FB1 - DC-DC Converters Feedback and Control

DC-DC Converters Feedback
and Control
Agenda
‰ Feedback generalities
‰ Conditions for stability
‰ Poles and zeros
‰ Phase margin and quality coefficient
‰ Undershoot and crossover frequency
‰ Compensating the converter
‰ Compensating with a TL431
‰ Watch the optocoupler!
‰ Compensating a DCM flyback
‰ Compensating a CCM flyback
‰ Simulation and bench results
‰ Conclusion
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2
Agenda
‰ Feedback generalities
‰ Conditions for stability
‰ Poles and zeros
‰ Phase margin and quality coefficient
‰ Undershoot and crossover frequency
‰ Compensating the converter
‰ Compensating with a TL431
‰ Watch the optocoupler!
‰ Compensating a DCM flyback
‰ Compensating a CCM flyback
‰ Simulation and bench results
‰ Conclusion
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3
What is Feedback?
‰ A target is assigned to one or several state-variables, e.g. Vout
= 12 V.
‰ A circuitry monitors Vout deviations related to Vin, Iout, T° etc.
‰ If Vout deviates from its target, an error is created and fed-back
to the power stage for action.
‰ The action is a change in the control variable: duty-cycle (VM),
peak current (CM) or the switching frequency.
Compensating for the converter shortcomings!
Input voltage
Vin
DC-DC
Output voltage
Vout
Input voltage
Vin
action
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4
control
Rth
Vth
Vout
The Feedback Implementation
‰ Vout is permanently compared to a reference voltage Vref.
‰ The reference voltage Vref is precise and stable over temperature.
‰ The error,ε = Vref − αVout, is amplified and sent to the control input.
‰ The power stage reacts to reduce ε as much as it can.
Power stage - H
Vout
Control
variable
d
Error amplifier - G
Rupper
+
-
Vin
α
-
Modulator - GPWM
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5
+
Vp
Vref
Rlower
Agenda
‰ Feedback generalities
‰ Conditions for stability
‰ Poles and zeros
‰ Phase margin and quality coefficient
‰ Undershoot and crossover frequency
‰ Compensating the converter
‰ Compensating with a TL431
‰ Watch the optocoupler!
‰ Compensating a DCM flyback
‰ Compensating a CCM flyback
‰ Simulation and bench results
‰ Conclusion
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6
Positive or Negative Feedback?
‰ Do we want to build an oscillator?
The « Plant »
ε
Vin(s) +
Vout(s)
H(s)
Error voltage
−
G(s)
Vout ( s )
Vin ( s )
=
H (s)
1+ H (s)G (s)
Open-loop gain T(s)
⎡
⎤
H (s)
Vout ( s ) = lim ⎢
Vin ( s ) ⎥
Vin ( s ) → 0 1 + G ( s ) H ( s )
⎣⎢
⎦⎥
Sign is neg for:
ϕ = -180°
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7
=1
To sustain self-oscillations, as Vin(s)
goes to zero, quotient must go infinite
Conditions for Oscillations
‰ when the open-loop gain equals 1 (0 dB) – cross over point
‰ total rotation is -360°: -180° for H(s) and -180° for G(s)
¾ we have self-sustaining oscillating conditions
Total phase delay at fc:
180
-180° H(s) power stage
-180° G(s) opamp
Loop gain |H(s)|
Gain is 1
at fc
90.0
total = -360°
0 dB
0
Loop phase arg H(s)
ϕ = -180°
-90.0
21
22
-180
1
10
100
1k
frequency in hertz
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8
10k
100k
-180°
The Need for Phase Margin
‰ we need phase margin when T(s) = 0 dB
‰ we need gain margin when arg T(s) = -360°
T(s)
80.0
180
phase
gain
-90.0
-180
vdbout in db(volts)
Plot1
ph_vout in degrees
0
0
0°
0 dB
2
gain
margin
Crossover
frequency fc
T(s) = 0 dB
-40.0
1
-80.0
10
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arg T(s)
= -360°
40.0
90.0
Phase margin:
The margin before the loop
phase rotation arg T(s)
reaches -360° at T(s) = 0 dB
phase
margin
100
1k
10k
100k
Gain margin:
The margin before the loop
gain T(s) reaches 0 dB at a
freq. where arg T(s) = -360°
Agenda
‰ Feedback generalities
‰ Conditions for stability
‰ Poles and zeros
‰ Phase margin and quality coefficient
‰ Undershoot and crossover frequency
‰ Compensating the converter
‰ Compensating with a TL431
‰ Watch the optocoupler!
