Boost converter duty cycle in DCM

[Ntip]

TL;DR Summary

I am trying to find the duty cycle of a boost converter in DCM

Hi, This is not a homework problem, I am trying to analyze how various parameters affect a boost converter operating in DCM most of the time. The input and output voltage is fixed so I will eventually like to sweep parameters for D, L, T, Vi, and R. R is less important though. Here I am using traditional parameter definitions.
D=duty cycle while the switch is on
L=inductance
T=period
Vi=input voltage
Vo=output votlage
R=load resistance

I am trying to find the duty cycle for a boost converter operating in DCM. I've solved for D, where D is the duty cycle while the switch is on. I end up with D=sqrt(-2LVo/(Vi^2RT)), but that doesn't make sense to me because of the negative. I've reworked it multiple times and I keep getting the same result.

 

[berkeman]

By definition, when you are in discontinuous mode, there is no well defined duty cycle. Are you running any simulations?

 

[DaveE]

I would solve this using the energy balance (like the way we do flyback converters). Assuming the time constants are are large compared to the switching speed, the inductor is charged up with slope +Vi/L with duration DT. This gives you the peak current in L and thus the energy input per cycle; Epk = (1/2)⋅L⋅(D⋅T⋅Vi/L)². In DCM all of this energy is transferred to the load which is T⋅Vo²/Ro.

I just sketched this out from memory, I haven't actually figured it out, so there could be a mistake here. Ask if it's confusing or erroneous and I'll look at it more carefully

 

[Ntip]

By definition, when you are in discontinuous mode, there is no well defined duty cycle. Are you running any simulations?

I am running simulations but I would ideally like to put an equation behind what I'm seeing. I have other topologies in mind which would require me to modify the traditional boost converter equation, but understanding the boost converter is a good starting point I think. I remember that in CCM there are assumptions that allow you to develop the equations so they're load independent but in DCM that's no longer true and it is then also a function of the load.

 

[DaveE]

There is an equation. As I described. The result is D = (Vo/Vi)⋅√(2⋅L/(T⋅R)). The result in you're OP can't be correct the dimensions aren't right (plus the sign error, which I don't understand).

 

[berkeman]

I am running simulations but I would ideally like to put an equation behind what I'm seeing. I have other topologies in mind which would require me to modify the traditional boost converter equation, but understanding the boost converter is a good starting point I think. I remember that in CCM there are assumptions that allow you to develop the equations so they're load independent but in DCM that's no longer true and it is then also a function of the load.

Several of the buck converters that I've designed into products over the years ran in DCM at light loads. There was no issue with it, other than an increase in ripple voltage (still withing spec), and a slightly slower transient load response. If you can simulate such conditions and the converter still meets spec, are there still issues?

 

[Ntip]

I would solve this using the energy balance (like the way we do flyback converters). Assuming the time constants are are large compared to the switching speed, the inductor is charged up with slope +Vi/L with duration DT. This gives you the peak current in L and thus the energy input per cycle; Epk = (1/2)⋅L⋅(D⋅T⋅Vi/L)². In DCM all of this energy is transferred to the load which is T⋅Vo²/Ro.

I just sketched this out from memory, I haven't actually figured it out, so there could be a mistake here. Ask if it's confusing or erroneous and I'll look at it more carefully

I think this is pretty much the approach that I took. Am I missing something fundamentally?
$V_{o}=<v_{o}>$
$I_{in}=<i_{in}>$
$P_{in}=V_o^2/R$

$I_{in}=<i_{L}>$
since we know the inductor current starts at 0 A and ends up at 0 A, We can simply find the area of the triangle using
$I_{in}=<i_{L}>= \frac {1} {2} bh$

if we let D = the traditional duty cycle and $D_A$ = the percent of the period where the inductor is discharging,

$I_{in}=\frac {\frac{1} {2} (D+D_A)Ti_{pk}} {T}$

$i_{pk}=\frac{V_iDT}{L}$

finding $D_A$
$-i_{pk}=\frac {-V_{in}-V_oD_AT} {L}$

$D_A=\frac {-Li_{pk}} {(V_{in}-V_o)T}$

$=\frac {-L(\frac {V_{in}DT}{L})}{(V_{in}-V_o)T}$

$D_A=\frac {-V_{in}D} {(-V_{in}-V_o)}$

substitute

$I_{L}=\frac {1} {2} (D -\frac {V_inD}{V_in-V_o})$

$=\frac {V_iD^2T} {2L}(1 -\frac {V_in}{V_in-V_o})$

find common denominator and simplify

$=\frac {V_iD^2T}{2L}(\frac {-V_o}{V_in-V_o})$

substitute into power balance
$\frac {V_in^2D^2T} {2L} (\frac {-V_o} {V_in-Vo})=\frac {V_o^2} {R}$

