] (1) Problem 1. University of California, Berkeley Fall 2010 EE142, Problem Set #9 Solutions Prof. Jan Rabaey

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1 University of California, Berkeley Fall 00 EE4, Proble Set #9 Solutions Ain Arbabian Prof. Jan Rabaey Proble Since the ixer is a down-conversion type with low side injection f LO 700 MHz and f RF f IF 00 MHz. (a) The current I through a MOSFET in saturation is given as 800 MHz, so that I µc ox W L (V GS V th ) where µ is the obility, C ox the oxide capacitance, W, L the diensions of the transistor, V GS the gate source voltage and V th the threshold voltage of the transistor. Fro the expression we note that due to the Square law nature of the device a cross ter.v GS.V th is generated. If V GS is the LO input and V th is ade proportional to the RF input, we can get a ixing product corresponding to the IF ter at the output. (b) In order to nd the conversion gain, we need to rst write the threshold voltage as a function of the RF input. We know that V th V to + γ( Φ f + V SB Φ f ) According to the SPICE level odel, Φ Φ f. Therefore, V th V to + γ( Φ + V SB Φ) V th V to + γ Φ( + V SB Φ ) In the proble V SB A RF sin(ω RF t). Since the aplitude A RF 50 V is uch saller than Φ 600 V, we can use the Binoial expansion and neglect the higher order ters. Therefore, we have V th V to + γ Φ( + V SB Φ ) V th V to + γ V SB Φ V th V to + γ A RF sin(ω RF t) Φ Also, we have V GS V bias + A LO sin(ω LO t), where V bias 0.7 V. As the MOS is described by a second order syste, we can now copute the total current and not ake any additional approxiation for the LO side (in addition to what we have done for the RF side). Hence, the current through the MOSFET is [ W I µc ox V bias + A LO sin(ω LO t) V to γ A ] RF sin(ω RF t) L Φ W I µc ox L [A LOsin (ω LO t) + γ A RF sin (ω RF t) + (V bias V to ) + (V bias V to )A LO sin(ω LO t) 4Φ (V bias V to )γ A RF sin(ω RF t) A LO sin(ω LO t)γ A RF sin(ω RF t) ] () Φ Φ

2 Fro (), the ixing product is given as Hence, the conversion gain W I ix µc ox L A γ LOA RF Φ sin[(ω RF ω LO )t] G conv,i I ix A RF W µc ox L A γ LO Φ We could also dene the conversion gain in ters of output voltage divided by the input voltage for which the output low frequency pole needs to be taken into account. To calculate this, we rst nd the output voltage v o,if G conv,i + ω IF R L C L A RF Using kω, C L 800 ff, W 8 µ, L 0.5 µ, γ 0.5, A LO 00 V, Φ 0.6 V, µ 40 c /Vs and C ox 4ɛ0 t ox F/, we get 9 and G conv,i ( ) ( ) µS v o,if 63.98µS 000Ω + (π00mhz 000Ω 800f) 0.05 V.858 V Hence, the conversion gain dened in ters of voltage is given as G conv,v v o,if v i,rf (c) Fro (), the LO coponent of the current is Substituting the values Therefore, the LO leakage to the output.858 V 50 V I LO µc ox W L (V bias V to )A LO v o,lo I LO 79.9µA I LO + ω LO R L C L Using the values we get The RF coponent of current fro () is v o,lo.68 V I RF µc ox W L (V bias V to )γ A RF Φ Substituting the values Therefore, the LO leakage to the output I RF 6.398µA v o,lo I RF + ω RF R L C L

3 Using the values we get v o,rf.544 V (d) Figure shows the output current spectru of the ixer. The output current at the IF is 3.6µA for a 50 V signal. Hence, the conversion gain is G conv 3.6µA 0.05 V 6.3µS This value atches very well with our analytical calculations. Figure : Output current spectru Figure shows the output spectru of the ixer. The IF output voltage, RF and LO leakage all atch well with our calculations. Figure : Output voltage spectru (e) The agnitude of the voltage applied at the RF port is liited. If the aplitude exceeds the cuto voltage of the source-bulk diode, the device gets forward biased. Also, since the RF input is injected through the bulk, the conversion gain is liited. This is because g b g. 3

