MOSFET DC BIASING

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MOSFET DC BIASING Powered By Docstoc
					Analogue Electronics I
      EAB 2014



      Jan 2009



 Lecture 8 - MOSFETs
       DC Biasing

                         1
   Topics To Cover

• DC Biasing For Depletion-Type and
  Enhancement-Type MOSFETs




                                      2
     MOSFETs Biasing… why?
 We are interested in finding the quiescent
 point (Q-point) of the response in the
 circuit (current and/or voltage)
 Steps to obtain the Q-point
1. Plot the transfer characteristics curve
   (ID vs VGS)
2. Draw the DC load line (network characteristics)
   superimposed on the curve
3. Obtain the Q-point by finding the intersection
   point of the two plots.


                                                     3
 Depletion type MOSFETs
Use Shockley’s Equation for simpler
calculation @ shorthand method
    VGS versus ID                                      2
                                       ⎛    V GS   ⎞
   VGS           ID      I D = I DSS   ⎜1 −
                                       ⎜           ⎟
                                                   ⎟
    0           IDSS                   ⎝    VP     ⎠
  0.3 VP      IDSS / 2
  0.5 VP      IDSS / 4
    VP           0


                                                           4
     Example 1 (voltage-divider)
For the circuit given,
determine:
a)   IDQ and VGSQ
b)   VDS




                                   5
     Example 1 …
                                                     ID(mA)
                                                                 11
Step 1: Plot the Transfer Characteristics Curve.                 10
                                                                 9
                                                                 8
Using Shockley’s Equation:                                       7
                                                          IDSS   6
When VGS = 0, ID = IDSS                                          5

     ID = 0, VGS = VP                                            4
                                                                 3
When VGS = VP/2 = -1.5 V,                                        2
     ID = IDSS/4 = 1.5 mA         VGS(V)                         1

                                           -3   -2   -1               1
                                           VP


                                                                          6
          Example 1 …
                                                                 ID(mA)
                                                                             11
Consider when VGS = +1 V,                                                    10
                                                                             9
                                      2
                      ⎛    V GS   ⎞                                          8
  I   D   = I   DSS
                      ⎜1 −
                      ⎜           ⎟
                                  ⎟                                          7
                      ⎝    VP     ⎠                                   IDSS   6
                                          2
                    ⎛      1 ⎞
          = (6 mA ) ⎜ 1 −
                                                                             5
                              ⎟
                    ⎝     − 3 ⎠                                              4
                                                                             3
          = 10 . 67 mA
                                                                             2

                                              VGS(V)                         1

                                                       -3   -2   -1               1
                                                       VP


                                                                                      7
      Example 1 …
 Step 2: Draw the DC load line.
Find equation for VGS using
voltage divider relationship:   Characteristics curve
      ⎛     10 M     ⎞
 VG = ⎜              ⎟ ×18V = 1.5V
      ⎝ 10 M + 110 M ⎠                     DC Load Line
 VGS = VG − I D RS = 1.5V − I D (750 Ω )
                                                          Intersection
        Set I D = 0 mA                                      (Q-point)

            VGS = VG = 1.5V
       Set VGS = 0V
                VG
           ID =    = 2 mA
                RS

                                                                     8
   Example 1 …
Step 3: Obtain the Q-point values
a) From the plot:
   IDQ = 3.1 mA and VGSQ = – 0.8 V

b) VDS = VDD – ID (RD + RS)
       = 18 – (3.1m)(1.8k + 750)
       = 10.1 V




                                     9
       Example 2 (self-bias)
For the network given,
determine:
a)   IDQ and VGSQ
b)   VD




                               10
            Example 2 (self-bias)
            Simplify the self-bias network:       20 V
                     20 V



                6.2kΩ                         6.2kΩ
                                 Vo

                            IDSS=8mA                     IDSS=8mA
Vi                                                        V P= -8V
                             VP= -8V


     1M Ω               2.4kΩ                         2.4kΩ




                                                                11
  Example 2 (self-bias)
  Step 1: Plot the Transfer Characteristics Curve.
Using Shockley’s Equation:                                  ID(mA)
When VGS = 0, ID = IDSS                                              11
                                                                     10
       ID = 0, VGS = VP
                                                                     9
                                                              IDSS
When VGS = VP/2 = -4 V,                                              8
      ID = IDSS/4 = 2 mA                                             7
                                                                     6
Consider when VGS = +1 V,                                            5
                                 2
                 ⎛    V   ⎞                                          4
     I D = I DSS ⎜ 1 − GS ⎟
                 ⎜
                 ⎝     VP ⎟
                          ⎠                                          3

                             2                                       2
                 ⎛     1 ⎞
        = (6 m ) ⎜ 1 −   ⎟                                           1
                 ⎝     −8⎠       VGS(V)

