March 28, 2017

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By B. W. Allen (auth.)

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20 Hz to 1 MHz) amplifier might be : RLl: 5kQ RB3: 100 kQ RB4: 10 ill hie: 2 kQ C s (input capacitance of T2): 200 pF hfe: 100 Cc : 10 ~ The small signal h-parameter equivalentcircuit is used in order to derive an expression for the voltage gain of a stage of amplification. c, signals that are being considered. Also, although Cs is low in value, typically in the order of a few hundred picofarads, its effect at high frequencies will be significant. 8) shows the physical capacitor Cc, and the stray capacitance that forms C; As already mentioned, C, affects the performance of the amplifier at low frequencies.

It can be clearly seen that the output load resistance is made up of four resistors connected in parallel. 12) . 12 Simplified circuit 33 2. 2314n Note: Remember that the effective resistance of several resistors in parallel is always less than the lowest resistor value. e. , calculate: (a) the gain ofthe amplifier. (b) the stage gain with an identical stage added. e. in high band, the reactance of the input capacitance of the next stage , Cs, begins to fall. ib to affect the performance of the circuit.

JffiCs 1 Rp -:--Jffi C s Zp = --=----"1 Rp+-j co ( PRODUCT) SUM c, Multiply top and bottom of the expression by jwCs Zp= 1--' : L = _ j ro Rp c, Rp +1 hl~ Rp hie1 ( 1 + Jffi Cs Rp ) This expression should be used if high band gain is to be calculated directly from component values. However, if AM has already been calculated then 36 Analogue Electronics for Higher Studies -hee Rp hie! = A M may be used in the expression for AH. 8 IfAM = 100 L180° • RLl = 5 kQ , RB = 10 kQ , hia = 2 kQ and =100 pF, calculate the gain at 10 kHz, 100 kHz, 1 MHz and 10 MHz.

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