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UK EE 462G - The BJT Differential Amplifier

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II. IntroductionSee 2.1 – 2.6, 8.1, 8.2, and 8.3 in HorensteinEE 462: Laboratory Assignment 9The BJT Differential AmplifierbyDr. A.V. RadunDr. K.D. Donohue (3/27/07)Department of Electrical and Computer EngineeringUniversity of KentuckyLexington, KY 40506Simulation-only Lab # 9 Due at Next lecture periodI. Instructional Objectives- Simulate a simple differential amplifier in SPICE- Determine common mode and differential mode gains- Determine the common mode rejection ratioII. IntroductionSee 2.1 – 2.6, 8.1, 8.2, and 8.3 in HorensteinPrevious experiments have used circuits designed for a discrete circuit implementation. However, most modern circuits are built as integrated circuits on a single silicon chip. Integrated circuits provide lower cost, higher speed, and lower power dissipation. On the other hand there are limitations that an integrated circuit implementation imposes on the types and size of components that can be used. For instance, capacitors above ~50 pF would occupy too much area in an integrated circuit (chip); similarly, large valued resistors require prohibitively large amounts of space. Since transistors occupy little chiparea, they are used liberally in integrated circuits. In addition, transistors on the same chip are well matched with almost identical values of  and other fabrication parameters unlike the case of discrete transistors where it is very difficult to match the BJT’s . Thisfact makes it possible to implement certain multi-transistor circuits in an integrated circuit that would be very difficult to implement with a discrete transistor design. One such circuit is the differential amplifier. Two versions of the differential amplifier are shown in Fig. 1.VCCRCRCVEEREVO2Q2V2+-Q1VO1V1+-(a)VCCRCRCVEEIOVO2Q2V2+-Q1VO1V1+-(b)Fig. 1. Differential amplifiers emitter branch with (a) resistor for negative feedback, and (b) active low impedance bypass element to ground.In order to maintain a stable bias for changing values of , a single transistor amplifier requires significant negative feedback. This was accomplished by using an emitter resistor in the previous lab as shown in Fig. 1a. This same negative feedback reduces the amplifiers gain, so a bypass capacitor was included to short the emitter resistor to ground for the AC signals to be amplified. The use of this bypass capacitor prohibits the effectiveuse of the amplifier at low frequencies. In addition, the large resistor and capacitor values cannot be used in an integrated circuit. A solution to this problem is to add anotheractive device to provide the required low impedance bypass to ground. One example of such a configuration is shown Fig. 1b, where more transistors can be used to act as an effective current source. A properly designed differential amplifier will amplify the difference between the two input signals V1 and V2 (V1 - V2) by a large gain factor and suppress the average or common component of the input signals (V1 + V2)/2. Thus, the differential gain is large and the common mode gain is small, ideally zero. The current source, Io (in Fig. 1b), serves to bias (provide the DC component) both of the transistors. In reality, a current source must be synthesized from various components. The crudest, yet easiest, approximation to a current source is a resistor as shown in Fig. 1a.The two input voltages can be expressed as linear combinations of the differential input Vidm = V1 - V2, and common mode input Vicm = (V1 + V2)/2. These equations constitute a linear transformation from (V1, V2) to (Vidm, Vicm). This transform is invertible so that the above equations can be solved for the input voltages in terms of Vidm, and Vicm. Since for a linear system, superposition holds, the output voltage for any arbitrary combination of V1 and V2 can be found once the voltage gain expressions for a pure differential input and a pure common mode input are known. The output only consists of a differential output when V1 = -V2, which corresponds to Vicm = 0. The output only consists of a common mode output when V1 = V2 corresponding to Vidm = 0.The differential mode gain with respect to the output Vout1 (measured with respect to ground) is defined as Vout1 / Vidm with Vicm = 0V and is called Admse1 (gain differential modesingle ended 1), likewise, the differential mode gain Admse2 (gain differential mode single ended 2) with respect to the output Vout2 is defined as Vout2 / Vidm with Vicm = 0V.The common mode gain with respect to the output Vout1 is defined as Vout1 / Vicm with Vidm = 0V and is called Acmse1 (gain common mode single ended 1), likewise, the common mode gain Acmse2 (gain common mode single ended 2) with respect to the output Vout2 is defined as Vout2 / Vicm with Vidm = 0V. Either output may be found by combining these two gains (using superposition) as follows: 2211211111VVAVVAVAVAV cmedmseicmcmseidmdmseout (1) 2212212222VVAVVAVAVAV cmedmseicmcmseidmdmseout(2)The expressions for the differential and common mode gains can be found using a small signal analysis of the circuit as usual (see Microelectronic Circuits and Devices by Mark Horenstein for sample derivations).The overall differential mode gain for the differential output Vout1-Vout2 is Admdiff =Admse1- Admse2 (gain differential mode differential) and the overall common mode gain is Acmdiff =Acmse1- Acmse2 (gain common mode differential). The purpose of the differential amplifier is to amplify the differential mode component of the input and to suppress the common mode component of the input (make the differential mode gain as high as possible and the common mode gain as small as possible). An important characterizationof the differential amplifier is its common mode rejection ratio defined as CMRR = 20log|Admdiff / Acmdiff |.III. Simulation Exercises1. Choose values for RC and RE so that the bias current through RE is approximately 2.1 mA and both transistor’s VCE is approximately 4.9V (for quiescent analysis set V1 = V2= 0V). Use VCC= - VEE = 15V,  = 200, and VBEf = 0.6V. Show your work.2. Simulate the circuit using SPICE to determine the gain differential mode differential Admdiff , the gain common mode differential Acmdiff, and the CMRR. To model the imperfections of a real circuit, set one transistor  -value to 160 and the other to 240. You may have to make one of transistor


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