Download Document

EE462L, Fall 2012
PI Voltage Controller for DC-DC
Converters
1
PI Controller for DC-DC Boost Converter Output
Voltage
Vpwm
(0-3.5V)
PWM mod.
and MOSFET
driver
DC-DC
conv.
Vout
(0-120V)
Open Loop, DC-DC Converter Process
error
Vset
+
–
Hold to 90V
Vpwm
PI
controller
PWM mod.
and MOSFET
driver
DC-DC
conv.
Vout
(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
Vset
+
–
Hold to 90V
Vpwm
PI
controller
PWM mod.
and MOSFET
driver
DC-DC
conv.
Vout
(scaled down
to about 1.3V)
G(s)  GPI (s)  GPWM (s)  GDC  DC (s)
1
G PI ( s )  K P 
sTi
Proportional
Integral
1
Gconv ( s)  G PWM  G DC  DC ( s) 
1  sT
Our existing boost process
3
Theory, cont.
error e(t)
Vset
+
–
PI
controller
!
Vpwm
PWM mod.
and MOSFET
driver
VPWM (t )  K P e(t ) 
DC-DC
conv.
Vout
1
e(t )dt

Ti
• 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
Has some “memory”
4
E.g. Buck converter
• Vin = 24 V
• Vout = 16 V (goal)
• L = 200 uH, C = 500 uF, R = 2 Ohm
5
E.g. Buck converter
!
• Vin = 24 V
• Vout = 16 V (goal)
• L = 200 uH, C = 500 uF, R = 2 Ohm
1
VPWM (t )   e(t )dt
Ti
• Ki = 40, Kp = 0
iL
vC
d
e
6
E.g. Buck converter
!
• Vin = 24 V
• Vout = 16 V (goal)
• L = 200 uH, C = 500 uF, R = 2 Ohm
1
VPWM (t )   e(t )dt
Ti
• Ki = 10, Kp = 0
iL
vC
d
e
7
E.g. Buck converter
!
• Vin = 24 V
• Vout = 16 V (goal)
• L = 200 uH, C = 500 uF, R = 2 Ohm
VPWM (t )  K P e(t )
• Ki = 0, Kp = 1
iL
vC
d
e
8
E.g. Buck converter
!
• Vin = 24 V
• Vout = 16 V (goal)
• L = 200 uH, C = 500 uF, R = 2 Ohm
VPWM (t )  K P e(t )
• Ki = 0, Kp = 0.1
iL
vC
d
e
9
E.g. Buck converter
• Vin = 24 V
• Vout = 16 V (goal)
• L = 200 uH, C = 500 uF, R = 2 Ohm
!
1
VPWM (t )  K P e(t )   e(t )dt
Ti
• Ki = 10, Kp = 1
iL
vC
d
e
10
Theory, cont.
Ti  Ri Ci

1 
1
 
G( s)   K P 
sTi  1  sT

Response of Second Order System
(zeta = 0.99, 0.8, 0.6, 0.4, 0.2, 0.1)
0.1
1.8
0.2
1.6
Vout ( s)
G( s)

Vset ( s) 1  G( s)
1.4
1
KP 
1 
 s 

T 
Ti K P 
Vout ( s )

Vset ( s )
1 K p  1
2

s  s

 T  TTi
work!
0.8
0.6
0.2
0
0
2
Ti  0.8T
2 n 
K p  2 nT  1 
0.99
0.4
s 2  2 n s   n2
1
 n2 
TTi
0.4
1.2
TTi
T
 1  2
6
8
10
Recommended in PI
literature
  0.65 K p  0.45
1 K p
2T
4
T
1
Ti
From above curve – gives some
overshoot
11
Improperly Tuned PI Controller
Mostly Integral Control - Oscillation
90V
Figure 11. Closed Loop Response with Mostly Integral Control
(ringing)
Mostly Proportional Control – Sluggish,
Steady-State Error
90V
Figure 12. Closed Loop Response with Mostly Proportional Control
(sluggish)
12
!
Op Amps
V−
I−
–
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
13
!
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
14
Example 2. Inverting Amplifier
(used for proportional control signal)
Rf
Rin
Vin
–
Vout
+
!
V
Vout  K (0  V )   KV , so V   out .
K
V  Vin V  Vout

 0.
KCL at the – node is 
Rin
Rf
Eliminating V yields
V
V
 out  Vin  out  Vout
K
K

 0 , so
Rin
Rf
 1
1
1 
 Vout 



 KRin KR f R f 


Rf
 Vout Vin
Vin

. For large K, then
, so Vout  Vin
.
Rin
Rf
Rin
Rin
15
!
Example 3. Inverting Difference
(used for error signal)
R
R
Va
–
Vout
+
R
Vb
R
V

Vout  K (V  V )  K  b  V  , so
 2

V
V
V  b  out .
2
K
V  Va V  Vout

 0 , so
KCL at the – node is
R
R
V  Vout
.
V  Va  V  Vout  0 , yielding V  a
2
Eliminating V yields
 V  Va 
 V  Va 
 K
V
V
V  Vout 
.
 , or Vout 1    K  b
 , so Vout  K out  K  b
Vout  K  b  a
2
2


2
2




2
 2

For large K , then Vout  (Va  Vb )
16
Example 4. Inverting Sum
(used to sum proportional and integral control signals)
Vout  K (0  V )   KV , so V 
R
Va
Vb
R
–
Vout
R
+
!
 Vout
.
K
KCL at the – node is
V  Va V  Vb V  Vout


 0 , so
R
R
R
3V  Va  Vb  Vout .
 V 
3 
 1  Va  Vb .
Substituting for V yields 3 out   Va  Vb  Vout , so Vout 
K

 K 
Thus, for large K , Vout  (Va  Vb )
17
Example 5. Inverting Integrator
(used for integral control signal)
!
~
~
Using phasor analysis, Vout  K (0  V ) , so
Vin
~
Vout
~
. KCL at the − node is
V  
K
Ci
Ri
–
Vout
+
~
 Vout
~
Eliminating V yields
K
Ri
~
 Vin
~
~
~
~
V  Vin V  Vout

 0.
1
Ri
jC
  V~out ~ 
 jC 
 Vout   0 . Gathering terms yields
 K

~
1
  Vin
1
 ~
~   1
~  1
 jC   1  
 jRi C   1   Vin For large K , the
, or Vout 
Vout 
K
  Ri
K

K
 KRi
~
 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 .
18
Op Amp Implementation of PI Controller
Signal flow
–
αVout
+
–
Vset
+
error
Rp
–
+
–
Proportional
(Gain = −Kp)
Buffers
(Gain = 1)
The 500kΩ pot is marked “504” meaning 50 • 10 4 .
The 100kΩ pot is marked “104” meaning 10 • 10 4 .
Vpwm
+
Difference
(Gain = −1)
Ri is a 500kΩ pot, Rp is a 100kΩ pot, and all other
resistors shown are 100kΩ, except for the 15kΩ
resistor.
–
15kΩ
+
Summer
(Gain = −1
1)
Ci
Ri
–
+
Inverting Integrator
(Time Constant = Ti)
(Note – net gain Kp is unity when, in the open loop condition and with the integrator disabled,
Vpwm is at the desired value)
19