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Download Homework 4 Solutions Problem 1 1) Circuit schematic is shown as
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Homework 4 Solutions Problem 1 1) Circuit schematic is shown as the problem. W/L = 20um/0.35um, n = 4 Size W/L to set the overdrive voltage of M1-M4 and M6 to roughly 100mV. Set n as 3, since the threshold voltage is 490mV, the Vgs of M1-M4 and M6 should be around 590mV. DC simulation result is shown below. In order to make sure that transistor M1-M4 and M6 have the overdrive voltage of approximately 0.1V, the width of transistor has been selected as 20um. 2) Sweep n and determine what (integer) value of n works best. The plot of X, Y, Z is shown below. The plot of M1 overdrive voltage is shown below. n = 4 is a good choice for the circuit. Firstly, increasing n, the voltage of node X, Y and Z will increase, making the transistors more saturated. Then, n = 4 makes the overdrive voltage of M1 closer to 0.1V. If n is larger than 4, the voltage of node X will be high, reducing the output voltage headroom. (Note that, it is also reasonable to choose n = 3.) 3) Perform .op analysis, the small-signal parameters of M1 and M2 are shown below. Small-signal parameters of M1 gds = 83.02u gm = 1.168m region = 2 vds = 200m vgs = 593.6m vth = 490.3m Small-signal parameters of M2 gds = 21.93u gm = 1.255m region = 2 vds = 800m vgs = 578.2m vth = 490.1m Small-signal parameters of M3 gds = 90.42u gm = 1.157m region = 2 vds = 192.6m vgs = 593.6m vth = 490.3m Small-signal parameters of M4 gds = 30.18u gm = 1.225m region = 2 vds = 401m vgs = 585.6m vth = 490.2m Small-signal parameters of M5 gds = 238.7u gm = 458.9u region = 1 vds = 196.5m vgs = 778.2m vth = 490m Small-signal parameters of M6 gds = 24.45u gm = 1.24m region = 2 vds = 581.7m vgs = 581.7m vth = 490.2m 4) Calculate the Ro of M1 and M2 branch Ro ro1 ro 2 gm 2 ro1ro 2 where gds1 = 83.02u, gds2 = 90.42u, gm2 = 1.157m. Therefore, Ro = 746.3kΩ. 5) To test the output resistance Ro, the AC input signal with 1V magnitude is injected into the output of current mirror, and the current flowing out of the AC voltage source is shown below. So, the output impedance is, Ro = 1/1.339u = 746.3kΩ, which is the same as the hand calculation, and both of them are close to gmro2. 6) Plot of Io vs. Vo Plot of Ro vs. Vo For region 1, the output voltage is between 0 and Vov, M1 and M2 both work at the linear region, the output impedance is very low, and the output current is small. Here, the upper voltage limit for region 1 is approximately 0.18V. For region 2, the output voltage is between Vov to 2Vov, M1 goes into the saturation region, and M2 still works in the linear region. The voltage range for region 2 is from 0.18V to 0.36V. For region 3, the output voltage is larger than 2Vov, M1 and M2 both are in the saturation region, the output current and output resistance increase quickly. Here, the lower voltage limit for region 3 is 0.36V. Note that, Vov could change with varying the output voltage Vo. 7) Sweep the value of IB from 50uA to 200uA in 10uA step increment. Plot of the voltage of nodes X, Y and Z and the overdrive voltage of M1 is shown below. Based on the simulation results, the biasing condition of high-swing cascode structure could almost linearly track the master current variation. 8) Connect the bulk of transistors M2, M4 and M6 to ground, the plot of the voltage of node X, Y, Z is shown below. Compared with the simulation results in 7), it could be found that the body effect does not have the effects of the high-swing cascode current mirror. Problem 2 1. “Fractional” band-gap reference (BGR) circuit is shown below. 1) Derive a closed-form expression for the BGR output, VOUT. The voltage across R1 is, VR1 VBE1 VBE 2 VT ln n Where n = 8. So, the current flowing through R1 is, I1 I 2 I R1 VR1 VT ln n R1 R1 The current flowing through R2 is, IR2 VBE1 R2 Since M1, M2 and M3 have the same aspect ratio, the current flowing through M3 should be, I out I R1 I R 2 Then, output voltage is, VT ln n VT ln n R1 R2 V ln n VBE1 Vout I out R3 R3 T R R2 1 2) Determine the ratios of R3/R1 and R2/R1 to produce a 0.5V temperature independent VOUT. Output voltage is, Vout R3 R2 V ln n V T BE1 R2 R1 Where n = 8, VBE1 = 0.7V, VT = 25mV. The factor M is, M R2 ln n 23.5 R1 So, R2 11.30 R1 Besides, Vout