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CMOS BIAS GENERATORS Master – Constant gm with Temp – The early generation R = [1-2/3]{2/gm} VB1' = [2/3]{1/gm} =185 ohmsw gm = 8xWs/Ls kp Veff = 8x6/0.4x 2.5x60uA/V^2 x1/5 =3.6mS 1/8 Ws/Ls Veff Inter digitate 2 {8xWs}/Ls {(9/4)8xWs}/Ls VB1 kRxWL/LL 8xWL/LL 8xWL/LL 144 uA/leg Veff = 200mV 8xWL/LL 144 uA/leg VB3 8xWL/LL LL = 1 um, LLP = 1 um, LS = 0.4 um, WS = 6 um, WLP = 15 um, WL = 6 um, VB4 200 fF 8xWL/LL 8xWL/LL VB4 8xWL/LL 8xWL/LL Inter digitate 4 Lerr < 5.5% Werr < 0.24% gm = 10ms, kpn = 150uA/V^2 Ws/Ls >???? 1. Determine the value of ID for all legs of current in the bias generator. 2. Determine the value of VB1, VB3, and VB4. 3. Find a general equation describing the minimum supply voltage for the bipolar bias generator. 4. Show gm is independent of temperature if R can be composed of a temperature independent resistor. 5. R is composed of two poly resistors; Rpn(1 + kn T) and Rpp(1 - kp T) where kp = -1.94x10-4/C, kn -1.79 x10-4/C, Rshn =89.6 ohms/sq, and Rshn =402 ohms/sq. Determine the ratio of lengths Rn to Rp required to develop a temperature independent resistor. Assume both resistors are 20 w minimum. SLAVE – The early generation VDD 8xWLp/LLp Veff = 200mV Inter digitate 3 VB1 = VDD – VT – V 8xkRxWL/LL 8xkRxWL/LL 16xkRxWL/LL LL = 1 um, LLP = 1 um, LS = 0.4 um, WS = 6 um, VB1 VB2 = VDD – VT – 3V+.58V WLP = 15 um, WL = 6 um, VB2 VB2 2xkRxWS/LS 48xkRxWS/LS 24xkRxWS/LS 288 uA/leg Inter digitate 3 144 uA/leg VB4 = VSS + VT + V 144 uA/leg VB4 VB3 = VSS + VT + 2.83V 8xWL/LL 2xWL/LL VB3 Inter digitate 2 gm = 10ms, kpn = 160uA/V^2 Ws/Ls > ??? VSS Lerr < 5.5% Werr < 0.24% 1. Determine the value of ID for all legs of current in the bias generator where the input current is from the master bias generator. 2. Determine the value of VB1, VB3, and VB4. Master ALL in one – Constant gm with Temp - wide swing generation R = [1-2/3]{2/gm} = [2/3]{1/gm} =185 ohms VB1' w gm = 8xWs/Ls kp Veff = 8x6/0.4x 2.5x60uA/V^2 x1/5 {2xWL}/LL =3.6mS 1/8 Ws/Ls Veff Inter digitate 2 VB2 VB1 8xWL/LL 8xWL/LL VB1 VB2 kRxWL/LL VB2 {8xWs}/Ls VB2 {8xWs}/Ls {8xWs}/Ls VB4 Veff = 200mV VB3 {8xWs}/Ls VB3 {8xWs}/Ls 144 uA/leg {8xWs}/Ls VB3 144 uA/leg 2xWL/LL {8xWs}/Ls {8xWs}/Ls VB4 200 fF 8xWL/LL 8xWL/LL VB4 VB4 8xWL/LL 8xWL/LL 8xWL/LL Inter digitate 4 Lerr < 5.5% Werr < 0.24% LL = 1 um, LLP = 1 um, gm = 10ms, kpn = 150uA/V^2 LS = 0.4 um, WS = 6 um, Ws/Ls >???? WLP = 15 um, WL = 6 um, 1. Determine the value of ID for all legs of current in the bias generator. 2. Determine the value of VB1, VB2, VB3, and VB4. 