EECS 242 – p. 1/20
EECS 242 Prof. Niknejad University of California, Berkeley University of California, Berkeley Two-Port Power Gain EECS 242 – p. 2/20 S V S Z 0 in + Z ) Z ) 2 S S 2 in | | Γ Z L in − Γ Γ 1 + Γ 1 )= −| −| = in (1 (1 S 2 2 | | Z 0 0 + − Z Z 1 2 (1+Γ 2 2 V V | | + 1 V = = = L in P P − 1 V University of California, Berkeley 0 + Z using circuit theory + 1 in in V + 1 Γ V − 1 + Γ 1 = in Z
Two-Port Power and Scattering Parameters The power flowing into a two-port can be represented by The power flowing to the load is likewise given by We can solve for In terms of the input and source reflection coefficient • • • • EECS 242 – p. 3/20 ) in Γ − 2 − V L ) )(1 Γ S S 22 Γ 2 S | − L + 0 Γ Z )(1 S + 1 2 Γ L S V −| in )+(1+Γ Γ Γ in + 1 21 1 S Γ 22 S 2 V − 2 Γ | ˛ ˛ ˛ S 1 − 21 L = + − (1+Γ 1 1 S − Γ S V + 2 ˛ ˛ ˛ 1 S 22 V 2 )(1 V 2 V S | = in 22 − 21 = S − 2 S 1 | | + V + 1 V = + (1+Γ 1 V L P University of California, Berkeley 21 )= S in = − 2 V (1+Γ + 1 V + 1 V
Two-Port Incident Wave Solve for The voltage incident on the load is given by • • EECS 242 – p. 4/20 rt s-parameters and ) 2 ) | 2 | ∗ S 2 | Γ ∗ S ) in S 2 Γ Γ | Γ −| L − + 2 2 −| 1 Γ | | −| (1 1 V S S 1 2 (1 ˛ ˛ ˛ −| Γ Γ 0 a 2 S + 2 | Z 1 V − (1 2 −| L V 1 ˛ ˛ ˛ 2 Γ 1 | | = 2 22 = 21 | 0 ∗ S S S S Z | ∗ S 8 V =Γ − | =Γ in 1 | Γ = in ˛ ˛ ˛ Γ = | + 1 avs V in L in P P P P = University of California, Berkeley = a + = 1 V p avs G P
Operating Gain and Available Power The operating power gain canthe be load written reflection in coefficient terms of the two-po The available power can be similarly derived from • • EECS 242 – p. 5/20 , then we have 0 Z e ) 2 2 | | S L Γ Γ 22 −| S − )(1 2 1 | | L 2 | 2 Γ | S Γ 21 −| S in | (1 Γ 2 = | − T 21 1 | S G | = L is a function of the two-port s-parameters and the load avs P University of California, Berkeley P T G = T G and = 0 S
Transducer Gain = Γ L Γ and source impedance. The transducer gain can be easily derived Note that as expected, If the two port is connected to a source and load with impedanc • • • EECS 242 – p. 6/20 L G . This is the 2 0 | Z = 0 21 rk we can change S n of the load. | 12 hopefully lossless, so S S ed as the gain from the G = 2 L 2 M G | 2 L | Γ L Γ 22 ¸ S −| 22 0 2 − | 1 S 1 21 | S | 2 11 21 | S S 21 · S | 2 | 2 S | S 1 Γ S G M Γ 11 University of California, Berkeley S −| − 1 0 1 | Z = TU G s assumption , we can simplify the expression by just assuming − + v 0 ≈
Unilateral Gain 12 S unilateral If The gain partitions into three terms, which can be interpret In reality the source/load matchingthe network power are gain passive is and 1 or less, but by virtue of the matching netwo source matching network, the gain of the two port, and the gai the gain of the two-port. • • • EECS 242 – p. 7/20 h ) 2 | 22 S 2 2 −| | | 2 | 22 11 )(1 21 S S 1 1 2 S | | ∗ ∗ 22 11 −| −| 11 S S 1 1 S = = = = −| S L Γ Γ (1 , the maximum gain is infinity. This is the = S,max L,max G G is potentially unstable. = 1 | 1 22 University of California, Berkeley > S | | T U,max ii G of S | = 1 | 11 S |
Maximum Unilateral Gain unstable case since We know that the maximum gain occurs for the biconjugate matc Note that if • • EECS 242 – p. 8/20 uencies is shown ds C he output can deliver o r in v m g in + − v University of California, Berkeley , this circuit has infinite power gain. This is a trivial fact gs = 1 C | 11 S |
Ideal MOSFET The AC equivalent circuit forabove. a Since MOSFET at low to moderate freq since the gate capacitance cannot dissipate power whereas t real power to the load. • EECS 242 – p. 9/20 ds R culated. Assume we ds C e the unilateral ds R 2 ˛ ˛ ˛ ˛ 2 ˛ ˛ ˛ ˛ ds 1 S 1 R V V V ˛ ˛ ˛ ˛ 2 i m g 2 R i | ˛ ˛ ˛ ˛ in v S R ds 2 1 8 m V | R g 2 m = )= g ∗ L = V avs L P I in − + 1 2 v ( ℜ T U,max = G gs University of California, Berkeley C L i P R i R
Real MOSFET s conjugate match the input/output A more realistic equivalent circuitassumption, is then shown the above. input If and we output mak power can be easily cal − + v • EECS 242 – p. 10/20 T in we should f y the FET is given by 2 2 ) « gs 2 T f ω f i /C 4 S „ R m V 2 g i ( ds R gs R i ds 1 4 1 R R jωC = = = 1 V T U,max T U,max G G University of California, Berkeley and minimize the input resistance while maximizing the T f
