analog bicmos design practices and pitfalls phần 6

Emitter cặp cùng, còn được gọi là bộ khuếch đại sự khác biệt, có thể loại được sử dụng thường xuyên nhất của bộ khuếch đại trong thiết kế mạch tích hợp. Các cặp emitter-coupled cung cấp các đặc điểm khác biệt giữa đầu vào cần thiết cho tất cả các bộ khuếch đại hoạt động.
Nội dung trích xuất từ tài liệu:
analog bicmos design practices and pitfalls phần 6Figure 5.17 Redrawn small signal equivalent circuit for determining theoutput impedance.This current is also equal to −Ie . Current Ix2 flows in the controlledcurrent source, but this current is defined as Vx Ix2 = gm2 v2 = αIe = −α ro2Total current Ix = Ix1 + Ix2 such that Vx Vx (1 − α) ≈ Ix = ro2 βro2and Ro = βro25.8 Emitter-Coupled PairsEmitter-coupled pairs, also known as differential amplifiers, are probablythe most often used type of amplifier in integrated circuit design. Theemitter-coupled pair provides differential input characteristics requiredfor all operational amplifiers. Cascading of sequential stages can be ac-complished without need for impedance matching, and relatively highgains can be realized in a small area of circuitry, especially when com-bined with current mirror active loads. The MOS differential amplifieris called a source-coupled pair.Figure 5.18 Emitter-coupled pair. The schematic representation of the emitter-coupled pair is shown inFigure 5.18. Note that the biasing current source can be a transistorsource (current mirror) or a simple resistor. If a resistor is used, thecurrent source IEE becomes zero. If a transistor is used, the transistorequivalent circuit replaces IEE and the resistor. Let us first consider the large signal transfer characteristic of theemitter-coupled pair shown in Figure 5.18. For simplicity, we will assumethe bias current source output resistance and the output resistances ofQ1 and Q2 are all infinite. This assumption is valid for the large signalanalysis, but not for the small signal analysis. We can also assume thatQ1 and Q2 are identical transistors with the same saturation current IS .If we use Kirchoff’s Voltage Law on the loop containing VI 1 , VI 2 and thebase-emitter junctions of Q1 and Q2, we obtain VI 1 − Vbe1 + Vbe2 − VI 2 = 0Recalling that Vbe = VT ln[IC /IS ], we can rewrite the equation aboveand solve for the ratio of collector currents IC 1 and IC 2 : VI 1 − VI 2 IC 1 = exp IC 2 VTFigure 5.19 Collector currents as a function of the input voltage.Next we sum currents at the node where the emitters of Q1 and Q2 areconnected: IC 1 IC 2 −(IE 1 + IE 2 ) = + = IEE α αWe now solve for the collector currents αIEE IC 1 = Vdif f 1 + exp − VTand αIEE IC 2 = Vdif f 1 + exp VTThese currents are plotted as a function of Vdif f in Figure 5.19. Notethat the currents become independent of Vdif f for values greater than3VT , or about 75 mV . At this point, all of the current IEE is flowingin only one of the transistors. The current change is linear for a regionslightly less than about ±2VT . The output voltages are given as Vo1 = VCC − IC 1 RCand Vo2 = VCC − IC 2 RCHowever, it is often the case that the differential output voltage, Vodif f =Vo1 − Vo2 , is of most interest. Vodif f is plotted against Vdif f in Figure5.20. It is possible to extend the range of linear operation by the addition ofemitter degeneration resistors as shown in Figure 5.21. The linear regionFigure 5.20 Differential pair output voltage as a function of the differentialinput voltage.Figure 5.21 The emitter degeneration resistors, RE , extend the linear rangeof the emitter-coupled pair.is extended by about ±IEE RE , but the voltage gain will be decreasedas a result of adding degeneration. For emitter-coupled pairs, we are most often interested in the smallsignal analysis when the dc differential input voltage is zero. In thiscase, Vdif f represents the ac signal. In analyzing this circuit, we makethe following assumptions: r The magnitude of the input signal Vdif f is small enough that the amplifier operates in the linear region. r The equivalent resistance of the biasing circuitry is finite. r ro for the transistors is much larger than RC and can be ignored in our analysis.It is convenient to define the input signal as a sum of two components,a dc common-mode voltage and an ac differential-mode voltage. Thedifferential-mode signal is defined as the difference between the two in-puts, while the common-mode signal is the average of the two inputs.That is Vid = VI 1 − VI 2 = Vdif fand VI 1 + VI 2 Vic = 2We can redraw our circuit in Figure 5.22 to see the significance of thesedefinitions. We can similarly define differential-mode and common-mode outputsignals vod = vo1 − vo2and vo1 − vo2 voc = 2We can identify vo1 and vo2 in terms of vod and voc vod vo1 = + voc 2and vod vo2 = − + voc ...