CASCADE AMPLIFIER

CASCADE AMPLIFIER

• 1.
  1 GLACEVARGHESET EC-A ROLL NO:45 cascade amplifier
• 2.
  cascade amplifier 2 SIGNIFICANCEOF CASCADE AMPLIFIER ● 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 TO IMPROVE THE GAIN
• 3.
  Multistage Amplifiers 3 Multi-stage amplifiersare 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 …… cascade amplifier
• 4.
  Example 4 Find Vout Av1 =5 Av2 = 10 Av3 = 6 Avn = 8 Vin = 5 mV cascade amplifier
• 5.
  cascade amplifier 5 Acascade amplifier is any two-port network constructed from a series of amplifiers, where each amplifier sends its output to the input of the next amplifier in a chain. The complication in calculating the gain of cascaded stages is the non-ideal coupling between stages due to loading. CONSTRUCTION
• 6.
  cascade amplifier 6
• 7.
  Cascade Connection  smallsignal gain is: by determine the voltage gain at stages 1 & stage 2 therefore - the gain in dB  input resistance ))(//)(//( 22121 SIN IN LCCmm S o V RZ Z RRrRgg V V A + == π 121 //// πrRRZIN = 21 VVV AAA = )log(20)( VdBV AA = 7 cascade amplifier
• 8.
  cascade amplifier 8 ●Output resistance assume 0=SV ,so 021 == ππ VV also 0211 == ππ VgVg mm Therefore LCOUT IIRRZ 2=
• 9.
  Exercise 1: Draw theac equivalent circuit and calculate the voltage gain, input resistance and output resistance for the cascade BJT amplifier in above Figure. Let the parameters are: Ω==Ω==Ω==Ω== kRRkRRkRRkRR EECC 1,2.2,7.4,15 21214231 ∞===== 0)(21 ,7.0,200,20 rVVVV ONBEQQCC ββ 9cascade amplifier
• 10.
  Solution 1 (cont.): Acequivalent circuit for cascade amplifier 10cascade amplifier
• 11.
  Solution 1: DC analysis: AtQ1: At Q2: AC analysis: At Q1: At Q2: Sg kr m 153.0 307.1 1 1 = Ω=π mAI AI CQ BQ 979.3 89.19 1 1 = = µ Sg kr m 153.0 307.1 2 2 = Ω=π mAI AI CQ BQ 979.3 89.19 2 2 = = µ 11cascade amplifier
• 12.
  Solution 1 (cont.): Fromthe ac equivalent circuit: At Q1, the voltage gain is: Where is the o/p voltage looking to the Q1 transistor and is the i/p resistance looking into Q2 transistor Therefore The voltage gain at Q1 is: 2iR 10QV )//( 21 10 1 iCm Q VQ RRg Vi V A −== 06.102)36.957//2.2(153.01 −=−= kAVQ Ω=== 36.957307.1//7.4//15// 222 kkkrRR Bi π 12cascade amplifier
• 13.
  Solution 1 (cont.): Fromthe ac equivalent circuit: At Q2, the voltage gain is: Where is the i/p voltage looking into the Q2 transistor Therefore, the voltage gain at Q2 is: The overall gain is then, ** The large overall gain can be produced by multistage amplifiers!! So, the main function of cascade stage is to provided the larger overall gain 2iQV )( 2 2 0 2 Cm iQ VQ Rg V V A −== 6.336)2.2(153.02 −=−= kAVQ 353,34)6.336)(06.102(21 =−−== VQVQV AAA 13cascade amplifier
• 14.
  Solution 1 (cont.): Fromthe ac equivalent circuit: The i/p resistance is: The o/p resistance is: Ω= == 36.957 307.1//7.4//15//// 121 kkkrRRRi π Ω== kRR Co 2.22 14cascade amplifier
• 15.
  THANK YOU 15cascade amplifier