This document describes a PI voltage controller for a DC-DC boost converter. It discusses using a PI controller in a closed-loop feedback system to regulate the output voltage of the boost converter. The proportional and integral terms of the PI controller are implemented using op-amps, with the proportional term providing immediate correction and the integral term providing gradual correction to eliminate steady-state error. Tuning the gains and time constants of the PI controller is important to achieve stable, non-oscillating control of the output voltage.

1. EE462L, Spring 2012
PI Voltage Controller for DC-DC
Boost Converter
1

2. 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

3. The Underlying Theory
error Vpwm Hold to 90V
PI PWM mod. DC-DC
Vset controller and MOSFET conv. Vout
+ driver
–
(scaled down
to about 1.3V)
xisting boost process
Proportional Integral 3

4. 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 ) + ∫ e(t )dt
Ti
• Proportional term: Immediate correction but steady state error (V pwm 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

5. Theory, cont.
work!
Recommended in PI
Ti = 0.8T literature
ζ = 0.65 K p = 0.45
From above curve – gives some
overshoot
5

6. 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

7. !
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

8. !
Example 1. Buffer Amplifier
(converts high impedance signal to low impedance signal)
Vout = K (V+ − V− ) = K(V – V )
in out
–
Vout
Vin + Vout + KVout = KVin
Vout (1 + K ) = KVin
K
Vout = Vin •
1+ K
K is large
Vout = Vin
8

9. !
Example 2. Inverting Amplifier
(used for proportional control signal)
Rf
, so .
Rin
Vin –
Vout KCL at the – node is .
+
Eliminating yields
, so
. For large K, then , so .
9

10. 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
Vb V− − Va V− − Vout
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

11. !
Example 4. Inverting Sum
(used to sum proportional and integral control signals)
− Vout
R Vout = K (0 − V− ) = − KV− , so V− = .
K
R
Va –
Vb
R Vout KCL at the – node is
+
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

12. Example 5. Inverting Integrator !
(used for integral control signal)
~ ~
Using phasor analysis, Vout = K (0 − V− ) , so
Ci ~
Ri ~ = − Vout
V− . KCL at the − node is
Vin – K
Vout
+ ~ ~ ~ ~
V− − Vin V− − Vout
+ = 0.
Ri 1
j ωC
~
− Vout ~
− Vin ~
 − Vout ~ 
~ K + jω C  
Eliminating V− yields
Ri  K − Vout  = 0 . Gathering terms yields
 
~
~  −1 1   Vin ~  −1 1  ~
Vout 
 KR − j ωC  + 1   = , or Vout 
 − jωRi C  + 1  = Vin For large K , the

 i K   Ri
 K K 
~
~ ~ ~ − Vin
expression reduces to Vout (− jωRi C ) = Vin , so Vout = (thus, negative integrator action).
jωRi C
~
For a given frequency and fixed C , increasing Ri reduces the magnitude of Vout .
12

13. Op Amp Implementation of PI Controller
Signal flow
– error Rp
αVout
+
–
–
+ 15kΩ – Vpwm
– +
Difference +
Vset Proportional
+ (Gain = −1) 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