Lecture 47 Switch Mode Converter Transfer Functions: T vd (s) and T vg (s) A. Guesstimating Roots of Complex Polynomials( this section is optional). Quick Insight n=. n th order case. Cuk example 4. Forth Order Filter Problem 8.9 B. Buck Boost T vd (s) and T vg (s) Analysis. Circuit Model. Loop Equations. T vg (s) in standard form : Two poles 4. T vd (s) in standard form : RHP Zero and two poles 5. Specific Buck-Boost Design a. DC conditions : CCM b. AC Bode plots : CCM C. Right Half Plane Zeros. Transient Response. Origin of RH plane Zero in differing delays. T(s) of a single Zero insight 4. Phase Margin Limitations with RH Plane Zero
Lecture 47 Switch Mode Converter Transfer Functions: T vd (s) and T vg (s) A. Guesstimating Roots of Complex Polynomials : P(s). Quick Insight OK approximation only if Why bother? Left Side Roots hard to guess a s + a s+ = (+ a s)(+ a a s) a a = + s[a + a a ] + a s << a Right Side Roots are easy to guess s = - a s = - a a low f root coeff. if a is big high f root if a is big This is the Widely separated root approximation: X--------------------------------X s = /a Means s = a /a This is valid if a /a << a Can we generalize this to higher order polynomials? Can we avoid root finder programs? It s really academic as root finders are widely available on calculators.. n th order case a. ideal case for approximation R(s) = + a s + a s + a n s n All a s are real
Form is arbitrary but USEFUL Beat/Cajole/Force into standard form R(s) = ( + t s)(+t s)...(+t n s) easy to see by expanding out all the factors. t s or time constants are either real or complex a = t + t +... t n a = τ( τ + τ +... τ n) + τ ( τ + τ 4... τ n) a = ττ ( τ + τ 4 +... τ n) + τ τ ( τ 4 + τ 5+... τ n)... a n = t t t...t n easy to see by expanding out factors. We ask under what conditions can solutions be easily approximated? We can use above for approximating roots if the roots are real and well seperated. Then only the first term of each equation is dominant. t > t > t >...> t n This is a big if and the goal is to be clever in choosing the t s in descending order when one does the factoring by hand of polynomials. Few cases are ideal. If successful, a = t ------------- t = a a = t t ----------- t = a /a }Works for simple second order insight case a = t t t --------- t = a /a... a n = t t t n ---- t n =a n /a n- In summary this assumes no inequalities are violated. if a > a a > a a «...«a n and if an- the P(s) = + a s + a s +... a n s N
4 P(s) must be in this standard form then the polynomial can be approximately factored by : P(s) = (+ a s) (+ a a s) (+ a a s) lowest f second f third highest f pole pole pole b. What if one inequality is not met a ideal: P(s) = ( + a s)( + a s)... (+ a n n- s) a a a a a... an an- Non-ideal case where one inequality is not satisfied we leave the corresponding roots in quadratic form: ak ak+ / / ak- ak P(s) = (+ a s)(+ a ak s).. + s + a k+ s a ak- ak-... (+ a n ) an- The quadratic equation below + a k s + a k+ s ak- ak- we solve as a usual quadratic }Valid only for this pair of roots only }a k > a k- * a k+ When several inequalities are violated then we must use cubic or higher order roots and often the roots are close in value.