‰ Compensating a DCM flyback
‰ Compensating a CCM flyback
‰ Simulation and bench results
‰ Conclusion
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10
Poles and Zeros
‰ A plant (power stage) loop gain is defined by:
N (s)
H (s) =
D (s)
numerator
denominator
‰ solving for N(s) = 0, the roots are called the zeros
‰ solving for D(s) = 0, the roots are called the poles
Two zeros
H (s)
s + 5k )( s + 30k )
(
=
s + 1k
sz1 = −5k
sz2 = −30k
s p1 = −1k
One pole
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5k
= 796 Hz
2π
30k
= 4.77 kHz
f z2 =
2π
1k
= 159 Hz
f p1 =
2π
f z1 =
Poles and Zeros
‰ A pole lags the phase by -45°at its cutoff frequency
0
Vin
Vout
Cutoff frequency
-3 dB
20.0
R2
1k
0
-20.0
V1
AC = 1
-40.0
C1
10nF
-60.0
|Vout(s)|
-1 slope
-20 dB/decade
1
0
-20.0
Vout ( s )
1
1
=
=
Vin ( s ) 1 + sRC 1 + s
-60.0
ω0
-80.0
10
ω0 =
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-45° at
cutoff
-40.0
1
RC
argVout(s)
100
1k
10k
100k
1Meg
2
10Meg
-90° delay
for f = ∞
Poles and Zeros
‰ A zero boosts the phase by +45°at its cutoff frequency
0
0
40.0
0
plot1
vdbout in db(volts)
Plot1
vdb2 in db(volts)
30.0
+1 slope
+20 dB/decade
+1 slope
+20 dB/decade
20.0
1
|Vout(s)|
1
-20.0
20.0
-40.0
Cutoff frequency
-3 dB
10.0
|Vout(s)|
0
Plot2
ph_v2 in degrees
70.0
Cutoff frequency
-3 dB
90°
+45° at
cutoff
1k
frequency in hertz
90°
70.0
10k
The general form of a zero:
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1Meg
0°
10.0
100k
10Meg
+45° at
cutoff
10
Vin
100
C1
10nF
1k
Vout
s
ω0
100k
argVout(s)
30.0
argVout(s)
G (s) = 1 +
10k
frequency in hertz
50.0
0°
100
1k
90.0
30.0
10
100
2
50.0
10.0
10
plot2
ph_vout in degrees
90.0
-60.0
V1
AC = 1
R2
1k
10k
frequency in hertz
100k
1Meg
2
10Meg
s
Vout ( s )
ω0
sRC
=
=
Vin ( s ) 1 + sRC 1 + s
ω0
1
ω0 =
RC
The Right Half-Plane Zero
‰ In a CCM boost, Iout is delivered during the off time: I out = I d = I L (1 − D )
Id(t)
Id(t)
IL1
IL0
Vin
L
Vin
L
IL(t)
IL(t)
Id0
Id1
d̂
t
t
D0Tsw
Tsw
D1Tsw
Tsw
‰ If D brutally increases, D' reduces and Iout drops!
‰ What matters is the inductor current slew-rate
‰ Occurs in flybacks, buck-boost, Cuk etc.
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d VL ( t )
dt
The Right-Half-Plane-Zero
‰ With a RHPZ we have a boost in gain but a lag in phase!
1
Plot1
vdbout in db(volts)
40.0
20.0
|Vout(s)|
+1 slope
+20 dB/decade
0
LHPZ
-20.0
G (s) = 1 +
-40.0
Plot2
ph_vout in degrees
argVout(s)
G (s) = 1 −
0
-90°
-90.0
2
-180
1
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15
10
ω0
RHPZ
180
90.0
s
100
1k
frequency in hertz
10k
100k
1Meg
s
ω0
Agenda
‰ Feedback generalities
‰ Conditions for stability
‰ Poles and zeros
‰ Phase margin and quality coefficient
‰ Undershoot and crossover frequency
‰ Compensating the converter
‰ Compensating with a TL431
‰ Watch the optocoupler!
‰ Compensating a DCM flyback
‰ Compensating a CCM flyback
‰ Simulation and bench results
‰ Conclusion
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16
How much Margin? The RLC Filter
‰ let us study an RLC low-pass filter, a 2nd order system
R1
{R}
3
L1
{L}
2
Vin
Vout
1
C1
{C}
T (s) =
1
LCs 2 + RCs + 1
T (s) =
1
s2
ωr 2
+ 2ζ
ωr
+1
1
s2
ωr 2
+
s
+1
ωr Q
1
LC
parameters
ωr =
f0=235k
L=10u
C=1/(4*3.14159^2*f0^2*L)
w0=({L}*{C})^-0.5
Q=10
R=1/((({C}/(4*{L}))^0.5)*2*{Q})
C
ζ =R
4L
zeta
s
=
Q=
1
2ζ
ωr resonant freq.
ζ damping factor
Q quality coeff.
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The RLC Response to an Input Step
‰ changing Q affects the transient response
Q=5
Q < 0.5 over damping
Q = 0.5 critical damping
Q > 0.5 under damping
1.80
Q=1
Plot1
vout#6, vout#5, vout#4, vout#3, vout in volts
Q = 0.707
1.40
Overshoot = 65%
1.00
11
9
10
8
7
Asymptotically stable
600m
Fast response and no overshoot!
Q = 0.5
200m
Q = 0.1
5.00u
15.0u
ti
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25.0u
i
35.0u
d
45.0u
Where is the Analogy with T(s)?
‰ in the vicinity of the crossover point, T(s) combines:
ƒ one pole at the origin, ω0 and one high frequency pole, ω2
9 Link the closed-loop response to the open-loop phase margin:
180
T (s) =
T(s)
80.0
phase
gain
-90.0
0
T (s)
0°
0 dB
2
-40.0
Link open-loop ϕm
with closed-loop Q
-80.0
10
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100
1k
=
1
s2
ω0ω2
+
s
ω0
(CL)
+1
1
-1
-180
1+ T (s)
T (s)
-2
10k
1+ T (s)
100k
(OL)
Close the
loop
40.0
vdbout in db(volts)
0
ph_vout
in degrees
90.0
1
⎛ s ⎞⎛
s ⎞
1
+
⎜ ⎟⎜
⎟
ω
ω
2 ⎠
⎝ 0 ⎠⎝
=
1
s2
ωr 2
+
s
+1
ωr Q
Closed-Loop Q Versus Open-Loop ϕm
‰ a Q factor of 0.5 (critical response) implies a ϕm of 76°
‰ a 45° ϕm corresponds to a Q of 1.2: oscillatory response!
