This is only a preview of the December 2025 issue of Silicon Chip. You can view 35 of the 104 pages in the full issue, including the advertisments. For full access, purchase the issue for $10.00 or subscribe for access to the latest issues. Articles in this series:
Articles in this series:
Items relevant to "RGB LED Star Ornament":
Items relevant to "Earth Radio, Part 1":
Items relevant to "DCC Decoder":
Items relevant to "Digital Preamplifier, Part 3":
Purchase a printed copy of this issue for $14.00. |
By Andrew Levido
Power
Electronics
Part 2: Controlling DC-DC Converters
Last month, we introduced average value analysis and used it to analyse a range of
DC-DC converter topologies at a high level. We then looked at a practical example of a
buck converter. This time, we will delve into the control systems required to make these
converters work correctly.
D
C-DC converters almost always operate
in a closed-loop mode, where some
kind of feedback is used to maintain the output voltage at the desired
level, despite changes in load or source
voltage. Often, the detail of the control
loop is hidden from the user because
it is incorporated in the driver chip,
such as was the case for our buck converter example last month.
Even when this is the case, it is
handy to have at least a basic understanding of how they work.
We will continue using the example
of the buck converter we introduced
last time. It was designed to convert a
nominal 12V input (ranging from 10V
to 14V) to a 6V output, to drive a 10W
load for a maximum output current of
600mA. The switching frequency was
500kHz. We went through the process
of selecting the filter components in
some detail.
If you want to know why we need
to close the loop around our converter,
take a look at Fig.1. This is the result of
a simulation of the converter with no
feedback and a fixed 50% duty cycle.
The source voltage (blue trace) is
programmed to be 12V for the first
30ms, dropping to 10V thereafter. In
contrast, the load is programmed to be
10W for the first 10ms, increasing to
20W thereafter, with the resulting load
current shown by the orange trace. The
purple trace shows the output voltage.
There is a significant transient when
the load changes, but no change in
output voltage, since the duty cycle
is fixed. The transient peaks at about
±300mV, which is not great, and it
takes about 5ms to settle.
Things get even worse when the
input voltage reduces at the 30ms
mark. Not only does the output voltage drop because there is no regulation, but the ringing at the transition is
horrendous. Again, it takes about 5ms
for the result to settle into something
close to the steady state value.
A properly designed closed-loop
control system should greatly reduce
or eliminate these problems.
Control theory is a huge topic, and
one that can be very heavy on mathematics. I will therefore not be giving
a control theory tutorial in the classical sense, but rather a practical exploration of some of the techniques that
can be employed. If you are a control
theory expert, you are hereby warned
that I am going to skip some details in
the interest of space and simplicity!
Creating a mathematical
model
Fig.1: without a closed loop control system, the output of the DC-DC converter
(purple trace) is unstable with changes in load (at 10ms) and input voltage (at
30ms).
To create a stable and effective control system, we have to build a mathematical model to describe our DC-DC
converter under dynamic conditions.
We have already discussed two models
for our converter: the very high-level
periodic steady state (PSS) model we
used for average value analysis, and
the very low-level cycle-by-cycle simulation model we used to verify the
ripple voltages and currents.
We can’t use the periodic steadystate model for the control loop since
it is dynamic behaviour we want to
take care of. Also, we don’t need to
model the cycle-by-cycle behaviour of
the converter since the behaviour we
are concerned with occurs over much
longer time periods than a single 2µs
cycle. Instead, we need an intermediate model.
Fig.2 shows the block diagram of
our converter with feedback presented
in the classic control theory manner.
Right in the middle is the modulator,
which takes a control voltage proportional to the duty cycle (vc) and
produces the switching waveform,
whose average value, ‹vx›, is applied
to the filter.
The filter takes this voltage and produces the smoothed output voltage, vo.
Each block has a gain, G, which relates
its output signal to its input.
Australia's electronics magazine
siliconchip.com.au
30
Silicon Chip
In front of these two blocks is
Fig.2: this classic
a compensator, which we will
depiction of the
describe in more detail below.
