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1、 Current Sources Structure and Typical application Differential Amplifiers Structure of Differential Pairs DC Analysis AC Analysis Common-Mode Rejection Ratio Multistage Amplifiers Coupling Modes DC Analysis AC Analysis Application of Multistage AmplifiersCurrent Source circuits can provide constant
2、 current in Integrated Circuits.EXEBEEI+2I= I +2I = I +2=ICCBEEXXXV-VI =II=+2RTypical Current CircuitIx is determined by Ucc and Rx .Current Source circuits can be used as the active load of amplifier. DC Analysis:CCBEXXUUIR1211 2CCXIIIThe circuit doesnt need high voltage supply. If the Ucc and Rx a
3、re appropriately matched, the quiescent operation point of the circuit could be acquired.12/ceceouiirruAur AC Analysis:1. A fundamental building block of analog ICs.2. Identical transistor characteristics of the differential-pair3. Two possible inputs and two possible outputs:(1) Single input: If an
4、 input signal is applied to either input with the other input connected to ground, the operation is referred to as “single-input”.(2) Double input: If two opposite-polarity input signals are applied, the operation is referred to as “double-input” or “differential-mode input”.(3) Common-mode: If the
5、same input is applied to both inputs, the operation is call “common-mode input”.4. High gain, high input impedance, and low output impedancei1cd1u =uu2i2cd1u =uu2di1i2u =uuci1i21u =u +u2Differential InputsCommon InputsBasic differential amplifierDC Bias:0V0.7V0.7VEBBEUUU()0.7VEEEEEEEEUUUIRR 1212CCEI
6、II1212CCCCCCCCECUUUI RUI R121212121211221211 2 /2since 22/22iiibbbiib iidiibidioccbcoccbcooobccidioccdidieoccvidirrrIIIVVI rVVVIVrVI RI RVI RI RVVVI RRVrVRRAVrrVRRAVr 22222eoccvidierVRRAVrr1212121111111 0/2/2/222iiibbbib ib ibiiooccc iic ieoccviierrrIIIVI rI rIVrVVI RRVrRVrVRRAVrr AC network of the
7、differential amplifier Using the half-circuit method.AC Analysis of differential mode:1. For the double output situation: 1C,|(2)LLudLLieRRARRRrr The half-circuit voltage gain: For the whole circuit: 12111121122oododododududiiiiiuuuuuAAuuuuu2. For the single output situation: 1112odLudidiuRAur 2212o
8、dLudidiuRAurC|LLRRR Input impedance of differential-mode:Output impedance of differential-mode: 12idididiidbuuZriiC2odZR12CododZZRThe output impedance of double-end output is:The output impedance of single-end output is:AC Analysis of differential mode:210212121ib ibEibiEcioccbciEocciiEVI rI RVIrRRV
9、VI RI RrRVRAVrR AC Analysis of common mode: AC network of common-mode input mode For the double-output: The voltage gain:0ucA For the single-output, using half-circuit: 1121 2ocbLLucicEbiEuIRRAuRIrR 2221 2ocbLLvcicEbiEuIRRAuRIrR C|LLRRR2LERR121vcvcAAUsually therefore, . Any signal that is common to
10、both inputs will be cancelled. A measure of the ability to cancel out common signals is called .dCMRuucAAK(log)20lg()uducCMRAKdBA The output of one amplifier is the input to the next amplifier.The overall voltage gain is determined by the product of gains of the individual stages.uuuuu123nAAAAA1iiZZ
11、nooZZ, ,uuuu123nAAAAare loaded gains.Multistage Amplifiers 1. RC coupling mode:The DC bias circuits are isolated from each other by the coupling capacitorsThe DC calculations are independent of the cascadingThe AC calculations for gain and impedance are interdependentBut big capacitor is difficult t
12、o achieve in IC, RC-coupling mode can hardly used in IC. 2. Direct coupling mode:Direct coupling mode between the former and latter stage is directly or resistor connected, the dc networks of each stage are also connected. The operation point of each stage is influenced by each other and the zero in
13、put drifting is also easily produced.Direct connection coupling mode is the amplifying ability for both ac input signals and the dc signals or signals varying slowly. Dont need the capacitor, which is convenient for the circuit integration. For direct coupling mode:24VCCU1240kBR13.9kCR2500kCR4VZU 14
14、5240. 1120.7V+4V=4.7VCQCEQBQBEQZUUUUUDC Analysis: 111244.75mA3.9CCCCEQRCUUIR111240.70.1mA240CCBEQBQBUUIRFor direct coupling mode:24VCCU1240kBR13.9kCR2500kCR4VZU 145240. 11145 0.1mA=4.5mACQBQII1210.5mACBQRCQIII22240 0.5mA=20mACQBQII22214VOCQCCCQCVVVIR22214V4V=10VCEQCQEQVVVBii11e1Z = Z= R|rInput imped
15、ance, first stage:Output impedance, second stage:Voltage gain:oC2Z = R2Bi2e2Z= R|r211BuC1i2C1e21eeR|ZR|R|rA=rruC22e2RA= -ruuu12A = A AFor RC coupling mode:For RC coupling mode:AC Analysis: From DC Analysis,26mV26mV6.54.0mAeErI212|ieZRRr121|665.2102.36.5CeueRRRrAr 22.2k338.56.5CueRAr 12102.3 ( 338.5)
16、34600uuuAAA 34600 25Vs865mVouiuA u112|=953.6ieZRRr22.2koCZRThis example is a CECB combination. This arrangement provides high input impedance but a low voltage gain.The low voltage gain of the input stage reduces the Miller input capacitance, making this combination suitable for high-frequency applications. Cu
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