Input ripple for Buck with input LC filter

Thread Starter

BitsNBytes

Joined Mar 22, 2021
10
Hello,

It's me again! This is a buck with input filter. The question I have is how to determine the inductor ripple current of L2.

1619404788921.png

To summarize my work with volt-sec balance and amp-sec balance:

1. Volt-sec balance for L1 -> 0 = Vg - VC1, so VC1 = Vg
2. Volt-sec balance for L2 -> 0 = VC1*D - V, so V = VC1 * D and since VC1 = Vg from #1, then V = Vg * D

3. Amp-sec balance for C1 -> 0 = IL1 - IL2 * D
3a. IL1 = IL2 * D is the DC component of IL1
4. Amp-sec balance for C2 -> 0 = IL2 - V/R
4a. IL2 = V/R is the DC component of IL2
4b. Since V = Vg * D, then IL2 = (Vg * D)/R

And since IL1 = IL2 * D, and IL2 = (Vg * D)/R, then finally:
IL1 = (D^2 * Vg) / R

All those are correct, and I have attached my work.

Here is the issue. From an ideal buck converter with no input filter, the inductor ripple current is ((Vg - V) * D * Ts) / (2 * L). This should be the same for L2 in the above circuit. If VC1 = Vg then I would calculate the inductor ripple in L2 as:

Eq. 1 : ((Vg - V) * D * Ts) / (2 * L2)

It's the same as before! But this answer is wrong. The correct answer is:
e
Eq. 2 : (Vg * D * Ts) / (2 * L2)

I do not see how we get from Eq1 to Eq2. Why/How does the output voltage drop out?
 

Attachments

MrAl

Joined Jun 17, 2014
8,473
Hello,

It's me again! This is a buck with input filter. The question I have is how to determine the inductor ripple current of L2.

View attachment 236770

To summarize my work with volt-sec balance and amp-sec balance:

1. Volt-sec balance for L1 -> 0 = Vg - VC1, so VC1 = Vg
2. Volt-sec balance for L2 -> 0 = VC1*D - V, so V = VC1 * D and since VC1 = Vg from #1, then V = Vg * D

3. Amp-sec balance for C1 -> 0 = IL1 - IL2 * D
3a. IL1 = IL2 * D is the DC component of IL1
4. Amp-sec balance for C2 -> 0 = IL2 - V/R
4a. IL2 = V/R is the DC component of IL2
4b. Since V = Vg * D, then IL2 = (Vg * D)/R

And since IL1 = IL2 * D, and IL2 = (Vg * D)/R, then finally:
IL1 = (D^2 * Vg) / R

All those are correct, and I have attached my work.

Here is the issue. From an ideal buck converter with no input filter, the inductor ripple current is ((Vg - V) * D * Ts) / (2 * L). This should be the same for L2 in the above circuit. If VC1 = Vg then I would calculate the inductor ripple in L2 as:

Eq. 1 : ((Vg - V) * D * Ts) / (2 * L2)

It's the same as before! But this answer is wrong. The correct answer is:
e
Eq. 2 : (Vg * D * Ts) / (2 * L2)

I do not see how we get from Eq1 to Eq2. Why/How does the output voltage drop out?

Hello,

Your solution #1 probably assumes that C1 is very large, and i believe that is not the general case. There may be an included assumption however that it is and then i would think vC1 would behave more or less like Vg.

But i also find it hard to believe that if we start with 100v in and 10v out that we get a peak inductor current of i1 and then if we increase the output from 10v to 90v that current peak current does not decrease. This can be tested with some random component values and a circuit simulator, or write up the exact time domain solutions and test it. But increase that output to 99.9v and there is no change in the peak inductor current? That's hard to believe at least at first glance.
This leads me to believe that the output voltage must be included in the expression for peak iL. Note that i have not tested this yet though i will look into it today sometime i hope [see below]

Are there any other assumptions for this problem? For example, maybe L2=L1 and C2=C1 or something like that. It is important to know this, because for some problems it simplifies the analysis by 100 fold or more. This problem has both 2nd order and 4th order exact solutions so any simplifications at the start would be great. On the other hand, we could assume some numerical values and test the assumed correct expression and all we have to do is find one case that doesnt work and we disproved the supposed correct expression.

Maybe also you might mention where you got this problem from.
I have seen typos abound in both textbooks and certainly on the web. One typo and nothing works at all. It can be frustrating until you do a full analysis to find out what doesnt fly.