10
Q
7.5
( 1+ tan( φ) )
2
1
4
ϕm
5
tan( φ)
2.5
0.5
0
0
25
50
φ⋅
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20
360
2⋅ π
75
76°
100
Summary on the Design Criteria
‰ compensate the open-loop gain for a phase margin of 70°
‰ make sure the open-loop gain margin is better than 15 dB
‰ never accept a phase margin lower than 45° in worst case
PM = 10°
5.12
PM = 25°
PM = 45°
PM = 76°
Plot2
vout2#a, vout2, vout2#b, vout2#d in volts
5.06
5.00
5
1
3
2
f ( Cout , f c , ΔI out )
4.94
f ( PM )
4.88
300u
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21
900u
1.50m
time in seconds
2.10m
2.70m
Agenda
‰ Feedback generalities
‰ Conditions for stability
‰ Poles and zeros
‰ Phase margin and quality coefficient
‰ Undershoot and crossover frequency
‰ Compensating the converter
‰ Compensating with a TL431
‰ Watch the optocoupler!
‰ Compensating a DCM flyback
‰ Compensating a CCM flyback
‰ Simulation and bench results
‰ Conclusion
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22
DC-DC Output Impedance
‰ A DC-DC conv. combines an inductor and a capacitor
‰ As f is swept, different elements dominate Zout,OL
Rlf
10m
Lout
100u
4
1
Open-loop model
2
Z0
Rlf
20.0
Vout
2
⎛ Rlf ⎞
1+ ⎜
⎟
⎝ Z0 ⎠
2
f0
Zout (dBΩ)
I1
AC = 1
3
Cout
1000uF
plot1
vdbout in db(volts)
0
Resr
1m
Lout
-20.0
f (Hz)
Cout
-40.0
RLf
Resr
-60.0
A buck equivalent circuit
⎛
1 ⎞
Z out = ( sLout + RLf ) || ⎜ Resr +
⎟
sCout ⎠
⎝
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2
Crossover region
1
10
100
1k
frequency in hertz
10k
100k
To avoid stability issues,
fc >> f0
1Meg
Closing the Loop…
‰ At the crossover frequency Zout,CL ≈ Zout,OL
100
|T(s)|
fc
Plot1
vdbout#b, vdbout, vdberr in db(volts)
50.0
0
|Zout,OL|
2
-50.0
5
3
|Zout,CL|≈| Zout,OL|
|Zout,CL|
-100
1
10
100
1k
frequency in hertz
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10k
100k
Calculating the Output Impedance
‰ the closed-loop output impedance is dominated by Cout
Z out ,CL
1
1
1
≈
≈
2π f c Cout 1 + T ( s ) 2π f cCout
1
2 − 2 cos (ϕm )
2
Zout
improves
Zout
degrades
Open-loop phase
margin affects the
closed-loop output
impedance
1
1 + T ( fc )
1
ϕm °
0
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25
20
40
60
80
An Example with a Buck
‰ Let’s assume an output capacitor of 1 mF
‰ The spec states a 80 mV undershoot for a 2 A step
‰ How to select the crossover frequency?
ΔVout ≈
ΔI out
2π f c Cout
2
fc ≈
= 4 kHz
80m ×1m × 2π
fc ≈
ΔI out
ΔVout Cout 2π
1
Z Cout @ 4 kHz =
= 40 mΩ
2π × 4k ×1m
Select a 1000-µF capacitor featuring less than a 40-mΩ ESR
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26
Setting the Right Crossover Frequency
180
80.0
90.0
40.0
0
-90.0
vdberr in db(volts)
Plot1
ph_verr in degrees
‰ Compensate the converter for a 4 kHz fc
Compensated open-loop gain
Buck operated in voltage-mode
ϕm = 70°
phase
0
4
fc
-40.0
3
gain
-180
-80.0
10
100
1k
frequency in hertz
10k
4 kHz
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27
100k
Step Load the Output
a
c
rLf
10m
7
L1
100u
Vout
3
d
11
PWM switch VM
Vin
10
vout
R10
1m
p
I1
X3
PWMVM
L = 100u
Fs = 100k
16
C5
1mF
H(s)
vout
12
GAIN
X1
GAIN
K = 0.5
‰ the load varies
from 100 mA to 2.1 A
C2
{C2}
PWM
gain
R7
{R3}
C1
{C1}
R2
{R2}
Rupper
10k
13
8
C3
{C3}
5
1
6
Verr
X2
AMPSIMP
V2
2.5
Rlower
10k
G(s)
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Measure the Obtained Undershoot
5.00
5
ΔV ≈ 40m
Plot1
vout in volts
4.98
4.96
ΔV ≈ 40m ×
ΔI
2 − 2 cos (ϕm )
2
= 70 mV
1.14
4.94
70 mV
4.92
1.61m
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29
2.42m
3.23m
time in seconds
4.05m
4.86m
Agenda
‰ Feedback generalities
‰ Conditions for stability
‰ Poles and zeros
‰ Phase margin and quality coefficient
‰ Undershoot and crossover frequency
‰ Compensating the converter
‰ Compensating with a TL431
‰ Watch the optocoupler!
