EE462L_PI_Controller_PPT

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posted:: 8/16/2012
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							     EE462L, Spring 2012
PI Voltage Controller for DC-DC
        Boost Converter




                                  1
PI Controller for DC-DC Boost Converter Output
                     Voltage
                                                                                     !
          Vpwm                      PWM mod.           DC-DC        Vout
         (0-3.5V)                  and MOSFET           conv.     (0-120V)
                                       driver

                       Open Loop, DC-DC Converter Process



              error                Vpwm                         Hold to 90V

                         PI                PWM mod.    DC -DC
  Vset                controller          and MOSFET    conv.        Vout
         +                                    driver
          –
                                                                (scaled down
                                                                to about 1.3V)


         DC-DC Converter Process with Closed-Loop PI Controller
                                                                                 2
                         The Underlying Theory


                                    Vout ( s )   G ( s)
                                               
                                    Vset ( s ) 1  G ( s )

              error                Vpwm                                  Hold to 90V

                         PI                PWM mod.          DC -DC
Vset                  controller          and MOSFET          conv.           Vout
         +                                    driver
          –
                                                                         (scaled down
                                                                         to about 1.3V)

                      G ( s)  G PI ( s)  G PWM ( s)  G DC  DC ( s)

                       1                                                            1
   G PI ( s )  K P                       Gconv ( s)  G PWM  G DC  DC ( s) 
                      sTi                                                        1  sT

       Proportional        Integral                    Our existing boost process         3
                                      Theory, cont.                           !
                   error e(t)         Vpwm

                            PI                PWM mod.        DC -DC
       Vset              controller          and MOSFET        conv.   Vout
              +                                  driver
               –



                                                     1
                          VPWM (t )  K P e(t ) 
                                                     Ti  e(t )dt
• Proportional term: Immediate correction but steady state error (Vpwm equals
  zero when there is no error (that is when Vset = Vout)).
• Integral term: Gradual correction
                   Consider the integral as a continuous sum (Riemman’s sum)
                   Thank you to the sum action, Vpwm is not zero when the e = 0


                                                                              4
Ti  Ri Ci                           Theory, cont.
                                                              Response of Second Order System
                1      1                                     (zeta = 0.99, 0.8, 0.6, 0.4, 0.2, 0.1)
G( s)   K P 
                   
                sTi  1  sT
                                                                               0.1
                                            1.8
                                              1.6                     0.2
Vout ( s )   G ( s)
           
                                              1.4
                                                                      0.4
Vset ( s ) 1  G ( s )                        1.2
                                                  1
                                              0.8
    work!      KP         1 
                   s 
                                            0.6

                        Ti K P 
                                                                            0.99
Vout ( s)
          
               T                            0.4
                                              0.2
Vset ( s)        1 K p  1
            s  s
             2
                  T   TT
                                                 0
                                                      0         2            4          6           8       10
                               i
                                                                       Recommended in PI
                                                          Ti  0.8T    literature
s 2  2 n s  n
                 2

                           1 K p                           0.65 K p  0.45
      1
n 
 2
                 2 n 
     TTi                       T                      From above curve – gives some
                                                      overshoot
                         2T              T
 K p  2 nT  1              1  2      1                                                         5
                         TTi              Ti
                                  Improperly Tuned PI Controller


                                                               Mostly Proportional Control – Sluggish,
Mostly Integral Control - Oscillation                                    Steady-State Error




                                                 90V                                                               90V




Figure 11. Closed Loop Response with Mostly Integral Control    Figure 12. Closed Loop Response with Mostly Proportional Control
                         (ringing)                                                         (sluggish)




                                                                                                                       6
                                                                             !
                                Op Amps
                           I−
                     V−          –

                                               Vout
                           I+
                     V+          +


Assumptions for ideal op amp

      Vout = K(V+ − V− ), K large (hundreds of thousands, or one million)

      I+ = I− = 0

      Voltages are with respect to power supply ground (not shown)
      Output current is not limited


                                                                             7
                                                              !
               Example 1. Buffer Amplifier
(converts high impedance signal to low impedance signal)

                        Vout  K (V+  V− ) = K(Vin – Vout)
         –
                Vout
  Vin    +              Vout  KVout  KVin

                        Vout (1  K )  KVin

                                       K
                        Vout  Vin 
                                     1 K

                        K is large

                         Vout  Vin
                                                              8
                                                                                      !
                    Example 2. Inverting Amplifier
                 (used for proportional control signal)


