Buck converter duty cycle

Discussion in 'Homework Help' started by notoriusjt2, Oct 29, 2010.

1. notoriusjt2 Thread Starter Member

Feb 4, 2010
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is the duty cycle the same as the duty ratio?

if so then D=Vo/Vs
D=25/50
D=0.5

2. thatoneguy AAC Fanatic!

Feb 19, 2009
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Duty cycle is a percentage of $OnTime \over Period$

In the case of the 100kHz, period is $1 \over frequency$ or $1\over 100000$ or 10 uS.

How many of those 10uS is the switch closed? If 5 uS, then it would be 50%

-ETA: Hint: It is a function of the load current more than it is a function of the output voltage

Last edited: Oct 30, 2010
3. notoriusjt2 Thread Starter Member

Feb 4, 2010
209
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sorry for the MSpaint drawing but its all I have right now

this is a graph of the discontinuous inductor current

DT=5uS

D1 is found from this equation

D1=0.26
I would think (D1T+DT)/T would be equal to $OnTime \over Period$

but I am getting an answer of 0.76
what am I not doing right here?

4. notoriusjt2 Thread Starter Member

Feb 4, 2010
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the duty RATIO here would be equal to Vo/Vs = 0.5 correct?

5. thatoneguy AAC Fanatic!

Feb 19, 2009
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D is the ratio of input and output voltages.

Duty cycle is the ratio of on time to off time of the output.

6. notoriusjt2 Thread Starter Member

Feb 4, 2010
209
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right so if you add up all the time that current flows and then divide it by the period you would have a duty ratio

based on the graph

DT + D1 T = total on time
T = period

5uS + 2.6uS = 7.6uS
7.6uS/10uS = .76

total on time/period =

7. thatoneguy AAC Fanatic!

Feb 19, 2009
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7.6μS / 10μS = 76% duty cycle, yes.

8. t_n_k AAC Fanatic!

Mar 6, 2009
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The approach one might take would be to assume the circuit is operating just at the point of going into discontinuous mode. That is, the minimum inductor current would momentarily fall to zero amp.

At this condition the peak inductor current will be twice the mean load current. Since ΔI=Imax-Imin, Imin=0 and Imean=(Imax+Imin)/2

Use D=f*L*ΔI/(Vs-Vo) to find the duty D.

The actual D would need to be less than the calculated D to have discontinuous mode condition.

Edit:

No - that wont work.

Since the output voltage is 50% of the supply - i.e. 25V - the rise (from 0 to Imax) and fall times (from Imax to 0) for the inductor current must be the same and equal to DT.

The mean current is Imean=DTImax/T=DImax

In this case Imean=125/25=5A

So Imax=5/D Amp=ΔI

But

D=f*L*ΔI/(Vs-Vo)=f*L*5/(D*(Vs-Vo))

So D^2=f*L*5/25=f*L/5

D=(f*L/5)^0.5=(100,000*5u/5)^0.5=(.1)^0.5=0.316

Last edited: Nov 6, 2010
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9. notoriusjt2 Thread Starter Member

Feb 4, 2010
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thank you. my text book makes no reference whatsoever of that formula for finding D

Feb 4, 2008
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11. notoriusjt2 Thread Starter Member

Feb 4, 2010
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recalculating with D=0.316 gives me...

but that answer is incorrect. am I not using the right formula here?

Feb 4, 2008
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13. notoriusjt2 Thread Starter Member

Feb 4, 2010
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based on this graph off of wikipedia, it appears that on time is only from 0 to DT.

so would duty cycle actually be DT/T

which would be equal to 0.316?

14. mik3 Senior Member

Feb 4, 2008
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The duty cycle is duty cycle, it doesn't change. You set the duty cycle and the output voltage depends on the duty cycle. Now, the relationship of the output voltage to the duty cycle depends whether the converter operates in CCM or DCM.

15. t_n_k AAC Fanatic!

Mar 6, 2009
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Kindly permit me an observation or two ...

You've been tinkering with these problems for some time now and still seem to be struggling.

As Mik3 & others point out, the particular converter mode of operation and topology dictate the approach required. I'm sure you've grasped that fact.

You asked me some posts back whether I remember the formulas for any given situation. In brief - no I don't. Well that's not exactly correct - I recall the relationship between switching duty and output voltage in a continuous mode buck converter. Rather than remembering a formula which may not be relevant to a particular case, I try to understand what's happening physically and then analyse the circuit.

The present example is a case in point. I came momentarily came 'unstuck' because I didn't initially think it through carefully.

You may derive some benefit from putting the text book aside and systematically analyzing the various cases for yourself - or make sure you understand why a certain formula is relevant for a particular case. Then you may be able to apply the formulas with confidence.

16. notoriusjt2 Thread Starter Member

Feb 4, 2010
209
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thats a very good idea, i will take that into consideration

however, the graph that i posted from wikipedia and the graph that is in my textbook shows two different on/off concepts. if i would have seen the wikipedia version from the beginning i probably would have gotten this concept much earlier