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Small-Signal Modeling at Work with Power Converters Christophe Basso IEEE Senior Member 1 • Chris Basso – APEC 2014 Course Agenda Introducing the PWM Switch Model CCM, DCM and BCM in Voltage Mode Pulse Width Modulator Gain The PWM Switch Model in Current Mode PWM Switch at Work in a Buck Converter A Simplified Approach to Modeling a DCM Boost Transfer Function of a BCM Boost in Current Mode Small-Signal Model of The Active Clamp Forward 2 • Chris Basso – APEC 2014 Course Agenda Introducing the PWM Switch Model CCM, DCM and BCM in Voltage Mode Pulse Width Modulator Gain The PWM Switch Model in Current Mode PWM Switch at Work in a Buck Converter A Simplified Approach to Modeling a DCM Boost Transfer Function of a BCM Boost in Current Mode Small-Signal Model of The Active Clamp Forward 3 • Chris Basso – APEC 2014 Manipulating Linear Networks A switching converter is made of linear elements! rDS ( on ) Vout rL L L C Vin Rload Vin rd on time rL C Vout Rload off time The non-linearity or discontinuity is coming from transitions linear Cannot differentiate DTsw linear linear vDRV ( t ) 4 • Chris Basso – APEC 2014 (1 − D ) Tsw on time off time Singularity State Space Averaging (SSA) Despite linear networks, equation is discontinuous in time Introduced in 76, SSA weights on and off expressions weighted during (1-D)Tsw xɺ = A1 D + A 2 (1 − D ) x ( t ) + B1 D + B 2 (1 − D ) u ( t ) weighted during DTsw A is the state coefficient matrix B is the source coefficient matrix This equation is now continuous in time: singularity is gone However, it became a non-linear equation You need to linearize it by perturbation (or differentiation) If you add a new element, you have to restart from scratch! S.Ćuk, "Modeling, Analysis and Design of Switching Converters", Ph. D. Thesis, Caltech November 1976 5 • Chris Basso – APEC 2014 The PWM Switch Model in Voltage Mode We know that non-linearity is brought by the switching cell a c L C u1 R p a: active c: common p: passive Why don't we linearize the cell alone? a c a . . c d PWM switch VM Switching cell p Small-signal model (CCM voltage-mode) p V. Vorpérian, "Simplified Analysis of PWM Converters using Model of PWM Switch, parts I and II" IEEE Transactions on Aerospace and Electronic Systems, Vol. 26, NO. 3, 1990 6 • Chris Basso – APEC 2014 The Bipolar Small-Signal Model A bipolar transistor is a highly non-linear system Replace it by its small-signal model to get the response Rb1 Rc b Vout Vg Vin Ib Ic rπ β Ib Q1 Ie e Rb2 Re Ce Ebers-Moll model 7 • Chris Basso – APEC 2014 Replace the Switches by the Model Like in the bipolar circuit, replace the switching cell… a c L u1 C . . R p …and solve a set of linear equations! u1 8 • Chris Basso – APEC 2014 . . L C R c An Invariant Model The switching cell made of two switches is everywhere! a d p a buck a buck-boost d PWM switch VM PWM switch VM p c c boost c d Ćuk d c p c PWM switch VM a d PWM switch VM p p a PWM switch VM 9 • Chris Basso – APEC 2014 Course Agenda Introducing the PWM Switch Model CCM, DCM and BCM in Voltage Mode Pulse Width Modulator Gain The PWM Switch Model in Current Mode PWM Switch at Work in a Buck Converter A Simplified Approach to Modeling a DCM Boost Transfer Function of a BCM Boost in Current Mode Small-Signal Model of The Active Clamp Forward 10 • Chris Basso – APEC 2014 CCM, DCM and BCM Operations Three types of conduction modes exist I peak iL ( t ) iL ( t ) iL ( t ) I peak iL ( t ) Tsw I peak iL ( t ) iL ( t ) Tsw Discontinuous Conduction iL ( t ) Mode Tsw Boundary Conduction Mode Third event dead time iL ( t ) Continuous Conduction Mode iL ( t ) Tsw Tsw Tsw > I peak < I peak = I peak 2 2 2 Each mode has its own small-signal characteristics A model is needed for these three modes! 11 • Chris Basso – APEC 2014 CCM Common Passive Configuration The PWM switch is a single-pole double-throw model d c a ia ( t ) ic ( t ) d' vap ( t ) p vcp ( t ) Install it in a buck and draw its terminals waveforms ia ( t ) ic ( t ) d c a L d' Vin C R vap ( t ) vcp ( t ) p Vout CCM VM 12 • Chris Basso – APEC 2014 The Common Passive Configuration Average the current waveforms across the PWM switch ic ( t ) ic ( t ) 0 Tsw t ia ( t ) I a = DI c ic ( t ) Tsw Averaged variables t 0 DTsw ia ( t ) Tsw = Ia = 1 Tsw DTsw ∫ i ( t ) dt = D i ( t ) a c 0 = DI c Tsw CCM VM 13 • Chris Basso – APEC 2014 The Common Passive Configuration Average the voltage waveforms across the PWM switch vap ( t ) vap ( t ) 0 t vcp ( t ) vcp ( t ) t 0 DTsw vcp ( t ) Tsw Tsw = Vcp = 1 Tsw DTsw ∫ 0 vcp ( t ) dt = D vap ( t ) Vcp = DVap Tsw Tsw Averaged variables = DVap CCM VM 14 • Chris Basso – APEC 2014 A Two-Port Representation We have a link between input and output variables DI c d a Vap p Two-port cell c Ic DVap p It can further be illustrated with current and voltage sources Ia d a Vap c Ic Vcp DI c DVap p p CCM VM 15 • Chris Basso – APEC 2014 A Transformer Representation The PWM switch large-signal model is a dc "transformer"! Ia Ic a c I I a = DI c Ic = a . . D 1 D dc equations! Vcp Vcp = DVap Vap = D p It can be plugged into any 2-switch CCM converter c rL Vin L a . . D 1 p C R Dc bias point Ac response CCM VM 16 • Chris Basso – APEC 2014 The Discontinuous Case In DCM, a third timing event exists when iL(t) = 0 L iL ( t ) D1 SW D2 Vin L C R C R C R during D1Tsw D3 Vin C on R iL ( t ) off L iL ( t ) during D2Tsw iL ( t ) = 0 iL ( t ) = 0 during D3Tsw t 0 D1Tsw Tsw D2Tsw D3Tsw DCM VM 17 • Chris Basso – APEC 2014 The Same Configuration as in CCM Draw the waveforms in the "common passive" configuration ia ( t ) I peak a Ia D1 D2 c D3 t vap ( t ) p Vin t vcp ( t ) Vin Vout t D1Tsw Tsw D2Tsw 18 • Chris Basso – APEC 2014 D3Tsw C off t I peak Vout L on Vin ic ( t ) Ic R Average the waveforms: Ia = I peak Ic = I peak Ic = 2 2 D1 D1 + I peak 2 D2 = I peak 2 2 I a D1 + D2 D + D2 = Ia 1 D1 2 D1 ( D1 + D2 ) DCM VM Derive Vcp to Unveil the New Model The addition of the third event complicates the equations vcp ( t ) Vcp = Vap D1 + Vcp D3 Vap Vcp = Vap D1 + Vcp (1 − D1 − D2 ) Vcp t D1Tsw D2Tsw a D3Tsw 1 . . Vcp = Vap D1 D1 + D2 Control input Ic Ia D1 + D2 + D3 = 1 c Ndcm I a = N dcm I c Vap = p Vcp N dcm Ic = Ia N dcm Vcp = N dcmVap N dcm = D1 D1 + D2 f ( D1 ) DCM VM 19 • Chris Basso – APEC 2014 Finally, Get the D2 Value In DCM the inductor average voltage per cycle is always 0 Vcp = Vout What is the averaged inductor peak current? I peak = vL ( t ) vL ( t ) D1Tsw L D1Tsw D1Tsw Vac = L = Vac I peak D1Tsw The peak current uses a previous expression Ic = I peak 2 ( D1 + D2 ) I peak = D2 = 20 • Chris Basso – APEC 2014 2Ic D1 + D2 2 LFsw I c − D1 D1 Vac DCM VM Variable Frequency Operation The power supply operates in a self-oscillating mode Vbulk Demag detector Np:Ns Lp +- 65 mV + . . . Vout Cout Rload Resr S Q Q Vdd R + Rpullup - FB GFB Rsense CTR Critical conduction