circuit for cascade amplifier R0 V0 Solution 1: DC analysis: At Q1: I BQ1 19.89A I CQ1 3.979mA At Q2: I BQ 2 19.89A I CQ 2 3.979mA AC analysis: At Q1: r 1 1.307k g m1 0.153S At Q2: r 2 1.307 k g m 2 0.153S Why the Q-point values same for both Q1 & Q2 ? Solution 1 (cont.): From the ac equivalent circuit: At Q1, the voltage gain is: AVQ1 Where and V0Q1 Ri 2 V0Q1 Vi g m ( RC1 // Ri 2 ) is the o/p voltage looking to the Q1 transistor is the i/p resistance looking into Q2 transistor Therefore Ri 2 RB 2 // r 2 15k // 4.7k // 1.307k 957.36 The voltage gain at Q1 is: AVQ1 0.153(2.2k // 957.36) 102.06 Solution 1 (cont.): From the ac equivalent circuit: At Q2, the voltage gain is: AVQ2 V0 g m ( RC 2 ) ViQ2 Where ViQ 2 is the i/p voltage looking into the Q2 transistor Therefore, the voltage gain at Q2 is: AVQ 2 0.153(2.2k ) 336.6 The overall gain is then, AV AVQ1 AVQ 2 (102.06)( 336.6) 34,353 ** The large overall gain can be produced by multistage amplifiers!! So, the main function of cascade stage is to provided the larger overall gain Solution 1 (cont.): From the ac equivalent circuit: The i/p resistance is: Ri R1 // R2 // r 1 15k // 4.7k // 1.307k 957.36 The o/p resistance is: Ro RC 2 2.2k ii) Cascode Connection -A cascode connection has one transistor on top of (in series with) another -The i/p into a C-E amp. (Q1) is, which drives a C-B amp. (Q2) -The o/p signal current of Q1 is the i/p signal of Q2 -The advantage: provide a high i/p impedance with low voltage gain to ensure the i/p Miller capacitance is at a min. with the C-B stage providing good high freq. operation **refer page 223 ii) Cascode Connection (cont.) From the small equivalent circuit, since the capacitors act as short circuit, by KCL equation at E2: V 2 g m 2V 2 r 2 solving for voltage V 2 g m1V 1 r V 2 2 g m1VS 1 2 Where 2 gm 2r 2 the output voltage is Vo ( g m 2V 2 )( RC // RL ) or r 2 RC // RL )VS Vo g m1 g m 2 1 2 ii) Cascode Connection (cont.) Therefore the small signal voltage gain: r 2 V0 RC // RL AV g m1 g m 2 VS 1 2 From above equation shows that: r 2 2 g m 2 1 1 2 1 2 So, the cascode gain is the approximately AV gm1 RC // RL * The gain same as a single-stage C-E amplifier iii) Darlington Connection -The main feature is that the composite transistor acts as a single unit with a current gain that is the product of the current gains of the individual transistors -Provide high current gain than a single BJT -The connection is made using two separate transistors having current gains of and 1 2 So, the current gain D 1 2 If 1 2 The Darlington connection provides a current gain of D 2 Figure 1: Darlington transistor iii) Darlington Connection (cont.) Figure shown a Darlington configuration refer page 222 iii) Darlington Connection (cont.) The small current gain Ai I o / I i : Since V 1 I i r 1 Therefore g m1V 1 g m1r 1 I i 1 I i Then, V 2 ( I i 1 I i )r 2 The o/p current is: I 0 g m1V 1 g m 2V 2 1 I i 2 (1 1 ) I i The overall gain is: I0 Ai 1 2 (1 1 ) 1 2 Ii ** The overall small-signal current gain = the product of the individual current gains iii) Darlington Connection (cont.) The input resistance: Known that: Vi V 1 V 2 I i r 1 I i (1 1 )r 2 So, the i/p resistance is: Ri r 1 (1 1 )r 2 The base of Q2 is connnected to the emitter of Q1, which means that the i/p resistance to Q2 is multiplied by the factor (1 1 ) , as we saw in circuits with emitter resistor. So, we can write: Therefore r 1VT 2VT r 1 1 I CQ 2 I CQ1 and I CQ1 I CQ 2 2 1 r 2 The i/p resistance is then approximately Ri 21r 2 **The i/p resistance tends to be large because of the multiplication