4 Proble Vdd C L v IF, v IF, p v LO, p I I I 3 I 4 v LO, v LO, p 00Ω : C L L C RF L L L 3 L 3 V bias I 0 V bias Figure 3: Down-conversion double balanced Gilbert ixer Figure 3 shows the circuit diagra of the down-conversion ixer. Assuing that the LO transistors act like ideal switches (i.e. the LO drive is large) with a switching function, we can write the current through the transistors as a product of the switching function and the RF current. For exaple, for the current I, the current fro the RF transistors includes the DC coponent plus the sall + A eff g V s sin(ω RF t), where A eff is the eective gain fro the RF source to the base- I signal coponent i.e. 0 eitter voltage of the BJT, g the transconductance of the RF transistors and V s is the dierential input RF swing. This current is switched using a Square wave s(t) which has a Fourier series s(t) + π {sin(ω LOt) + 3 sin(3ω LOt) + 5 sin(3ω LOt) +...} Thus, or I s(t)[ I 0 + A eff g V s sin(ω RF t)] I [ + π {sin(ω LOt) + 3 sin(3ω LOt) + 5 sin(3ω LOt) +...}][ I 0 + A V s eff g sin(ω RF t)] () Siilarly, we can write the current through the other branches taking into consideration their switching functions. I 3 [ π {sin(ω LOt) + 3 sin(3ω LOt) + 5 sin(3ω LOt) +...}][ I 0 A eff g V s sin(ω RF t)] (3) I [ π {sin(ω LOt) + 3 sin(3ω LOt) + 5 sin(3ω LOt) +...}][ I 0 + A eff g V s sin(ω RF t)] (4) I 4 [ + π {sin(ω LOt) + 3 sin(3ω LOt) + 5 sin(3ω LOt) +...}][ I 0 A V s eff g sin(ω RF t)] (5) The output dierential voltage is given as v out [(I + I 3 ) (I + I 4 )]Z out 4

5 where Z out + (ωrl C L ) I I 4 [ + π {sin(ω LOt) + 3 sin(3ω LOt) + 5 sin(3ω LOt) +...}][A eff g V s sin(ω RF t)] I 3 I [ π {sin(ω LOt) + 3 sin(3ω LOt) + 5 sin(3ω LOt) +...}][A eff g V s sin(ω RF t)] Therefore, v out [ 4 π {sin(ω LOt) + 3 sin(3ω LOt) + 5 sin(3ω LOt) +...}][A eff g V s sin(ω RF t)] (6) + (ωrl C L ) The output voltage corresponding to the IF is given v IF π A eff g V s sin[(ω LO ω RF )t] + (ωif C L ) The output power is therefore The input power Therefore, the power gain is v IF P out P in (V S ) R S V S 6R S G P P out P in 4v IF V S. R S [ ] 4( π ) A eff g R L. R S + (ωif C L ) Here R S 50Ω, 00Ω, C L 0pF, G P 0 0 db and Therefore, This gives g 0.5I 0 V T 0.5 5A S [ ] 0 4( π ) A eff (96.5 S) (π00mhz 00 0pF) 00 A eff We now need to nd A eff in ters of the atching network paraeters. For this consider Figure 4. Here, we assue that the L in the second part of the circuit is used to resonate out the reactive coponents. This leaves us with a real ipedance which is atched to 50Ω using an L-atch circuit consisting of C and L 3. There are any other ways to design this atching network part. For e.g. the L-atch can also be absorbed into the second stage. Any design which atches the input to 50Ω satisfying the given power gain specications will receive full credit. In Figure 4(a), we consider the rst part of the circuit. The resistance values at resonance (GHz) are shown. Hence, we have v y V S 4. With a atching network quality factor Q M, the voltage v x at resonance is v x v y ( + jq M ) V S 4 ( + jq M ) 5