‰ Compensating a DCM flyback
‰ Compensating a CCM flyback
‰ Simulation and bench results
‰ Conclusion
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30
How do we Stabilize the Converter?
Select the crossover frequency fc (assume 4 kHz)
Provide a high dc gain for a low static error and good input rejection
Shoot for a 70° phase margin at fc
Evaluate the needed phase boost at fc to meet (3)
Shape the G(s) path to comply with 1, 2 and 3
40.0
180
20.0
90.0
0
ph_voutin degrees
Plot1
vdboutin db(volts)
1.
2.
3.
4.
5.
Gain
Phase
-20.0
-90.0
-40.0
-180
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Asc ,CL ( s ) =
0
10
31
Open-loop Bode plot of the power stage, H(s)
Arg H(s) @ fc
|H(s)| @ fc
100
1k
frequency in hertz
10k
1
100k
Asc ,OL ( s )
1+ T ( s)
First, Provide Mid-Band Gain at Crossover
1. Adjust G(s) to boost the gain by +21 dB at crossover
¾ Create the so-called mid-band gain
180
40.0
90.0
0
ph_voutin degrees
Plot1
vdboutin db(volts)
20.0
Gain
Push the
gain up.
0 [email protected]
0
Tailor G(s) to
exhibit a gain
of +21 [email protected] fc.
Phase
|H(s)|= -21 dB
-20.0
-90.0
Arg H(s)= -175°
-40.0
-180
10
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100
1k
frequency in hertz
4 kHz
10k
100k
Second, Provide High Gain in DC
2. An integrator provides a high dc gain but rotates by -270°
¾ This is the origin pole
C1
100n
60 dB
360
60.0
1
R1
10k
2
E1
1k
Vout
4
180
0
-30.0
p in unknown
Plot1
vdbout in db(volts)
30.0
-20 dB per decade
slope -1
V1
AC = 1
0
-180° by inverting
op amp
-180
-90° by pole
at the origin
-60.0
10
-360
100m
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8
1
10
100
frequency in hertz
1k
10k
100k
-270°
Third, Evaluate the Phase Boost at fc
Plot1
ph_vout#a in degrees
0
-40.0
arg H(s) at 4 kHz
arg H(s)
-80.0
-120
-160
arg H(s)
-175°
18
arg G(s)
Plot3
p in unknown
0
-90.0
-113°
arg G(s)
-180
+155°
Plot2
p in unknown
0
Phase boost at fc
-90.0
ϕm = 70°
arg H(s)G(s)
-180
-270
ϕm
-360
10
100
1k
frequency in hertz
1
10k
arg H ( f c ) − 270° + BOOST − ϕm = −360°
BOOST = ϕm − arg H ( f c ) − 90° = 70° + 175 − 90 = 155°
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11
-270
-360
+
100k
How do We Boost the Phase at fc?
‰ The phase boost is created by combining zeros and poles
⎛
⎜1 +
G ( jω ) = ⎝
⎛
⎜⎜1 +
⎝
j
ω ⎞
⎟
ω z1 ⎠
ω ⎞
j
⎟
ω p1 ⎟⎠
⎛
⎜1 +
arg G ( jω ) = boost = arg ⎝
⎛
⎜⎜1 +
⎝
j
ω ⎞
⎟
ω z1 ⎠
ω ⎞
j
⎟
ω p1 ⎟⎠
⎛ fc ⎞
⎛ fc ⎞
arg G ( f c ) = arctan ⎜
⎟⎟
⎟ − arctan ⎜⎜
f
f
⎝ z1 ⎠
⎝ p1 ⎠
Assume 1 zero placed at 705 Hz, 1 pole at 22 kHz and a 4-kHz crossover frequency:
⎛ 4k ⎞
⎛ 4k ⎞
arg G ( 4 kHz ) = arctan ⎜
arctan
−
⎟
⎜
⎟ = 80 − 10.3 ≈ 70°
⎝ 705 ⎠
⎝ 22k ⎠
‰ If poles and zeros are coincident, no phase boost!
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40.0
360
20.0
180
0
p in unknown
plot1
vdbout in db(volts)
How do We Boost the Phase at fc?
Gain
|G(s)|
G100 Hz = 38 dB
Gain at
fc = 21 dB
33
0
fp = 22 kHz
fz = 705 Hz
-20.0
-180
-270°
-40.0
32
-360
10
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Phase boost
at fc = 71°
Phase
Arg G(s)
100
1k
frequency in hertz
4 kHz
10k
100k
Type 2
How do We Boost the Phase at fc?
‰
‰
‰
‰
The type 1 configuration
No phase boost, pure integral term
Permanent phase lag of -270°
Ok if argH(fc) < -45° for a ϕm of 45°
C1
10n
1
R1
10k
2
G (s) =
E1
10k
Vout
4
V1
AC = 1
Vout ( s )
Vin ( s )
=
1
1
=
s
sR1C1
1
ω p1 =
R1C1
ω0
1 pole at the origin
Type 1
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How do We Boost the Phase at fc?