DC-DC converter
The input to the compensator is
closed-loop
an error, e, that is the difference
system is useful
between the reference setpoint vr
to understand
and the feedback signal vfb. The
the circuit
feedback signal is derived from the
blocks we need
output voltage we want to control
to model.
by some sensor, with a gain of H.
Fig.3: the DCThese blocks are shown in a
DC converter
more familiar way in Fig.3. The
arranged
modulator includes the Mosfet and
according to the
diode switches, plus the circuitry
block diagram
that converts a control voltage to
in Fig.2 is the
pulse-width-modulation to drive
starting point for
the Mosfet. At its simplest, this is
the development
just a comparator with an output
of a control
that is high whenever the control
model of the
voltage exceeds a ramp waveform
converter. The
compensator is
at the switching frequency.
the missing piece
The modulator therefore takes a
of the puzzle at
control voltage, vc, and produces
this stage.
the average voltage, ‹vx›. The modulator has a gain of Gm and it is easy
to see that this must be the unitless negative input for the feedback. This
quantity Gm = Vsrc/Vramp. If we assume anticipates that the compensator will
the ramp voltage is 1V, the modulator probably involve op amps.
gain simplifies to the source voltage,
Simplifying the complexwhich is 12 in our example.
It is worth noting that there is no frequency domain
need for any switching to occur in this
Describing the filter and compenmodel of the modulator. The modu- sator gains requires us to dip our toes
lator produces an output equal to the into something called the complex-
average voltage over one switching frequency domain, also known as the
period. This voltage may vary with s-domain. Analysis of the behaviour
time, but the time scale we are con- of capacitors and inductors involves
cerned about is much longer than the differential equations, which we usu2µs switching period.
ally describe in the time domain – for
Because this modulator controls the example, the relationship between
average voltage applied to the filter, voltage and current in an inductor is
this converter is said to be employ- v = L di/dt.
ing voltage mode control. There is
We can describe the same equations
another way to do it, which we will in the s-domain, which makes the
come to later.
mathematics much simpler, because
The sensor gain is trivial to calcu- differentiation is reduced to multilate. It is simply the divider ratio rep- plication and integration is reduced
resented by the two resistors, so Hs = to division. The conversion from the
R1 ÷ (R1 + R2), or 0.204 in this case. time domain to the s-domain requires
The buffer is there to isolate the com- something called a Laplace transform.
pensator from the divider.
You don’t need to know how to do
We will ignore the compensator for this (unless you are an electrical engithe moment, but notice that I have neering student) because you can write
absorbed the error summing junc- the s-domain expression for most comtion into the black box and added a mon components very easily without
positive input for the reference and a having to do any maths.
Table 1 – passive component complex impedances
Component
Time Domain
s-Domain
Complex Impedance Z(s)
Inductor
v = L di/dt
v = iLs
Ls
Capacitor
i = C dv/dt
i = vCs
1/Cs
Resistor
v = Ri
v = ir
R
siliconchip.com.au
Australia's electronics magazine
The “s” in s-domain is a complex
frequency, which is just a way of
describing a unit-amplitude sinewave
with a given frequency and phase. We
won’t, however, have to deal with any
complex numbers in this explanation
– we can simply treat s like any other
algebraic variable.
To transform the inductor mentioned above to the s-domain, the time
differential (d/dt) is simply replaced by
s to get the s-domain equivalent relationship v = i • L • s, which I hope
you can see is kind of analogous to
Ohm’s law. The complex impedance
of the inductor, indicated by Z(s), is
therefore simply Ls.
Using a similar logic, we can arrive
at the s-domain expressions for the
voltage/current relationships of common parts, and hence their complex
impedances, shown in Table 1.
To show just how useful this transformation is, we will use these complex impedances to calculate the transfer function of a simple RC low-pass
filter shown at the top of Fig.4. If you
substitute in the complex impedances
from the table and use the normal
voltage divider equation, you get the
s-domain transfer function 1 ÷ (RCs +
1) after a little bit of algebraic manipulation. It’s that easy.
Poles and zeros
You can see that there must be a
particular value of complex frequency
s that makes the denominator of the
December 2025 31
expression equal zero, and therefore
the transfer function becomes undefined (but very large). This value, s = -1
÷ (RC), is called a “pole” because you
can visualise it as a spike rising to infinity on the two-dimensional s-plane.