5 c. Cubic Polynomial: In general for a cubic polynomial this may occur two ways + a s + a s + a s. First and Second Close Roots Case First guess (+ a s)(+ a a s)(+ a a s) ---X--------X-------------------------------------X--------- s s s close pair isolated root (+ a s)(+ a a s) solution not valid (+ a a s) still valid approximation Use: ( + a s+a s ) s = - a a to solve for unknown s and s just solve quadratic quadratic validity (+ a s)(+ a a s) + (a + a a ) s + a s k = a > a a Summary: a k > a k- a k+ _ a > a oa a > a + a s + a s + a a s s and s s. Second and Third Roots Close Case ---X-------------------------------X------X--------- s s s
6 Original equation + a s + a s + a s (+a s) (+ a a s) (+ a a s) Isolated root Invalid First guess valid approximation Close roots Not valid (+ a a s)(+ a a s) Use: + ( a a ) s + (a a ) s + ( a a + a a ) s + a a s Use this quadratic to solve for s and s only if provided a a a k > a k- a k+ a a a > a a k = Summary: a > a a ( + a s) + a a s + a a s s = - /a yields s and s. Cubic Polynomial Case for CuK CuK converter with D, D left out for simplicity Go G(s) = + s( L + L R ) + s L C+ s L L C R a a a How to factor D(s) depends on circuit values. We want (but not always valid) D(s) = (+ a s)(+ a a s)(+ a a s)
T(s) = + s( L + L R 7 ) + s RC L L L + s L R a. a > a > a ideal case of well separated roots a a If L is chosen >> L the approximate factorization is OK because: L R > RC > L for the ideal case of well separated roots R + s L [ ] R + s RC + s L R easy to get roots from circuit a elements. Consider D(s) = ( + a s) ( + a s) (+ a a s) Whereas if no approximations are made: + s( L + L RCL ) + s + s L R L + L R a a What if you find the second inequality is violated a / a a due to component choices! L + L L» RC R L + L but not L / / R This will require a different approach. b. Second and Third Terms Roots Close Then for easy root solving we break T(s) into two terms: one with a single real root and the second with two possibly complex roots. Still valid Use this Quadratic + s( L + L R ) + src L L L + s C* L L
Again this is valid only if both L >> L and L /R >RC + s L [ ] R + src+ s CL If also we find low Q for the quadratic then we get two well separated roots from the quadratic term as well c. First and Second Terms Have Close Roots This is a case when the first inequality is violated. L + L RC L L / / R L + L R The inequalities are the same as L >L and RC> L /R. It is no longer required that L /R >>RC Return to Quadratic (+ s L R ) form for first and second roots s and s. We still have a valid approx. for s = - a /a =L /R and the quadratic is simply (+ s( L + L R ) + s L C) = (+s L /R +s L C) Solve this quadratic for s and s 8 4. Fourth Order Polynomial from a two section L-C Filter Erickson Pbm. 8.9. a. Two Section L-C Filter A two-section L-C filter has the following transfer function: G(s) = + s L + L + L L C s ( L (C + C )) + s + s 4 (ll CC ) R R The element values are: R 50 mw C 680 mf C 4.7 mf
9 L 500 mh L 50 mh G(s) = + s L + L + L L C s 4(C + C ) + L C + s + 4 s L L C C R R a a a a 4 [ ] ( ) R = 50 mw L = 500 mh C = 680 mf L = 50 mh C = 4.7 mf (a) Factor G(s) into approx. real and quadratic poles. Compute a a a a 4 and t t t t 4 a = 0.0 t = a = 0.0 a = 4.6 x 0-9 t = a a a.5 x 0-6 Non ideal sequence a = 40 x 0 - t = = 99.4 x 0-6 we must use a a quadratic to solve s, s a 4 = 80 x 0-8 t 4 = a4 a = 5. x 0-9 a a Check: a >> but a a is not > a a a a >> a >> aa4 a4 Valid to break out one quadratic a a (=) 0.0-6.5x0-6 -9.8x0 5.x0 G(s) = ( + a s)( + a a s + a s )( + a 4 s) a a
s s ( + )( + Qw + s w ) + ( + s w ) wo Compare coefficients, we get w o = o 0.0 = 90.9 rad/ s f = 4.47 Hz w = a a Q w = a a w = a a4 0 = 5687.96 rad/ s f = 905.Hz Q = 5.65db = 4559 rad/ s f = 677.KHz Justification: Actual exp. G(s) = a o + as+ as + as + a4 s A o A A A A 4 Approx. exp. G(s) = + ( a a + a + a 4 a ) s+ ( a +a + a a 4 a a ) s + (a + a 4 a + a 4 a a s ) 4 A o A A A A 4 Numerical values: Pole Finder Actual Factoring Approx. A o A. x 0 -.0 x 0 - A.4 x 0-7.76 x 0-7 A.4 x 0-0.4008 x 0-0 A 4 7.99 x 0-7 7.99 x 0-7
G 0db fo f f f 0 fo f f f /G 45 5 5 We wish to reduce Q from 5.65 db to db without changing break frequencies. Is this possible? (c) Now Q = a From, = a Qw, a a = _ w = a w a a Q = a = a a but C >> C a a a L >> L approx. a L R a L C L a L C R a 4 L L C C
L LLC Then Q = R R = L LC R C w o = = R a L a w = = a L C a w = = a4 RC So Q depends only on R, L, C. Let s change L and C. Can t change R as it changes w o R = 50 mw, Q = L = 0.0 C and C = 0 w = 0 5688 =.76mF L = 0.0 C = 7.58 mh Check: w = a L + L = a LLC = 5687.96 rad/ s f = 905. Hz same B. Buck-Boost T vd (s) and T vg (s) Analysis This is the crucial material for this lecture as part A can be replaced by using a root solver on a basic calculator.. Equivalent ac Small Signal model The Buck-Boost small signal circuit model was developed earlier.