Mode (CRM) Boundary Conduction Mode Borderline Conduction Mode (BCM) 21 • Chris Basso – APEC 2014 Benefits of Quasi-Resonance Wait for core demagnetization and valley voltage A iD ( t ) 2.10 1.50 0.9 0.3 -0.3 V 230 170 Valley 110 Core is reset 50 vDS ( t ) 0 2.039m 2.052m 2.066m 2.080m 2.093m BCM VM 22 • Chris Basso – APEC 2014 Idealized Waveforms Draw the PWM switch waveforms in a buck configuration ia ( t ) I peak t I peak ic ( t ) ic ( t ) Tsw t vcp ( t ) Vap t BCM VM 23 • Chris Basso – APEC 2014 Derive Founding Equations Average current in terminal C is straightforward I peak ic ( t ) 0 Tsw = I peak 2 The off time depends on the voltage across the inductor toff = L I peak Vcp I peak = 2 I c toff = 2 LI c Vcp Period and duty ratio come easily as ton is imposed ton + toff = Tsw D= ton Tsw Need to transform the error voltage into time 24 • Chris Basso – APEC 2014 BCM VM Generating the On Time The on time modulator works as a PWM block Vdd ton (Verr ) = I Verr Verr C I 2 PWM 5 1 C 4 d C ton (Verr ) = dVerr I DRV The modulator small-signal gain is a simple coefficient C = 100 pF I = 20 µA C = 5u I Verr Assuming 1 V = 1 µs ton 5 BCM VM 25 • Chris Basso – APEC 2014 Final PWM Switch Model in BCM Just add the on-time modulator to the VM PWM switch Verr k= Vac I peak = Vcp toff = C I Vac ton L L I peak Vcp ton a c 1 . . D= D I peak ton ton + toff toff ton p toff duty-cycle a Fsw (kHz) Ip c vc PWM switch BCM p X1 PWMBCMVM C. Basso, "Switch Mode Power Supplies: SPICE Simulations and Practical Design", McGraw-Hill 2008 26 • Chris Basso – APEC 2014 BCM VM Course Agenda Introducing the PWM Switch Model CCM, DCM and BCM in Voltage Mode Pulse Width Modulator Gain The PWM Switch Model in Current Mode PWM Switch at Work in a Buck Converter A Simplified Approach to Modeling a DCM Boost Transfer Function of a BCM Boost in Current Mode Small-Signal Model of The Active Clamp Forward 27 • Chris Basso – APEC 2014 What About the PWM Block? In voltage mode, the duty ratio depends on Verr v (t ) verr ( t ) = Vp Vp + vPWM ( t ) - D (t ) verr ( t ) 60 100 30 10 28 • Chris Basso – APEC 2014 ton ( t ) Tsw = 1 Vp t ton ( t2 ) ton ( t1 ) D (t ) = Tsw verr ( t ) t % ton ( t ) d (t ) t d 1 D (Verr ) = GPWM = dVerr Vp Including the PWM Contribution In a simulation fixture, insert a gain block after Verr a D c PWM switch VM GPWM = PWM switch model in VM d p 1 Vp Modulator gain Verr 29 • Chris Basso – APEC 2014 Considering Feedforward A perturbation disturbs operations and must be fought In a power supply, the input voltage is a perturbation iˆout Zˆ out ,OL GVin vˆin Vref - vˆerr - Plant vˆout The perturbation must affect the output to trigger action Why not reacting before the perturbation hits? This is the principle of feedforward 30 • Chris Basso – APEC 2014 Input Contribution in a Buck Converter The transfer function of a CCM buck converter includes Vin V H ( s ) = in Vp 1+ 1+ s sz1 s s + ω0 Q ωo 2 Let's make Vp a function of Vin: Vp (Vin ) = kFFVin H (s) = Vin k FF Vin 1+ 1+ s sz1 s s + ω0 Q ωo 2 = 1 k FF 1+ 1+ s sz1 s s + ω0 Q ωo 2 The transfer function no longer depends on Vin 31 • Chris Basso – APEC 2014 How to Make Vp a Function of Vin? Make the sawtooth capacitor current depend on Vin Vin vC ( t ) R Verr Verr PWM HL LL IC t C DRV HL = hi line LL = lo line 32 • Chris Basso – APEC 2014 Ramp Amplitude and Input Voltage We can neglect the sawtooth ramp amplitude IC ≈ Vin R The peak value, Vp, is linked to the time constant τ Vp = IC Vin V Tsw = Tsw = in Cramp Cramp Rramp τ Fsw In the time domain, the peak value will change vramp ( t ) = V p V t t = in Tsw τ Fsw Tsw At t = ton, the error voltage equals Vramp Verr = Vin ton V = in D τ Fsw Tsw τ Fsw D (Verr ) = Verr τ Fsw Vin 33 • Chris Basso – APEC 2014 An In-Line Equation to Include Feedforward Differentiate the expression to get the small-signal gain GPWM = ∂D (Verr ) ∂Verr = τ Fsw Vin = 1 k FF Vin k FF = 1 Fswτ The feedforward block requires an ABM source in a R = 82 kΩ C = 470 pF Fsw = 500 kHz c d Vin B1 Voltage PWM switch VM p V(err)/(V(in)*{kFF}) parameters: k FF = kFF=52m err 1 = 52m 500k × 470 p × 82k ABM: Analog Behavioral Model 34 • Chris Basso – APEC 2014 Course Agenda Introducing the PWM Switch Model CCM, DCM and BCM in Voltage Mode Pulse Width Modulator Gain The PWM Switch Model in Current Mode PWM Switch at Work in a Buck Converter A Simplified Approach to Modeling a DCM Boost Transfer Function of a BCM Boost in Current Mode Small-Signal Model of The Active Clamp Forward 35 • Chris Basso – APEC 2014 Peak Current Mode Control In voltage-mode, the loop controls the duty ratio In current-mode, the inductor peak current is controlled PWM latch clock vc ( t ) R Q vr ( t ) iL ( t ) Ri + vr ( t ) + iL ( t ) Ri Slope comp. a Ri D c D' p An artificial ramp is added for stabilization purposes 36 • Chris Basso – APEC 2014 We Want the Average Current Definition The value Ic is the inductor current at half the ripple ic ( t ) a I peak ic ( t ) S a Ri ∆I L b Tsw S2 0 t DTsw (1 − D ) Tsw Tsw Current at point b is that of a minus half the inductor ripple ic ( t ) Tsw = Ic = Vc S a S D ' Tsw − DTsw − 2 Ri Ri 2 CCM CM 37 • Chris Basso – APEC 2014 Define the Converter off-Slope Use a buck configuration to see voltages at play ia ( t ) ic ( t ) d c a L d' Vin C R vap ( t ) vcp ( t ) p Vout The downslope depends on the output voltage Vout: S2 = − Vout L The inductor average voltage is 0 at steady-state Vcp = Vout 38 • Chris Basso – APEC 2014 S2 = − Vcp L CCM CM A Current Mode Generator Update the previous equation to obtain final definition Vc T S − Vcp (1 − D ) sw − a DTsw Ri 2 L Ri Ic = Peak current setpoint Half inductor ripple Group 2nd and 3rd terms I µ = Vcp (1 − D ) Tsw S a + DTsw 2 L Ri Compensation ramp Inductor ripple and compensation ramp alter peak value c c ∆I L 2 Vc Ri Vc Ri Sa DTsw Ri p Iµ p CCM CM 39 • Chris Basso – APEC 2014 CM or VM Lead to Similar Input Currents Average the current waveforms across the PWM switch ic ( t ) ic ( t ) 0 t ia ( t ) ic ( t ) dTsw Tsw = Ia = 1 Tsw dTsw ∫ i ( t ) dt = D i ( t ) a 0 I a = DI c Tsw Ia = t 0 ia ( t ) Tsw c Tsw = DI c D= Vcp Vac Ic Vcp Vap CCM CM 40 • Chris Basso – APEC 2014 The PWM Switch Model in Current Mode The final model associates three current sources c a Ia Vcp Vin L Ic Vap Vc Ri Ic Iµ C R p This is the large-signal current-mode PWM switch model V. Vorpérian,"Analysis of Current-Controlled PWM Converters using the Model of the PWM Switch", PCIM Conference ,1990 41 • Chris Basso – APEC 2014 Final Model Includes Subharmonic Effects A simple capacitor is enough to mimic instability Ia c a L Ic Vcp Vin Vap Vc Ri Ic Iµ Cs resonant tank C R p As the instability is placed at half the switching frequency: Fsw 1 = 2 2π LCs 42 • Chris Basso – APEC 2014 Cs = 1 L ( Fsw π ) 2 CCM CM The PWM Switch in DCM The PWM CM can work in discontinuous mode ic ( t ) I peak ic ( t ) Vc Ri Current