Real MOSFET (cont) output resistance. maximize the device At the center resonant frequency, the voltage at the input of This can be written in terms of the device unity gain frequenc The above expression is very insightful. To maximum power ga • • • EECS 242 – p. 11/20 acheived for a atching. This is ibility if we back-off ad impedances that case, though, we’d like esign. We can trade off University of California, Berkeley bandwidth noise gain flatness linearity etc. Design for Gain • • • • • single point on the Smithfrom Chart, the we peak will gain. find that a lot more flex So far we have onlypossible discussed when power the gain device using is bi-conjugateto unconditionally m design stable. with In a many potentially unstable device. Moreover, we would like to introduce more flexibility in the d We can make this tradeoffcan by realize identifying a a given range value of of source/lo power gain. While maximum gain is gain for • • • EECS 242 – p. 12/20 , 1 dB . 0 a device. ible error. ng inequality y unilateral, or else we 2 ) | 2 2 ) | 22 U S 22 1 | S − 2 | (1 −| 11 S | < )(1 2 | 2 | T TU 21 11 G S G | S 2 | < −| 12 2 is compared to the power gain under the ) (1 S | T U 1 G can be used to test the validity of the unilateral = . If the inequality is tight, say on the order of m (1 + University of California, Berkeley U TU G
Unilateral Design unilateral figure of merit then the amplifier can be assumed to be unilateral with neglig unilateral assumption No real transistor is unilateral.use But cascades most of are devices predominantl (such as theThe cascode) to realize such assumption It can be shown that the transducer gain satisfies the followi Where the actual power gain • • • • EECS 242 – p. 13/20 S g ˛ ˛ ˛ ˛ ˛ ) 2 | 11 S 1) + 1 − −| S (1 g ( S g 2 | − 11 1 S | √ corresponding to the maximum gain. ˛ ˛ ˛ ˛ ˛ S L ∗ 11 G = G S,max L,max . One can show that a fixed value of S ˛ ˛ ˛ ˛ 1 G G = ≤ = = S L S L C g 1) + 1 g g S − ≤ g . A similar equation can be derived for the load. 0 S plane ∗ 11 , and g S ( S S R 2 Γ and | University of California, Berkeley = 0 = 1 11 S | S S R ≤ | , C S − g − S = 1 ≤ S Γ ˛ ˛ ˛ ˛ S Γ 0 | g
Gain Circles Note that for represents a circle on the We now can plot gain circles for the source and load. Let By definition, More simply, • • • EECS 242 – p. 14/20
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Potentially Unstable Unilateral Amplifier real part less than the critical value. identical to before as long as we avoid source or load reflecti For a unilateral two-port with The same is true of the load impedance if resistance is larger than negative real part. Thus we can still design a stable amplifie • • EECS 242 – p. 17/20 to find ∗ 11 . /S . 1 ) , the amplifier 1 in = 5dB Z > ( S | ℜ G 11 > S | ) ◦ S Z 60 ( ∠ the negative real input resistance. is potentially unstable. We begin by plotting Since Now any source inside this circle is stable, since ℜ We also draw the sourcecircle for gain 00 = 0 . • • • 12 = 5 S
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> — 0.0 • EECS 242 – p. 18/20 . In 5 dB of j . Note the real part is he BW) but with as 35 . ∗ 11 236 . /S 1 3 + 0 ) . = 0 ) ) j 2 j − ) 4 | . 4 ) 2 . | 0 = 11 S g S 11 ) − S S − 4 . g −| 0 75 + 0 (1 −| . − − 2 | (1 ∗ (1 15(1 11 . S 11 S 2 g | S S = 50( = 50(0 g = 3 − 11 −| S S 1 S in 1 Z g √ Z −| 1 = University of California, Berkeley S = R S C gain circle is calculated as follows = 5dB S
Amp Example (cont) G interpreted as negative The input impedance is read off the Smith Chart from The We can select any point on this circle and obtain a stable gain particular, we can pick alarge point of near a the real origin impedance (to as maximize possible: t • • • EECS 242 – p. 19/20 , on p G p g 2 | mpedaance. 2 p 21 g S 2 | | 2 . The transducer gain is ˛ ˛ ˛ = 21 p p S ) 2 g | G 2 | 2 12 ) L | S ∆ Γ | 11 ∆ | S 22 + −| ∗ S + ) p 2 ∆ 2 | p g | plane. The radius and center are − | g − 22 L 1 2 L 21 | | S Γ | Γ S ∗ 22 ( « 22 S 2 p 12 −| S ( ˛ ˛ ˛ g S p | L (1 L g −| Γ 2 K | 1+ 1 ∆Γ 2 22 − − 21 S = − ˛ ˛ ˛ S − 11 | 1 1 L S ˛ ˛ ˛ C q University of California, Berkeley − is a circle on the = 1 p L „ g R = p G
Bilateral Amp Design given by not used since the source impedance is a function of the load i the other hand, is only a function of the load. In the bilateral case, we will work with the power gain It can be shown that • • EECS 242 – p. 20/20 at this occurs is in the stable ∗ in = 0 = Γ ” S 1 ” Γ 1 − 2 p,max − g 2 2 2 | K K 21 p S p − 12 − S K | K “ + “ | that is in the stable region and not too ˛ ˛ ˛ ˛ 21 L S 21 12 and check to see if Γ 1 S S 12 p,max ˛ ˛ ˛ ˛ in S g | Γ | or compromise. = 21 L = S Γ 12 S p,max University of California, Berkeley | G p,max K g 2 − 1 p , or g = 0 L R region. If not, iterate on close to the stability circle. Bilateral Amp (cont) 1. Specify 2. Draw operating gain3. circle. Draw load stability circle. Select 4. Draw source stability5. circle. To maximize gain, calculate We can also use thiswhen formula to find the maximum gain. We know th The design procedure is as follows • •