3. Find a general equation describing the minimum supply voltage for the bipolar bias generator. Master – Constant gm with Temp - wide swing generation R = [1-2/3]{2/gm} = [2/3]{1/gm} =185 ohms VB1' w gm = 8xWs/Ls kp Veff = 8x6/0.4x 2.5x60uA/V^2 x1/5 {2xWL}/LL =3.6mS 1/8 Ws/Ls Veff Inter digitate 2 VB2 VB1 8xWL/LL 8xWL/LL VB1 {(9/4)8xWL}/LL VB2 kRxWL/LL 8xWL/LL VB2 8xWL/LL VB2 8xWL/LL VB4 Veff = 200mV VB3 VB3 {8xWs}/Ls 8xWL/LL 8xWL/LL 144 uA/leg VB3 144 uA/leg 2xWL/LL 8xWL/LL VB4 200 fF 8xWL/LL 8xWL/LL VB4 VB4 8xWL/LL 8xWL/LL 8xWL/LL Inter digitate 4 Lerr < 5.5% Werr < 0.24% LL = 1 um, LLP = 1 um, gm = 10ms, kpn = 150uA/V^2 LS = 0.4 um, WS = 6 um, Ws/Ls >???? WLP = 15 um, WL = 6 um, 1. Comment on the difference between the two wide swing master bias generators above. Be specific on the advantages of each. Use the geometries to guide your discussion. SLAVE - wide swing generation 2. Using the ideas presented above with wide swing mirrors develop a slave bias generator to develop VB1, VB2, VB3, and VB4. Assume the slave generator current per leg is ½ that of the master. Insert Visio drawing here. ALL in ONE VB1 w R = [4/3]{1/gm} WLp/LLp gm =8xWs/Ls kp Veff 8xWLp/LLp 1/8 Ws/Ls Veff Inter digitate 2 {(9/4)8xWs}/Ls VB2 VB1' VB1 kRxWL/LL 16xWL/LL 2xWL/LL Veff = 200mV VB3 L = 2um, LLP = 1um, LS = 0.4um, WS = 6um, WLP = 18um, WL = 6um, VB4 200 fF 8xWL/LL 8xWL/LL VB4 8xWL/LL 8xWL/LL 75 uA/leg Inter digitate 4 Lerr < 5.5% Werr < 0.24% gm = 10ms, kpn = 170uA/V^2 Ws/Ls > 16fx4.25/0.4 BIPOLAR bias generator 25 uA/leg kRxWL/LL kRxWL/LL kRxWL/LL kRxWL/LL VB1 10 pf M1 M2 VB1 VB1 VB2 =VCM kRxWL/LL 0.5Ws/Ls Q3 Q4 Q1 Q2 A:1 where A = 8 VB4 1 pF I =UT/RBLn(A ratio) 10 pf gm =1/RBLn(A ratio) VB4 WL/LL WL/LL WL/LL = 2WS/2LS 1. Show by derivation that the bias current generated by the bipolar bias generator above is proportional to temperature. 2. Find a general equation describing the minimum supply voltage for the bipolar bias generator. 3. Find the geometry of M1 and M2 in terms of the bias resistor and bipolar ratio, veff and KPp. BIPOLAR (Con’t) 25 uA/leg Ic = Is e^(VBE/nUt) kRxWL/LL kRxWL/LL VBE3 + VBE2 = IRB + VBE4 + VBE3 VB1 Assuming IB << IC RB = Ut/I LN[(A1 A4)/(A2 A3)] I = Ut/R LN[(A1 A4)/(A2 A3)] Q3 Q1 I PTAT A3 A1 gm = I/ Ut Q4 Q2 gm = k/RB LN[(A1 A4)/(A2 A3)] A4 A2 I =UT/RBLn(A ratio) Constant for All Temp RB gm =1/RBLn(A ratio) gm = K/R 1. Using the above ideas from bipolar generators develop an equivalent MOS subthreshold generator. Assume each leg is to have an ID of 100nA. BIPOLAR (Con’t) kR WL/LL VB1 Rc Vin- Vin+ Re Vo- Vo+ Vcm = VB2 Rs = 500 ohm 8.64 mA/leg 1.08 mA/leg 4 WL/LL 4 WL/LL ½ WL/LL ½ WL/LL VB4 VB4 w Vcmfb gm = k/R LN[(A1 A4)/(A2 A3)] ADiff = RC/(1 + gmRS) Constant for All Temp ADiff = RC/(1 + K RS/RB) gm = K/R 1. Derive the transconductance for the amplifier above showing all work. 2. Derive the differential gain in terms of the bias resistor, RB above showing all your work.

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posted: | 12/19/2010 |

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