‰ The type 2 configuration
‰ Phase boost up to 90°
‰ Ok if argH(fc) < -90°
C2
62pF
C1
2nF
1
R1
10k
1 + sR2C1
G (s) = −
⎛
⎡ C1C2 ⎤ ⎞
sR1 ( C1 + C2 ) ⎜⎜1 + sR2 ⎢
⎥ ⎟⎟
+
C
C
2 ⎦⎠
⎣ 1
⎝
R2
116k
2
4
E1
10k
If C2 << C1
Vout
3
V1
AC = 1
ω po =
1
1
1
ω p1 =
ω z1 =
R2C2
R2C1
R1C1
1 pole at the origin
1 zero
1 pole
Type 2
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38
How do We Boost the Phase at fc?
‰ The type 3 configuration
‰ Phase boost up to 180°
‰ Ok if argH(fc) < -180°
G (s) = −
C2
350pF
C1
11nF
1
C3
22nF
R3
321
R2
20k
2
If C2 << C1 and R3 << R1
4
E1
10k
ω z1 =
5
Vout
3
V1
AC = 1
R1
10k
sC3 ( R1 + R3 ) + 1
sR2C1 + 1
⎛
C C ⎞ ( sR3C3 + 1)
sR1 ( C1 + C2 ) ⎜1 + sR2 1 2 ⎟
C1 + C2 ⎠
⎝
1
R2C1
ω p1 =
1
R1C3
1
=
R2C2
ωz2 =
1
ω p2
R3C3
1 pole at the origin
2 zeros
2 poles
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39
ω po =
1
R1C1
Type 3
Finally, We Test the Open-Loop Gain
5.
6.
Given the necessary boost of 155°, we select a type-3 amplifier
A SPICE simulation can give us the whole picture!
a
c
rLf
10m
7
L1
100u
Vout
d
11
PWM switch VM
Vin
10
vout
3
R10
1m
p
X3
PWMVM
L = 100u
Fs = 100k
R11
1
16
C5
1mF
Buck stage
vout
12
GAIN
X1
GAIN
K = 0.5
C2
{C2}
R7
{R3}
C1
{C1}
2
R2
{R2}
Rupper
10k
13
8
C3
{C3}
5
LoL
1kH
1
6
CoL
1kF
Verr
X2
AMPSIMP
V2
2.5
Rlower
10k
9
Vstim
AC = 1
Type 3
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40
1 pole at the origin
2 zeros at 500 Hz
2 poles at 50 kHz
Finally, We Test the Open-Loop Gain
80.0
360
40.0
180
0
p in unknown
plot1
vdberr in db(volts)
An ac simulation gives us the open-loop Bode plot
fc = 4 kHz
0
-40.0
-180
-80.0
-360
10
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41
Gain
T(s)
15
ϕm = 70°
Phase
Arg T(s)
14
100
1k
frequency in hertz
10k
100k
Agenda
‰ Feedback generalities
‰ Conditions for stability
‰ Poles and zeros
‰ Phase margin and quality coefficient
‰ Undershoot and crossover frequency
‰ Compensating the converter
‰ Compensating with a TL431
‰ Watch the optocoupler!
‰ Compensating a DCM flyback
‰ Compensating a CCM flyback
‰ Simulation and bench results
‰ Conclusion
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Type 2 with a TL431
‰ Litterature examples use op amps to close the loop.
‰ Reality differs as the TL431 is widely implemented.
‰ How to convert a type 2 to a TL431 circuit?
K
R
K
R
TL431A
A
2.5V
A
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43
A shunt regulator!
R
A
K
Type 2 with a TL431
‰ A TL431 implements a two-loop configuration
FB signal
Rpullup
FB
D2
MBR20100CT
L1
2.2u
Vout
solution A
Vdd
Rbias
RLED
Rupper
Gnd
slow
lane
fast
lane
Vcc
FB
FB signal
Rpulldown
Gnd
www.onsemi.com
C3
100uF
Czero
X1
TL431A
solution B
44
C2
1mF
Rlower
Adding a Pole for a Type 2 Circuit
‰ The pole is a simple capacitor on the collector
Rpullup
FB
Vdd
Vdd
FB
Rpulldown
Cpole
⎛ sRupper C zero + 1 ⎞ ⎛
VFB ( s )
1
G (s) =
= −⎜
⎜ sRupper C zero ⎟⎟ ⎜⎜ 1 + sR pullup C pole
Vout ( s )
⎝
⎠⎝
f po =
1
2π Rupper Czero
Pole at the origin
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45
fz =
1
2π Rupper Czero
Low frequency zero
G=
R pullup
RLED
CTR
Mid-band gain
Cpole
Or on the
emitter
⎞ R pullup
CTR
⎟⎟
⎠ RLED
fp =
1
2π R pullup C pole
High frequency pole
The Type 2 Final Implementation
‰ The LED resistor fixes the mid-band gain
Vout
RLED
Vdd
Rupper
1
3
U2A
Rpullup
Czero
2
Cpole
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U2B
X1
TL431
Rlower
What TL431?
‰ The TL431 is available under several grades
ƒ TL431AI, 2.495 V, ± 2.2% TA = -25 °C to +85 °C
ƒ TL431AC, 2.495 V, ± 1.6% TA = -25 °C to +85 °C
ƒ TL431BI, 2.495 V, ± 0.8% TA = -25 °C to +85 °C
• BV = 37 V, IK,max = 100 mA and IK,min = 1 mA
‰ The TLV431 can regulate to a lower output
ƒ TLV431A, 1.24 V, ± 2% TA = -25 °C to +85 °C
ƒ TLV431B, 1.24 V, ± 1% TA = -25 °C to +85 °C
• BV = 18 V, IK,max = 20 mA and IK,min = 100 µA
NCP100 down to 0.9 V
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Agenda
‰ Feedback generalities
‰ Conditions for stability
‰ Poles and zeros
‰ Phase margin and quality coefficient
‰ Undershoot and crossover frequency
‰ Compensating the converter
‰ Compensating with a TL431
‰ Watch the optocoupler!