[LATER]
Ok it looks like that second equation is correct. It looks like the first inductor has an averaging effect on the current so it kind of regulates the current in the second inductor. We will have to look deeper to find out exactly what is happening. The 2nd inductor peak to peak current stays the same with load, but the 2nd inductor average current changes with load.
My guess for now is that you have to do some time domain analysis to get the correct result.

[LATER LATER]
(NOTE: E=Vg here)
Ok using a time domain approach seems to work.
Using the derivative of the 2nd inductor current and assuming Vout=E*D we get the same result which must be correct (second equation).
The reasoning behind this is probably because the output is considered constant and the output voltage is inherent in the input voltage through the duty cycle so no need for Vout directly.

Simple averaging techniques sometimes dont work because the phase of the variable is not in step with the switch. In those cases i think you always have to work it out in the time domain one way or another. The simpler time domain technique is to assume some things are constant (that's where that 2nd equation comes from), the complete exact solution assumes nothing other than what was assumed all along such as ideal components, unless of course you care to include parasitic effects. The exact solution requires solving a 4th degree polynomial so it makes it difficult to find symbolic solutions, even the numerical time equations are a bit difficult to reach, although it is possible. Sit down, roll up your sleeves, and get to work :)
 
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Thread Starter

BitsNBytes

Joined Mar 22, 2021
10
Ok using a time domain approach seems to work.
Using the derivative of the 2nd inductor current and assuming Vout=E*D we get the same result which must be correct (second equation).
The reasoning behind this is probably because the output is considered constant and the output voltage is inherent in the input voltage through the duty cycle so no need for Vout directly.
This is a problem from UC Boulder, specifically from the textbook. The only "clue" is that the components are "sufficiently large to ensure the ripple is small." The presumptions are listed below and funny enough, this question has been asked multiple times across other sites:

https://electronics.stackexchange.c...sign-of-a-buck-converter-with-lc-input-filter

I agree with the small ripple approximations from the above thread, but I do not agree with the equations for capacitor voltage ripple and inductor current ripple.

From your explanation above, I am still not seeing the answer. The "derivative of the 2nd inductor current", so writing the canonical form of the current through an inductor and taking the derivative yields:

d iL(t) / dt = delta-v(t) / L

I am still just seeing delta-v as (Vg - V) and not just Vg.

Funnier still, the answer provided:

(Vg * D * Ts) / 2 * L

is the inductor ripple current of a Boost converter (pg. 28, Fundamentals of Power Elec, 2nd edition). Must be a coincidence.
 

MrAl

Joined Jun 17, 2014
8,473
This is a problem from UC Boulder, specifically from the textbook. The only "clue" is that the components are "sufficiently large to ensure the ripple is small." The presumptions are listed below and funny enough, this question has been asked multiple times across other sites:

https://electronics.stackexchange.c...sign-of-a-buck-converter-with-lc-input-filter

I agree with the small ripple approximations from the above thread, but I do not agree with the equations for capacitor voltage ripple and inductor current ripple.

From your explanation above, I am still not seeing the answer. The "derivative of the 2nd inductor current", so writing the canonical form of the current through an inductor and taking the derivative yields:

d iL(t) / dt = delta-v(t) / L

I am still just seeing delta-v as (Vg - V) and not just Vg.

Funnier still, the answer provided:

(Vg * D * Ts) / 2 * L

is the inductor ripple current of a Boost converter (pg. 28, Fundamentals of Power Elec, 2nd edition). Must be a coincidence.
I dont think you have to "know" the inductor voltage for all time.
For example:
L*di/dt=vC
That is one way to express the simplified time domain solution (vC is the voltage across the output cap, noting that the inductor is in parallel with that cap during the 'off' period).
Note that is for the "off" time not the "on" time as one terminal of the inductor is considered grounded through the (presumably zero Ohm) diode when it is conducting.
The principle of Continuity of States says that the ripple going "up" will be the same as the ripple ramping "down" when steady state has been reached. Therefore we should be able to get the solution for either the 'on' time or the 'off' time, with one slope being positive and the other negative but the amplitude will be the same.
Try that see what you can come up with. Be sure to include D and watch the time t can be a little tricky because we are only doing the 'off' time not both 'on' and 'off'.
Also, recall that vC=D*E. where vC is the output cap voltage.
Once you solve this i can give you the entire ODE set then you can solve for lots of stuff if you feel up to it :)
 
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Thread Starter

BitsNBytes

Joined Mar 22, 2021
10
At least I am in good company, because I have shown this to a few people and no one can get it.

It has been a long time since I had to go back to first principles with ODE's, so I attached my work for the first interval to prove to myself that the time domain solution is equivalent to the small ripple approximation. I think my work is correct.

I also attached this for interval 2, but I can see that there will be a D' or (1-D) factor.

