UNIVERSITY OF CALIFORNIA AT BERKELEYCollege of EngineeringDepartment of Electrical Engineering and Computer SciencesDiscussion Notes #7EE 105 Spring 2007Prof. Wu1 Frequency Analysis of a Common Source Amplifier1.1 MotivationSo far, when discussing gain, we’ve been talking about DC gain. That means if we slowly apply a small signalvin, we know how to get voutbased on the equations we’ve der ived so far, such asvoutvin= −gm(ro||rD) fora common source amplifier. However, when designing amplifiers, we almost always are dealing with signalswith 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 materialwill come out as it originally was, just much louder than before, which is exactly what we want from anaudio amplifier (though we may use an equalizer to adjust the exact frequency response of our amplificationsystem).Thus, we have to worry about the poles and zeroes in an amplifier to ensure the frequency response meetsthe specifications we need for whatever the amplifier is being built for, such as audio signals.1.2 DerivationConsider the small signal model of a common source amplifier shown in figure 11.19(c) of Razavi (the one onthe right—the left one is for the bipolar equivalent). Although the derivation on page 519 uses the genericterms in Figure 11.19(d), I’m going to use the MOS-specific terms in this document since at this point, we’reonly concerned with MOS transistors (and it would also be pointless to re-copy the derivation from page519 if I didn’t change something). Also note that we could replace RLwith (ro||RL) at any point withoutchanging the validity of the solution—we’re leaving out rofor brevity (you could also assume that ro>> RLto 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− VoutZGD+VGZGS+VG− VinRS= 0 (1)Vout− VGZGD+VoutZDB+VoutRL+ gmVG= 0 (2)Note that VGS= VGsince the source is grounded. We can simplify these equations by noting thatZC=1sC, where s represents the complex number s = σ + jω (if you want to think of it in terms of whatyou learned in EE40, just treat s as if s = jω). If we plug this into our equations, we get:(VG− Vout) sCGD+ VGsCGS+VG− VinRS= 0 (3)(Vout− VG) sCGD+ VoutsCDB+VoutRL+ gmVG= 0 (4)1We now have a system of two equations with two unknowns: VGandVoutVin. Let’s solve this system. We’llstart by solving Equation (4) for VG:Vouts (CGD+ CDB) +1RL= VG(sCGD− gm) (5)⇒ VG= Vouts (CGD+ CDB) +1RLsCGD− gm(6)Let’s also separate VGto one side of Equation (3):VGs (CGD+ CGS) +1RS= Vouts (CGD+ CGS) +VinRS(7)Plugging the result from Equation (6) into Equation (7) gives:Vouts (CGD+ CDB) +1RLsCGD− gms (CGD+ CGS) +1RS= Vouts (CGD+ CGS) +VinRS(8)Dividing both sides by Vingives:VoutVins (CGD+ CDB) +1RLsCGD− gms (CGD+ CGS) +1RS=VoutVins (CGD+ CGS) +1RS(9)VoutVin"s (CGD+ CDB) +1RLsCGD− gms (CGD+ CGS) +1RS− s (CGD+ CGS)#=1RS(10)Getting a common denominator on the left gives:VoutVin1sCGD− gms (CGD+ CDB) +1RLs (CGD+ CGS) +1RS− s2(CGD+ CGS) (CGD− gm)=1RS(11)Now we are in a position to solve for Vout/Vin:VoutVin=(sCGD− gm) RLas2+ bs + 1(12)a = RSRL(CGSCGD+ CDBCGD+ CGSCDB) (13)b = (1 + gmRL) CGDRS+ RSCGS+ RL(CGD+ CDB) (14)I’ve left out some messy algebra that Razavi skips over as well. It should be obvious how to get thesolution above, though actually solving for a and b requires multiplying out the terms of Equation (11) andmatching a with the s2coefficients and b with the s coefficients (after dividing by the constant term to ensureit is 1). At this point we only care about analyzing the result, so let’s ignore any further algebra.First, we can note there is a zero at ωz=gmCGD. Razavi mentions that this is usually at a very highfrequency, out of the range that would affect our frequency response. Consider if gm∼ 10−3and CGD∼10−15, then ωz∼ 1012, or in the terahertz range, which is very high frequency.21.3 Finding the PolesSince the denominator of the transfer function is quadratic in s, we know we will have 2 poles. Let’s call themωp1and ωp2. We should be able to factor the denominator such that as2+bs+1 =sωp1+ 1sωp2+ 1. Nowconsider if one pole dominates—that is, assume ωp1<< ωp2. Then we can further simplify this expressionto:as2+ bs + 1 =sωp1+ 1sωp2+ 1(15)=s2ωp1ωp2+sωp1+sωp2+ 1 (16)≈s2ωp1ωp2+sωp1+ 1 (17)⇒ b ≈1ωp1(18)a ≈1ωp1ωp2≈bωp2(19)Now we can s olve for the dominant pole, ωp1, and secondary pole, ωp2:ωp1≈1b(20)≈1(1 + gmRL) CGDRS+ RSCGS+ RL(CGD+ CDB)(21)ωp2≈ba(22)≈(1 + gmRL) CGDRS+ RSCGS+ RL(CGD+ CDB)RSRL(CGSCGD+ CDBCGD+ CGSCDB)(23)2 Example 11.9This example is already solved in the book, but it’s useful to go over it anyway. We just need to findhow CGD, CGS, CDB, and RLare changed by the new circuit topology. Drawing out all of the parasiticcapacitances gives the circuit in Figure 11.21(b) (Razavi asks why CSB1, CGS2, and CSB2don’t come intoplay—the answer: the sources are all connected to the bodies, so CSB1and CSB2are gone; CGS2is attachedbetween two DC voltages Vband 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. CGSrepresented thecapacitance from the input to ground, so that remains as CGS1. CGDrepresented the capacitance frominput to output, which in this circuit is just CGD1. CDBwas the capacitance from the output to ground,which now has 3 parallel capacitances, so CDBis replaced by CDB1+ CDB2+ CGD2.We can summarize this as follows:CGS−→ CGS1CGD−→ CGD1CDB−→ CDB1+ CDB2+ CGD2We also note that in this case, RL= (ro1||ro2). Thus we can find ωp1and ωp2by plugging these newvalues into Equations (21) and
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