Page 1 of 3 ELEC 312: ELECTRONICS II : ASSIGNMENT-3 Department of Electrical and Computer Engineering Winter 2012 1. A common-emitter amplifier that can be represented by the following equivalent circuit, has C π = 10 pf, C µ = 0.5 pf, C L = 2 pf, g m = 20 ma/v, β = 100, r x = 200 Ω, R L / = 5 kω and R sig = 1 kω. Find (i) the mid band gain A M, (ii) the frequency of the zero f Z, and (iii) the approximate values of the pole frequencies f P1 and f P2. Hence estimate the 3-dB frequency f H. Note that R sig is the equivalent Thevenin resistance looking towards the signal source and includes the effects of R sig, r x and r π. For approximate estimates, you may use OCTC method. + V sig / V o - (i) A M = -r π (g m R L )/( R sig + r x + r π ); (ii) f Z = g m /(2π C µ ) (iii) f P1 = 1/[2π{(C π + C µ (1+ g m R L )) R sig +(C L + C µ )R L }]; f P2 = [(C π + C µ (1+ g m R L ))R sig +(C L + C µ )R L ]/[2π{C π (C L + C µ )+C L C µ )} R sig R L ]; f P1 << f P2 & f P1 << f Z, hence f H f P1 2. Analyze the high-frequency response of the CMOS amplifier shown below. The dc bias current is 100 µa. For Q 1, µ n C ox = 90 µa/v 2, V A = 12.8 V, W/L = 100 µm/1.6 µm, C gs = 0.2 pf, C gd = 0.015 pf. For Q 2, C gd = 0.015 pf, C gs = 36 ff and V A = 19.2 V. Assume that the resistance of the input signal generator is negligibly small. Also, for simplicity assume that the signal voltage at the gate of Q 2 is zero. Find the low-frequency (i.e., at DC) gain, the frequency of the pole, and the frequency of the zero. You may use nodal analysis. Note: ff=10-15 F, pf=10-12 F. Assignment 3 ELEC 312/Winter 12 R.Raut, Ph.D.
Page 2 of 3 DC gain = - g m (r 01 // r 02 ), where g m = [2µ n C ox I D W/L], r 0 =V A / I D and Small-signal gain, v o / v i = (g m -sc gd1 )/[1/ r 01 +1/ r 02 +s(c L +C gd1 )] where C L = C gd2 f Z = g m /(2π C gd1 ); f p = (1/2π)[( 1/ r 01 +1/ r 02 )/(C L +C gd1 )] 3. A CG amplifier is specified to have C gs = 2 pf, C gd = 0.1 pf, C L = 2 pf, g m = 5 ma/v, χ = 0.2, R sig = 1 kω and R L / = 20 kω. Neglecting the effects of r o, find the low-frequency gain v o / v sig, the frequencies of the poles f P1 and f P2 and hence an estimate of the 3-dB frequency f H. For a CG amplifier you can use g mb = χg m. Use ac equivalent circuit. From the small-signal equivalent circuit, v o /v i = [{1/(g m +g mb )}/{R S +1/(g m +g mb )}](g m +g mb )R L ; f p1 = 1/[2π C gs {R sig //(1/(g m +g mb ))}]; f p2 = 1/[2π(C gd +C L )R L ]. f p2 <<f p1, f p2 is the dominant pole and f H f p2 4. (a) Consider a CS amplifier having C gd = 0.2 pf, R sig = R L = 20 kω, g m =5 ma/v, C gs = 2 pf, C L (including C db ) = 1 pf, and r o = 20 kω. Find (i) the low-frequency gain A M, and (ii) estimate f H using open-circuit time constants. Hence determine the gain-bandwidth (GBW=mid-freq. gain times f H ). A M = g m R L ; f H = 1/(2πτ H ) where τ H = C gs R gs + C gd R gd + C L R L, R gs = R sig, R gd = R sig (1+g m R L )+ R L ; GBW = A M f H 5. Consider the following circuit for the case: I = 200 µa and V OV = 0.25 V, R sig = 200 kω, R D = 50 kω, C gs = C gd = 1 Pf (for both transistors). Find the dc (i.e., lowfrequency) gain, the high-frequency poles, and an estimate of f H. (hint: need to find g m from I and V OV data!). V G1 = V S. [(2/g m )/((2/g m )+R S )], I = V G1 /(2/g m ), V O = IR D hence, A O = V O /V S = g m R D /(2+ g m R S ); f P1 = 1/[2π R S (C gs /2+C gd )]; f P2 = 1/(2πR D C gd ) Assignment 3 ELEC 312/Winter 12 R.Raut, Ph.D.
Page 1 of 5 ELEC 312: ELECTRONICS II : ASSIGNMENT-4 Department of Electrical and Computer Engineering Winter 2012 1. (a) Consider a CS stage having C gd = 0.2 pf, R sig = 20 kω, g m =5 ma/v, C gs = 2 pf, and r o = 20 kω. (b) A CG stage is connected in totem-pole configuration with the CS transistor in (a) to create a cascode amplifier. The ac parameters of this stage are identical with those of the CS stage. Regarding the body-effect in the CG stage assume χ = 0.2. Further R L = 20 kω, and is shunted by a load capacitance C L =1 pf. Show a schematic diagram of the system using NMOS transistors. Show the ac equivalent circuit. Find (i) the low-frequency gain AM, and (ii) estimate the gain-bandwidth of the system. You may use OCTC method to determine the dominant high frequency pole f H of the system. For the cascade amplifier: AC Equivalent circuit:
Page 2 of 5 V gs2 = V g V s2 = 0 V s2 V bs2 = V b V s2 = 0 V s2 For low frequency gain, ignore all C gs and C gd Consider the 2- node system and derive V o / V sig + + = + + = Here = = = Using the values: - 98.37 v/v For Dominant role calculation, note: For C gd1, the Miller effect amplifications are : i) At input (1 + K 1 ) C gd1, K 1 = = =0.8. ii) At input (1 + C gd1 = (1+ 1.2) C gd1
Page 3 of 5 C gd2 does not have miller effect So the AC equivalent circuit is Ignoring r and r as was done in the class lecture. CE-CB Cascade The time constants are: τ 1 = 2.36 10-12 20 10 3 = 4.72 10-8 sec τ 2 =.. = 4.07 10-10 sec τ 3 = 1.2 10-12 20 10 3 sec τ 1, τ 3 are close enough,so dominant time constant principle may not apply
Page 4 of 5 We will take τ H = τ 1 + τ 2 + τ 3 = f H =. =. MHZ GBW = -98.37 2.2 10 6 = 218.6 MHZ 2. For the following circuit, let the bias be such that each transistor is operating at 100-µA collector current. Let the BJTs have h fe = 200, f T = 600 MHz, and C µ = 0.2 pf, and neglect r o and r x. Also, R sig = RC = 50 kω. Show the ac equivalent circuit. Find (i) the low-frequency gain, (ii) the high-frequency poles, and (iii) an estimate of the dominant high frequency pole f H of the system. Now find the GBW (gain-bandwidth) of the system. You may use half-circuit technique. 3. In the following circuit assume both transistors operate in saturation and λ 0. For each transistor you can assume the parasitic capacitances as C gsi, C gdi, (i=1,2).
Page 5 of 5 Draw the ac equivalent circuit, analyze and derive the expression for the dominant pole frequency.