So I am almost happy with the inconsistency. Alas.
 

Attachments

MrAl

Joined Jun 17, 2014
8,473
At least I am in good company, because I have shown this to a few people and no one can get it.

It has been a long time since I had to go back to first principles with ODE's, so I attached my work for the first interval to prove to myself that the time domain solution is equivalent to the small ripple approximation. I think my work is correct.

I also attached this for interval 2, but I can see that there will be a D' or (1-D) factor.

So I am almost happy with the inconsistency. Alas.
Hello again,

That looks good, and notice that it does NOT match the original equation #2.
What it looks like to me now is that the original #2 was for the MAXIMUM peak to peak inductor current, and that of course comes with a fractional duty cycle of 1/2 (0.50).
In fact, if you equate #2 with #1 and solve for the duty cycle, you would get 0.50 also. That means the only time #2 matches your new formula is when the duty cycle is exactly 0.50, and that makes me wonder why they included the duty cycle D in equation #2 because if you make it anything other than 0.50 it becomes very much in err of the true result of the peak to peak 2nd inductor current.
I tested this in a simulator as well as a numerical ODE solver program and both results came up very close to the right value when the equation #2 came up just plain wrong when the duty cycle was anything other than 0.50. In one case, for D=0.25, equation #2 came up with ipp=3 amps while the better formula came up with 1.5 amps, and all simulations show 1.5 amps not anything even close to 3.0 amps.

So the conclusion must be that the original equation #2 is only valid for D=1/2 and also the new equation like yours is valid for any duty cycle. Of course during the design phase of these circuits we usually look at D=1/2 mostly anyway because we want to know worst case, but for me that does not give them any artistic license to include D in the formula if it is only valid when D=1/2.
What i suspect however is that they may have shown that formula AFTER explaining that D=1/2 because the design procedure would use D=1/2 as a target duty cycle for the sake of understanding what the worst case inductor current could be. Note however that does not include the start up inductor current if there is no soft start and the start up current could be 10 times higher :)

It is kind of funny i followed the advice i gave you about the inductor and i came up with the same formula (v=D*E in mine) and so i had to look at it several times because it did not match equation #2 of the first post. After some investigation it becomes clear what was going on.
 
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Thread Starter

BitsNBytes

Joined Mar 22, 2021
10
I call shenanigans on the stated answer of

(Vg * D * Ts) / 2L

I hate to believe that the answer is a typo after all these years of the course work being covered. I have no idea what other assumptions are being made. I spent enough time on this particular problem. I could ask the TA and see what they say.

I need to review other materials for the course and prepare for the final, so thank you for your help.
 

MrAl

Joined Jun 17, 2014
8,473
I call shenanigans on the stated answer of

(Vg * D * Ts) / 2L

I hate to believe that the answer is a typo after all these years of the course work being covered. I have no idea what other assumptions are being made. I spent enough time on this particular problem. I could ask the TA and see what they say.

I need to review other materials for the course and prepare for the final, so thank you for your help.
Yes i second that nomination :)
As i was saying though, in the write up for that there could have been additional text like:
"... and so we end up with this expression where D=1/2 ..."
so either before or after that expression they might have explained that D must be 1/2 for it to work. If someone pulled that out of context that could have led to this misunderstanding. Either that or they just calculated it wrong. Or maybe they have it online and have the expression shown on a page where they also have background circus music playing :)
 

MrAl

Joined Jun 17, 2014
8,473
Hey, don't you owe me the full set of ODE equations?
Oh yeah we finished with this so here are the two sets, one for ON and one for OFF modes.
Note the following:
i1=L1 current, left to right
i2=L2 current, left to right
v1=C1 voltage, positive up
v2=C2 voltage, positive up
R3=load resistor
and of course there are the time derivatives of each of them.
If you wanted to you could solve for the exact time domain solutions from these two sets.
I named R3 "R3" because i like to leave room for adding inductor ESR (R1) and capacitor ESR (R2) later if i want to see how that changes things.
An assumption is that the circuit operates in continuous mode meaning the diode always conducts when the switch is off. This is typical. The diode drop when on is zero volts also, but that can also be changed.
Also E1 is the constant DC input source voltage.

OFF:
di1/dt=(E1-v1)/L1
dv1/dt=i1/C1
di2/dt=-v2/L2
dv2/dt=(i2*R3-v2)/(C2*R3)

ON:
di1/dt=(E1-v1)/L1
dv1/dt=(i1-i2)/C1
di2/dt=(v1-v2)/L2
dv2/dt=(i2*R3-v2)/(C2*R3)
 
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