UNIVERSITY OF CALIFORNIA AT BERKELEY College of Engineering Department of Electrical Engineering and Computer Sciences Discussion Notes 7 EE 105 Prof Wu 1 1 1 Spring 2007 Frequency Analysis of a Common Source Amplifier Motivation So far when discussing gain we ve been talking about DC gain That means if we slowly apply a small signal gm ro rD for vin we know how to get vout based on the equations we ve derived so far such as vvout in a common source amplifier However when designing amplifiers we almost always are dealing with signals with high frequencies Consider a typical audio amplifier The human ear has a range of hearing from about 20Hz to 20kHz When designing an audio amplifier we d like the gain to be constant across this range of frequencies Otherwise when listening to a piece of music we d have some notes being amplified much more than others leading to poor sounding music If all notes are amplified equally approximately then our source material will come out as it originally was just much louder than before which is exactly what we want from an audio amplifier though we may use an equalizer to adjust the exact frequency response of our amplification system Thus we have to worry about the poles and zeroes in an amplifier to ensure the frequency response meets the specifications we need for whatever the amplifier is being built for such as audio signals 1 2 Derivation Consider the small signal model of a common source amplifier shown in figure 11 19 c of Razavi the one on the right the left one is for the bipolar equivalent Although the derivation on page 519 uses the generic terms in Figure 11 19 d I m going to use the MOS specific terms in this document since at this point we re only concerned with MOS transistors and it would also be pointless to re copy the derivation from page 519 if I didn t change something Also note that we could replace RL with ro RL at any point without changing the validity of the solution we re leaving out ro for brevity you could also assume that ro RL to justify this choice as well Let s start by writing KCL at the gate and at the output node This gives the following equations VG VG Vin VG Vout 0 ZGD ZGS RS Vout Vout Vout VG gm VG 0 ZGD ZDB RL 1 2 Note that VGS VG since the source is grounded We can simplify these equations by noting that 1 where s represents the complex number s j if you want to think of it in terms of what ZC sC you learned in EE40 just treat s as if s j If we plug this into our equations we get VG Vin 0 RS 3 Vout gm V G 0 RL 4 VG Vout sCGD VG sCGS Vout VG sCGD Vout sCDB 1 We now have a system of two equations with two unknowns VG and start by solving Equation 4 for VG Vout Vin 1 Vout s CGD CDB VG sCGD gm RL VG Vout Let s solve this system We ll s CGD CDB 5 1 RL 6 sCGD gm Let s also separate VG to one side of Equation 3 1 Vin VG s CGD CGS Vout s CGD CGS RS RS 7 Plugging the result from Equation 6 into Equation 7 gives Vout s CGD CDB sCGD gm 1 RL 1 Vin s CGD CGS Vout s CGD CGS RS RS 8 Dividing both sides by Vin gives 1 1 Vout Vout s CGD CDB RL s CGD CGS s CGD CGS Vin sCGD gm RS Vin 1 Vout s CGD CDB RL 1 s CGD CGS s CGD CGS Vin sCGD gm RS 1 RS 9 1 RS 10 Getting a common denominator on the left gives 1 Vout Vin sCGD gm 1 s CGD CDB RL 1 s CGD CGS RS 2 1 RS 11 s CGD CGS CGD gm Now we are in a position to solve for Vout Vin sCGD gm RL Vout Vin as2 bs 1 a RS RL CGS CGD CDB CGD CGS CDB b 1 gm RL CGD RS RS CGS RL CGD CDB 12 13 14 I ve left out some messy algebra that Razavi skips over as well It should be obvious how to get the solution above though actually solving for a and b requires multiplying out the terms of Equation 11 and matching a with the s2 coefficients and b with the s coefficients after dividing by the constant term to ensure it is 1 At this point we only care about analyzing the result so let s ignore any further algebra m First we can note there is a zero at z CgGD Razavi mentions that this is usually at a very high frequency out of the range that would affect our frequency response Consider if gm 10 3 and CGD 10 15 then z 1012 or in the terahertz range which is very high frequency 2 1 3 Finding the Poles Since the denominator of the transfer function is quadratic in s we know we will have 2 poles Let s call them s s 2 p1 and p2 We should be able to factor the denominator such that as bs 1 p1 1 p2 1 Now consider if one pole dominates that is assume p1 p2 Then we can further simplify this expression to 2 as bs 1 b a s s 1 1 p1 p2 s s s2 1 p1 p2 p1 p2 s s2 1 p1 p2 p1 1 p1 b 1 p1 p2 p2 15 16 17 18 19 Now we can solve for the dominant pole p1 and secondary pole p2 p1 1 b 20 1 1 gm RL CGD RS RS CGS RL CGD CDB b a 1 gm RL CGD RS RS CGS RL CGD CDB RS RL CGS CGD CDB CGD CGS CDB p2 2 21 22 23 Example 11 9 This example is already solved in the book but it s useful to go over it anyway We just need to find how CGD CGS CDB and RL are changed by the new circuit topology Drawing out all of the parasitic capacitances gives the circuit in Figure 11 21 b Razavi asks why CSB1 CGS2 and CSB2 don t come into play the answer the sources are all connected to the bodies so CSB1 and CSB2 are gone CGS2 is attached between two DC voltages Vb and VDD which both go to AC ground eliminating that capacitor as well Looking at Figure 11 21 b we can see how the capacitances are transformed CGS represented the capacitance from the input to ground so that remains as CGS1 CGD represented the capacitance from input to output which in this circuit is just CGD1 CDB was the capacitance from the output to ground which now has 3 parallel capacitances so CDB is replaced by CDB1 CDB2 CGD2 We can summarize this as follows CGS CGS1 CGD CGD1 CDB CDB1 CDB2 CGD2 We also …
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