                       Rf                                                       V
                                          Vout  K (0  V )   KV , so V   out .
           Rin                                                                    K
     Vin           –                                             V  Vin V  Vout
                            Vout          KCL at the – node is                      0.
                   +                                               Rin         Rf
                                          Eliminating V yields

                                           V           V
                                           out  Vin  out  Vout
                                            K          K           0 , so
                                             Rin          Rf

        1     1    1       Vin                      Vout Vin                   Rf
 Vout                      . For large K, then            , so Vout  Vin     .
        KRin KR f R f      Rin                      Rf     Rin                  Rin
                      


                                                                                      9
                       Example 3. Inverting Difference
                                                                                                       !
                           (used for error signal)

                                                                        V        
               R           R                    Vout  K (V  V )  K  b  V  , so
          Va                                                             2       
                       –                             V     V
                                   Vout         V  b  out .
               R       +                              2     K
                                                                      V  Va V  Vout
          Vb
                                                KCL at the – node is                      0 , so
                   R                                                     R           R
                                                                                        V  Vout
                                                V  Va  V  Vout  0 , yielding V  a          .
                                                                                            2
                                                Eliminating V yields

         V   V  Vout               V         V  Va             K        V  Va 
Vout  K  b  a        , so Vout  K out  K  b       , or Vout 1    K  b      .
          2     2                     2         2                  2      2 


                               For large K , then Vout  ( a  Vb )
                                                           V

                                                                                                  10
                                                                                                !
              Example 4. Inverting Sum
 (used to sum proportional and integral control signals)

                                                                                      Vout
                         R                    Vout  K (0  V )   KV , so V           .
                                                                                       K
             R
       Va            –
             R                 Vout           KCL at the – node is
       Vb            +

                                              V  Va V  Vb V  Vout
                                                                       0 , so
                                                 R       R        R

                                              3V  Va  Vb  Vout .

                             V                              3 
Substituting for V yields 3 out   Va  Vb  Vout , so Vout    1  Va  Vb .
                             K                               K    


                             Thus, for large K , Vout  ( a  Vb )
                                                          V

                                                                                            11
                   Example 5. Inverting Integrator                                                      !
                  (used for integral control signal)
                                                                             ~             ~
                                                      Using phasor analysis, Vout  K (0  V ) , so

                        Ci                                   ~
                                                      ~      Vout
            Ri
                                                      V        . KCL at the − node is
   Vin            –                                           K
                             Vout
                  +                                   ~     ~   ~    ~
                                                      V  Vin V  Vout
                                                                         0.
                                                         Ri        1
                                                                  jC
                           ~
                          Vout         ~
                                       Vin              ~
                                                       Vout ~ 
            ~                                  j C           
                                                      K  Vout   0 . Gathering terms yields
Eliminating V yields        K
                                 Ri                            
                               ~
 ~  1             1     Vin       ~  1             1     ~
Vout 
      KR    j  C   1  
                                , or Vout 
                                                 jRi C   1   Vin For large K , the
                                                                 
         i         K     Ri            K             K    
                                                            ~
                                                           Vin
expression reduces to Vout ( jRi C )  Vin , so Vout 
                       ~                 ~        ~
                                                                (thus, negative integrator action).
                                                         jRi C
                                                                              ~
For a given frequency and fixed C , increasing Ri reduces the magnitude of Vout .
                                                                                                       12
                 Op Amp Implementation of PI Controller
                                       Signal flow



                –                                         error            Rp
    αVout
                +
                                              –
                                                                       –
                                              +              15kΩ                                      –             Vpwm
                –                                                      +
                                             Difference                                                +
      Vset
                +                           (Gain = −1)                     Proportional
                                                                                                            Summer
                                                                           (Gain = −Kp)
                                                                                                           (Gain = −1
                                                                                                                    1)
                 Buffers
                (Gain = 1)
                                                                                Ci
                                                                  Ri
Ri is a 500kΩ pot, Rp is a 100kΩ pot, and all other
                                                                       –
resistors shown are 100kΩ, except for the 15kΩ
resistor.                                                              +
                                                                        Inverting Integrator
The 500kΩ pot is marked “504” meaning 50 • 10 4 .                      (Time Constant = Ti)
The 100kΩ pot is marked “104” meaning 10 • 10 4 .


                (Note – net gain Kp is unity when, in the open loop condition and with the integrator disabled,
                                               Vpwm is at the desired value)
                                                                                                                         13