Homework 2 Solutions. Perform.op analysis, the small-signal parameters of M1 and M2 are shown below.

Transcription

1 Problem 1 Homework 2 Solutions 1) Circuit schematic Perform.op analysis, the small-signal parameters of M1 and M2 are shown below. Small-signal parameters of M1 gds = 9.723u gm = 234.5u region = 2 vds = vgs = 1.256

2 vth = 488m Small-signal parameters of M2 gds = 9.723u gm = 234.5u region = 2 vds = vgs = vth = 488m M1 and M2 both are in the saturation region. 2) For M1, Small-signal gain is, A G R G 234.5u VM, 1 m out m out ds1 1 So, small-signal gain of M1 is For M2, Small-signal gain is, A G R G 234.5u VM, 1 m out R 1/ g R 9.11k m out ds1 1 So, small-signal gain of M2 is Total voltage gain is, AV AV, M1 AV, M R 1/ g R 9.11k 3) Plot of voltage gain vs. output voltage

3 In order to plot the voltage gain versus output voltage, we need to create the feedback loop and sweep DC output voltage, the schematic is shown below. With the above schematic, we could sweep DC output voltage. Assume that the differential amplifier is the ideal VCVS with very large gain, and its maximum and minimum output voltage is 2.5V and 0V respectively. Therefore, DC output voltage is equal to the input of amplifier and sweeping it is considered as sweeping output voltage. The input voltage of M1, Vi, should change with sweeping output voltage, so that we could calculate the voltage gain. (In calculator, select derive from the special function, for example, use deriv(vo)/deriv(vi) to calculate the small signal gain). The simulation results are shown below.

4 4) Plot of small-signal gain vs. frequency

5 Since the AC magnitude of input voltage source is 1, output voltage is directly equal to the voltage gain. Based on the simulation results, the voltage gain is also 4.565, which is almost the same as that of calculation results in problem Output impedance is, R 1/ g R 9.11k o ds Schematic to extract output impedance Ro, Ro = Vx/Ix, if we set AC magnitude Vx=1, then output impedance Ro = 1/Ix. Based on the schematic shown above, the simulation result of Ix is shown below.

6 Ix = uA, therefore, Ro = 1/109.72u = 9.11kΩ, which is the same as the calculated output impedance in problem 5. 7) Large-signal swing plot,

7 The upper limit of large-signal swing is the voltage across the resistor R2, the maximum value for output voltage swing is V DD, when M2 is cut-off. The low limit of large-signal swing is the overdrive voltage of M1. In order to make sure that M1 is in the saturation region, output voltage should be larger than V ov,m1. Problem 2 1) Perform.op analysis, the small-signal parameters of M1 and M2 are shown below. Small-signal parameters of M1 gds = 31.98u

8 gm = 1.561m gmbs = 235.8u region = 2 vds = vgs = 801.1m vth = 721.6m Small-signal parameters of M2 gds = 5.255u gm = 272.3u gmbs = 55.1u region = 2 vds = vgs = 796m vth = 579.8m M1 and M2 are both in the saturation region. 2) For M1, Small-signal gain is, A G R G 1.561m VM, 1 m out out m1 mb1 ds1 1 m R 1/ g 1/ g 1/ g R 0.518k So, small-signal gain of M1 is

9 For M2, Small-signal gain is, A G R G 272.3u VM, 2 m out out m2 mb2 ds2 2 m R 1/ g 1/ g 1/ g R 2.31k So, small-signal gain of M2 is Total voltage gain is, A A, 1 A, V V M V M 3) Plot of voltage gain vs. output voltage 4) Plot of small-signal gain vs. frequency

10 Since the AC magnitude of input voltage source is 1, output voltage is directly equal to the voltage gain. Based on the simulation results, the voltage gain is 0.509, which is the same as that of calculation results in problem 2. 5) Output impedance is, R 1/ g 1/ g 1/ g R 2.31k o m2 mb2 ds2 2 6)

11 Ro=Vx/Ix, if we set AC magnitude as Vx=1, then the output impedance Ro=1/Ix. Based on the schematic shown above, the simulation result of Ix is shown below.

12 Ix=432.53u, therefore, Ro=1/432.53u=2.31kΩ, which is the same as the calculated output impedance in problem 5. 7) Large-signal swing plot

13 The upper limit of large-signal swing is the overdrive voltage of M1 and the Gate-to-Source voltage of M2, the maximum value for output voltage is V DD -V ov,m1 -V GS,M2. (Note that, if we assume that the maximum input voltage is V DD, the maximum output voltage should be V DD -V GS,M1 -V GS,M2.) The low limit of large-signal swing is very close to the ground, when M2 is cut-off. Connect the bulk to the source for both two transistors, and repeat above 1)

14 (Note, the size of M1 and M2 are changed, however, if the size is not changed in the homework, the answer is still acceptable as long as the simulation is rational.) 1) Perform.op analysis, the results are shown as the following part. Small-signal parameters of M1 gds = 13.48u gm = 597.4u region = 2 vds = vgs = 799.2m vth = 489.6m Small-signal parameters of M2 gds = 3.986u

15 gm = 198.1u region = 2 vds = vgs = 794.7m vth = 488.1m 2) For M1, Small-signal gain is, A G R G 597.4u VM, 1 m out R 1/ g 1/ g R 1.41k m out m1 ds1 1 So, small-signal gain of M1 is For M2, Small-signal gain is, A G R G 198.1u VM, 2 m out R 1/ g 1/ g R 3.31k m out m2 ds2 2 So, small-signal gain of M2 is Total voltage gain is, A A, 1 A, V V M V M 3) Plot of voltage gain vs. output voltage

16 4) Plot of small-signal gain vs. frequency Since the AC magnitude of input voltage source is 1, output voltage is directly equal to the voltage gain. Based on the simulation results, the

17 voltage gain is 0.551, which is almost the same as that of calculation results in problem 2. 5) Output impedance is Ro gm 1/ g R 3.31k 6) 1/ 2 ds2 2

18 The analysis part is the same as the last question. Ix=302.02u, therefore, Ro=1/302.02u=3.31kΩ, which is the same as the calculated output impedance in problem 5. 7) Large-signal swing plot

19 The analysis part is the same as the previous part.