ECE343 Test 1: Feb 10, :008:00pm, Closed Book. Name : SOLUTION


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1 ECE343 Test : Feb 0, 00 6:008:00pm, Closed Book Name : SOLUTION C Depl = C J0 + V R /V o ) m C Diff = τ F g m ω T = g m C µ + C π ω T = g m I / D C GD + C or V OV GS b = τ i τ i = R i C i ω H b Z = Z A Z = Z /A. pts) The circuit below shows a simple commonsource amplifier with a sourcedegeneration resistor. Use a smallsignal analysis to find the lowfrequency gains A o = v out / and A = v /. Give your answer in terms of R in,, R D, and any applicable smallsignal parameters. Assume that γ = 0, λ = 0, and C DB = C SB = 0 Solution: The small signal model is illustrated on the right above. The controlled current source sets the current through R s, and determines v : v = g m v gs )R s = g m R s v ) v + g m R s ) = g m R s A = v = Once v is determined, the output voltage is found using Substituting the above value of v gives [ v out = g m R D g mr s + g m R s v out = R D g m v gs ) = g m R D v ) g mr s + g m R s ] [ = g m R D g ] mr s = g mr D + g m R s + g m R s A o = v out = g mr D + g m R s
2 . a) 8 pts) For the circuit of problem, examine the impact of the parasitic capacitors C GD and C GS. Use the Miller theorem to redraw the circuit showing the equivalent capacitors between the various circuit nodes and ground. Give your component values in terms of C GD, C GS, A o = v out / and A = v /. Solution: Miller s theorem relates the equivalent capacitance to the gain between the nodes that the capacitor is tied to. Use A to deal with C GS, and A o for C GD. C in = C GS A ) + C GD A o ) C o = C GD /A o ) C = C GS /A ) b) 8 pts) If is chosen sufficiently large, and R D, then the gains from problem can be approximated as A o = v out R D A = v Using these approximations, argue that the bandwidth of the amplifier is primarily determined by C GD and not C GS ). Solution: If A, then all of the above C GS terms vanish. Essentially, no current will flow through C GS if the two terminal voltages are equal, so it drops out of the circuit. Only the C GD term remains... and the contribution to the input capacitance is potentially large. The above reduce to: C in C GD + R ) D C o C GD + R ) S R D C 0 C GD
3 3. a) pts) Considering only C GD, and using the approximations given in problem b, use the opencircuit time constant method to give an approximate expression for the amplifier bandwidth. Solution: The open circuit time constants are found by scaling each capacitor by the DC) resistance measured at the capacitor terminals. At the input, this resistance is just R in. The resistance at the output node is R D in parallel with the resistance seen looking into the drain of the transistor. In this case r o = is given, so the result is just R D. τ = R in C in = R in C GD + R ) D τ = R D C o = R D C GD + R ) S = R D + )C GD R D The 3 dbfrequency is approximately the reciprocal of the sum of all time constants: ω H τ + τ = ) ] in rad/sec) [R in + R D RS + R D + C GD b) 3 pts) Assume that g m = 0 ma/v, R D = 5 kω, R in = 00 kω, C GD = pf. Evaluate the approximate) amplifier gain, 3dB bandwidth in khz), and gainbandwidth product for = 500 Ω. Solution: A o = R D = 0 ω H = krad/sec [00 kω)) + 5 kω) Ω)] pf) f H = ω H = khz π GBW = 0 f H =.44 MHz c) 3 pts) Repeat problem 3b for = kω. Solution: A o = R D = 5 ω H = 650 krad/sec [00 kω)6) + 5 kω) + kω)] pf) f H = ω H = 6.6 khz π GBW = 5 f H =.3 MHz
4 4. The circuit below shows a folded cascode CMOS amplifier utilizing a simple current source M, supplying current I, and a cascoded current source M 4, M 5 ) sinking current I. All transistors are saturated. Assume, for simplicity, that all transistors have equal smallsignal parameters g m and r 0, and that g m r 0. Neglect the body effect, and neglect all parasitic capacitors. a) 8 pts) Give approximate expressions for all of the resistances indicated: R o, R o, R o3, R o4, R o5, R in3 and R out. Give your answers in terms of g m and r o. Solution: Start with the easy cases: R o = R o = R o5 = r o The impedance looking into the drains of M 4 and M 3 have an additional term given by the resistance attached at their source terminals, scaled up by their intrinsic gains: R o4 = r o4 + + g m4 r o4 )r o5 g m r o R o3 = r o3 + + g m3 r o3 )R o R o ) = r o + + g m r o )r o /) g mr o The resistance into the source of M 3 is /g m3 plus and additional term given by the resistance attached to the drain terminal, scaled down by the intrinsic gain of M 3 : R in3 = R o4 + + g mro = + r o r o g m3 + g m3 r o3 g m g m r o g m Finally, R out = R o3 R o4 ) gmr gm o ro) g mr o + g m r o = g mr o 3
5 b) pts) Reevaluate R in3 for the case when the output node is shorted to ground, and use current division to find the shortcircuit transconductance G m. Show that G m g m. Solution: For the output shorted, the resistance attached to the drain of M 3 changes to 0 instead of R o4 ), so R in3 = g m3 If it helps, redraw the smallsignal model for the case in which the output is shorted to ground as shown below. With some practice, you should get to the point where you can work directly from the original circuit diagram.) The controlled current source output from M is divided between R o R o = r o / and R in3 = /g m. The output current is the current through R in3 : ) ) r o / gm r o i out = g m = g m r o / + /g m3 g m r o + G m = i ) out gm r o = g m short g m r o + g m for g m r o Large) c) 6 pts) Evaluate the voltage gain v out / for the case g m = ma/v, and r 0 = 0 kω. Solution: Here, the intrinsic gain is g m r o = 0. The amplifier voltage gain is v out = G m R out g mr o 3 = 0) 3 = 33.3
6 5. 8 pts) Using V DD =.8 V and a pair of MOSFETs, design the current sink shown below) to provide an output current i O of 00 µa nominal value. To simplify matters, assume that the nominal value of the output current is obtained at v O V GS. It is further required that the circuit operate for v o in the range of 0.0 V to V DD and that the change in i O over this range be limited to 3% of the nominal value of i O. Find the required value of R and the device dimensions of M. For the fabricationprocess technology utilized, µ n C ox = 50 µa/v, V A = 0 V/µm, and V t = 0.6 V. Solution: The required current sink characteristic is shown on the right above. The length of M is found using the required output impedance. r o 0 V/µm)L 66.6 kω 00 µa L.66 µm µa =.6 V = 66.6 kω 6 µa I ll use the minimum required device length L =.66 µm.) Since the current source must operate at v o = 0. V, the overdrive voltage of M and M ) is set to 0. V, giving V G = 0.8 V for both MOSFETS. The width of M is set to give the desired 00 µa output current: 00 µa = 50 µa/v W L ) 0. V) W L = 40 W = 40L = µm) = 06.4 µm. The current through M will be one fourth of that in M because of the different value of W /L. So the design should set I REF = 50 µa. Using the known gate voltage of V G = 0.8 V, the value of R is determined: R = V 50 µa = 0 kω.
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