vg(t) ig I d. :D. i(t) L (Vg-V)d(t). D':. I d C v(t) R The buck-boost small signal equations follow. We aim to develop these equations to yield two separate transfer functions as described below G vg and G vd : This is our goal to extract these two functions for the buck-boost from the equivalent circuit. Setting one input to zero will result in great simplifications to the loop equations of the full circuit.. Loop Equations: $ i (t) = D $ i(t) + I d(t) $ (for the first circuit loop on left above) g L d$ i dt = D v (t) + D v(t) + (v - v) d $ (middle circuit loop) g g
4 C dv = - D $ i - $v(t) dt R + I$ d(t) (output loop) Next do the Laplace transform for the small signal output voltage. $v(s) G vd (s) d(s) $ + G vg (s) $v g(s) $v(s) $v(s) $d(s) vg $ (s)=0 v$ g (s) d(s)=0 $ Simple but tedious algebra leads to: Or in the form, that we will use later, after putting DC analysis in. Vg V sli $ $ DD' v $ g(s) v(s) = D' s L d(s) - R s LC D' s L R s + + + + LC yields T vd (s) T vd (s) = $v(s) $d(s) yields T vg (s) $v(s) T v g (s) = v (s) g
5. Line to Output Transfer Function In standard form, such that coefficients of s ο are unity, The line (mains) to output transfer function is: D T (s) ( D' ) vg = s L D' R s LC Pole location in the dynamic transfer + + D' function depend on the choice of DC operating point, D. L eff (D) L/(D ) = f(d, L, C, R). In standard quadratic pole pair form: go T vg (s) = T (constant low f gain) s + Qw + ( s with Q in db units o w ) o Identify like terms T go - D Buck-boost Gain @ dc D w o Leff C = D The dynamic transfer function rollover LC frequency depends on the choice of DC operating point f(d ) Qw = L eff R o Q = D R c Note also that the second order Q also depends on L
6 the DC operating point chosen for the converter Q = f(d ) 4. Control to Output Transfer Function: T vd (s) $v(s) $d(s) $v g(s) = 0 T vd (s) = f(d, V g,v, I, L, C, R) All DC voltages/currents as well as D and component values of L and C effect the dynamic response. In standard form such that s o coefficients are unity: T vd(s) = Vg D' LI s V Vg V + s L + R s L C eff eff or in the standard form below, Poles of T vd (s) are the same as poles of T vg (s), but the zero is unique. It is a right half plane zero. We effect w z location by choice of all DC values for V g, V, I etc. w z (RHP ZERO) = D R/DL we will show below. Compare to standard. quadratic form: T do(constant low f gain)[- s/ w z] s + Qw + ( s = T VD (s) o w ) o It s easy to design the DC gain for Buck-Boost given V(out) and
the duty cycle. Vg V T do(dc) = D' Vg V = = D' DD' 7 V Recall for Buck-Boost DC conditions: V = - D g D and I = - V R D Be sure to note that I(DC) is in inner loop of the buck-boost circuit model on page. Then we find from the DC conditions the zero location: g w z = V - V LI = D R right-half plane D L Next we try to get more specific and quantitative. V = -.6.4 5. Specific Buck-Boost Design D = 0.6 R = 0 V g = 0V L = 60mH } f sw = C = 60mF a. We know DC operating conditions 0 = - 45V
I = - 45 =.5A in middle loop of model, on page. 0.4 We also find I out = - 4.5A in the outer loop of the page circuit. Recall that I(middle) = I out + I in? for CCM conditions b. AC values in std. transfer functions (using db for plots) for these DC conditions. First a T VG (s) having only poles is considered. G go = D D' =.5 or.5 in db units 8 o f o = w π = D' π LC = 400 Hz Q = D' R C L = 4 in normal units or in db units T vg(s) = - D =.5 =.5db D s + ()( f o) + ( s ) π π fo Q 400 Hz low resonance Q = D R C/ L = 4 = db Now for a T VD (s) having a right hand plane zero we have a new set of values. V G do = = 469 or 5.4 db D(D') f z = D' R = 6.6 khz πdl s T vd (s) = G do w z s s + + ( ) ()π f o π f o
