# EE462L_PI_Controller_PPT

Document Sample

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

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

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)).
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

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 views: 6 posted: 8/17/2012 language: Norwegian pages: 13