The complex frequency domain
uses radians per second as the unit of
frequency, but converting to Hertz (via
the relationship that one cycle is 2π
radians), this pole is at a frequency of
1 ÷ (2πRC)Hz. This should be familiar
as the corner frequency of an RC filter.
If we turn to the RC high-pass filter
at the bottom of Fig.4 and go through
the same process, we arrive at the
s-
domain transfer function RCs ÷
(RCs + 1). This too has a single pole,
at –1 ÷ RC, but the s in the numerator means there is also a value of s
where the numerator of the fraction
becomes zero, so the transfer function
is equal to zero.
It may come as no surprise to know
that this is called a “zero”. In this case,
the zero occurs at s = 0, so the LC highpass filter has a transfer function with
one pole at 1 ÷ (2πRC) and one zero at
the origin. By origin, we really mean
the origin of the s-plane, but you can
think of this zero as being at “0Hz”.
So, you can easily work out the poles
& zeroes of any network by calculating
the transfer function in the s-domain
as a fraction. Poles are the values of
s (frequency) when the denominator
is zero, and zeroes are the values of s
when the numerator is zero.
Poles & zeros have a specific meaning in the frequency domain. When a
signal encounters a pole, the slope of
the gain reduces by −20dB/decade and
the phase shifts by −90°. Conversely,
when a signal encounters a zero, the
slope of its gain increases by 20dB/
decade and the phase shifts by +90°.
Fig.5 shows this in action for the
two filters we just analysed. In the
case of the low-pass filter, the signal
is unattenuated and has a phase shift
of zero until it encounters the pole at
which point it turns down to a slope of
−20dB per decade. At the same time,
the phase shifts by −90°.
Fig.4: finding the s-domain
transfer functions for
these filters is as simple as
substituting the expressions
from Table 1 into the voltage
divider equation and
performing some elementary
algebra.
In the case of the high-pass filter, the
zero at the origin means that the gain is
increasing by +20dB per decade from
‘zero’ and the phase is already shifted
by +90°. When the pole is encountered,
the −20dB/decade shift puts the gain
back to flat, and the −90° shift brings
the phase back to zero.
In practice, this means that you can
work out the transfer function, the
pole and zero locations and therefore
the gain/phase versus frequency characteristic of any network of resistors,
capacitors and inductors with a bit of
high-school algebra. No wonder engineers like the s-domain. Thanks, Monsieur Laplace!
Buck converter output filter
Armed with this new knowledge,
we are ready to work out the transfer
function, Gf, of the buck converter’s
output filter to fill out our control system model. This is shown on the lefthand side of Fig.6.
If you have a go at this yourself, you
will see I have made a small approximation in the denominator, where an
RCs term is dropped because it is very
much smaller that the dominant RLs2
term. This kind of simplification is
common in complex frequency analysis, so watch out for them.
The resulting transfer function has
an s2 term in the denominator. This
signifies that the transfer function has
two poles; in this case, both are at 1
÷ (2π√LC). This means the roll-off at
this frequency is −40dB/decade, and
the phase shifts by −180°.
There is also one zero at 1 ÷ 2πRC
due to the ESR of the of the capacitance.
This pulls the gain back up to −20dB/
decade and the phase back up to −90°.
This double-pole is a problem for
our converter. Gain ‘peaking’, like that
shown dotted due to the Q of the LC
filter, and the rapid phase transition
can cause instability like the ringing
we saw in Fig.1. If we compare this
plot to the simulation on the righthand side of Fig.6, you can see that the
peaking extends to about 15dB, which
is potentially quite serious.
Designing a compensator
Fig.5: the poles and zeros obtained from the s-domain transfer functions
in Fig.4 can be easily transferred to the frequency response plots. A pole
introduces a −20dB/decade slope change and a −90° phase shift. A zero
32introduces
Silicon
Chip
Australia's
electronics
a +20dB/decade
slope change and
a +90° phase
shift. magazine
Finally, we come to the design of the
compensator. There are several ways
to go about this, but they all come
down to manipulating the poles and
zeros of the closed-loop transfer function. Their number, position and relationship can be tweaked to ensure the
siliconchip.com.au
Fig.6: the DC-DC converter output
filter has two poles at a frequency
determined by the inductance and
capacitance, and a zero at the
frequency of the capacitance and its
ESR. The resulting frequency response
contains a nasty peak & steep phase
shift responsible for the ringing in
Fig.1.