6 R where Q M H RS. Therefore, v x V S + Q M 4 V S 4 RH R S R s R H R s R s v y C v x L R s v y C vx R s v s L 3 v s L 3 (a) R s R H R s v y C v x L C eff L v s L 3 + v gs - R H (b) Figure 4: Matching network circuit Now consider Figure 4(b) and consider the second part of the circuit. Here, C eff C gs + C gd g τ f + C jeo + C jc 96.5S 6.4ps + 0fF + 60fF 95.36fF We get this approxiation fro Figure 5(a). Note we can also use the Miller approxiation as shown in Figure 5(b). This changes the answer slightly. 6

7 /g /g /g v i C gd v y v i v i i y C gs g (v i -v x ) C gd C gs +C gd g (v i -v x ) C gs +C gd g (v i -v x ) v x v y ~-v i +v x i y ~sc gd (v i +v i -v x ) v x v x L L L (a) /g /g /g v i C gd v y v i v i i y C gs g (v i -v x ) C gd C gs g (v i -v x ) C gs g (v i -v x ) v x v y ~-v i +v x i y ~sc gd (v i +v i -v x ) v x v x L L L (b) Figure 5: Capacitance approxiation Hence, Also, we can neglect r π because at GHz, and ω T g C eff 96.5S 95.36fF rad/s r π β g S.56kΩ ω RF C eff πghz 95.36fF 74Ω For the second part of the circuit if we denote the quality factor as Q, then (L + L )ω RF R H Q Therefore, v gs v x Q V S 4 At the RF frequency, L + L resonates with C eff. RH R S (L + L )ω RF R H 7

8 Therefore, Fro this expression, v gs V S RH 4 R S V S 4 ω RF R H C eff RS R H ω RF C eff A eff v gs V S R S R H ω RF C eff With ω RF π.ghz, C eff 95.36fF, R S 50Ω, we get For atching, Also, since L + L resonates at RF, R H 439.3Ω L R H nH ω T L + L ωrf C eff (πghz) 95.36fF 7.67 nh This gives L 3.49nH RH Q M R S R H L 3 ω RF Q M (πghz) nh Therefore, L 3 C L 3 + Q. nh M ω RF L 3.4 pf Assuing a drop of 0.5 V across the current source (because in practice it will be ipleent by another transistor) and a drop of 0.7 V across the RF transistor, we choose V bias. V. Assue a V CE 0.6 V across the RF transistor and a noinal V BE 0.7 V, choose V bias.8 V. (b) For siulation an LO aplitude of 500 V was used to switch the transistors. The RF aplitude unless entioned was set at 0 V. Figure 6 shows the output and input power of the ixer. The power gain fro the RF to IF is found to be 53.4 ( 6.7) 9.46 db close to our required value of 0 db. 8

9 Figure 6: Down conversion power gain of the ixer. Plot shows the output and input power. (c) Figure 7 and 8 show the input ipedance and the reection coecient at the RF port respectively. The RF port is well atched to 00Ω with an S 7 db at GHz. Figure 7: Input ipedance at the RF port 9

10 Figure 8: S at the RF port (d) Figure 9 shows the spectru of the voltage and current with an LO aplitude 500 V. The large signal LO input ipedance is found to be Z LO 0.5 V 864.7µA 578Ω Please note that your answer ay vary slightly depending on the LO aplitude you have used. All correct approaches will receive full credit. Figure 9: Spectru at the LO input - voltage and current (e) Fro (6), we can see that both the RF and LO leakage for this topology is zero. That is why it is called a Double Balanced ixer. Figure 0 shows the output voltage spectru. It is clear that the output levels of the RF and LO are very low. 0

11 Figure 0: Output voltage spectru (f) In order to calculate the noise fro the devices, we assue that the LO transistor are hard switches so that they donot contribute to the output noise. Referring to Figure, we have noise sources fro the base and collector currents of the RF transistors and fro the input port resistance. The noise of the load resistances are neglected because these are considered in the stages subsequent to the ixer. Vdd C L v IF, v IF, p v LO, p I I I 3 I 4 v LO, v LO, p 00Ω : C L Q0 Q L C i c i c RF i b L L L 3 L 3 i b V bias I 0 V bias Figure : Circuit diagra for noise calculation

