1

EE462L, Fall 2012PI Voltage Controller for DC-DC

Converters

2

Vpwm (0-3.5V)

PWM mod. and MOSFET

driver

DC-DC conv.

Vout (0-120V)

Vset Vout

(scaled down to about 1.3V)

PI controller

PWM mod. and MOSFET

driver

DC-DC conv.

error

+ –

Open Loop, DC-DC Converter Process

DC-DC Converter Process with Closed-Loop PI Controller

PI Controller for DC-DC Boost Converter Output Voltage

Hold to 90VVpwm

!

3

Vset Vout

(scaled down to about 1.3V)

PI controller

PWM mod. and MOSFET

driver

DC-DC conv.

error

+ –

The Underlying Theory

Hold to 90VVpwm

)()()()( sGsGsGsG DCDCPWMPI

iPPI sT

KsG 1)(

Proportional Integral

)(1

)()()(

sGsG

sVsV

set

out

sT

sGGsG DCDCPWMconv 1

1)()(

Our existing boost process

4

Vset Vout

PI controller

PWM mod. and MOSFET

driver

DC-DC conv.

error e(t)

+ –

Vpwm

1( ) ( ) ( )PWM Pi

V t K e t e t dtT

Theory, cont.

• 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”

!

5

E.g. Buck converter• Vin = 24 V

• Vout = 16 V (goal)

• L = 200 uH, C = 500 uF, R = 2 Ohm

6

E.g. Buck converter• Vin = 24 V

• Vout = 16 V (goal)

• L = 200 uH, C = 500 uF, R = 2 Ohm

• Ki = 40, Kp = 0

!

iL

vC

d

e

1( ) ( )PWMi

V t e t dtT

7

E.g. Buck converter• Vin = 24 V

• Vout = 16 V (goal)

• L = 200 uH, C = 500 uF, R = 2 Ohm

• Ki = 10, Kp = 0

!

iL

vC

d

e

1( ) ( )PWMi

V t e t dtT

8

E.g. Buck converter• Vin = 24 V

• Vout = 16 V (goal)

• L = 200 uH, C = 500 uF, R = 2 Ohm

• Ki = 0, Kp = 1

!

iL

vC

d

e

( ) ( )PWM PV t K e t

9

E.g. Buck converter• Vin = 24 V

• Vout = 16 V (goal)

• L = 200 uH, C = 500 uF, R = 2 Ohm

• Ki = 0, Kp = 0.1

!

iL

vC

d

e

( ) ( )PWM PV t K e t

10

E.g. Buck converter• Vin = 24 V

• Vout = 16 V (goal)

• L = 200 uH, C = 500 uF, R = 2 Ohm

• Ki = 10, Kp = 1

!

iL

vC

d

e

1( ) ( ) ( )PWM Pi

V t K e t e t dtT

11

iii CRT

sTsTKsG

iP

111)(

i

p

Pi

P

set

out

TTT

Kss

KTs

TK

sVsV

11

1

)()(

2

22 2 nnss

Response of Second Order System(zeta = 0.99, 0.8, 0.6, 0.4, 0.2, 0.1)

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

0 2 4 6 8 10

0.99

0.1

0.4

0.2

in TT

12

T

K pn

12

121212

iinp T

TTTTTK

TTi 8.0

65.0 45.0pK

Recommended in PI literature

From above curve – gives some overshoot

Theory, cont.

)(1

)()()(

sGsG

sVsV

set

out

work!

12

Improperly Tuned PI Controller

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

Mostly Proportional Control – Sluggish, Steady-State Error

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

Mostly Integral Control - Oscillation

90V 90V

13

Op Amps

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

– + V+

Vout

V− I−

I+

!

14

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

– + Vin

Vout

) = K(Vin – Vout)( −+out VVKV

inoutout KVKVV

inout KVKV )1(

KVV inout

1

K

K is large

inout VV

!

15

Example 2. Inverting Amplifier(used for proportional control signal)

– +

Rf

Rin Vin

Vout

KVVKVout )0( , so KV

V out .

KCL at the – node is 0

f

out

in

inRVV

RVV

.

Eliminating V yields

0

f

outout

in

inout

R

VKV

R

VKV

, so

in

in

ffinout R

VRKRKR

V

111 . For large K, then in

in

f

outRV

RV

, so in

finout R

RVV .

!

16

Example 3. Inverting Difference(used for error signal)

– +

Vout

Va

Vb R

R R

R

VV

KVVKV bout 2

)( , so

KVV

V outb 2.

KCL at the – node is 0

RVV

RVV outa , so

0 outa VVVV , yielding 2outa VV

V

.

Eliminating V yields

22outab

outVVV

KV , so

22about

outVV

KV

KV , or

221 ab

outVV

KKV .

For large K , then baout VVV

!

17

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

– +

Vout Va

Vb

R

R

R

KVVKVout )0( , so KV

V out .

KCL at the – node is

0

RVV

RVV

RVV outba , so

outba VVVV 3 .

Substituting for V yields outbaout VVVKV

3 , so baout VVK

V

13 .

Thus, for large K , baout VVV

!

18

Example 5. Inverting Integrator(used for integral control signal)

– +

Ci Ri

Vin Vout

Using phasor analysis, )~0(~ VKVout , so

KV

V out~

~ . KCL at the − node is

01

~~~~

Cj

VVRVV out

i

in

.

Eliminating V~ yields 0~

~~~

outout

i

inout

VKV

CjR

VKV

. Gathering terms yields

i

in

iout R

VK

CjKR

V~

111~

, or iniout V

KCRj

KV ~111~

For large K , the

expression reduces to iniout VCRjV ~~ , so CRj

VV

i

inout

~~

(thus, negative integrator action).

For a given frequency and fixed C , increasing iR reduces the magnitude of outV~ .

!

19

(Note – net gain Kp is unity when, in the open loop condition and with the integrator disabled,

Vpwm is at the desired value)

Ri is a 500kΩ pot, Rp is a 100kΩ pot, and all other resistors shown are 100kΩ, except for the 15kΩ resistor. The 500kΩ pot is marked “504” meaning 50 • 10 4 . The 100kΩ pot is marked “104” meaning 10 • 10 4 .

– +

– +

– +

– +

– +

– +

error

Summer (Gain = −11)

Proportional (Gain = −Kp)

Inverting Integrator (Time Constant = Ti)

Buffers (Gain = 1)

Vset

αVout

Vpwm

Rp

Ci Ri

15kΩ

Difference (Gain = −1)

Op Amp Implementation of PI ControllerSignal flow