Fig.7: we create a compensator
characteristic (▪) to transform the
filter response into the dominant pole
characteristic (▪). The compensator
requires two poles & two zeroes.
system is stable and to optimise the
transient response.
For example, if we are most worried about overshoot, we can set up a
highly damped system that will eliminate it at the expense of response time
and settling time. On the other hand,
we might be concerned about getting
to the new voltage quickly, and can
therefore tolerate a bit of overshoot.
I’m not going to dig into all of that
here. Instead, I will demonstrate a simple technique called dominant-pole
compensation that focuses on the
open-loop poles and zeros. This is
the same technique used to ensure op
amps are stable in closed-loop applications.
The aim is to make the open-loop
frequency response look like a single-
pole filter that rolls off the gain at a
steady −20dB/decade until it reaches
0dB at some crossover frequency, fc.
This roll-off is required to ensure the
gain drops to unity well before the
phase shift reaches −180°.
The difference between the phase
shift at the crossover frequency and
−180° is the ‘phase margin’, a measure of the stability of the circuit and
its tolerance to external disturbances.
We can achieve this roll-off by adding a compensator with the characteristics shown in Fig.7. The modulator
siliconchip.com.au
and the sensor gains are not frequency-
dependent, so they don’t impact the
shape of the gain/frequency characteristic of the open-loop converter, and
can be ignored. They do impact the
absolute value of the gain, but we are
not concerned with that here.
If we want to transform the filter
gain/frequency characteristic (in red)
into the dominant pole characteristic
in blue, we require the compensator to
have the green characteristic.
It should have a pole at the origin,
to set the roll-off to −20dB/decade.
We then need two zeroes at the corner
frequency of the LC filter to cancel the
double-pole in the filter response. The
Fig.8: this Type III compensator has
the necessary poles and zeroes. The
transfer function is calculated in
exactly the same way as for the filter;
the component values are discussed in
the text.
Australia's electronics magazine
compensator should have another pole
at the frequency of the output capacitance/ESR zero to cancel the corresponding zero in the filter.
So in total, we need a compensator
with two zeroes and two poles, one of
which is at the origin.
This is known as a Type III compensator, and a suitable circuit is shown in
Fig.8. The reference voltage is applied
to the non-inverting input of the op
amp, and the feedback signal is applied
to the inverting input via R3/C3.
You can work out the compensator’s
transfer function in the s-domain in
exactly the same way as we did for the
filter. This time, we just assume that
vr is zero, and use the usual equation
for the gain of an inverting amplifier:
G = −Zf /Zi, where Zf is the impedance
of C2 in parallel with R1 plus C1, and
Zi is the impedance of R3 in parallel
with C3.
The resulting transfer function
(again with a minor simplification)
is shown below the circuit diagram.
There are two zeros, one formed by
R1C1 and one by R3C3, and two poles,
one at the origin and one formed by
R1C2.
The component values shown are
calculated by setting the frequency
of both zeroes, 1 ÷ (2πR1C1) and 1 ÷
(2πR3C3), to be the same as the LC
December 2025 33
seems likely the chip uses a similar
compensator internally.
Current-mode control
Fig.9: compare this closed-loop response to that of Fig.1. The disturbances when
the load or supply voltage change are much smaller and settle much faster.
The lower graph shows the open loop gain and phase with the compensator
included. Contrast this with the filter characteristic in Fig.6.
filter’s 1 ÷ (2π√LC), which is 1.56kHz.
I chose 100nF and 1kW for nice, round
values.
The non-origin pole at 1 ÷ (2πR1C2)
should be at the same frequency as the
zero formed by the output capacitor
and its ESR (20.4kHz). Since we have
already chosen R1, we can calculate
C2 to be 7.8nF. This is not a standard
value, so I used 8.2nF, the nearest one.
