I am struggling with moving average model of dc-dc converter as below. The image is my modified version from the lecture here (page 18).

My question is written in the image. Can anyone explain that?
Sorry for the bad writing.

enter image description here

  • \$\begingroup\$ It is an approximation assuming that the voltage changes for one switching cycle are small. It is usually a good approximation because the filter capacitors are specifically chosen such that the ripple voltages are small. \$\endgroup\$ – rioraxe Dec 16 '16 at 2:24
  • \$\begingroup\$ If so, I don't see how this model differs from the one in chapter 2 in the link below (as in the page 20). ecee.colorado.edu/copec/book/slides/Ch2slide.pdf \$\endgroup\$ – anhnha Dec 18 '16 at 0:20
  • \$\begingroup\$ This is a Buck Boost model using an average over period Ts with an example for D=0.5 while the Ch2slide 20 is for Boost only subinterval 1 (same) but subinterval 2 is not same with choke in series with source rather than shunt switched from source to cap. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 18 '16 at 1:11
  • \$\begingroup\$ @anhnha I can't figure out what you are trying to point out. Nevertheless, on page 20 of your link, it says "Small ripple approximation" which clearly implies it is an assumption going forward. \$\endgroup\$ – rioraxe Dec 18 '16 at 4:05
  • \$\begingroup\$ @rioraxe: I am confused about how the integral in red above is equal to the right expression in my first post. You said that it is an approximation assuming that the voltage changes for one switching cycle are small. If we look at the inductor voltage VL(t). It is a constant <vg(t)> over dTs and equal to another constant <v(t)> over (1-d)Ts. So how does this model differs from the model in chapter 2? In chapter 2, we ignore ripple so the voltage over dTs is constant and (1-d)Ts is also constant. Where the model differs from the one in chapter 2? \$\endgroup\$ – anhnha Dec 18 '16 at 5:50

The notation is pretty confusing, but after studying the slides my interpretation is:

d is the fraction of "on" time during the cycle. d' is the fraction of off time during the cycle. Averaged across the cycle, VL will approximately equal input voltage VG for d fraction of the time and will approximately equal output voltage V for d' fraction of the time. So the average voltage will be d*VG + d'*V.

Since everything is a function of time (hopefully slowly changing), VG and V are replaced by averages over the cycle. And d and d' are replaced by d(t) and d'(t). (This is how the model preserves the low frequency components and removes the high frequency components.) This gives the right hand side of the equation you're looking at. Because these are changing very slightly over a cycle, it's not exact equality but approximate equality so the ≈ symbol is used.

Note that in the integrals, d is used for its differential meaning, while in the other equations d indicates the PWM fraction, which makes things harder to understand. And d' is not the derivative of d but 1-d.

Edit to include some of the comment discussion:

Consider a real-life example such as a power supply where the input voltage Vg(t) is 170V + 60 Hz ripple + 30 kHz switching ripple. You can simplify this by making Vg constant at 170V (which is what chapter 2 does). Or you can make a more realistic simplification by averaging out the ripple and getting Vg(t) = 170V + 60 Hz ripple (which is what chapter 7 does).

One tricky part is that in approximating the integral in the original question, you're assuming that Vg(t) is approximately constant over the switching time period. But after that point in the analysis, you treat the moving averages as functions of t, not constants.

  • \$\begingroup\$ You said that VG and V are replaced by averages over the cycle. Yes, but the average here is moving average <vg(t)> and <v(t)> and they are functions of time not a constant. So the how the integral in red above is equal to the right expression? If you say that it is an approximation, assuming that <vg(t)> and <v(t)> are constant over dTs and (1-d)Ts, respectively then how does this model differs from the model given in chapter 2 where ripple is ignored and the inductor voltage are also constant over the range dTs and (1-d)Ts respectively? \$\endgroup\$ – anhnha Dec 18 '16 at 5:55
  • \$\begingroup\$ Page 18 shows the two sides are approximately equal (≈) while your red text has equal (=). They aren't equal, just approximately equal. The key idea is the averages change very slowly, so over one cycle you can consider them just about constant even though they aren't exactly constant (as in chapter 2). If Vg changed rapidly compared to the cycle time, then the equation wouldn't work. But if these things change at a low frequency compared to the cycle frequency, then you can treat them as constant over the cycle and the two sides are approximately (but not exactly) equal. \$\endgroup\$ – Ken Shirriff Dec 18 '16 at 7:16
  • \$\begingroup\$ As you see in the page 16-17, the approximation sign is because VL(t) is not equal to <Vg(vt)> during D and v(t) during 1-D. \$\endgroup\$ – anhnha Dec 18 '16 at 7:22
  • \$\begingroup\$ To see the difference in models, look at chapter 2 page 12. The Vg and V lines are horizontal, while in chapter 7 they are slightly sloped. Chapter 2 makes the assumption these are constant, while chapter 7 does not. That's why chapter 7 uses the moving average - the dotted line for the moving average is horizontal since the average is assumed to be constant over short times (but really a function of t). \$\endgroup\$ – Ken Shirriff Dec 18 '16 at 7:30
  • \$\begingroup\$ As you said, in chapter 7, Vg and V are slightly sloped. I can see it from the picture. However, during the calculation, the moving average voltages <Vg(t)> and <v(t)> are assumed to be constant. Then I don't see any different from the model in chapter 2. \$\endgroup\$ – anhnha Dec 18 '16 at 7:36

