tailieunhanh - 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. | Figure Redrawn small signal equivalent circuit for determining the output impedance. Vx x a r o2 Vx fro2 This current is also equal to -Ie. Current Ix2 flows in the controlled current source but this current is defined as Ix2 gm2V2 ale Total current Ix Ix1 Ix2 such that Ix Vx 1 - a r o2 and Ro fro2 Emitter-Coupled Pairs Emitter-coupled pairs also known as differential amplifiers are probably the most often used type of amplifier in integrated circuit design. The emitter-coupled pair provides differential input characteristics required for all operational amplifiers. Cascading of sequential stages can be accomplished without need for impedance matching and relatively high gains can be realized in a small area of circuitry especially when combined with current mirror active loads. The MOS differential amplifier is called a source-coupled pair. Figure Emitter-coupled pair. The schematic representation of the emitter-coupled pair is shown in Figure . Note that the biasing current source can be a transistor source current mirror or a simple resistor. If a resistor is used the current source I EE becomes zero. If a transistor is used the transistor equivalent circuit replaces I EE and the resistor. Let us first consider the large signal transfer characteristic of the emitter-coupled pair shown in Figure . For simplicity we will assume the bias current source output resistance and the output resistances of Q1 and Q2 are all infinite. This assumption is valid for the large signal analysis but not for the small signal analysis. We can also assume that Q1 and Q2 are identical transistors with the same saturation current Is. If we use Kirchoff s Voltage Law on the loop containing V 1 VI2 and the base-emitter junctions of Q1 and Q2 we obtain Vịi - Vbei Vbe2 - Vị2 0 Recalling that Vbe VT ln Ic Is we can rewrite the equation above and solve for the ratio of collector currents IC1 and IC2 Ici Vị 1 - Vị2 Figure Collector currents as a function of the .