This small error won’t make a huge
difference, since the values for output capacitance and ESR we are using
(102.4µF and 76mW) are hardly precision values themselves.
Simulation
I simulated this control circuit to
see how it will perform. The result is
shown at the top of Fig.9. The stimulus
is exactly the same as for Fig.1 – the
load resistance changes from 10W to
20W at the 10ms mark, and the source
34
Silicon Chip
voltage drops from 12V to 10V at the
30ms mark.
The difference is dramatic. There
is just a small blip at the 10ms mark,
which peaks at about +20mV (+0.3%),
and lasts for just 150µs. At the 30ms
mark, the voltage dips about 130mV
when the input changes, but it recovers
without overshoot in around 600µs.
I ran a second simulation to plot
the open-loop gain of the entire circuit, which is shown at the bottom of
Fig.9. While you can clearly see the
expected -20dB/decade roll-off from
100Hz to 100kHz, a small amount of
gain peaking is still visible, but it is
much less pronounced. The simulation also calculates the crossover frequency, which is 4.36kHz, and the
phase margin, which is 66°.
You may recall that the crossover
frequency we calculated from the
TPS5410 data sheet was 4.5kHz, so it
Australia's electronics magazine
We mentioned above that there is
a second type of modulator that can
be used in DC-DC converters. This is
known as current-mode control, and
it is shown in Fig.10. In this case, the
modulator controls the average current
through the inductor, ‹il›, instead of
the average voltage at the filter input.
At the beginning of every switching
period T, the latch is set, so the Mosfet switches on and the inductor current begins to increase. This current is
monitored by a current sensor, usually
a resistor, and the resulting voltage is
compared to the control voltage, vc.
When the current reaches the level
defined by the control voltage, the
latch is reset and the Mosfet switches
off until the next cycle.
The result is that the peak current
is determined by the control voltage
and the value of the sense resistor. If
the current ripple is small, this turns
out to be a reasonable approximation
to the average current. This approximation is not perfect, so most real-life
implementations use something called
slope compensation to avoid a potential instability at very high duty cycles.
This is not relevant for the discussion below, so I will not go into it here.
Editor’s note: see the LED Dazzler
project (February 2011; siliconchip.
au/Article/899) for a practical implementation of slope compensation in a
switch-mode regulator.
The transfer function of a current
mode modulator is therefore approximated by Gm = 1 ÷ Rsense. Note that this
gain is a transconductance (a voltage to
current conversion). Because the output of the modulator is now a current,
we have to revisit the transfer function
of the output filter and the compensator, since the filter is no longer the
voltage divider we modelled earlier.
Fig.11 shows the calculation of the
filter transfer function, which is performed in exactly the same way as
before. What is different is that the
inductor is effectively irrelevant, since
the modulator current passes through
it (you can think of the inductor as part
of the modulator if that helps).
The output voltage is now dependent on the impedance of the filter
capacitor, including its ESR, and the
impedance of the load.
The upshot of this is that the transfer
siliconchip.com.au
function is simpler, with a single pole
at a frequency dependent on the output capacitance and the load resistance, and a zero at the frequency
related to the capacitance and its ESR,
like before.
It follows that our compensator can
be simpler, as it now needs only two
poles and one zero, as shown at the
bottom of Fig.11. This is known as a
Type II compensator, and an example
is shown in Fig.12.
The component values of R1 & C1
are calculated so that the frequency of
the zero (1 ÷ 2πR1C1) is the same as
the zero formed by the output capacitor and the load, at 1 ÷ 2πRloadC,
which is 156Hz.
The pole at 1 ÷ 2πR1C2 should be at
the same frequency as the zero formed
by the output capacitor and its ESR
(20.4kHz). Given that R1 is 10kW, the
closest standard capacitor value comes
out to be 820pF. The value of R3 should
be very low compared to R1, and can
even be zero.
I simulated this converter as well,
although in this case, a change in input
voltage was not modelled. Fig.14 on
page 36 shows the results with a much
magnified scale. The peak excursion
and recovery time are very similar to
the voltage mode controller, as we
would expect.
