By B. W. Allen (auth.)
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Additional resources for Analogue Electronics for Higher Studies
At that point, the output rapidly changes state and goes high. The time delay is determined by the time constant of the C-R circuit. g. the input to a stage of amplification) can be related to the upper cut-off frequency, f2, of an amplifier. 7, the series C-R circuit is often referred to as a Thevenin 's arrangement, and the parallel circuit is known as a Norton's circuit. It is possible to equate the two circuits via Thevenin to Norton transformation, hence the series and parallel circuits are equivalent.
17 Current dividing network 41 2. Cascaded Amplifiers In the amplifier under consideration, ZI = RLI and Z2 = XCc + hie2. ---- i 3 = hfe ib ( Ru + Rin2+ j 1 w ec ) and V3 = - ts Rin2 -hfe lo RL 1 Rin2 ( RL1 + Rin2 + j w1Cc ) To make the final expression simpler, it is helpful if the formula can be expressed in terms of R p ( R p was the term used for all resistances when C c had no effect on the circuit. It is R LI II Rin2 ). Divide top and bottom by (R LI + R in2) . - h fe lb [ As RL1 ( RL1 + Hn2 ) + RL1 Rin2 ( RL1 + Rin2 ) Rin2 + Rin2) (RL1 RLI + Rin2 ( R LI + Rin2 ) ( RLI + Rin2 ) = +.
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. _1 + [s» Cs Rp Rp =RLl . 25/(D. At 10kHz AH = - - - - - - - . : . 55' Note: It is clear that 10 kHz is, in fact, in mid band for this amplifier. At 100 kHz Instead ofre-working the complete expression for AH, a short-cut may be employed. The frequency for this calculation is a factor of 10 larger than for the last, and therefore the imaginary part or 'j' value will also be 10 times larger, hence 2.