Simplify and solve for D
$D^2= \frac {-2LV_o} {V_i^2RT}$
$D=\sqrt(\frac {-2LV_o} {V_i^2RT})$

 

[Ntip]

Several of the buck converters that I've designed into products over the years ran in DCM at light loads. There was no issue with it, other than an increase in ripple voltage (still withing spec), and a slightly slower transient load response. If you can simulate such conditions and the converter still meets spec, are there still issues?

I think it's easier with a buck converter because you don't have to worry about blowing up components. With a boost converter, the gain is a lot higher which is my concert. For instance, simulating a 100 V input with a 50% duty cycle would be 200 V output in CCM. In DCM, it reaches steady state at about 825 V so figuring out how to control this with the duty cycle would be ideal.

 

[berkeman]

I think it's easier with a buck converter because you don't have to worry about blowing up components. With a boost converter, the gain is a lot higher which is my concert. For instance, simulating a 100 V input with a 50% duty cycle would be 200 V output in CCM. In DCM, it reaches steady state at about 825 V so figuring out how to control this with the duty cycle would be ideal.

Can you post a Spice simulation of that scenario please? I'd like to simulate it a bit more to help my understanding. Thanks.

 

[DaveE]

I think it's easier with a buck converter because you don't have to worry about blowing up components. With a boost converter, the gain is a lot higher which is my concert. For instance, simulating a 100 V input with a 50% duty cycle would be 200 V output in CCM. In DCM, it reaches steady state at about 825 V so figuring out how to control this with the duty cycle would be ideal.

Yes, you have to be careful with boost and flyback (boost-buck) converters to avoid no-load operation. In practice, what usually happens is that the magnetics saturate as their current grows. You keep putting energy in but it can't get out. However, that is the light load regime, which is usually easily avoided with either a pre-load or feedback control that responds to excess voltage by reducing the duty cycle.

Once you've applied the easy fixes, the residual problem is what @berkeman said, the response is slow and messes up your dynamic response. It's actually even worse that that since you end up with a RHP zero, which is a real PIA for stability.

The result is that you tend to see converters that either operate in CCM, with low current ripple, or DCM as circuits that deliver packets of energy. Designers usually avoid crossing the CCM-DCM boundary, unless they are comfortable with an overcompensated (slow) response.

 

[DaveE]

finding $D_A$
$-i_{pk}=\frac {-V_{in}-V_oD_AT} {L}$

$D_A=\frac {-Li_{pk}} {(V_{in}-V_o)T}$

$=\frac {-L(\frac {V_{in}DT}{L})}{(V_{in}-V_o)T}$

$D_A=\frac {-V_{in}D} {(-V_{in}-V_o)}$

Sign errors in this section. Ipk = (Vi-Vo)⋅D_A⋅T/L & D_A = Vi⋅D/(Vi-Vo) is the correct version.

The inductor down slope is (Vi-Vo)/L. This is a negative number since Vi<Vo.

Now: I've already had 1 1/2 beers tonight. So, don't expect me to do algebra for you. Look at it again and and complain if you think I'm wrong.

 

[Ntip]

Can you post a Spice simulation of that scenario please? I'd like to simulate it a bit more to help my understanding. Thanks.

Here you go. I've only simulated out to 100 ms but it can obviously be changed. I think steady state for the no load condition is ~5 s-ish. You can still see that by 100 ms it has reached ~330 V instead compared to 200 V in CCM.

**edit**
nevermind, it will not let me upload LTspice *.asc files. Here is a screenshot. I used L=30 mH, C=5 uF, Rload= 50 Ohm for CCM and 1 MOhm for DCM. Ton=5 us, T=10 us. Vs=100 V. I used the generic nmos with 8000 in parallel (m=8000).