12 Noise fro the RF Port resistance The voltage gain fro the RF port is given as A V π A eff g + (ωif C L ) S π Note that the input noise is bandpass ltered so there will not be any excess noise added due to noise folding. Therefore, the output spectral density due to the input resistance : S vo,rs 4k B T (R s ) (A V ) V /Hz Noise fro the Collector current of the RF transistors The total noise at the output ust be calculated taking into account noise folding. If we split the RF transistor noise sources as ic ic and ic ic for analyzing it dierentially, the total noise current at the output is given as : i o [ s(t)( i c i c ) + s(t)( i c i c ) s(t)( i c i c ) s(t)( i ] c i c ) where s(t) [ + π {sin(ω LOt) + 3 sin(3ω LOt) + 5 sin(3ω LOt) +...}] s(t) [ π {sin(ω LOt) + 3 sin(3ω LOt) + 5 sin(3ω LOt) +...}] We have the factor of in front because only the collector current goes to the output as the ixer is degenerated with inductors. Upon siplifying we get the total noise at the output as S vo,ic 4( [ π ) + ( 3 ) + ( ] [ ] S 5 ) +... ic + (ωif C L ) 4 Each of the ters, ( 3 ), ( 5 ),... indicates the gain fro the sidebands. The current noise spectral density due to the collector current is given as : S i c Noise fro all the sidebands : qi c A 8 0 A /Hz S vo,ic Noise fro the st sideband alone is found to be V /Hz S vo,ic, V /Hz Noise fro the st and nd sideband alone is found to be S vo,ic, V /Hz Noise fro the st, nd and 3rd sideband is found to be S vo,ic, V /Hz

13 Noise fro the Base current of the RF transistors In order to nd the noise fro the base current we need to nd the transfer function fro i b to the output. The transfer function between the collector current and i b is : i c g [R H + jω(l + L )] + jωl (g + jωc π ) + jωc π (R H + jωl ) i b Note that this transfer function is low pass in nature and hence only the noise fro the rst sideband reaches the output. At the RF frequency of GHz, substituting the values : i c 9i b The current noise spectral density due to the collector current is given as : S i b qi b q I c β A A /Hz 50 The output noise due to i b and i b is given as S vo,ib 4( π ) [ V /Hz + (ωif C L ) ] (9) S ib (g) Figure shows the siulated output noise of the ixer with one sideband, two sideband and three sideband. Note that the sidebands occur at frequencies f lo, 3f lo and 5f lo. We notice very little change between the siulations. This because the noise fro the input port doinates as shown in Figure 3 and is also apparent fro our calculations. It is also worth noticing that the calculated values are slightly ore than those fro siulation. The reasons for these can be attributed to the fact that in our calculations we assued a Square wave LO drive and also we did not consider ltering at the interediate nodes (like the drain of the RF transistors) which reduce the total noise reaching the output. Figure 4 shows the siulated noise gure of the ixer. The noise gure is found to be 3.3 db at 00 MHz. Figure : Siulated noise of the ixer with dierent sidebands 3

14 Figure 3: Siulated noise contribution fro various sources. R4 is the input PORT resistance, Q0 and Q the RF transistors 4

15 Figure 4: Siulated noise gure of the ixer (h) Figure 5 shows the output spectru for three dierent aplitudes. The RF tones had a frequency of GHz ± MHz. Figure 6 shows the IIP3 of the ixer. The IIP3 is found to be 5 dbv 77.8 V. Figure 5: Siulated output spectru for RF aplitudes of 0 V,5 V and.5 V 5

16 40 0 Output Aplitude per tone (dbv) Fundaental Distortion X: 5.0 Y: Input Aplitude per tone (dbv) Figure 6: IIP3 of the ixer 6