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- Inductor-calculation in Buck/Setp-Down
- 10-05-2007
- ole.tetzschner
October 5, 2007, 9:54 am
Hi
I am wondering about the calculation of the inductor in a buck-design.
There is a lot of examples and equations out on the Internet, and they
seems to work :)
But i can't understand (or see the logic in), that if I design a
powersupply to deliver 5 Amp. and calculate the inductor to be 4uH...
but if I put 1 Amp in the same equation, I got 16uH ???
Why is the inductor getting larger when the load on the powersupply is
getting lower??? should there be a minimum load??? and should the
inductor-calculation be based upon that???.
With kind regards
Ole Tetzschner
I am wondering about the calculation of the inductor in a buck-design.
There is a lot of examples and equations out on the Internet, and they
seems to work :)
But i can't understand (or see the logic in), that if I design a
powersupply to deliver 5 Amp. and calculate the inductor to be 4uH...
but if I put 1 Amp in the same equation, I got 16uH ???
Why is the inductor getting larger when the load on the powersupply is
getting lower??? should there be a minimum load??? and should the
inductor-calculation be based upon that???.
With kind regards
Ole Tetzschner
Re: Inductor-calculation in Buck/Setp-Down
zero each cycle, it becomes discontinuous and the behavior changes. The
change is still within reach of an error amplifier, so a variable regulated
supply isn't impossible or anything.
Tim
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Re: Inductor-calculation in Buck/Setp-Down
Consider dI/dt = V/L. For a given switcher frequency,
i.e. a given inductor-charging time, thinking simply,
if you need I to be 5x smaller, you'll want L to be
5x larger. This implies sqrt 5 = about 2.2x as many
turns, and for 5x lower current you can use 5x smaller
wire cross-section area, not to mention core area, so
the physical size of the inductor can be smaller.
Think another way - as you go to higher currents, the
impedances must go lower. The implies ot only low R
but low L. At very very high currents your wires may
be thick bars of metal and an inductor may be a huge
special pieces of ferrite clamped around the bar (i.e.
made with one turn).
Re: Inductor-calculation in Buck/Setp-Down
Thanks for the replies :)
But what if my designed power supply, calculated to handle a load of 5
amp, suddenly got a lighter load... let's say 100mA, what would then
happen??? i know the duty-cycle changes, but what about the output???
would there be more ripple or what???
Re: Inductor-calculation in Buck/Setp-Down
When the load becomes so light that the inductor current ramps all the way to
zero before the next switching cycle begins, you've entered so-called
"discontinuous mode." From a stability point of view, the system has changed
so in a poorly-designed (or perhaps just super-cheaply-designed where it was
"known" that there'd always be at least a minimum load) power supply the
system might go unstable and you'll get an oscillating output, but in most
cases you'd expect the power supply would still be stable to really nothing at
all "interesting" happens. Further decreases in load will decreate the switch
duty cycle, although in general this won't be your worst case of ripple
because the load is so very light in the first place. Many controller ICs for
switchers have special modes that they enter at light loads that are meant to
improve overall efficiency (when the load itself is light, suddenly all the
losses in the rest of the switcher become a much bigger factor in the power
supply's overall efficiency), such as switching to "burst" mode where they
switch for a few cycles at a higher duty cycle and then shut off completely
for a bit.
You might want to download a copy of LTspice and play around with the various
example circuits. It also sounds as though you might want to pick up a book
on switching power supply design
---Joel
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