An intuitive approach to small signal analysis
Understanding and designing analog circuits requires a good grasp of small signal analysis methods. Small signal analysis is used so frequently in analog circuits that we hope to find a faster way to complete circuit performance analysis. In CMOS analog circuits, there is a very simple small signal analysis method, which we call the intuitive analysis method. This method is based on the circuit diagram of the CMOS circuit and does not require drawing a small signal model. It superimposes AC changes onto DC variables. This technique identifies the transistor that converts input voltage into current. We call these tubes transconductance transistors. The current generated by the transconductance transistor flows through the resistor to AC ground. Multiplying the current by this resistance gives you the voltage at this node. This method is both fast and can be used to check small signal analysis using small signal models.
Let us use this method to analyze the differential amplifier in Figure 5.2-5.
Figure 5.2-11 is a redraw of the differential amplifier shown in Figure 5.2-5 with the AC voltage and current determined. NOTE: AC current can flow against DC. This means that the current actually decreases rather than changes direction.
VDDM3M4 is shown in Figure 5.2-11 and works as a differential mode. We see that the currents of M1 and M2 are 0.5gm1νd and 0.5g2νM respectively.
If we still remember a few knowledge points we have learned, the small signal observation analysis method described above is very useful. These knowledge points are: the small signal output resistance of the cascode structure is approximately equal to the r of the common source transistor multiplied by the gm /m of the common gate transistor. This relationship can be expressed as: row (cascode) ≈ rω (common source) × 8crta. (common gate) (5.2-34) In addition to this relationship, it will be very useful to examine the conditions in Figure 5.2-10.
The source of the mid-transconductance transistor has a resistor connected to ground. Under this condition, we can use equation (5.2-26) to express the effective transconductance gm (effective), that is: where gn (41 where) = gn1 + gnR (5.2-35)
g In the formula, g is the transconductance of the tube, and R is the small signal resistance from source to ground. For equation (5.2-26), R=2Tas, gm=gm; using equation (5.2-34) and equation (5.2-35), designers will be able to analyze almost all circuits in the back of this textbook with the help of intuitive analysis. However, this method is not suitable for the analysis of small signal frequency response, and only some simple conclusions can be applied (typically, the pole of the MOSFET circuit is equal to the reciprocal of the product of the AC resistance and capacitance from the node to the AC ground).
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