R3 R2 VT ln n VBE1 0.5 R2 R1 R3 0.388 R2 So, R3 R3 R2 4.38 R1 R2 R1 3) Consider the finite r0 effect of M1, M2 and M3. Suppose the R1 = 1kΩ and ro = 0.1MΩ, determine the numerical value of VOUT. The current flowing through M1, M2 and M3 is, I D1 I D 2 I D3 I R1 I R 2 VT ln n VBE1 R1 R2 Where R1 = 1kΩ, R2 = 11.3 kΩ, n = 8. So, I D1 I D 2 I D3 VT ln n VBE1 0.114mA R1 R2 Considering the channel-length modulation, for a single MOS transistor, ro 1 ID 1 0.088 ro I D For M1, M2 and M3, the current determined by VGS should be the same. Assume that, I D ,VGS 1 p cox VGS VTH 2 2 Therefore, for M1, I D1 I D,VGS 1 VDS 0.114mA Where |VDS|=1.8V. Then, I D,VGS 0.098mA For M3, I D ,VGS [1 (VDD VOUT )] VOUT R3 Where R3 = 4.38kΩ. Therefore, considering the finite ro effect, output voltage is, VOUT 0.505V 4) Explain intuitively how the two feedback loops formed by the opamp and the left and right arms of the VT-reference circuit can be stable. Based on the schematic show above, the loop from node X to node Y is the positive feedback for the op-amp, and the loop from node X to node Z is the negative feedback for the op-amp. For the positive feedback, the magnitude of gain is, | AV , pos | g m,M 1 (R 2 || 1 g m,Q1 ) Where, gm,M1 is the transconductance of M1, gm,Q1 is the transconductance of Q1. For the negative feedback, the magnitude of gain is, | AV ,neg | g m,M 2 (R 2 || ( 1 g m,Q1 R1 )) Where gm,M1=gm,M2, gm,Q1=gm,Q2. Therefore, the negative feedback gain is larger than the positive feedback, which could make sure that the system could be stable. In this analysis, R1 is critical. 2. Brokaw band-gap reference circuit is shown below. 1) Derive a closed-form expression for the BGR output, VOUT. The voltage across R1 is, VR1 VBE1 VBE 2 VT ln N Where N = 16. So, the current flowing through R1 is, I1 I 2 I R1 VR1 VT ln N R1 R1 The current flowing through R2 is, I R 2 2 I R1 2VT lnN R1 Then, output voltage is, Vout I R 2 R2 VBE1 VBE1 2 R2 ln N VT R1 If M = 23.5, then, M 2 R2 ln N 23.5 R1 Therefore, R2 4.24 R1 2) Still ignore ∆R but include VOS, re-derive the expression for Vout. Considering the input offset voltage VOS of op-amp, the current flowing through Q2 is, I1 VDD VX VR1 R R1 And the current flowing through Q1 is, I2 VDD VX VOS V I1 OS R R Then, the voltage across R1 is, VR1 VBE1 VBE 2 VT ln( N V VX VOS I2 ) VT ln( N DD ) I1 VDD VX Therefore, the output voltage is, Vout VBE1 R2 I1 I 2 VBE1 V VX VOS 2 R2 R VT ln( N DD ) 2 VOS R1 VDD VX R V VX VOS VBE1 VT ln DD RI S 2 Where, IS2 is the saturation current for bipolar transistor Q1. Then, 1 2 R2 / R1 Vout VT VOS VDD VX VOS R2 R In order to minimize the dependence of VOUT on VOS, the ratio of R2/R should be minimized, and the ratio of R2/R1 is determined by the M factor to achieve the temperature-independent output voltage. Besides, the voltage VX should be reduced. 3) Include both ∆R and VOS, re-derive the expression for VOUT. Considering the input offset voltage VOS of op-amp and ∆R, the current flowing through Q2 is, I1 VDD VX VR1 R R1 And the current flowing through Q1 is, (assume that ∆R/R<<1) I2 VDD VX VOS I1R VOS V I1 OS R R R R R R Then, the voltage across R1 is, I V VX VOS R VR1 VBE1 VBE 2 VT ln N 2 VT ln N DD VDD VX R R I1 Therefore, the output voltage is, (assume that ∆R/R<<1) Vout VBE1 R2 ( I1 I 2 ) VBE1 V VX VOS 2 R2 R2 R VT ln( N DD ) VOS R1 VDD VX R R R R VBE1 V VX VOS 2 R2 R2 VT ln( N DD ) VOS R1 VDD VX R R VBE1 VT ln( VDD VX VOS 1 ) R R IS 2 Where, Is2 is the saturation current for bipolar transistor Q1. 4) Explain intuitively how the two feedback loops formed by the opamp and the left and right arms of the VT-reference circuit can be stable. The loop form node X to node Y is the positive feedback for the opamp, and the loop from node X to node Z is the negative feedback. Ignore the early effect, and assume that β>>1, for the positive feedback, the magnitude of the gain is, | AV , pos | g m,Q 2 R 1 g m,Q 2 (R1 R2 ) For the negative feedback, the magnitude of the gain is, | AV ,neg | g m,Q1 R 1 g m,Q1 R2 Assume that, the current flowing through Q1 and Q2 is the same, then gm,Q1 = gm,Q2, therefore, the magnitude of negative feedback gain is larger than that of positive feedback gain, and the system is stable. R1 is critical to make sure that the system is stable, since it could reduce the positive feedback gain with emitter degeneration.