Ee462 l pi_controller_ppt
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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.
Ee462 l pi_controller_ppt
- 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