 

[DaveE]

OK, IMO, this is not the sort of generic problem that is not helped by simulation. You won't really understand it without getting the basic operation plus algebraic manipulations right. You are definitely on the right track, I can tell you understand the basic, essential information here. You just need to be more careful with your algebra. Draw some pictures, like a graph of inductor current, to help you keep your signs straight. You are closer to the answer (and, BTW, the right approach) than you may realize.

 

[Baluncore]

nevermind, it will not let me upload LTspice *.asc files.

The LTspice.asc and LTspice.plt files are ascii files. If you change their names by attaching the extension .txt they can be attached to the post by drag and drop as ...
LTspice.asc.txt and LTspice.plt.txt

Look in LTspice dropdown menu Tools. "Copy bitmap to clipboard"; is clearer than screen shot. Attach the picture as a .jpg or .png to reduce size and maintain readability.

 

[Ntip]

****Ignore this and I'll reply again below for ease of reading. I edited the comment and for some reason all of the latex formatting went away. I'll poste again below with formatting.***

Sign errors in this section. Ipk = (Vi-Vo)⋅D_A⋅T/L & D_A = Vi⋅D/(Vi-Vo) is the correct version.

The inductor down slope is (Vi-Vo)/L. This is a negative number since Vi<Vo.

Now: I've already had 1 1/2 beers tonight. So, don't expect me to do algebra for you. Look at it again and and complain if you think I'm wrong.

Thanks, I will work through it again but I think the negative sign is correct. I drew the graph out for the inductor current but just didn't upload it. It does help a lot.

Here's where I get $-i_{pk}$ in that step.

$V= \frac {L\Delta i}{\Delta t}$

rearranging
$\Delta i = \frac {V\Delta t}{L}$

for $DT<t<D_A$

$V=V_{in}-V_o$
$\Delta t = D_AT$

finding $\Delta i$
$i_o=i_{pk}$
$i_f=0$

$\Delta i=(i_f-i_o)$

$=0-i_{pk}$

$\Delta i=-i_{pk}$

substituting into $V= \frac {L\Delta i}{\Delta t}$ and solving for $D_A$

$(V_{in}-V_o)= \frac {L(-i_{pk})}{D_AT}$

$D_A=\frac {-Li_{pk}}{(V_{in}-V_o)T}$$D_A=\frac {-V_{in}D} {(V_{in}-V_o)}$

We can substitute $i_{pk}$ , which is positive, giving
$D_A=\frac {-V_{in}D} {(V_{in}-V_o)}$so I think anyway. I know that's a lot of math so I'll go through and check again but I keep getting negative. I hope it's positive because that would make life easy and give me a solution, but just don't see how it is positive.

 

[Ntip]

The LTspice.asc and LTspice.plt files are ascii files. If you change their names by attaching the extension .txt they can be attached to the post by drag and drop as ...
LTspice.asc.txt and LTspice.plt.txt

Look in LTspice dropdown menu Tools. "Copy bitmap to clipboard"; is clearer than screen shot. Attach the picture as a .jpg or .png to reduce size and maintain readability.

Ok I see. I've attached it here.

 

[Ntip]

Sign errors in this section. Ipk = (Vi-Vo)⋅D_A⋅T/L & D_A = Vi⋅D/(Vi-Vo) is the correct version.

The inductor down slope is (Vi-Vo)/L. This is a negative number since Vi<Vo.

Now: I've already had 1 1/2 beers tonight. So, don't expect me to do algebra for you. Look at it again and and complain if you think I'm wrong.

Thanks, I will work through it again but I think the negative sign is correct. I drew the graph out for the inductor current but just didn't upload it. It does help a lot.

Here's where I get $-i_{pk}$ in that step.

$V= \frac {L\Delta i}{\Delta t}$

rearranging
$\Delta i = \frac {V\Delta t}{L}$

for $DT<t<D_A$

$V=V_{in}-V_o$
$\Delta t = D_AT$

finding $\Delta i$
$i_o=i_{pk}$
$i_f=0$

$\Delta i=(i_f-i_o)$

$=0-i_{pk}$

$\Delta i=-i_{pk}$

substituting into $V= \frac {L\Delta i}{\Delta t}$ and solving for $D_A$

$(V_{in}-V_o)= \frac {L(-i_{pk})}{D_AT}$

$D_A=\frac {-Li_{pk}}{(V_{in}-V_o)T}$$D_A=\frac {-V_{in}D} {(V_{in}-V_o)}$

We can substitute $i_{pk}$ , which is positive, giving
$D_A=\frac {-V_{in}D} {(V_{in}-V_o)}$so I think anyway. I know that's a lot of math so I'll go through and check again but I keep getting negative. I hope it's positive because that would make life easy and give me a solution, but just don't see how it is positive.