setpoint Sa Ri S1 Theoretical value Sa = 0 S2 Ic Tsw D3Tsw 0 t D1Tsw D2Tsw Tsw The average current Ic is somewhere in the downslope S2 I peak = Vc − D1Tsw Sa Ri Ic = Vc − D1Tsw Sa − α D2Tsw S 2 Ri DCM CM 43 • Chris Basso – APEC 2014 Derive the Inductor Average Current We must now obtain the value of α to get Ic I peak − I c α I peak S2 Ic Tsw Ic t D2Tsw Ic is the area under the triangle divided by the switching period Ic = I peak D1 2 + I peak D2 2 = I peak α I peak = I peak − I peak 44 • Chris Basso – APEC 2014 D1 + D2 2 D1 + D2 2 α = 1− D1 + D2 2 DCM CM Adopt the CCM Structure for DCM Substitute and rearrange to get the inductor current Ic = Vcp Vc D1Tsw Sa − − D2Tsw Ri Ri L D1 + D2 1 − 2 If we stick to the original CCM architecture Ic = Vc − Iµ Ri Iµ = with Vcp D1Tsw Sa + D2Tsw Ri L D1 + D2 1 − 2 c V D + D2 D1Tsw Sa + D2Tsw cp 1 − 1 Ri L 2 Vc Ri p DCM CM 45 • Chris Basso – APEC 2014 Discontinuous Waveforms Let's have a look at the PWM switch voltages in DCM ia ( t ) I peak ia ( t ) Tsw t vap ( t ) Vin t ic ( t ) DCM t vcp ( t ) Vap D3Tsw 0 t D1Tsw 46 • Chris Basso – APEC 2014 Vout D2Tsw DCM CM Derive the Duty Ratios From the DCM voltage-mode PWM switch we have: Vcp = Vap D1 D1 + D2 D1 = D2Vcp Vap − Vcp From the operating DCM waveforms Ia = I peak D1 I peak = 2 2I a D1 I c = I peak D1 + D2 2 I a = Ic D1 D1 + D2 Almost there, just need to express D2 I peak = Vac D1Tsw L I peak = and 2Ic D1 + D2 Vac 2Ic D1Tsw = L D1 + D2 D2 = 2 LFsw I c − D1 D1 Vac DCM CM 47 • Chris Basso – APEC 2014 The DCM Model is Complete! We can use this model for DCM simulations L a c Ia Ic Vc Ri Ia Vin Iµ C R p I a = Ic D1 D1 + D2 Iµ = V D + D2 D1Tsw Sa + D2Tsw cp 1 − 1 Ri L 2 DCM CM 48 • Chris Basso – APEC 2014 Current Mode Borderline In CM, the error voltage sets the peak current I peak = Vc Ri The peak current also depends on duty ratio D iL ( t ) I peak DTsw I peak = iL ( t ) Vcp L Vac L Tsw Tsw Vac DTsw L D= Vc L Ri VacTsw ton = Vc L Ri Vac BCM CM 49 • Chris Basso – APEC 2014 Operating Points of BCM Current Mode The off-time duration depends on Ipeak too I peak = Vcp L (1 − D ) Tsw toff = Verr L Ri Vcp Switching frequency comes easily Tsw = ton + toff = Vc L 1 1 + Ri Vap Vcp The average inductor current Ic is straigthforward Ic = I peak 2 = Vc 2 Ri The relationship between Ia and Ic is always the same I a = DI c BCM CM 50 • Chris Basso – APEC 2014 This Completes the BCM CM Model The model is really simple, two current sources L a c Ia Ic Vc DI c 2 Ri Vin C R p Other ABM sources will compute D and Tsw BCM CM 51 • Chris Basso – APEC 2014 Course Agenda Introducing the PWM Switch Model CCM, DCM and BCM in Voltage Mode Pulse Width Modulator Gain The PWM Switch Model in Current Mode PWM Switch at Work in a Buck Converter A Simplified Approach to Modeling a DCM Boost Transfer Function of a BCM Boost in Current Mode Small-Signal Model of The Active Clamp Forward 52 • Chris Basso – APEC 2014 A Buck Converter in Current Mode Identify the diode and switch position in a buck CM a a duty-cycle c c vc PWM switch CM p p Current mode Replace switches by the small-signal PWM switch model 53 • Chris Basso – APEC 2014 A Small Signal Model The model includes current sources and conductances Ic a c 1 gi vˆc ki vˆcp g r vˆap g f vˆc ko 1 go Cs p ko = 1 Ri ki = D Ri g f = Dg o − DD ' Tsw 2L I gi = D g f − c Vap go = − gr = Tsw Sa 1 D' + − D L Sn 2 Ic − go D Vap V. Vorpérian,"Analysis of Current-Controlled PWM Converters using the Model of the PWM Switch", PCIM Conference ,1990 54 • Chris Basso – APEC 2014 Plug the Model and Simplify Plug, simplify and rearrange: test in between! Ic a Vin 1 gi 1 go vˆc ko vˆc ki vˆcp g r p L c C Cs R rC vˆap g f We want the control-to-output function, remove input stimulus vˆin = 0 vˆap g f = 0 The input contribution can also disappear, no interest in Zin 55 • Chris Basso – APEC 2014 Time to Call FACTS! End-up with a simpler and less ugly sketch c L Vout ( s ) 1 C Vc ( s ) ko 1 go Cs 2 R rC There are 3 storage elements: 3rd-order system FACTS: Fast Analytical Circuit TechniqueS 56 • Chris Basso – APEC 2014 Don't Use Brute-Force Algebra! You can use brute-force analysis… Vout ( s ) L Z th 1 1 Vth = Vc ( s ) k0 g o sCs C R Vth Z (s) 1 1 Z th = g o sCs rC …or consider FACTS to write a 3rd-order system TF H ( s) = H0 N (s) D (s) = H0 N (s) 1 + a1s + a2 s 2 + a3 s 3 57 • Chris Basso – APEC 2014 Start with the Dc Gain Consider dc and high-frequency states for L and C impedance ZC = C impedance 1 sC Z L = sL L Dc state Z C = ∞ Cap. is an open circuit HF state Z C = 0 Cap. is a short circuit Dc state Z L = 0 Inductor is a short circuit Inductor is an open circuit HF state Z L = ∞ In the circuit, open capacitors and short inductors Vout ( s ) Vc ( s ) ko 1 go 1 H 0 = k0 R g0 R 10 s! 58 • Chris Basso – APEC 2014 Identify The Zeros Zeros prevent the excitation from reaching the output Vin ( s ) Z (s) = ∞ Vout ( s ) rC + response rC Z (s) = 0 excitation sr C + 1 1 = C sC sC srC C + 1 = 0 R C ω z1 = Z (s) = 0 1 rC C We have the zero expression, half of the work is done H ( s ) = H0 1+ s sz1 1 + a1s + a2 s 2 + a3 s3 59 • Chris Basso – APEC 2014 Finding the Poles The poles are linked to the time constants of the system These time constants solely depend on the structure Remove the excitation signal to isolate the structure short circuit voltage source current source open circuit The denominator order depends on the storage elements 1 storage element 1st-order 2 storage elements L C C 2nd-order Not always! Consider individual state variables. 60 • Chris Basso – APEC 2014 Start by Identifying the Time Constants The excitation is zero, elements are in their dc states 3 storage elements, 3 time constants, 3 drawings 1 g0 R? τ 1 = f ( Cs ) τ 1 = Cs R τ 2 = f ( L) τ2 = τ 3 = f (C ) τ 3 = C rC + rC R? 1 g0 1 R go R rC R? R 1 g0 rC L 1 +R g0 1 R g0 61 • Chris Basso – APEC 2014 First Coefficients a1 and a2 FACTs tell us that a1 sums up all time constants a1 = τ 1 + τ 2 + τ 3 Dimension is time For a2, we multiply combined-time constants a2 = τ1τ 12 + τ 1τ 31 + τ 2τ 32 Dimension is time2 What is this new time constants definition,τ 2 ? 1 Cs (HF) τ 12 Cs (HF) τ 31 L C (dc) R? L (HF) τ 32 L (dc) C R? C Cs (dc) R? V. Vorpérian, “Fast Analytical Techniques for Electrical and Electronic Circuits”, Cambridge Press, 2002 62 • Chris Basso – APEC 2014 Coefficient a2 Mixes Time Constants Drawings are key to avoid mistakes Cs (HF) τ 12 R? 1 g0 L C (dc) R τ 12 = L R rC R? Cs (HF) R? τ 31 L (dc) C R 1 g0 τ 31 = rC C rC R? 63 • Chris Basso – APEC 2014 (Carefully) Mixing Time Constants The last drawing completes the a2 expression L (HF) τ 32 R? C Cs (dc) R 1 g0 τ 32 = ( rC + R ) C rC R? a2 coefficient is there! 