How about this:

enter image description here

Notice there is a slight variation in the notation of the result.

Edit for clarification and further point:

From the above equations we get:

\begin{equation} \ <v_{L}(t) > _{T_{s}} = <v_{g}(t) > _{T_{s}} + <v(t) > _{T_{s}} \end{equation}

This equation implies that the contents are continous for the whole period. Now everybody agrees on this, right? We further continue the equation and get:

\begin{equation} \ <v_{g}(t) > _{T_{s}} = d(t) <v_{g}(t) > _{dT_{s}} + d'(t) <v_{g}(t) > _{d'T_{s}} \end{equation}

\begin{equation} \ <v(t) > _{T_{s}} = d(t) <v(t) > _{dT_{s}} + d'(t) <v(t) > _{d'T_{s}} \end{equation}

but (as can be seen on graph) we assumed this:

\begin{equation} \ v_{g}(t) = 0 \ (d<t<T_{s}) \\ \ v(t) = 0 \ (0<t<d) \end{equation}


\begin{equation} \ <v_{g}(t) > _{T_{s}} = d(t) <v_{g}(t) > _{dT_{s}} \end{equation}

\begin{equation} \ <v(t) > _{T_{s}} = d'(t) <v(t) > _{d'T_{s}} \end{equation}

This makes sense because:

\begin{equation} \ \frac{d(t)}{T_{s}} < 1 \end{equation}

\begin{equation} \ \frac{d'(t)}{T_{s}} < 1 \end{equation}

The average of those signals taken over a longer time within Ts would drop because they are zero. The maximum average occurs within the zones they are non-zero.

Therefore, what we finally get is:

\begin{equation} \ <v_{L}(t) > _{T_{s}} = d(t) <v_{g}(t) > _{dT_{s}} + d'(t) <v(t) > _{d'T_{s}} \end{equation}

Which is not the result found in the slides but I am not sure where I made a mistake in the math... Maybe the slides just have a minor mistake in notation? (It happens)

  • \$\begingroup\$ Can you prove the expression after "so:" above? Also if you use the result after "so:", you can not get the final formal as you wrote. You made mistake substituting them. \$\endgroup\$ – anhnha Dec 18 '16 at 6:03
  • \$\begingroup\$ The parameter encircled ( <Vg(t)>|Ts and <v(t)>|Ts ) also includes (1/Ts) in their formula. So you get: (<Vg(t)>|Ts + <v(t)>|Ts ) on the upper encircled equation, right? For the bottom encircled equation, you get: (d(t)<Vg(t)>|dTs + d'(t)<v(t)>|d'Ts) using the same idea. Since: (<Vg(t)>|Ts + <v(t)>|Ts ) = (d(t)<Vg(t)>|dTs + d'(t)<v(t)>|d'Ts) it follows that from: (<Vg(t)>|Ts + <v(t)>|Ts ) = (<v_L(t)>|Ts) you get: (<v_L(t)|Ts>) = (d(t)<Vg(t)>|dTs + d'(t)<v(t)>|d'Ts) \$\endgroup\$ – user132236 Dec 18 '16 at 14:38
  • \$\begingroup\$ Yes, but your final expression is not same as the result in the lecture or in my first post. \$\endgroup\$ – anhnha Dec 18 '16 at 16:05
  • \$\begingroup\$ Well to be honest, I don't agree with the original result. Because, if Vg(t) is averaged between t and t+dTs, why write <Vg(t)>|Ts ? Technically, it is not even continous from t+dTs to Ts in the original result. That's why I added "0" to make it continous. But perhaps I'm wrong here... \$\endgroup\$ – user132236 Dec 18 '16 at 16:46
  • \$\begingroup\$ Yes, also do you agree with the discussion between me and Ken Shirriff above? \$\endgroup\$ – anhnha Dec 18 '16 at 16:58

I prefer to analyze using the energy storage function E=1/2LI^2 where V1*dt1=V2*dt2 [volt-second] in continuous mode.

