# CMOS BIAS GENERATORS

Document Sample

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 – 3V+.58V               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 + 2.83V

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.

DOCUMENT INFO
Shared By:
Categories:
Stats:
 views: 25 posted: 12/19/2010 language: English pages: 11
pptfiles
About