1 L 1 L a2 = C s R + Cs R rC C + ( rC + R ) C 1 go R go +R g0 64 • Chris Basso – APEC 2014 …And a3 is ? For a3, we multiply by a third time-constant a3 = τ 1τ 12τ 31,2 Dimension is time3 What is this new time constant definition? Cs (HF) L (HF) τ 31,2 R? R 1 g0 C (dc) τ 31,2 = ( rC + R ) C rC R? The final coefficient has been identified 1 L a3 = Cs R ( rC + R ) C go R 65 • Chris Basso – APEC 2014 Time to Check Results A Mathcad® sheet can be built to verify these calculations H ( s ) = G0 ω z1 = 1+ s sz1 1 + a1s + a2 s + a3 s 2 3 1 G0 = k0 R g0 1 rC C 1 1 L a1 = Cs R+ + C rC + R 1 go g0 +R g0 1 L 1 L a2 = Cs R + Cs R rC C + ( rC + R ) C 1 g R g o o +R g0 1 L a3 = Cs R ( rC + R ) C go R 66 • Chris Basso – APEC 2014 5 V/1 A buck Vin = 10 V, Fsw = 100 kHz, Ri = 0.25Ω, Se = 2.5 kV s C = 100 µF, rC = 0.1Ω, L = 100µH, Cs = 101nF, Vc = 1.28 V I c = 4.94 A ki = 2 Ω −1 k0 = 4 Ω −1 g 0 = 0.01 Ω −1 g f = −7.5 mΩ −1 g r = 0.49 Ω −1 g i = −250 mΩ −1 G0 = 12 dB f z1 = 15.9 kHz See What SPICE is Saying Use the large-signal model, SPICE linearizes it for you a {Se}*V(D)/({Ri}*{Fsw}) + v(c,p)*(1-V(D))*({1/Fsw}/(2*{L})) L1 VIC {L} 4.95V c c 5 a B3 Current V(Vc)/{Ri} C1 100uF 0V -2.50mV V1 10 p Vout 4.95V C2 B4 Current {Cs} B2 Current V(D)*I(VIC) 10.0V parameters Fsw=100kHz L=100u Cs=1/(L*(Fsw*3.14)^2) Ri=250m Se=2.5k 4.95V 1 495mV 1.28V R3 1m R2 B1 100m Voltage v(c,p)/v(a,p) Vstim 1.28 AC = 1 p R1 1 4 D vc Then compare results with those of Mathcad® 67 • Chris Basso – APEC 2014 Excellent Agreement! Superimposed curves mean transfer functions are identical dB ° 20 0 10 − 50 0 −100 − 10 −150 H(f) − 20 10 100 68 • Chris Basso – APEC 2014 3 1´ 10 4 1´ 10 5 1´ 10 10 arg H ( f ) 100 5 3 1´ 10 4 1´ 10 1´ 10 Rearranging Expressions The denominator is not really in a low-entropy form D ( s ) = 1 + a1s + a2 s 2 + a3 s 3 This is a third-order polynomial form that can be factored H(f) D ( s ) ≈ 1 + a1s D ( s) ≈ 1+ Low frequency High frequency a2 a s + 3 s2 a1 a1 69 • Chris Basso – APEC 2014 Final Lap! The transfer function can now unveil peaking and damping H ( s) ≈ H0 1+ s ωz 1 1 1+ s s s H0 = 2 ωp 1+ ω Q + ω n n 1 ωp = 1 ωn = T 1 + sw mc (1 − D ) − 0.5 RC LC π Q= Tsw S mc = 1 + e Sn ωz = 1 π mc (1 − D ) − 0.5 1 R 1 Ri 1 + RTsw m 1 − D − 0.5 ) c( L 1 rC C 20 10 0 − 10 Artificial ramp Inductor on slope − 20 10 H(f) 100 10 3 R. B. Ridley, “A new Continuous-Time Model for CM Control”, IEEE Transactions of Power Electronics, Vol. 6, April 1991 70 • Chris Basso – APEC 2014 10 4 10 5 Course Agenda Introducing the PWM Switch Model CCM, DCM and BCM in Voltage Mode Pulse Width Modulator Gain The PWM Switch Model in Current Mode PWM Switch at Work in a Buck Converter A Simplified Approach to Modeling a DCM Boost Transfer Function of a BCM Boost in Current Mode Small-Signal Model of The Active Clamp Forward 71 • Chris Basso – APEC 2014 A DCM Boost Converter The goal is to regulate the LED string current D L Vout Vf Cout Vin SW rC Ri I out Rsense PWM control The LED current is sensed via a shunt element 72 • Chris Basso – APEC 2014 Characterize the LED String Evaluate forward drop and dynamic resistance n LEDs in series A A A A IF n vf rd rLEDs ∑r VT0 vZ ∑V i =1 di n K i =1 T0i K v f = I F rd + VT0 K K Lab. measurements require simple V and A-meters rLEDs = V f1 − V f 2 I F1 − I F2 = 27.5 − 26.4 = 55Ω 0.1 − 0.08 VZ ≈ V f1 − RLEDs I F1 = 27.5 − 0.1× 55 = 22 V 73 • Chris Basso – APEC 2014 A Simplified Approach The LED string is driven by a current source I out Vout Rac = rLEDs + Rsense Rac Vc Vout = Rac I out + Vz I out VZ Vout ( s ) = Rac I out ( s ) VZ sets the operating point, Rac sets the ac response iˆo ( s ) Z (s) vˆo vˆc ( s ) rC I out Cout 74 • Chris Basso – APEC 2014 Rac Current Source Model It is convenient and fast to consider 1st-order models Use the converter transfer function in DCM Vout = Vin 1+ 1+ 2Tsw D 2 Rdc L 2 We purposely consider an instantaneous power response if D changes, Pout immediately translates this is not true for boost or buck-boost converters: RHPZ high-frequency phenomena are also lost 1st-order rC R I out D C 75 • Chris Basso – APEC 2014 Define the Duty Ratio In the previous expression, D is unknown vc ( t ) Ri Se Ri Sn = I peak Vin L iD ( t ) DTsw I peak V S = c − e DTsw Ri Ri I peak = 76 • Chris Basso – APEC 2014 DTswVin L D= Vc L SeTsw L + RiTswVin Update the Output Current Equation Massage the dc transfer equation to unveil Iout Vout = Vin 1+ 1+ 2Tsw D 2 Vout I out L 2 Inject the duty ratio definition, solve for Iout I out = 2Vout LVc 2 1 2 Tsw 2V out − 1 − 1 ( Se L + RiVin )2 Vin There are two modulated variables, Vc and Vout vˆin = 0 77 • Chris Basso – APEC 2014 A Simple Model You obtain two small-signal sources: iˆout ∂I (V , V ) Vin 2Vc L g1 = out c out = 2 ∂Vc Tsw (Vout − Vin )( Se L + RiVin ) g2 = ∂I out (Vc ,Vout ) ∂Vout =− = g1vˆc + g 2 vˆout Vin 2Vc 2 L 2Tsw (Vin − Vout ) The circuit is quite simple 2 ( Se L + RiVin )2 iˆo ( s ) vˆo rC vˆc ( s ) vˆo ( s ) Rac Cout 78 • Chris Basso – APEC 2014 ∂I out ∂Vc vˆout ∂I out ∂Vout vˆc vˆc vˆo A Current Driving an Impedance The second term is a simple resistance I R V I= Update the final model 2Tsw (Vin − Vout ) ( Se L + RiVin ) 2 V R R1 = 2 Vc 2Vin 2 L Z (s) vˆo rC R1 vˆc ( s ) Rac Vout ( s ) = Vc ( s ) g1Z ( s ) Cout 79 • Chris Basso – APEC 2014 Use FACTS to Get the Impedance Obtain the impedance (transfer function) in a snapshot Vres ( s ) Response Ω I exc ( s ) rC R1 Rac Z (s) = Vres ( s ) I exc ( s ) = Z0 Cout Excitation For dc, open the capacitor R1 80 • Chris Basso – APEC 2014 rC Rac Z 0 = R1 Rac 1+ s s z1 1+ s s p1 No Algebra to Get the Result! At the zero frequency, the response disappears Cout Short circuit rC rC + sr C + 1 1 = C out sCout sCout srC Cout + 1 = 0 ω z1 = 1 rC Cout Remove the excitation and look at the resistance driving Cout Z ( s ) = R1 Rac rC R1 Rac R? 1 + srC Cout 1 + sCout ( rC + R1 Rac ) τ = rC + ( R1 Rac ) Cout 81 • Chris Basso – APEC 2014 Final Expression Associate the impedance expression with g1 Vout ( s ) Vc ( s ) ω z1 = 1+ = H0 1+ s ω z1 s H0 = ω p1 1 rC Cout ω p1 = Vin 2Vc L Tsw (Vout − Vin )( Se L + RiVin ) 2 ( Rac R1 ) 1 Cout ( rC + R1 Rac ) However, we want the control-to-output current expression Vout ( s ) s Vsense ( s ) rLEDs Rsense 82 • Chris Basso – APEC 2014 Vsense ( s ) Vc ( s ) 1+ ω z1 Rsense = H0 s Rsense + rLEDs 1+ ω p1 Checking our Model Response Ac simulation of the current source-based approach parameters Vout 2 R3 4m Rd 55 R1 Ri=0.25 {R1} vc=0.4 Se=100k B1 6 1 Vc Current Fsw=1Meg Vsense {k1}*V(Vc)/{k2} C1 L=3.3u Rsense 2.2uF Tsw=1/Fsw 11 Vin=12 Vout=32.85 k1=Vin^2*vc*L k2=Tsw*(Vout-Vin)*(Se*L+Ri*Vin)^2 R1=2*Tsw*(Vin-Vout)^2*(Se*L+Ri*Vin)^2/(Vc^2*(Vin^2)*L) V2 {Vc} AC = 1 83 • Chris Basso – APEC 2014 Simplified Approach vs PWM Switch The comprehensive model with the PWM switch L1 3.3u 12.0V 10 12.0V c 9 p 32.6V Vout Rd 55 400mV 0V 11 Vin 12 vc a 8 PWM switch CM 405mV dc duty-cycle 5 X3 PWMCM L = 3.3u Fs = 1Meg Ri = Ri Se = 100k Shunt model R3 4m 7 23.8V X2 IRF530M 32.6V 2 5.86V 1 23.8V 6 3 1.77V 4 X1 AMPSIMP parameters Ri=-0.25 vc=0.4 V2 {Vc} AC = 1 C1 2.2uF Vsense Rsense 11 84 • Chris Basso – APEC 2014 V1 22 Final Results The RHPZ effect is not modeled in the simplified approach 50.0 ( dB) 30.0 10.0 Vout ( f ) -10.0 Vc ( f ) -30.0 15.0 (°) -15.0 Simple model -45.0 -75.0 arg -105 1 Vout ( f ) Vc ( f ) PWM switch 10 100 1k 10k 100k 1Meg 85 • Chris Basso – APEC 2014 Real Measurement vs Model Deviation occurs, as expected, because of the RHPZ 3.3uH Control to Output (R29 current sense) 10 0 model prototype 0 -20 -40 Gain (dB) (Calculated) -10 -60 Gain (dB) (Vin=12V) Phase (o) (Calculated) Phase (o) (Vin=12V) -80 -20 -100 -30 RHPZ -120 100 1,000 10,000 100,000 C. Basso, A. Laprade, "Simplified Analysis of a DCM Boost Converter Driving an LED String", www.how2power.com 86 • Chris Basso – APEC 2014 Course Agenda Introducing the PWM Switch Model CCM, DCM and BCM in Voltage Mode Pulse Width Modulator Gain The PWM Switch Model in Current Mode PWM Switch at Work in a Buck Converter A Simplified Approach to Modeling a DCM Boost Transfer Function of a BCM Boost in Current Mode Small-Signal Model of The Active Clamp Forward 87 • Chris Basso – APEC 2014 Why Bordeline Operation? More converters are using variable-frequency operation This is known as Quasi-Square Wave Resonant mode: QR Valley switching ensures extremely low capacitive losses DCM operation saves losses in the secondary-side diode Easier synchronous rectification The Right Half-Plane Zero is pushed to high frequencies Smooth signals iD ( t ) Less noise vDS ( t ) 88 • Chris Basso – APEC 2014 Low CV² losses id ( t ) vDS ( t ) What is the Principle of Operation? The drain-source signal is made of peaks and valleys A valley presence means: The drain is at a minimum level, capacitors are naturally discharged The converter is operating in the discontinuous conduction mode V 480 360 Vin valley 240 120 toff ton DT vDS ( t ) 0 2.061m 2.064m 2.066m 2.069m 2.071m Flyback structure BCM = Borderline or Boundary Conduction Mode 89 • Chris Basso – APEC 2014 A QR Circuit Does not Need a Clock The system is a self-oscillating current-mode converter L Vout 50 mV + - Vin C S Q d (t ) Vout R + - + 2.5 V 90 • Chris Basso – Huawei Technical Seminar 2012 vc ( t ) vsense ( t ) Ri R A Winding is Used to Detect Core Reset When the flux returns to zero, the aux. voltage drops Discontinuous mode is always maintained vaux ( t ) Vbulk V Core is reset 800 . . . delay 400 200 iL p ( t ) 0 vDS ( t ) vaux ( t ) = − N vDS ( t ) 600 V delay A dϕ ( t ) 10 400m dt 0 vaux ( t ) iL p ( t ) 20 800m 0 -10 -400m ϕ =0 -20 -800m 2.251m 2.255m 2.260m 2.264m 2.269m set 91 • Chris Basso – Huawei Technical Seminar 2012 Test the Large-Signal Model Response Insert the PWM BCM CM model in the boost converter L1 {L} 10.0V X1 PWMBCMCM Vout Fsw (kHz) 11.0V vc L=22u Ri=-0.1 PWM switch BCM p 33.4V parameters a Vin 10 ton c 10.0V 15.8V C1 470u 500mV V3 0.5 AC = 1 92 • Chris Basso – APEC 2014 R1 10 SPICE Gives Us the Response SPICE linearizes the model for us around the bias point dB 40.0 pole G0 20.0 0 -20.0 -40.0 -1 slope Vout ( f ) Vc ( f ) zero ° 0 -40.0 RHPZ? -80.0 -120 arg -160 1 Vout ( f ) Vc ( f ) 10 100 1k 10k 100k 1Meg 93 • Chris Basso – APEC 2014 Derive a Small-Signal Model A Quasi Resonant model is built with the PWM switch model Vc L 1 1 + Ri Vac Vcp V Ic L 3 variables Ia = c Ic = Ri VacTsw 1 1 Vac + Vac Vcp Vc 1 variable Ic = 2 Ri D= c a Vc 2 Ri DI c p Vc L Ri VacTsw Tsw = These are large-signal equations that need linearization I c = f (Vc ) iˆa = iˆc = ∂I a (Vcp , Vac , I c ) ∂Vcp 94 • Chris Basso – APEC 2014 ∂I c (Vc ) ∂Vc vˆcp + I c ,Vac vˆc 1 iˆc = vˆc 2 Ri ∂I a (Vcp , Vac , I c ) ∂I c = vˆc kc iˆc + Vap ,Vac kc = 1 2 Ri ∂I a (Vcp , Vac , I c ) ∂Vac vˆac I c ,Vcp Large to Small-Signal Final steps before the small-signal model iˆa = vˆcp I c 0Vac 0 (V ac 0 kcp = (V + Vcp 0 ) 2 + I c 0Vac 0 ac 0 + Vcp 0 ) Vcp 0 Vcp 0 + Vac 0 kic = 2 iˆc − vˆac Vcp 0 I c 0 (V + Vcp 0 ) ac 0 Vcp 0 Vcp 0 + Vac 0 kac = 2 Vcp 0 I c 0 (V ac 0 + Vcp 0 ) 2 A small-signal model can now be assembled a c Vcp kcp Vac kac I c kic conductance Vc kc conductance p conductance 95 • Chris Basso – APEC 2014 Always Check the Small-Signal Response Always verify if coefficients are well derived! R2 10m parameters 33.4kV L1 {L} 9.97V Fsw ton 7 Vg 10 Vin=10 Vout=15.6 M=Vout/Vin Ic=-2.5 Vac=-10 Vap=-Vout Vcp=Vin-Vout kcp=Ic*Vac/(Vac+Vcp)^2 kic=Vcp/(Vcp+Vac) kac=Vcp*Ic/(Vac+Vcp)^2 kc=1/(2*Ri) 500m V Vc V3 0.5 AC = 1 15.8V a V(Vc)*{L}/({Ri}*(V(a,c)+1u)) B6 Voltage R1 10 C1 470u Rdum 1u c a B1 Current V(c,p)*{kcp} Vc 9.97V B2 Current I(Vc)*{Kic} B3 Current V(a,c)*{kac} p 96 • Chris Basso – APEC 2014 Vout p c 11.0uV 4 10.0V L=22u Ri=-0.1 9.97V 1/((V(Vc)*{L}/{Ri})*(1/V(a,c)+1/V(c,p))) B5 Voltage 10 B4 Current V(Vc)*{kc} A Simple Intermediate Sanity Check Same response as with the non-linear model: good to go dB 40.0 20.0 0 -20.0 -40.0 Vout ( f ) Vc ( f ) ° 0 -40.0 -80.0 -120 arg -160 1 Vout ( f ) Vc ( f ) 10 100 1k 10k 100k 1Meg 97 • Chris Basso – APEC 2014 The Model at Work in the BCM Boost Replace the switch/diode in the boost configuration c L Vc kc p Vcp kcp Vin Vc a I c kic Vac kac C R In the original model, Ic leaves node c. It enters it in a boost. 98 • Chris Basso – APEC 2014 Identify Static Parameters in the Coefficients Depending on configurations, update variables in coefficients kcp = Vcp 0 = Vin − Vout Ic0 = − I outVout Vin I outVout Vin Vin ( −Vin + Vin − Vout )2 kac = − Vac 0 = −Vin kic = kcp = 1 R kac = I outVout (Vin − Vout ) Vin ( −Vin + Vin − Vout )2 Vin − Vout Vin − Vout − Vin M 1 − R R kic = 1 − 1 M kc = 1 2 Ri 99 • Chris Basso – APEC 2014 First Step is Dc Gain Open the output capacitor, short the inductor Vc kc Vin Vout ( s ) p c Vc ( s ) I out ( s ) Vcp kcp I c kic Vac kac R a Write output voltage expression I out ( s ) = Vc ( s ) kc + Vcp ( s ) kcp − I c ( s ) kic − Vac ( s ) kac ( ) ( ) I out ( s ) = Vc ( s ) kc + V( c ) ( s ) − V( p ) ( s ) kcp − Vc ( s ) kc kic − V( a ) ( s ) − V( c ) ( s ) k ac 100 • Chris Basso – APEC 2014 Dc Gain Derivation There is no contribution from the source, Vin(s) = 0 I out ( s ) = Vc ( s ) kc − Vout ( s ) kcp − Vc ( s ) kc kic = Vc ( s ) kc (1 − kic ) − Vout ( s ) kcp I out ( s ) = Vout ( s ) R 1 Vc ( s ) kc (1 − kic ) = Vout ( s ) kcp + R Vout ( s ) kc (1 − kic ) = 1 Vc ( s ) kcp + R H0 = R 4 MRi We have the first term of our transfer function s sz1 Vout ( s ) = H0 s Vc ( s ) 1+ s p1 1+ 101 • Chris Basso – APEC 2014 Deriving the Zero Position For the zero, the stimulus does not reach the output current in the resistance is 0, node p voltage is also 0 p c Vc kc VL ( s ) L Vc ( s ) Vout ( sz ) = 0 i p ( sz ) = 0 Vcp kcp I c kic Vac kac R a All the inductor ac current is absorbed before reaching R Vc ( s ) kc + V( c ) ( s ) − V( p ) ( s ) kcp = Vc ( s ) kc kic + V( a ) ( s ) − V( c ) ( s ) kac 102 • Chris Basso – APEC 2014 What is the Root to N(s) = 0? The voltage at node