Buck converter question - Page 1
Forum section: electronics / projects

author: gxti
author_rank: Frequent Contributor
timestamp: August 15, 2011, 04:26:31 am
content: I'm currently working on a simple but beefy DC-DC converter, that takes 12VDC and outputs 0-12VDC. It's intended to be part of a modular bench supply, so the minimum load current is unknown, but it's specified for up to 10A out yet should be able to sustain 10mA loads without too much ripple.Here's my specifications (they are flexible):12VDC in, 0-12VDC outOutput load 0-10ASwitching freq 100kHzCurrent ripple: 20%Voltage ripple: 1%Synchronous rectifier topology, so forward losses are around 0.1VSo here's the question: If I design for 10A load, my calculator tells me to use a 15uH inductor. But if I lower the current, that valueincreasesproportionally. I can find 10A, 15uH inductors, but not 10A, 1500uH inductors. What do I choose, and what kind of ill effects can I expect if I use the lower-value inductor with lower load currents? Does the larger output capacitor compensate well enough, at least when operating in constant-voltage mode?

author: Psi
author_rank: Super Contributor
timestamp: August 15, 2011, 04:51:19 am
content: As i understand it, you should spec the inductor for the max output.Here's a good app note about choosing an inductor if you really want to get into the detailhttp://www.national.com/an/AN/AN-1197.pdf

author: amspire
author_rank: Super Contributor
timestamp: August 15, 2011, 04:58:42 am
content: The 15uH should work fine - the switcher will just stop switching if is has to when the current is low or zero.You just have to look at the voltage rise on the output capacitors resulting from a minimum width switching pulse.Say your minimum ON time was 1 uSec with no load. The current into the 15uH inductor would rise from 0 to 12/15e-6 * .000001 secs = .8 AThis is an energy of 1/2 * L * I^2 = 4.8 uJNow this is the minimum amount of energy that capacitors have to be able to absorb without an ugly amount of ripple.So say you had 10,000uF of capacitance with 12 V across it. The energy already in the capacitor is already 1/2*C *V^2 = .72 JSo in comparison, the 4.8 uJ is nothing - it would equate to a voltage increase of under 3 mV.The moral of this is that the only ripple you really have to worry about is at maximum current. You definitely don't want to use a 1500uH inductor if you want that 10A output.Richard

author: Zero999
author_rank: Super Contributor
timestamp: August 15, 2011, 04:42:48 pm
content: A buck regulator is unsuitable for this application because the output voltage will be equal to the input voltage when it's set to 12V so the oscillator will have to stop and turn the transistor on continuously. There will be no regulation and some voltage loss in the transistor and inductor so the output voltage will be slightly under 12V.

author: gxti
author_rank: Frequent Contributor
timestamp: August 16, 2011, 03:53:09 am
content: Quote from: Psi on August 15, 2011, 04:51:19 amThis will keep me busy for a while, thanks!Quote from: amspire on August 15, 2011, 04:58:42 amThis is what my intuition told me, but my electronics intuition isn't as good as my programming intuition so I had to ask.Quote from: Hero999 on August 15, 2011, 04:42:48 pmThis makes sense, I will accept a lower output voltage. This is the first component of a modular bench power supply, the mains part comes later so I'm planning to run this component off an ATX power supply until it's done. Once I get started on the mains part (getting ahead of myself here) I'll design for a higher intermediate voltage so that the second stage (the subject of this thread) can keep regulation at 12 volts.

author: allanw
author_rank: Frequent Contributor
timestamp: August 16, 2011, 04:08:57 am
content: It's difficult to design the control loop of a buck regulator for wildly varying currents. As you turn down the current, the converter will go into DCM mode which radically changes the voltage output vs duty cycle curve from linear to something pretty weird. This will mess up the transient response of your voltage regulator, since the controller will have to deal with a weird condition. A variable output voltage also changes the closed loop behavior. If you don't care about power efficiency, it might be a good idea to sink a fake load just so that the buck converter has a constant current to worry about. (I just made that last sentence up so I have no idea if that's what people actually do or not in these situations)Or, they probably design controllers that handle this condition explicitly. Also, I think a synchronous buck converter wouldn't have this problem. It might be a good idea to use one anyway since this is high power.I'm sure if you search for buck converter CCM and DCM control you'll find a lot of information about it.source: sat near people designing 100A GPU power supplies

author: NiHaoMike
author_rank: Super Contributor
timestamp: August 17, 2011, 01:38:08 am
content: Hysteresis mode control is very easy to implement and would work properly whether you're running in DCM or CCM. The main disadvantage is output ripple, but a linear post regulator can fix that. That also fixes the problem of going all the way down to 0V, which would require an impractically low duty cycle for an output just above 0V. Just have the buck regulate 300mV or so above the output or some value like 1V, whichever is higher.You can modify the ATX power supply to vary the output voltage so your buck converter wouldn't have to operate over such a wide ratio. Every one I have encountered will go up to 14.4V (charging voltage of a car battery) quite easily. They should also be able to go down to 8V or less quite easily.

author: Zero999
author_rank: Super Contributor
timestamp: August 17, 2011, 04:56:36 pm
content: Quote from: gxti on August 16, 2011, 03:53:09 amWhy not use a SEPIC or forard converter so the input voltage can be above or below the input voltage?

author: gxti
author_rank: Frequent Contributor
timestamp: August 17, 2011, 10:24:24 pm
content: Because those are more complicated. I'm trying to minimize the chances of making a dud. Also, SEPIC seems like a poor choice for a 10A supply due to the series capacitor.

author: NiHaoMike
author_rank: Super Contributor
timestamp: August 19, 2011, 04:38:50 am
content: A SEPIC (one with a coupled inductor, at least) is actually more or less a flyback converter with a 1:1 ratio transformer and a coupling capacitor that also acts as a "lossless" snubber.

author: Zero999
author_rank: Super Contributor
timestamp: August 19, 2011, 06:22:41 pm
content: 10A seems a bit high for a fly-back.I think a forward or push-pull converter are probably the best options.