‰ Compensating a DCM flyback
‰ Compensating a CCM flyback
‰ Simulation and bench results
‰ Conclusion
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The Optocoupler is the Treator Here!
‰ You need galvanic isolation between the prim. and the sec.
‰ An optocoupler transmits light only, no electrical link
LED
Creepage path
a
c
Silicone
dome
Detector
IF
Ic
500 µm
Detector
e
CTR =
k
Ic
× 100
IF
Current Transfer Ratio
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Clearance
LED
French
specimen
Luigi Galvani, 1737-1798
Italian physician and physicist
The Internal Pole should be Known
‰ The photons are collected by a collector-base area.
‰ This area offers a large parasitic capacitance: opto pole!
Vdd
Vout
Vdd
Rpullup
Rpullup
RLED
RLED
VFB
C
CTR
VFB ( s )
Vout ( s )
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=−
R pullup CTR
RLED
1
1 + sR pullup C
If fp is above 5 times fc, its effect is negligible
If fp is close to fc, phase margin degradation
Assess the CTR Variations
‰ CTR changes with the operating current!
‰ Try to select collector bias currents around 2-5 mA
CTR between 0.63 and 1.25
Normalized to 1 (0 dB)
0.63 gives –4 dB
1.25 gives +2 dB
fc
Watch out for
crossover frequency
changes and phase margin
at CTR extremes!
SFH-615
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51
Changing the Pullup Affects the Pole Position
‰ A low pullup resistor offers better bandwidth!
10 kHz
30 kHz
Rpullup = 1 kΩ
Rpullup = 4.7 kΩ
‰ Changing the bias point affects the CTR
CTR
5 kHz
Rpullup = 15 kΩ
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VFB ( s )
Vout ( s )
=−
R pullup
RLED
CTR
‰ If Rpullup = RLED, then |G0| = 0 dB…?
Agenda
‰ Feedback generalities
‰ Conditions for stability
‰ Poles and zeros
‰ Phase margin and quality coefficient
‰ Undershoot and crossover frequency
‰ Compensating the converter
‰ Compensating with a TL431
‰ Watch the optocoupler!
‰ Compensating a DCM flyback
‰ Compensating a CCM flyback
‰ Simulation and bench results
‰ Conclusion
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53
Stabilizing a DCM Flyback Converter
‰
‰
‰
‰
‰
We want to stabilize a 20 W DCM adapter
Vin = 85 to 265 Vrms
Vout = 12 V/1.7 A
Fsw = 60 kHz
Selected controller: NCP1216
1.
2.
3.
4.
5.
Obtain a power stage open-loop Bode plot, H(s)
Look for gain and phase values at cross over
Compensate gain and build phase at cross over, G(s)
Run a loop gain analysis to check for margins, T(s)
Test transient responses in various conditions
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Stabilizing a DCM Flyback Converter
‰ Capture a SPICE schematic with an averaged model
DC
6
vc
a
duty-cycle
389mV
90.0V
X2x
XFMR
RATIO = -166m
3
p
2
PWM switch CM
839mV
-76.1V
c
Vin
90
AC = 0
D1A
mbr20200ctp
12.0V
vout
4
12.6V
R10
20m
0V
X9
PWMCM
L = Lp
Fs = 65k
Ri = 0.7
Se = Se
13
L1
{Lp}
8
V(errP)/3 > 1 ?
1 : V(errP)/3
12.0V
1
C5
3mF
B1
Voltage
Coming from FB
‰ Look for the bias points values: Vout = 12 V, ok
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vout
Rload
7.2
Stabilizing a DCM Flyback Converter
‰ The feedback portion includes the optocoupler pole
parameters
Vdd
5
5.00V
errP
vout
2.52V
5
K
Rled
{Rled}
Rpullup
{Rpullup}
S+A
X4
POLE
FP = pole
K=1
2.52V
2.52V
11.7V
Verr
Rupper2
{Rupper}
LoL
1kH
err
18
11
X7
Optocoupler
Cpole = 1/(6.28*pole*pullup)
CTR = CTR
9
2.52V
14
CoL
1kF
0V
Cpole2
{Cpole}
10
10.7V
Czero1
{Czero}
15
Vstim
AC = 1
2.49V
X10
TL431_G
Rlower2
{Rlower}
Vout=12
Ibridge=250u
Rlower=2.5/Ibridge
Rupper=(Vout-2.5)/Ibridge
Lp=450u
Se=100m
fc=1k
from
pm=60
Bode
Gfc=-24
pfc=-77
G=10^(-Gfc/20)
boost=pm-(pfc)-90
pi=3.14159
K=tan((boost/2+45)*pi/180)
Fzero=fc/k
Fpole=k*fc
Rpullup=20k
RLED=CTR*Rpullup/G
Czero=1/(2*pi*Fzero*Rupper)
Cpole=1/(2*pi*Fpole*Rpullup)
CTR=1.5
Pole=6k
Automated compensation
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Stabilizing a DCM Flyback Converter
‰ Get the open-loop power stage transfer function, H(s)
60.0
Gain at 1 kHz
-22.7 dB
30.0
0
-30.0
1
|H(s)|
-60.0
180
Phase at 1 kHz
-79 °
90.0
0
-90.0
argH(s)
-180
10
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2
100
1k
10k
100k
Stabilizing a DCM Flyback Converter
‰ Boost the gain by +22 dB, boost the phase at fc
4
80.0
GM
35 dB
40.0
0
-40.0
5
|T(s)|
-80.0
Cross over
1 kHz
180
90.0
0
Margin at 1 kHz
60°
-90.0
argT(s)
-180
10
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100
1k
10k
4
100k
Stabilizing a DCM Flyback Converter
‰ Test the response at both input levels, 90 and 265 Vrms
‰ Sweep ESR values and check margins again
12.04
Vout(t)
Hi
line
12.00
6
4
11.96
11.92
100
mV
Low
line
11.88
200 mA to 2 A in 1 A/µs
3.00m
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59
9.00m
15.0m
21.0m
27.0m
Agenda
‰ Feedback generalities
‰ Conditions for stability
‰ Poles and zeros
‰ Phase margin and quality coefficient
‰ Undershoot and crossover frequency
‰ Compensating the converter
‰ Compensating with a TL431
‰ Watch the optocoupler!