Maybe the book explains it better, enter image description here

I also had a hard time following his notation as well.


You are focusing on the math as opposed to realizing why the math must work.

For any switcher operating in continuous mode, the area of the ON cycle must equal the OFF cycle. The area of the yellow rectangle must equal the orange rectangle for the switcher to be in steady-state mode.

enter image description here

During ON time current comes from the source, but during the OFF time current comes from the inductor forward biasing the diode. \$I_{MAX}\$ and \$I_{MIN}\$ must meet up or capacitor voltage changes. It's why switchers require a minimum current.

Wikipedia - Buck converter

  • \$\begingroup\$ Sorry but your answer has nothing to do related to my question above. You are talking about small ripple approximation considering Vin and Vout are exactly constant during D and 1-D periods. However, my question is about moving average and why that integral works when the function under the integral sign is not constant over the lower to upper limits range. \$\endgroup\$ – anhnha Dec 20 '16 at 2:19
  • \$\begingroup\$ The math still has to hold because the underlying physics is defined by the math. \$\endgroup\$ – user132236 Dec 22 '16 at 22:10

enter image description here

From that picture it is important to notice that < v_g(t) > is the dotted line not the one that has a slope. What you are doing is calculating the average of inductor voltage over one switching cycle. So even though the averages change in time, for one specific time period the average notated by < v_g(t) > is constant because it is defined that way.

In other words you messed up notations. In the integral that you made yourself in hand writing you shouldn't have used the over swithcing cycle averages < v_g(tau) > and < v(tau) > but just plain v_g(tau) and v(tau). That is because you don't know the averages for that time period yet but you are only calculating them. They are defined to be the answer of those time integrals. < v_g(t) > for the on time and < v(t) > for the off time.

Also then the interal derived by you only works if calculated from the beginning of the switching cycle but the equation that is on the left side from it works always but of course you knew that and just calculated it from beginning of swithing cycle for convience.



$$ \begin{align} d(\tau) &= 1\text{ for } t\leq \tau \leq t+dT_z\\ &= 0 \text{ for } t+dT_s < \tau \leq t+T_s \end{align} $$


$$ \begin{align} d'(\tau) &= 0\text{ for } t\leq \tau \leq t+dT_z\\ &= 1 \text{ for } t+dT_s < \tau \leq t+T_s \end{align} $$


$$ \begin{align} \left< v_L(t)\right>_{T_s}&= \frac{1}{T_s}\int_t^{t+T_s}v_L(\tau)d\tau =\\ &= \frac{1}{T_s} \left[1 \times\int_t^{t+dT_s} v_g(\tau)d\tau +0\times\int_{t+dT_s}^{t+T_s} v_g(\tau)d\tau+\\ + 0 \times\int_t^{t+dT_s} v(\tau)d\tau +1\times\int_{t+dT_s}^{t+T_s} v(\tau)d\tau \right] \end{align} $$ Appliying the inverse of the distributive property and knowing that $$ \int_t^{t+dT_s} x(\tau)d\tau + \int_{t+dT_s}^{t+T_s} x(\tau)d\tau = \int_t^{t+T_s} x(\tau)d\tau $$ then, $$ \begin{align} \left< v_L(t)\right>_{T_s}&= \frac{1}{T_s}\left[d(\tau)\times\int_t^{t+T_s}v_g(\tau)d\tau+d'(\tau)\int_t^{t+T_s}v(\tau)d\tau \right]\\ &=d(\tau)\times\frac{1}{T_s}\int_t^{t+T_s}v_g(\tau)d\tau+d'(\tau)\times\frac{1}{T_s}\int_t^{t+T_s}v(\tau)d\tau\\ &= d(\tau)\times\left< v_g(t)\right>_{T_s}+d'(\tau)\times\left< v(t)\right>_{T_s} \end{align} $$ thus obtaining the awaited result.

  • \$\begingroup\$ Thank you for the help. However, I think that your derivation in the last three lines are not correct. I don't see how you can apply distributive property to get what you showed in the last three lines above. \$\endgroup\$ – anhnha Dec 24 '16 at 15:46
  • \$\begingroup\$ It is not the distributive, it is the inverse of the distributive. And the d and d' functions are piecewise linear, therefore you can multiply by them any sum (such as an integral) the same way as any other function. \$\endgroup\$ – Marcos Judewicz Dec 27 '16 at 3:00

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