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ELE 1110D Lecture review Common-emitter amplifier

ELE 1110D Lecture review Common-emitter amplifier. Some functions of transistors Current-source Emitter Follower Common-emitter amplifier. ELE 1110D Lecture review Common-emitter amplifier. Vout is biased at middle of V+ AC Gain: G = - R c / R E. (V+)/2 + G*0.1. (V+)/2. 0.1. 0.

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ELE 1110D Lecture review Common-emitter amplifier

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  1. ELE 1110D Lecture reviewCommon-emitter amplifier • Some functions of transistors • Current-source • Emitter Follower • Common-emitter amplifier

  2. ELE 1110D Lecture reviewCommon-emitter amplifier • Vout is biased at middle of V+ • AC Gain: G = - Rc / RE (V+)/2 + G*0.1 (V+)/2 0.1 0 -0.1 (V+)/2 - G*0.1

  3. re=50 re=25 re=12.5 ELE 1110D Lecture reviewCommon-emitter amplifier • Intrinsic emitter resistance • Gain: G = -Rc / re = -gmRc • Gain is varied with VBE !!! • Add RE to reduce the effect of re

  4. ELE 1110D Lecture reviewCommon-emitter amplifier • Temperature effect • If it is heated up • Ic increase, then VBE increase • VBE increase, then Ic increase further • Ic increase will cause further heat up • Add RE for compensation • Ic increase will not cause VBE decrease • VBE decrease drive Ic decrease • High AC gain by adding a parallel Capacitor

  5. ELE 1110D Lecture reviewCurrent Mirror • Act as programmable current source • Q1 and Q2 are twins • VBE are the same • Ic1 = Ic2 • Temperature effect • Temperature increase, cause Ic1 increase • Voltage drop across 15K resistor increase • VBE decrease, drive the Ic1 to decrease

  6. ELE 1110D Lecture review Differential Amplifier • A differential amplifier is a two-input device. • Only amplify the difference in signal. • By-pass the common signal. • For example: • Input signal : 5V, 3V • Common signal : 4V, 4V • Differential signal : 1V, -1V • Ideal differential amplifier: • Large differential gain. • Zero common mode gain.

  7. ELE 1110D Lecture review Differential Amplifier • Consider no input signals i.e. DC quiescent point • VA = -0.6V. • ITAIL = (-0.6 – (-15)) / 7.5k = 2mA • ITAIL is quite constant • Change in VA is small • Large negative voltage • Simple current source • By symmetry, IC on both transistor = 1mA.

  8. ELE 1110D Lecture review Differential Amplifier • Consider only the differentialsignals • For example, • dVIN/2 = 1V • Assume the simple current source has large impedance (open circuit) • Current across the branch = 2 / (100 + 100) = 10mA • VA = 1- 10mA*100 = 0V • In general, • Due to symmetric of 2 RE, • VA = 0V. • Therefore, practically, the differential signals have no effect at A.

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