c depends on the inductance V( c ) ( s ) = −VL ( s ) = − I c ( s ) sL = −Vc ( s ) kc sL Substitute in the previous equations and simplify Vc ( s ) kc − Vc ( s ) kc sLkcp = Vc ( s ) kc kic + Vc ( s ) kc sLkac Solve for s, this is the zero position 1 − kic = sL ( kac + kcp ) sz1 = 1 − kic L ( kac + kcp ) Substitute kic, kac, kcp The root is positive, this is a Right Half Plane Zero sz1 = R LM 2 103 • Chris Basso – APEC 2014 For the Pole, Reduce Excitation to 0 Excitation is Vc: all current sources f (Vc) are open p c V( c ) ( s ) = 0 R? L Vcp kcp Vac kac R a Look for the resistance “seen” by the capacitor C The source Vcpkcp can be reworked vˆ( c ) = 0 Vcp ( s ) kcp = −V p ( s ) kcp = − 104 • Chris Basso – APEC 2014 Vout ( s ) R Replace by a resistance R A Really Simple Circuit Difficult to beat in terms of problem solving p c R? L R R a The pole due to the capacitor comes immediately R ? = R || R = R 2 τ= R C 2 s p1 = 2 RC 105 • Chris Basso – APEC 2014 The Complete Expression is Ready With a few steps, we have our transfer function s 1− ω z1 Vout ( s ) = H0 s Vc ( s ) 1+ H0 = ωp R 4 MRi ωp = 1 2 RC ωz = 1 R LM 2 1 We can easily add the capacitor ESR contribution rC + rC sr C + 1 1 = C sC sC srC C + 1 = 0 C Z (s) = 0 106 • Chris Basso – APEC 2014 ω z2 = 1 rC C s s 1 − 1 + ω ω Vout ( s ) z1 z2 = H0 s Vc ( s ) 1+ ωp R Pole changes: τ = + rC C 2 1 Time to Confront Mathcad® with SPICE! If equations are correct, curves must superimpose perfectly ° dB 40 H(f) 100 20 0 − 20 − 40 1 0 arg H ( f 10 ) − 100 100 3 10 4 10 5 10 6 10 The BCM boost response in CM is first-order with RHPZ 107 • Chris Basso – APEC 2014 Course Agenda Introducing the PWM Switch Model CCM, DCM and BCM in Voltage Mode Pulse Width Modulator Gain The PWM Switch Model in Current Mode PWM Switch at Work in a Buck Converter A Simplified Approach to Modeling a DCM Boost Transfer Function of a BCM Boost in Current Mode Small-Signal Model of The Active Clamp Forward 108 • Chris Basso – APEC 2014 Active Clamp Forward in Voltage Mode An Active Clamp Forward (ACF) is a forward converter… 1: N L Cclp Lmag C R Q2 Vin Cclp Q1 Low-side GND referenced …featuring a controlled-upper-side switch for ZVS operations 109 • Chris Basso – APEC 2014 Control Strategy A deadtime is inserted to let VDS swing towards grounds vDS ( t ) ZVS vQ2 ( t ) DT vQ1 ( t ) DT Quasi-ZVS can be implemented on the drain voltage 110 • Chris Basso – APEC 2014 Active Clamp Forward in Voltage Mode Energy is stored in the magnetizing inductance at turn on L 1: N Cclp VCclp Q2 iL ( t ) Lmag imag ( t ) C R NiL ( t ) Vin imag ( t ) + NiL ( t ) Q1 We need to offer a path at turn off to demagnetize the core 111 • Chris Basso – APEC 2014 All Parasitic Capacitors are Charged At turn off, the current charges the lump capacitor L 1: N Cclp VCclp Q2 Lmag iL ( t ) imag ( t ) imag ( t ) Vin imag ( t ) C Q1 Clump The drain voltage increases until it touches Vclamp 112 • Chris Basso – APEC 2014 R Magnetizing Current Resonates The upper-side MOSFET switches on with delay: ZVS imag ( t ) Cclp VCclp L 1: N Lmag iL ( t ) C R imag ( t ) Q2 Vin imag ( t ) Q1 Clump Magnetizing current decreases to 0 then reverses 113 • Chris Basso – APEC 2014 Deadtime Gives Quasi-ZVS Operation The deadtime lets magnetizing current reach –Imag,peak imag ( t ) Cclp VCclp Q2 L 1: N Lmag iL ( t ) C R imag ( t ) imag ( t ) Vin imag ( t ) imag ( t ) Q1 Clump imag ( t ) At this moment, Q2 opens and current discharges Clump 114 • Chris Basso – APEC 2014 Full ZVS at Nominal Power is Difficult True ZVS is difficult to reach, quasi-ZVS is usually obtained Vclamp max vDS ( t ) 0 0 imag ( t ) − I mag , peak Q2 opens True ZVS Q1 closes 2 1 1 Lmag ( I mag , peak − − NI L ) > ClumpVin 2 2 2 115 • Chris Basso – APEC 2014 Building an Ac Model A model can be built following different methods Write large-signal equations of voltages and currents Assemble sources to build a large-signal model Linearize expressions and derive transfer functions o Reveal the presence of the PWM Switch model o Implant its already-available small-signal model o Solve for the transfer function expression Both approaches have pros and cons Going along both paths helps to cross-check results 116 • Chris Basso – APEC 2014 Identify PWM Switches Pair A pair of switches appear in primary and secondary sides L Cclp C R Ic Vin c Ip p a Ic c a p Ip 117 • Chris Basso – APEC 2014 Primary-Side PWM Switch Model Identify currents and voltages during transitions Cclp Vds − on 0 Vin Cclp Lmag Lmag ron 2 I mag ron1 ( I mag + NI out ) I mag Vds − off 0 Vclamp I mag + NI out During the on-time DTsw vDS − on ( t ) DTsw 118 • Chris Basso – APEC 2014 = ( I mag + NI out ) ron1 During the off-time (1 − D ) Tsw vDS − off ( t ) (1− D )Tsw = Vin + VC + ron 2 I mag The First PWM Switch Model is Here Connect the PWM Switch the right way, check polarities Cclp Cclp Lmag Lmag vDS ( t ) ron 2 I c ron 2 I c Vloss (1 − D ) I c Vin Vin V pa V pa (1 − D ) a V pa (1 − D ) a d PWM switch VM Ic Vap ( D − 1) I c = I mag c DVap − DI c −Vap c Vclamp (1 − D ) I c p p Tsw Vloss 119 • Chris Basso – APEC 2014 Secondary-Side PWM Switch Model The forward converter is a buck-derived topology N (Vin − Vloss ) Vloss 2 a 3 . .I out DN (Vin − Vloss ) c Vout 5 4 D D d 6 PW M switch VM p X2 PWMCCMVM 1: N 1 Vin I p = DNI out Vloss = ( I mag + NI out ) ron1 Ip Vloss = I mag + ron1 D Primary-side loss NI out 120 • Chris Basso – APEC 2014 Check the Dual-PWM Switch Response Compare ac response with that of the large-signal version {ron1}*(I(VLP)+(I(V5)/V(D))) B4 Voltage in X2 PWMCCMVM 47.7V 48.0V 7.95V 17 Iin V1 48 C2 0.2u 91.7V D out 3.30V 11 Vout 6 3.80V 3.80V 48.0V 10 VIL {Vf} L1 0.5u c 16 X1 XFMR RATIO = N L2 50u a d 0V p PWM switch VM C1 500u R1 100m 1 5 {ron2}*I(VLP) Vclp R2 50m VLP B2 Voltage 48.0V 0V 2 clp 4 B5 Voltage ({ron1}*(I(VLP)+(I(V5)/V(D))))*V(D) 47.8V V5 18 c p 91.7V PWM switch VM parameters 478mV L3 1kH 14 C3 1kF 0V 478mV out 12 478mV 3.30V 13 15 d V6 3.3 Vstim AC = 1 a N=1/6 ron1=60m ron2=60m Vf=0.5 E1 10k R5 100m D X3 PWMCCMVM D Check first if dc operating points are similar Use this fixture to also run transient simulations 121 • Chris Basso – APEC 2014 Large-Signal Equations with SPICE Non-linear equations are linearized by SPICE V(D)*{N}*(V(in)-{ron1}*(I(VLP)+I(VIL)*{N}))-{Vf} L1 0.5u 3.30V in 48.0V 0V 10 Iin V1 48 B4 Current I(VIL)*V(D)*{N} L2 50u C2 0.2u 48.0V 4 5 Vclp clp VLP 91.7V VIL 3.30V out Vout 6 3.30V C1 500u B3 Voltage 0V R1 100m 1 R2 50m V5 48.0V B1 Current (1-V(D))*I(VLP) 2 B2 Voltage parameters (V(in)+V(clp,in)+{ron2}*I(VLP))*(1-V(D))+V(D)*{ron1}*(I(VLP)+(I(V5)/V(D))) D E1 10k R5 100m 478mV L3 1kH 14 C3 1kF 0V 478mV N=1/6 ron1=60m ron2=60m Vf=0.5 out 12 478mV 13 15 3.30V Vstim AC = 1 V6 3.3 Proposed by Dr. José Capilla, ON Semi, November 2012 122 • Chris Basso – APEC 2014 Curves Perfectly Superimpose Plot1 vdbout#b, vdbout 30.0 Vout ( f ) D( f ) 20.0 10.0 0 -10.0 Plot3 vdbclp#a, vdbclp 100 Vclamp ( f ) D( f ) 80.0 Control-to-clamp-voltage transfer function 60.0 40.0 8 3 20.0 60.0 Plot4 idbin, idbin#a I in ( f ) D( f ) 50.0 Input current response 40.0 30.0 Plot2 ph_vout#a, ph_vout 20.0 Vout ( f ) D( f ) 10 9 20.0 -20.0 -60.0 7 2 -100 -140 Control-to-output transfer function argument 10 100 1k frequency in hertz 10k 100k 123 • Chris Basso – APEC 2014 Compare Transient Responses Cycle-by-cycle results are similar to that of average model Plot1 vout#a, vout in volts 4.50 3.50 12 2 2.50 vout ( t ) 1.50 Plot3 ilmag#a, ilmag in amperes 500m imag ( t ) 4.00 2.00 0 13 11 -2.00 -4.00 vclamp ( t ) 130 Plot2 va, vclp in volts ∠ 6 1 Control-to-output transfer function magnitude 110 14 7 90.0 70.0 50.0 200u 124 • Chris Basso – APEC 2014 600u 1.00m time in seconds 1.40m 1.80m Simulate with Small-Signal Sources Plug the small-signal models of the PWM Switch in B8 Voltage V(a2,p2)*V(dac) {ron1}*(I(VLP)+I(VIL)*{N}) in 48.0V 47.7V 7.94V 13 V1 48 Iin Imag C2 0.2u L2 50u 48.0V 191.7V 0 L1 0.5u VIC2 3.80V c2 11 c2 VIL {Vf} 3.80V out 3.30V Vout 6 3.80V X1 XFMR RAT IO = N X5 dcXFMR B7 Current I(VIC2)*V(dac) -33.0nV C1 500u 0V R1 100m 1 d VLP 48.0V 0V 2 91.7V Vclp 9 5 B2 Voltage {ron2}*I(VLP) B6 Voltage 3.80V a2 a2 47.8V D p2 4 B5 Voltage R3 1n R2 50m p2 p2 V5 V(D)*{ron1}*(I(VLP)+{N}*I(VIL)) parameters N=1/6 ron1=60m ron2=60m Vf=0.5 Iout=33 c1 c1 p1 p1 0V Vstim AC = 1 VIC1 VDbias 478.279m 47.8V 478mV 20 d D D dac X4 dc XFMR 176fV 19 B3 Current I(VIC1)*V(dac) B1 Voltage (V(a1,p1)*V(dac))/V(D) a1 176fV a1 Ac modulation R4 1n Dc bias Check if ac response is ok before proceeding! Simplify circuitry to concentrate on control-to-output only 125 • Chris Basso – APEC 2014 Same Response Between Fixtures Plot1 vdbout#a,vdboutin db(volts) 20.0 14.0 8.00 2.00 Vout ( f ) D( f ) -4.00 38 36 Plot2 ph_vout#a,ph_voutin degrees 20.0 -20.0 -60.0 ∠ Vout ( f ) D( f ) 39 37 -100 -140 10 126 • Chris Basso – APEC 2014 100 1k 10k 100k Make the Circuit Look Friendlier Re-arrange sources to simplify the electrical circuit Mag. current generator B8 Voltage {ron1}*{N}^2*(I(VLP)/{N}+I(VIL)) (V(4,0)*V(dac))/{D} in 15 V1 48 Iin C2 0.2u B6 Voltage B7 Current I(VIL)*V(dac) X1 XFMR RATIO = N Lmag 50u 10 5 VIL {Vf} Lout 0.5u 14 Vout 17 3 C1 500u 4 B10 Current I(VIL)*{D} B11 Voltage V(9,0)*{D} R1 100m 1 R2 50m VLP B2 Voltage {ron2}*I(VLP) 9 Isolated buck converter 2 6 B5 Voltage {D}*{ron1}*(I(VLP)+{N}*I(VIL))+{ron1}*{Iout}*{N}*V(dac) B4 Voltage Vclp 13 V(7,6)*{D} parameters B9 Current dac I(VLP)*{D} Vstim AC = 1 7 B3 Current I(VLP)*V(dac) B1 Voltage (V(6)*V(dac))/{D} N=1/6 ron1=60m ron2=60m Vf=0.5 Iout=33 D=478.279m dac = dˆ D = D0 Run a sanity check further to any simplification 127 • Chris Basso – APEC 2014 Circuit is Looking Simpler Now Look for ways to get simpler equations, final arrangement VIL {Vf} Lout 0.5u in 4 Vout 17 3 V1 48 C2 0.2u Lmag 50u B12 Voltage 10 5 R1 100m 1 R2 50m VLP B2 Voltage {ron2}*I(VLP) C1 500u 2 6 B5 (({N}*V(in)-{ron1}*{N}^2*(I(VLP)/{N}+I(VIL)))*(1+V(dac)/{D}))*{D} Voltage {D}*{ron1}*(I(VLP)+{N}*I(VIL))+{ron1}*{Iout}*{N}*V(dac) B3 Voltage Vclp 13 parameters V(6)*((V(dac)/{D})+1)*{D} B9 Current I(VLP)*{D} dac Vstim AC = 1 Final ac check is mandatory! 128 • Chris Basso – APEC 2014 Secondary-side contribution – neglected for simplicity N=1/6 ron1=60m ron2=60m Vf=0.5 Iout=33 D=478.279m Simpler Circuit Does Not Distort Response We are good to start analyzing the equivalent circuit Plot1 vdbout#a, vdbout in db(volts) 30.0 20.0 10.0 0 18 15 -10.0 Plot2 ph_vout#a, ph_vout in degrees 20.0 -20.0 -60.0 19 17 -100 -140 10 100 1k frequency in hertz 10k 100k 129 • Chris Basso – APEC 2014 Start with the Clamp Circuitry For the dc transfer function, open caps and short inductors 48.0V V1 48 6 Vin Vclp VLP Imag 48.0V 1 92.0V c Vclamp B9 Current I(VLP)*{D} B5 Voltage V(c)*{D} parameters Vclp = V( c ) D = Vclamp D N=1/6 ron1=60m ron2=60m Vf=0.5 Iout=33 Vin=48 D=478.279m Vclamp = Vin + Vclp Vclp = (Vin + Vclp ) D Vclp = Vin dac 0V Vstim AC = 1 D 0.478 = 48 ≈ 44 V 1− D 1 − 0.478 Vclamp = Vin + Vclp = 92 V This is a buck-boost dc transfer function 130 • Chris Basso – APEC 2014 We Need the Magnetizing Current Use KVL and KCL to get the mag. current expression Vin 48V ac = 0 C2 0.2u iˆmag 3 VLmag Lmag 50u 4 B1 Voltage {ron2}*imag ( B2 Voltage {D}{ron1}*Imag The secondary side contribution has been purposely neglected, Iout and îout 1 B5 Voltage V(c)*(V(dac)+{D}) iˆmag (1 − D ) c iˆmag D iˆmag B9 Current imag*{D} ) VLmag = V(C ) − V( C ) dˆ + D0 + D0 ron1iˆmag 2 VC 131 • Chris Basso – APEC 2014 Collect Terms and (Carefully) Simplify Use KVL and KCL to get the mag. current expression VLmag = V ( 2 ) − Vin with ( ) V ( 2 ) = V( C ) − V(C ) D0 + dˆ + D0 ron1iˆmag The voltage at node (c) has a dc and an ac component ( VLmag + vˆLmag = V(C ) + vˆ( C ) − V( C ) + vˆ(C ) ) ( D + dˆ ) + D r 0 iˆ 0 on1 mag − Vin − vˆin VLmag + vˆLmag = V(C ) + vˆ( C ) − V(C ) dˆ − V(C ) D0 − vˆ( C ) dˆ − vˆ(C ) D0 + D0 ron1iˆmag −Vin − vˆin ≈0 0 Sort out an ac and a dc equation VLmag Tsw =0 Vin = V(C ) (1 − D0 ) dc VLmag = V( C ) − V(C ) D0 − Vin ac vˆLmag = vˆ(C ) (1 − D0 ) − Vclamp dˆ + D0 ron1iˆmag 132 • Chris Basso – APEC 2014 Vclamp = Vin (1 − D0 ) Express the Magnetizing Current imag The clamp capacitor ac voltage depends on imag 1 vˆ(C ) = iˆmag (1 − D0 ) sCclp 1 + iˆmag ron 2 = iˆmag (1 − D0 ) sCclp + ron 2 substitute iˆmag = − vˆLmag sLmag =− vˆ(C ) (1 − D0 ) − Vclamp dˆ + D0 ron1iˆmag sLmag Solve for imag I mag ( s ) = D ( s ) Vclamp sCclp D0 − 2 D0 + 1 + sCclp ( ron 2 + D0 ron1 − D0 ron 2 ) + s 2 Lmag Cclp 2 133 • Chris Basso – APEC 2014 Identify Second-Order Coefficients Identify terms with a second-order polynomial form I mag ( s ) D (s) = Vclamp (1 − D0 ) sCclp 2 r (1 − D0 ) + D0 ron1 2 Lmag Cclp 1 + sCclp on 2 +s 2 2 (1 − D0 ) (1 − D0 ) Develop and rearrange: I mag ( s ) D (s) M0 = sCclp = M0 1+ 134 • Chris Basso – APEC 2014 s s + ω0M QM ω0M Vclamp (1 − D0 ) 2 = Vin (1 − D0 ) 3 ω0M = 2 QM = 1 − D0 Cclp ron 2 (1 − D0 ) + D0 ron1 Lmag 1 − D0 Lmag Cclp Re-Write the Expression Nicely A tuned network offers the following transfer function A0 Q ( dB ) 20 dB/decade -20 dB/decade H ( s ) = A0 H(f) 90° ∠H ( f ) 0° 1 ω s 1+ 0 + Q s ω0 -90° ω0 I mag ( s ) D (s) A0 = I mag ( s ) sCclp = M0 1+ s s + ω0M QM ω0M D (s) 2 Vclamp = A0 1 ω s 1 + 0M + ω s 0M A0 (ω0M ) = 67.713 dB ron 2 (1 − D0 ) + D0 ron1 135 • Chris Basso – APEC 2014 Time for an ac Check SPICE can obtain the ac response in a snap-shot 48.0V parameters 6 V1 48 C2 0.2u 92.0V 2 Lmag 50u 48.0V 5 B1 Voltage {ron2}*I(VLP) VLP Imag 48.0V 1 B2 Voltage {D}*{ron1}*I(VLP) B5 Voltage 92.0V c 348.0V V(c)*(V(dac)+{D}) B9 Current I(VLP)*{D} 136 • Chris Basso – APEC 2014 N=1/6 ron1=60m ron2=60m Vf=0.5 Iout=33 Vin=48 D=478.279m dac 0V Vstim AC = 1 QM Curves Perfectly Superimpose It is important to obtain similar plots, otherwise: error! 