 

[Ntip]

Sign errors in this section. Ipk = (Vi-Vo)⋅D_A⋅T/L & D_A = Vi⋅D/(Vi-Vo) is the correct version.

The inductor down slope is (Vi-Vo)/L. This is a negative number since Vi<Vo.

Now: I've already had 1 1/2 beers tonight. So, don't expect me to do algebra for you. Look at it again and and complain if you think I'm wrong.

I think I see what you mean now about the slope being negative which i agree with. I think the signs should work out without just flipping "because we know" the output is larger. By keeping them negative, ##D_A=\frac {-V_{in}D} {(V_{in}-V_o)} works out to be positive since you have a negative divided by a negative.

Are I on track with what you were talking about, or am I still missing something?

 

[DaveE]

This may (or may not help). In any case R.W. Erickson's stuff is great and online (although not easy to find if you don't know here to look). He uses slightly different nomenclature. But I'm posting it because it's a great reference. Bookmark it. He was a grad student of my prof. R.D. Middlebrook, and he has the best documentation online of those methods. Look around page 20-40, or so.

http://altor1.narod.ru/Books_Docs/Fundamentals_of_Power_Electronics.pdf

 

[Ntip]

This may (or may not help). In any case R.W. Erickson's stuff is great and online (although not easy to find if you don't know here to look). He uses slightly different nomenclature. But I'm posting it because it's a great reference. Bookmark it. He was a grad student of my prof. R.D. Middlebrook, and he has the best documentation online of those methods. Look around page 20-40, or so.

http://altor1.narod.ru/Books_Docs/Fundamentals_of_Power_Electronics.pdf

Thanks, this is really good actually really good. I have the book and have found individual chapters, but not all chapters together. I'll take a look and hopefully get this figured out.

 

[DaveE]

Wait, time-out. My formula was the one with the sign errors. Specifically, this one D_A = ...
The correct one is your version D_A = Vi⋅D/(Vo-Vi). This is obvious if you consider that Vo > Vi and D>0, D_A>0.

I'll blame the beer, but the truth is I'm probably just lazy, LOL.

 

[Ntip]

Wait, time-out. My formula was the one with the sign errors. Specifically, this one D_A = ...
The correct one is your version D_A = Vi⋅D/(Vo-Vi). This is obvious if you consider that Vo > Vi and D>0, D_A>0.

I'll blame the beer, but the truth is I'm probably just lazy, LOL.

Yes that is correct now. I kept the negative sign there, you factored it out by flipping the order of Vo and Vi. Both ways work. Personally, I like your arrangement better without the negative sign but both are now the same.

lol hey beer will do that. What's the saying...don't drink and derive? It was just off by a minus sign so it was close enough.

 

[alan123hk]

Summary:: I am trying to find the duty cycle of a boost converter in DCM

I am trying to find the duty cycle for a boost converter operating in DCM. I've solved for D, where D is the duty cycle while the switch is on. I end up with D=sqrt(-2LVo/(Vi^2RT)), but that doesn't make sense to me because of the negative. I've reworked it multiple times and I keep getting the same result

It does not seem easy to derive the duty cycle equation directly. If you don’t mind, there are other simpler methods. You can first derive the voltage transfer function, and then find the duty cycle from it. The clever predecessors have obtained the transfer function of the DCM boost converter, you can refer to Wikipedia:-

https://en.wikipedia.org/wiki/Boost_converter

The duty cycle equation of DCM boost converter seems to be more complicated than expected.

 

[Ntip]

It does not seem easy to derive the duty cycle equation directly. If you don’t mind, there are other simpler methods. You can first derive the voltage transfer function, and then find the duty cycle from it. The clever predecessors have obtained the transfer function of the DCM boost converter, you can refer to Wikipedia:-

https://en.wikipedia.org/wiki/Boost_converter

The duty cycle equation of DCM boost converter seems to be more complicated than expected.

Thanks. I think this follows is the same method presented in Erickson's book. I'm able to derive it now...let's just see how well this works. Fingers crossed.