Lecture 7 - 1 Lecture 7 Introduction to electronic analog...

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Lecture 7: Introduction to electronic analog circuits 361-1-3661 1 5.4. Differential amplifier Our aim is to find a new circuit that is able to amplify a difference between two voltage signals, each relative to the ground (see Fig. 1). (Note that the elementary single-transistor amplifiers have only one input and are unable to amplify the difference between two voltage signals, each relative to the ground). We will call such a circuit differential amplifier. We will define (see Fig. 1) the small-signal voltages at the amplifier inputs as v a and v b , the differential small-signal voltage as v ε = v a v b (1) and the common-mode small-signal voltage as 2 b a cm v v v + = . (2) An ideal differential amplifier should amplify only the differential signal v . Practical amplifiers, however, have also an undesirable common-mode voltage gain. The ratio of the small-signal differential voltage gain, A v , and the small-signal common-mode voltage gain, A vcm , is used to describe the capability of a differential amplifier not to amplify common- mode signals. We will refer to this ratio, measured in dB, as the common-mode rejection ratio vcm v A A CMRR 10 log 20 = . (3) It will also be important for us to improve the linearity of the mutual transconductance gain, g m , of the new circuit compared to that of the elementary CE amplifier. Since we need two high impedance inputs, and we use the double-transistor configuration shown in Fig. 2. We will see below, that the architecture of the differential amplifier is based on its symmetry; hence, we chose identical transistors and define their static state with the help of a current mirror. Static-state The static state of the differential amplifier in Fig. 2 is defined by the static current source I N , R N =r oS , representing the slave transistor of a current mirror circuit. The static voltages of the transistor base terminals are zero, and their base-to emitter voltages approximately equal 0.7 V. A v v O v a v b v o =A v ( v a v b ) v v a v b v cm 0 V A v v O v a v b v o =A v ( v a v b ) v v ε /2 v cm v /2 v cm v /2 v /2 Fig. 1. Differential amplifier. Therefore, the voltage drop on the static current source I N , r oS is V CC + 0.7. Having this voltage and r oS , one can easily find I N that provides the needed currents of the transistor emitters. Amplified by α F , the emitter currents give us the collector currents. Amplified by R C , the collector currents give us the voltage drop on the load resistors. Subtracted from V CC , the voltage drop on the load resistors gives us the collector potentials V C . Adding 0.7 V, gives us the collector-to emitter voltages V CE . Divided by β F , for a given V CE
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Lecture 7 - 1 Lecture 7 Introduction to electronic analog...

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