60 50 20⋅ log Hclp ( i⋅ 2π ⋅ f k ) 40 , 10 A ( 20 ) arg H clp ( i⋅ 2π ⋅ f k ) ⋅ 0 180 π − 50 0 4 1×10 1×10 5 fk Max SPICE is 63.710 dB Max Mathcad® is 63.713 dB "Can do!" 137 • Chris Basso – APEC 2014 Run Another Round of Rearrangement Now concentrate on the buck output stage only B8 Voltage (V(4,0)*V(dac))/{D} {ron1}*{N}^2*(I(VLP)/{N}+I(VIL)) 8 Vin X1 XFMR RATIO = N 5 9 B6 Voltage B7 Current I(VIL)*V(dac) 1 3 Vin 48V 7 7 B6 Voltage B7 Current I(VIL)*V(dac) B8B8 Voltage Voltage V(9,0)*V(dac) V(7,0)*V(dac) 1 1 B11 Voltage V(9,0)*{D} V(7,0)*{D} dc 138 • Chris Basso – APEC 2014 ( V ( 7 ) D + dˆ ac 6 B10 Current I(VIL)*{D} R1 100m 6 R2 50m {ron1}*{N}^2*(I(VLP)/{N}+I(VIL)) 5 C1 500u B11 Voltage V(9,0)*{D} B10 Current I(VIL)*{D} I mag V ( 7 ) = NVin − ron1 N + I out N 8 Vout 2 reflect 4 2 X1 XFMR RATIO = N VIL {Vf} Lout 0.5u ) F ( s) VIL {Vf} Lout 0.5u Vout 2 3 C1 500u R1 100m 4 R2 50m Further Simplification is Necessary Extract the ac current from the equation, iˆmag V( 7 ) = NVin − ron1 N 2 + I out + iˆout N Develop V(1) keep ac terms only ( imag ( t ) Tsw =0 iˆmag V(1) = NVin − ron1 N 2 + I out + iˆout N ) ( ) ( D + dˆ ) ˆˆ r − DI N 2 r − DN 2 iˆ r − I N 2 dr ˆ − N 2 di ˆˆ r DNVin + NVin dˆ − DNiˆmag ron1 − Ndi mag on1 out on1 out on1 out on1 out on1 dc ≈0 dc ≈0 Rearranging the result, we obtain: ron1 ≪ 1 ˆ NVin dˆ − DNiˆmag ron1 − DI out N 2 ron1 − DN 2 iˆout ron1 − I out N 2 dr on1 ( dˆ ≪ 1 ) 2ˆ ˆ ˆ dˆ NVin − I out N 2 dr on1 − DNimag ron1 − DN iout ron1 N2 ≪1 Neglecting small terms, we finally obtain: ˆ ˆ vˆ(1) = dNV in − DNimag ron1 139 • Chris Basso – APEC 2014 Add the Second-Order Response of LC Filter The magnetizing current definition is: I mag ( s ) D (s) = M0 sCclp s s 1+ + ω0M QM ω0M 2 = M (s) Once re-injected in to the previous definition, we have: V1 ( s ) = D ( s ) NVin − D0 Nron1 D ( s ) M ( s ) V1 ( s ) = D ( s ) NVin − D0 Nron1 M ( s ) The source is filtered by the 2nd-order LC filter: 1 F ( s ) = F0 2 Lout rL 3 Cout rC ω0F = 140 • Chris Basso – APEC 2014 1+ Rload 4 F (s) 1+ F0 = 1 Lout Cout rL + Rload rC + Rload QF = s szF s s + ω0F QF ω0F RLoad Rload + rL ω zF = 2 1 rC Cout Lout Cout ω0F ( rC + RLoad ) Lout + Cout ( rL rC + Rload ( rL + rC ) ) We Have the Final Transfer Function The transfer function reveals the mag. current contribution Vout ( s ) D (s) = F ( s ) ( NVin − D0 Nron1 M ( s ) ) YES! The mag. current substracts and explains the notch sCclp Vout ( s ) = F0 N Vin − D0 ron1 M 0 2 2 D (s) s s s s + + 1+ 1+ ω0FQF ω0F ω0M QM ω0M s szF 1+ This is the control-to-output transfer function of the ACF 141 • Chris Basso – APEC 2014 Final Sanity Check - Magnitude Compare the analytical ac response with that of SPICE 20 10 H 1( i⋅ 2π ⋅ f k ) V 20⋅ log , 10 0 − 10 Vout ( f ) − 20 10 D( f ) 100 3 4 1×10 1×10 1×10 fk Magnitude curves superimpose perfectly 142 • Chris Basso – APEC 2014 5 1×10 6 Final Sanity Check - Phase Compare the analytical ac response with that of SPICE 0 arg H i⋅ 2π ⋅ f ⋅ 180 ( 1( k)) π − 100 ∠ Vout ( f ) D( f ) 10 3 1×10 100 1×10 4 5 6 1×10 1×10 fk Argument curves are in excellent agreement too 143 • Chris Basso – APEC 2014 The Bench is the Final Referee Build a hardware using simple gates arrangements D1A m br2545ctp 20 L1 74uH 1 Vout 2 14 D1B m br2545ctp C5 100u Vin 51V C1 480uF V15 V1 15V V15 U1A CD4093B 4V 50 kHz R7 4.7k 0 5 U1B CD4093B V15 V15 Vgen 23 13 VEE 3 15 QA IRF9530 Q4 2N2907 U1E CD4093B V15 U1C CD4093B V15 C8 1uF 17 21 V15 R2 100k 16 R3 14k U1D CD4093B C2 330p Q3 2N2222 Out M 7 R4 1k U2A LM393T C7 0.1uF R1 1k 26 6 C6 220n QM SPP04N60C3 V15 V15 VCC 25 rC = 76m XFMR RATIO = 0.25 V15 C9 0.1uF V15 ac input Out A 11 9 8 R5 1k C3 330p U1F CD4093B R10 1k Q1 2N2222 10 C4 Q2 0.1u 2N2907 12 D2 1N4148 R6 10k This active clamp forward converter delivers 5 V/5 A 144 • Chris Basso – APEC 2014 R9 1 The Prototype Hardware The circuit is assembled using available components Make sure caps. are well characterized before soldering With the kind help of Yann Vaquette, Application engineer 145 • Chris Basso – APEC 2014 Typical Prototype Waveforms voutA ( t ) voutM ( t ) vDS ( t ) N-channel 146 • Chris Basso – APEC 2014 Quasi-ZVS is Ensured on the Drain voutA ( t ) voutM ( t ) 140 ns 100 V Vin = 51 V vDS ( t ) 28 V N-channel 147 • Chris Basso – APEC 2014 ZVS is Also Ensured on Clamp Switch voutA ( t ) voutM ( t ) 240 ns Body diode vDS ( t ) P-channel 148 • Chris Basso – APEC 2014 −100 V Current in the Clamping Network voutA ( t ) voutM ( t ) iCclamp ( t ) 149 • Chris Basso – APEC 2014 SPICE Test Fixture We have used the following SPICE simulation fixture {ron1}*(I(VLP)+(I(V5)/V(D))) X2 PWMCCMVM B4 Voltage in 51.0V 49.8V 12.5V 17 Iin Vin 51 C2 0.23u 99.8V L2 {Lmag} 51.0V 10 Vclp L1 74u c 20 X1 XFMR RATIO = N VIL {Vf} R3 60m 5.20V 11 6.16V D PWM switch VM 6.16V Vout 5.85V 0V 2 4 B5 Voltage ({ron1}*(I(VLP)+(I(V5)/V(D))))*V(D) V5 18 c parameters p 99.8V X3 PWMCCMVM R5 100m 1.98V D GAIN X4 GAIN K = 1/4 a d PWM switch VM N=0.25 Lmag=580uH ron1=0.9 ron2=0.6 Vf=0.65 19 C3 1kF 0V L3 1kH 14 E1 10k 1.98V out 12 1.98V 5.20V 13 15 Vstim AC = 1 D Check dc points versus hardware values: ok 150 • Chris Basso – APEC 2014 R1 1 1 R2 76m 51.0V 50.4V C1 480uF 0V p VLP B2 Voltage clp out 6 d 495mV 5 {ron2}*I(VLP) a 16 V6 5.2 Magnitude Response There is a slight shift but overall agreement is good SPICE measured H(f) cap. ESR is often the offender in loop measurement 151 • Chris Basso – APEC 2014 Phase Response Good agreement between curves, expecially peaking measured SPICE ∠H ( f ) The notch Q depends on resistive elements rDS(on) etc. 152 • Chris Basso – APEC 2014 Comparison with Simplis Response Excellent agreement between curves, almost no shift Simplis measured H(f) 153 • Chris Basso – APEC 2014 Comparison with Simplis Response Despite a slightly lower peaking, phase agreement is ok Measured ∠H ( f ) 154 • Chris Basso – APEC 2014 Simplis Simulation also Confirms Damping! Inserting a 10-Ω resistance damps the notch nicely simulated H(f) measured simulated ∠H ( f ) measured 155 • Chris Basso – APEC 2014 A Practical Case The model helped to stabilize this 3.3-V/30-A converter ∠T ( f ) f c = 31.9 kHz ϕm = 62.9° 156 • Chris Basso – APEC 2014 T(f) ϕm Transient Response Test Step-load transient response confirms stability margins Overshoot = 42.2 mV vOUT ( t ) 3.3 V Undershoot = 48.2 mV VIN = 36 V, 15 A to 22.5 A - Slew rate 1 A / µs 157 • Chris Basso – APEC 2014 Conclusion The PWM switch model is an essential tool for modeling We have seen how to derive it in different operating modes Small-signal modeling using the PWM switch is simple and fast When modeling converters, always proceed step by step Always perform intermediate sanity checks (SPICE, Mathcad®…) Analytical analysis does not shield you against lab. experiments Analysis, simulation and bench: the path to success! Merci ! Thank you! Xiè-xie! 158 • Chris Basso – APEC 2014