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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9
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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11
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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12
-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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13
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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14
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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17
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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18
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:
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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19
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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23
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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24
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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28
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 dB@fc
0
Tailor G(s) to
exhibit a gain
of +21 dB@ fc.
Phase
|H(s)|= -21 dB
-20.0
-90.0
Arg H(s)= -175°
-40.0
-180
10
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32
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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33
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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34
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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35
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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36
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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37
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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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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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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42
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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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
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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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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
TL431AI, 2.495 V, ± 2.2% TA = -25 °C to +85 °C
TL431AC, 2.495 V, ± 1.6% TA = -25 °C to +85 °C
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
TLV431A, 1.24 V, ± 2% TA = -25 °C to +85 °C
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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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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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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55
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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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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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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65
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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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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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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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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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
•
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Semiconductor at www.onsemi.com
•
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the design of highly efficient power supplies at
www.onsemi.com/powersupplies
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