‰ Compensating a DCM flyback
‰ Compensating a CCM flyback
‰ Simulation and bench results
‰ Conclusion
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60
Stabilizing a CCM Flyback Converter
‰
‰
‰
‰
‰
We want to stabilize a 90 W CCM adapter
Vin = 85 to 265 Vrms
Vout = 19 V/4.8 A
Fsw = 60 kHz
Selected controller: NCP1230
1.
2.
3.
4.
5.
Obtain a power stage open-loop Bode plot, H(s)
Look for gain and phase values at cross over
Compensate gain and build phase at cross over, G(s)
Run a loop gain analysis to check for margins, T(s)
Test transient responses in various conditions
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Stabilizing a CCM Flyback Converter
6
a
90.0V
X2x
XFMR
RATIO = -0.25
-78.8V
p
2
PWM switch CM
DC
duty-cycle
467mV
vc
‰ Capture a SPICE schematic with an averaged model
4
vout
vout
19.7V
19.0V
c
Vin
90
AC = 0
3
D1A
mbr20200ctp
R10
14.4m
0V
X9
PWMCM
L = Lp
Fs = 65k
Ri = 0.25
Se = 0
13
L1
{Lp}
Rload
4
19.0V
786mV
8
V(errP)/3 > 1 ?
1 : V(errP)/3
1
C5
6600u
B1
Voltage
‰ Look for the bias points values: Vout = 19 V, ok
‰ Vsetpoint < 1 V, enough margin on current sense
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Stabilizing a CCM Flyback Converter
‰ Capture a SPICE schematic with an averaged model
parameters
Vdd
5
5.00V
errP
2.36V
vout
5
K
Rpullup
{Rpullup}
S+A
X4
POLE
FP = pole
K=1
Rled
{Rled}
2.36V
2.36V
18.7V
Verr
Rupper2
{Rupper}
LoL
1kH
err
18
11
R7
1
X8
Optocoupler
Fp = Pole
CTR = CTR
2.36V
14
CoL
1kF
0V
15
17.7V
9
2.49V
Vstim
AC = 1
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Cpole2
{Cpole}
X10
TL431_G
10
Czero1
{Czero}
Rlower2
{Rlower}
Vout=19
Ibridge=250u
Rlower=2.5/Ibridge
Rupper=(Vout-2.5)/Ibridge
Lp=350u
Se=20k
fc=1k
from
pm=60
Bode
Gfc=-22
pfc=-71
G=10^(-Gfc/20)
boost=pm-(pfc)-90
pi=3.14159
K=tan((boost/2+45)*pi/180)
Fzero=fc/k
Fpole=k*fc
Rpullup=20k
RLED=CTR*Rpullup/G
Czero=1/(2*pi*Fzero*Rupper)
Cpole=1/(2*pi*Fpole*Rpullup)
CTR=1.5
Pole=6k
Stabilizing a CCM Flyback Converter
‰ Capture a SPICE schematic with an averaged model
32.0
Gain at 1 kHz
-22 dB
16.0
Sub harmonic
poles
0
-16.0
4
180
Phase at 1 kHz
-71 °
90.0
Inject ramp
compensation
0
-90.0
argH(s)
-180
10
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64
100
1k
10k
ramp
|H(s)|
-32.0
6
100k
Stabilizing a CCM Flyback Converter
‰ The easiest way to damp the poles:
¾ Calculate the equivalent quality coefficient at Fsw/2
¾ Calculate the external ramp to make Q less than 1
1
Q=
⎛
π ⎜D'
⎝
Se =
⎞
Se 1
+ − D⎟
Sn 2
⎠
=
1
=8
3.14 × ( 0.5 − 0.46 )
Sn ⎛ 1
90 × 0.25
⎞ Vin Ri ⎛ 1
⎞
⎛ 1
⎞
− 0.5 + 0.46 ⎟ = 36 kV s
⎜ − 0.5 + D ⎟ =
⎜ − 0.5 + D ⎟ =
⎜
D'⎝π
⎠ Lp D ' ⎝ π
⎠ 320u × (1 − 0.46 ) ⎝ 3.14
⎠
2.3 Vpp
Rramp
18 kΩ
DRV
CS
NCP1230
internal
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Se 36k
=
= 51%
Vin Ri
On-time slope
S n 70k
Lp
2.3
S ramp =
= 153 kV s
15u
M S R
0.51× 70k × 18k
Rcurrent = r n ramp =
= 4.1 k Ω
S ramp
153k
Mr =
Rcurrent
Ri
Stabilizing a CCM Flyback Converter
‰ Boost the gain by +22 dB, boost the phase at fc
11
Cross over
1 kHz
80.0
40.0
GM
20 dB
0
-40.0
10
|T(s)|
-80.0
180
90.0
0
Margin at 1 kHz
60°
-90.0
argT(s)
-180
10
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66
100
1k
10k
11
100k
Stabilizing a CCM Flyback Converter
‰ Test the response at both input levels, 90 and 265 Vrms
‰ Sweep ESR values and check margins again
19.11
Vout(t)
Hi
line
19.03
12
11
18.95
112 mV
18.87
Low
line
18.79
1.80m
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67
5.40m
9.00m
time in seconds
12.6m
16.2m
Agenda
‰ Feedback generalities
‰ Conditions for stability
‰ Poles and zeros
‰ Phase margin and quality coefficient
‰ Undershoot and crossover frequency
‰ Compensating the converter
‰ Compensating with a TL431
‰ Watch the optocoupler!