Discontinuous current
Before we finish with control theory
for now, we have to cover one more
topic. So far, we have assumed that
the current through the converter is
continuous. That is, we have assumed
the peak-to-peak current is small compared to the average. But what happens as the average inductor drops
to the point it is close to the peak-topeak ripple?
Fig.13 shows our old friend, the buck
converter, where we have plotted the
inductor current and the voltage vx that
we have used in all of our analyses.
On the left is the case when the
inductor current is high enough that
the negative excursion of the (exaggerated) current ripple remains positive.
This is known as the continuous current case, and has been the assumption
we have used to derive the modulator
models for both voltage mode and current mode examples above.
We could, however, imagine a situation when the load and therefore
inductor current reduces to the point
that the inductor current ramps down
siliconchip.com.au
Fig.10: a
current-mode
modulator
controls the
average inductor
current instead
of the average
voltage as in
Fig.3. This
change flows on
to the filter and
compensator
transfer
functions.
Fig.11: the current-mode filter
transfer function is simpler, with just
one pole set by the capacitor and
the load resistance and one zero due
to the capacitor and its ESR. The
compensator therefore only needs
two poles and one zero.
Fig.12: this Type II compensator
has the requisite poles (one at the
origin) and one zero to compensate
the current-mode controller. See the
text for the component values.
Fig.13: discontinuous mode occurs when the current reaches zero before the
end of the switching period. This changes the modulator transfer function,
and thus the compensator is required for optimum performance.
Australia's electronics magazine
December 2025 35
Fig.14: the resulting closed-loop response to a load change is similar to that of
the voltage mode controller (albeit with a different scale).
Silicon Chip kcaBBack Issues
$10.00 + post
$11.50 + post
$12.50 + post
$13.00 + post
$14.00 + post
January 1997 to October 2021
November 2021 to September 2023
October 2023 to September 2024
October 2024 onwards
September 2025 onwards
All back issues after February 2015 are in stock, while most from January 1997 to
December 2014 are available. For a full list of all available issues, visit: siliconchip.com.
au/Shop/2
PDF versions are available for all issues at siliconchip.com.au/Shop/12
We also sell photocopies of individual articles for those who don’t have a computer
Compact HiFi Headphone Amplifier
Complete Kit
SC6885: $70
December 2024
& January 2025
siliconchip.au/Series/432
This kit includes everything required to build the Compact HiFi Headphone Amplifier. The case is
included, but you will need your own power supply.
36
Silicon Chip
Australia's electronics magazine
to zero at some time (Zt) before the end
of the switching period, as shown on
the right. This is known as discontinuous current operation.
When the inductor current falls to
zero, the inductor voltage must also
be zero, so the point vx will be at the
output voltage, Vload.
It’s pretty easy to see that the modulator transfer function will not be the
same as for the continuous conduction
case. The fact that the transfer function changes in discontinuous mode
has implications for our control loop.
You can design an optimised control loop for a switching converter
operating in either mode, but compromises are required if the converter will
operate in both modes. Converters are
therefore generally designed to work
in one mode or the other. High-power
converters usually use continuous current mode, and thus have a specified
minimum output current.
Low power converters (say, ≤50W)
often operate exclusively in discontinuous current mode. Of course, depending on what you are powering, it may
be necessary to operate in both modes,
in which case compromises much be
made to ensure stability while keeping reasonably good load and line
regulation.
Conclusion
Whew! This has been a pretty heavy
article, but I think it is important to
have a basic understanding of the principles of control theory as applied to
DC-DC converters.
You may never have to completely
design a compensator from scratch,
as the manufacturers do a lot of the
heavy lifting for us, but it is important to understand what’s going on for
those occasions when things don’t go
to plan.
At the very least, we have added
yet another tool to our growing set of
ways to look at and understand power
electronics systems.
As the number of models at our disposal grows, I am reminded of the saying, “all models are wrong, but some
models are useful”. Knowing just
which model or tool to use in what circumstance is the mark of a true expert,
and comes only with experience.
Next time, we will look at isolated DC-DC converters and reverse-
engineer a commercial converter to
see what we can learn from the proSC
fessionals.
siliconchip.com.au
|