        = 10 . 13 mA                      -8 -7 -6 -5 -4 -3 -2 -1
                                          VP
                                                                          1   2



                                                                                  12
   Example 2 (self-bias)
Step 2: Draw the DC load line.
From the input loop:
 VGS = − I D RS = − I D (2.4 kΩ )   Characteristics curve

  Set I D = 0 mA
       VGS = 0V
                                     DC Load Line           Intersection
                                                              (Q-point)

 Set       VGS = −6V
                VG
       ID = −      = 2.5 mA
                RS


                                                                      13
    Example 2 (self-bias)
Step 3: Obtain the Q-point values

      a) From the plot:
         IDQ = 1.7 mA and VGSQ = – 4.3 V

      b) VD = VDD – IDRD
            = 20 – (1.7m)(6.2k)
             = 9.46 V




                                           14
Enhancement-Type MOSFETs
Different equation is used:
          I D = k (V GS − V T               )2
                            I D ( on )
          k =
                   (V   GS ( on )   − VT   )2



             VGS versus ID
          VGS                              ID
         VGS(on)                         ID(on)
           VT                              0

                                                  15
     Example 3 (feedback-biasing)

For the circuit given,
determine:
a)   IDQ and VDSQ




                                16
     Example 3 …
      Simplify the self-bias network:
                  12 V                         12 V



               2kΩ                           2kΩ

                                        Vo
     10M Ω               I D(on)= 6mA                 I D(on)= 6mA

                         VGS(on)= 8V                  VGS(on)= 8V
Vi
                         VGS(th)= 3V                  VGS(th)= 3V




                                                                17
       Example 3 …
       Step 1: Plot the Transfer Characteristics Curve.
                                      I D (on )
Find value of k:        k =
                              (V   GS (on ) − V T   )
                                                    2


                             6m
                         =
                           (8 − 3 )2
                         = 0 . 24 × 10 − 3 A / V        2

Select a few points:
For VGS = 6 V,                     For VGS = 10 V,
ID = k(VGS – VGS(T))2              ID = (0.24m) x (10 – 3)2
  = (0.24m) x (6 – 3)2                 = 11.76 mA
   = 2.61 mA

                                                              18
Example 3 …




              19
     Example 3 …
     Step 2: Draw the DC load line.


                                 Characteristics curve
From output circuit:
VDS = VGS
    = VDD – IDRD
                                  DC Load Line
                                                         Intersection
ID = 0                                                     (Q-point)
VGS = VDD = 12 V

VGS = 0
ID = VDD / RD = 6 mA


                                                                        20
 Example 3 …
Step 3: Obtain the Q-point values

 a) From the plot:
    IDQ = 2.75 mA and VDSQ = 6.4 V




                                     21
     Example 4 (voltage-divider)
For the circuit given,
    determine:
a)   IDQ
b)   VGSQ
c)   VDSQ




                                   22
   Example 4 (voltage divider)
    Step 1: Plot the Transfer Characteristics Curve.
                                       I D (on )
Find value of k:        k =
                              (V GS (on ) − V GS (T ) )2
                                 3m
                          =
                              (10   − 5 )2
                          = 0 . 12 × 10 − 3 A / V     2
Select a few points:
For VGS = 15 V,                 For VGS = 20 V,
ID = k(VGS – VGS(T))2           ID = (0.12m) x (20 – 5)2
  = (0.12m) x (15 – 5)2              = 27 mA
  = 12 mA

                                                           23
Example 4 (voltage divider)
  ID(mA)




   40


   30


   20


   10
 ID(on)

                                                   V GS(V)
             5        10       15   20   25   30
           VGS(th)   VGS(on)


                                                             24
    Example 4 (voltage divider)
 Step 2: Draw the DC load line.
                ⎛ R2      ⎞
         VG   =⎜⎜R +R     ⎟ V DD
                          ⎟
                ⎝ 1   2   ⎠
              = 18 V
From the input circuit:            At ID = 0,
VGS = VG – IDRS                    VGS = 18 V
    = 18 – ID(0.82k)
                                   At VGS = 0,
                                   ID = 21.95 mA

                                                   25
Example 4 (voltage divider)

              Characteristics curve



         DC Load Line



                                      Intersection
                                        (Q-point)




                                                     26
   Example 4 (voltage divider)
Step 3: Obtain the Q-point values
 a) From the plot:
     IDQ = 6.7 mA and VGSQ = 12.5 V
 b) From the output circuit:
   VDS = VDD – ID(RS + RD)
         = 40 – (6.7m)(0.82k + 3k)
         = 14.4 V

                                      27
Question & Answer




                    28

				
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Description: MOSFET DC Biasing for Depletion and Enhancement. Cover Voltage Divider, Self-Bias, and Feedback Bias.