EECE 310 Homework 8 with Solution, Fall 2007

American University of Beirut Department of Electrical and Computer Engineering EECE 310 – Electronics Fall 2007 – 2008 Homework 8 The MOSFET in the circuit has Vt = 1.2 V, k’(W/L) = 1 mA/V2, and VA = 1/λ = 50 V. RG +9 V RD RS -9 V a) Find the values of RS, RD, and RG so that ID = 1 mA, the largest possible value of RD is used while a maximum signal swing at the drain of ± 2.5 V is possible keeping the MOSFET in the saturation region, and the input resistance at the gate is 5 MΩ. b) Find the values of gm and ro at the bias point. c) If the amplifier input (at the gate) is connected to a source with a 100 KΩ source resistance, and to a load (at the drain) with a resistance of 50 KΩ, find the voltage gain and the power gain from signal source to output. d) To increase the voltage gain, a capacitor is used to bypass part of the resistor RS. How much of RS should be bypassed to obtain a voltage gain of –5 from signal source to load? e) Verify your results using PSpice. Assume that the source and load are coupled to the MOSFET gate and drain, respectively, using capacitors. All capacitor values are 500μF, including the one used to partially bypass RS. Use a sinusoidal source with a frequency of 1 KHz. Show the voltage gain, current gain, and maximum voltage swing at the drain. e) For unbypassed Rs M1 menhn W = 100um L = 10um -1.000mA 0A 1.000mA C2 500uF V 0V 0V RS 6.44k RL 50k 0A Vo: red, Vi in green. Av=Vo/Vi= ‐0.897.248/1= ‐0.897.248 out Rg 5000K 0A Rd 7.7k 0V 0V V in C1 500uF Vs FREQ = 1KHz VAMPL = 1mV VOFF = 0 0A V1 9Vdc 9.000V 1.000mA V2 9Vdc 1.303V 1.000mA R7 100K Time 5.0ms 5.5ms 6.0ms 6.5ms 7.0ms 7.5ms 8.0ms V(IN) V(OUT) -1.0mV -0.5mV 0V 0.5mV 1.0mV (5.7487m,897.248u) Io in red and Ii in green Time 5.0ms 5.5ms 6.0ms 6.5ms 7.0ms 7.5ms 8.0ms -I(Rg) -I(RL) -20nA -10nA 0A 10nA 20nA (5.7450m,17.945n) (5.7450m,-195.983p) Ai=Io/Ii = ‐17.945n/195.983p =91.564 A/A 1 pt Po in red and Pi in green Time 5.0ms 5.5ms 6.0ms 6.5ms 7.0ms 7.5ms 8.0ms W(Rg) W(RL) 0W 5pW 10pW 15pW 20pW (5.2450m,16.101p) (5.2450m,192.045f) Ap=Po/Pi = 16.1p/192f =83.85 W/W To find the maximum swing at the drain we can change Vs until VD=VG‐ Vt Note: The maximum swing here is less than ±2.5V since when we designed the circuit and calculated the value of RD, we neglected the swing at the gate (swing at drain /Gain). VD‐2.5 > VG + (2.5/Av) ‐ Vt However Av turned out to be less than 1 and the swing of 2.5 requires Vg to change by more than 2.5V which can no longer be neglected. That is why the maximum swing at the drain here is less than 2.5V. Trace of VG‐ Vt and VD Time 5.0ms 5.5ms 6.0ms 6.5ms 7.0ms 7.5ms 8.0ms V(M1:d) V(M1:g)-1.2 -4.0V -2.0V 0V 2.0V 4.0V (6.2498m,134.531m) (5.7498m,2.4677) We find that for Vs=1.3V (amplitude) VD starts to become equal to VG‐ Vt for some values. So for larger values of Vs we will have distortion. So the maximum swing will be ±1.17V e) For bypassed Rs Vo: red, Vi in green. Av=Vo/Vi= ‐4.993/1= ‐4.993 V/V V2 9Vdc out RL 50k 0A 1.303V V1 9Vdc 1.000mA M1 menhn W = 100um L = 10um -1.000mA 0A 1.000mA Vs2 FREQ = 1KHz VAMPL = 1mV VOFF = 0 0A 9.000V R7 100K 0A C1 500uF 0V V Rd 7.7k Rsb 5.903k in 0V -3.099V 0V C2 500uF 0V V C3 500uF RS 537 Rg 5000K 0A Time 5.0ms 5.5ms 6.0ms 6.5ms 7.0ms 7.5ms 8.0ms V(IN) V(OUT) -5.0mV 0V 5.0mV (5.7450m,4.9938m) Io in red and Ii in green Time 5.0ms 5.5ms 6.0ms 6.5ms 7.0ms 7.5ms 8.0ms 1 -I(RL) 2 I(Vs2) -100nA -50nA 0A 50nA 100nA 1 -1.0nA -0.5nA 0A 0.5nA 1.0nA 2 >> (5.7450m,195.983p) (5.7450m,99.877n) Ai=Io/Ii = 99.877n/195.983p =510 A/A Po in red and Pi in green Time 5.0ms 5.5ms 6.0ms 6.5ms 7.0ms 7.5ms 8.0ms 1 W(Rg) 2 W(RL) 0W 50fW 100fW 150fW 1 200fW 0W 200pW 400pW 2 600pW >> (5.2450m,498.394p) (5.2472m,192.045f) Ap=Po/Pi = 498.4p/192f =2595 W/W To find the maximum swing at the drain we can change Vs until VD=VG‐ Vt Note: The maximum swing here is slightly less than ±2.5V since now the gain is ‐5 and the swing at the gate can be neglected Trace of VG‐ Vt and VD Time 5.0ms 5.5ms 6.0ms 6.5ms 7.0ms 7.5ms 8.0ms V(M1:d) V(M1:g)-1.2 -2.0V 0V 2.0V 4.0V (5.2453m,-791.196m) (5.7421m,3.2833) We find that for Vs=0.41V (amplitude) VD starts to become equal to VG‐ Vt for some values. So for larger values of Vs we will have distortion. So the maximum swing will be ±2.04V