1 EE462L, Fall 2012 PI Voltage Controller for DC-DC Converters.

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

KsG1

)(

Proportional Integral

)(1

)(

)(

)(

sG

sG

sV

sV

set

out

sTsGGsG 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 P

i

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( ) ( )PWM

i

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( ) ( )PWM

i

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 P

i

V t K e t e t dtT

11

iii CRT

sTsTKsG

iP

1

11)(

i

p

Pi

P

set

out

TTT

Kss

KTs

T

K

sV

sV

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

121

212

iinp T

T

TT

TTK

TTi 8.0

65.0 45.0pK

Recommended in PI literature

From above curve – gives some overshoot

Theory, cont.

)(1

)(

)(

)(

sG

sG

sV

sV

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 K

VV out .

KCL at the – node is 0

f

out

in

in

R

VV

R

VV.

Eliminating V yields

0

f

outout

in

inout

R

VK

V

R

VK

V

, so

in

in

ffinout R

V

RKRKRV

111. For large K, then

in

in

f

out

R

V

R

V

, 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

K

VVV outb

2.

KCL at the – node is 0

R

VV

R

VV outa , so

0 outa VVVV , yielding 2outa VV

V

.

Eliminating V yields

22outab

outVVV

KV , so

22about

outVV

KV

KV , or

221 ab

outVV

KK

V .

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 K

VV out

.

KCL at the – node is

0

R

VV

R

VV

R

VV outba , so

outba VVVV 3 .

Substituting for V yields outbaout VVVK

V

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

K

VV out

~~ . KCL at the − node is

01

~~~~

Cj

VV

R

VV out

i

in

.

Eliminating V~

yields 0~

~~~

outout

i

inout

VK

VCj

R

VK

V

. Gathering terms yields

i

in

iout R

V

KCj

KRV

~1

11~

, or iniout VK

CRjK

V~

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 • 104 . The 100kΩ pot is marked “104” meaning 10 • 104 .

– +

– +

– +

– +

– +

– +

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