In summary: Q in db 400 Hz 9 If L goes to small values(dcm) what happens to the zero location? In G vd (f) Phase Plots we will find Two poles cover 80 o over a frequency range, where now Q is in real units and not in db lower - Q + Q f = 0 f = 0 (400) = 00Hz upper o o /8 /8 f = 0 f = 0 400 = 5Hz In G vd (f) Gain Plots we will find: We have a double pole at 400 Hz showing a resonant Q peak. Gvd Q effects the peaking Q = 4 or db The RHP zero will add amplitude above f z = 6.6 khz but will cause additional phase lag threatening stability if we use feedback. f
0 Below are plotted Gvd and Gvd Note carefully the unity gain line at 0 db. RH plane zero starts to contribute phase angle at 660 Hz and LIMITS phase margin at unity gain. Additional phase contribution starts at f z /0 @ 660 Hz. The G vg amplitude and phase is much simpler as shown below. Here the f margin at unity gain is also small but not as bad as T vd.
C. Right Half Plane Zero Issues and Indicators of This The above two switch mode circuits have right-half plans zeros.. Transient Response Indicators of a RHP zero a. Typical of nd order poles Vo(final) Vg t versus b. Typical of (RH Plane/zero) d d Vo t Vo(initial) Vo(initial) Vo(final) This RH plane zero occurs for several PWM dc-dc converters boost buck-boost flyback It does not occur for buck converter. WHY?
. Origin of RH Plane Zero Differing time delays for D which changes V o immediately & i m which changes V o only slowly due to switching over several switch cycles, causing initial V o to fall for all converters with output driven by i m. We have therefore the case that: (i m ) stays the same over short t time periods but D and D change fast. D chnages instantaneously by the control circuit knob, this instantaneously lower D and thus V o goes down initially and only later increases when the effect of bigger D makes it to the output. * i Lm cannot change instantaneously * time interval d = -d can change instantaneously which is the duration i m is delivered to the output. This causes a phase lag and momentary drop in V o followed by subsequent increase in V o sudden change in d that builds up i m slowly. i L varies slowly i D duration changes instantaneously but i Lm does not V o then id(t) <id(t)>ts il(t) t d=0.4 d=0.6 t v(t) d=0.4 d=0.6 t
. T(s) of a single right half plane zero insight Vo z V ~ (- s/ w )} A RHP zero acts as two parallel paths to V o d uin(s) + - uout(s) s/wz Overall transient response: former initial change final change 4. Phase Margin Limitations of Any Closed Loop with RHP Zero Present AoL A CL = Want (Aβ) = + A β Then < Aβ far from 80 o How far? 76 o away (for conservative designer), 60 for average designer and 0 away for poor designer! Flyback converter bode plots: Oddity is gain (low f) is non-linear w.r.t. V d f range with -40db/decade drop depends on low f gain. Not enough φ margin come close to instability at unity gain φ 80 o
Note the unity gain location at 0 db. Flyback operating DCM does not have right half plane zero! Without LH plane zero above the phase margin would be nil. Moreover, RH plane zero returns phase to 80 o. 4