‰ Compensating a DCM flyback
‰ Compensating a CCM flyback
‰ Simulation and bench results
‰ Conclusion
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Testing a UC3843 Converter
‰ A 19 V/3 A converter is built around an UC3843
‰ The converter operates in CCM or DCM
T1
86H-6232
0.18 : 1 : 0.25
HV-bulk
R19
47k
R13
47k
R3
47k
C2
10n
D5
MBR20100
-
+
400V
.
IC4
KBU4K
C11
100p
R6
6k
Vref
C10
470n X2
R23
1Meg
R7
10k
1
CMP
Ref
2
FB
Vcc
7
3
CS
DRV
6
8
C5a
1.2mF
C5b
1.2mF
C7
220uF
25V
25V
25V
C13
2.2nF
Type = Y1
R17
47k
IN
L1
2 x 10mH
Schaffner RN122-1.5/02
C4
100uF
.
.
D2
MUR160
L2
2.2u
D8
1N4937
R1
330
Gnd
R8
1k
R14
4.7k
M1
400V
R11
10k
4
Rt
GND
U1
UC3843
5
R12
10k
SPP11N60S5
R16
10
Vref
R18
47k
R10
56k
U3B
R24
1Meg
U3A
R5
1k
85-260 Vac
R6a
1
R15
4.7k
C12
220p
C16
4.7nF
C15
10nF
Vout
C3
220uF
C6
100n
R6b
1
IC2
TL431
R9
10k
Gnd
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Full Load Leads to CCM Operation
2
simulated
2
CCM operation, Rload = 6.3 Ω
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Reduce the Load to Enter in DCM
simulated
DCM operation, Rload = 20 Ω
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From the Open-Loop Bode Plot, Compensate
‰ The TL431 is tailored to pass a 1 kHz bandwidth
Vref
Vout
RLed
Rupper
66 kΩ
Cpole
RLED =
R pullup CTR
10
18
20
=
4.7 k × 0.45
= 266 Ω
7.94
We place a zero at 300 Hz:
1 kΩ
CTR
= 45%
Calculate mid-band gain: +18 dB
Czero
C zero =
Rpulldown
Rlower
10 kΩ
1
2π f zero Rupper
=
1
= 8 nF
6.28 × 300 × 66k
We place a pole at 3.3 kHz:
C pole =
1
1
=
= 10 nF
2π f pole R pulldown 6.28 × 3.3k × 4.7 k
k factor method
“Switch-Mode Power Supplies: SPICE Simulations and Practical Designs”, McGraw-Hill
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72
Verify in the Lab. the Open-Loop Gain
‰ Sweep extreme voltages and loads as well!
Simulated
CCM operation, Rload = 6.3 Ω, Vin = 150 Vdc
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Verify in the Lab. the Open-Loop Gain
Simulated
0
100
1k
10k
CCM operation, Rload = 6.3 Ω, Vin = 330 Vdc
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100
Verify in the Lab. the Open-Loop Gain
Simulated
DCM operation, Rload = 20 Ω, Vin = 330 Vdc
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As a Final Test, Step Load the Output
‰ Good agreement between curves!
Simulated
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Vin = 150 V
CCM
2 to 3 A
1 A/µs
As a final test, Step Load the Output
‰ DCM operation at high line is also stable
Simulated
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77
Vin = 330 V
DCM
0.5 to 1 A
1 A/µs
Conclusion
‰ DC-DC loop compensation cannot be overlooked
‰ It is important to understand the impact of phase margin
‰ The crossover frequency affects the output impedance
‰ Current mode CCM or DCM is ok with a TL431-based type 2
‰ Make sure the optocoupler is characterized, watch the pole!
‰ Use SPICE before going to the bench: NO trial and error!
‰ Once the simulation is stable, build the prototype
‰ Simulations and laboratory debug: the success recipe!
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For More Information
•
View the extensive portfolio of power management products from ON
Semiconductor at www.onsemi.com
•
View reference designs, design notes, and other material supporting
the design of highly efficient power supplies at
www.onsemi.com/powersupplies
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