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Registered Member #3943
Joined: Sun Jun 12 2011, 05:24PM
Location: The Shire, UK
Posts: 552
Hi all,
I recently downloaded a calculator for the MC34063 DC-DC converter IC and have a question about the peak switch current in this type of converter.
I had always assumed that in a buck converter the input current is always less than the output current, but apparently this is not the case. Or at least partially.
Below I have entered 10v in and 5v 1 amp out, so that should give a maximum switch current of 500mA since that is equal to 5 watts. But the calculator tells me otherwise...
What causes this 2 amps peak switch current? I don't understand why any of the input current is larger than the actual output current.
Anyone who could explain this to me would be greatly appreciated. Thanks!
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
Yes, in a buck, or step-down converter, the mean input current is less than the output current. However, the input current is not steady, so you need to keep the peak current within the switch limitations.
When the switch goes on, you have 5v across the inductor, so the current will slew at 5v/21uH = 240kA/s. In half a cycle of your switching clock, 10uS, current will increase by 2.4 amps. The solution will be to increase the value of the inductor, or increase the switch clock frequency, so the input delta I is less, then your mean input current + swing will stay within 2A.
In fact even in the limit of a very large inductor, driving the ripple down to a very small value, the peak input and switch current will always be more than the output current.
Registered Member #3943
Joined: Sun Jun 12 2011, 05:24PM
Location: The Shire, UK
Posts: 552
Dr. Slack wrote ...
When the switch goes on, you have 5v across the inductor, so the current will slew at 5v/21uH = 240kA/s.
Thanks, I think I am starting to understand now.
I don't really know how best to describe this, but does the current get "added" to what is already flowing through the inductor in continuous mode since the current never drops to zero?
But in discontinuous mode the peak currents would be much lower because the inductor current returns to zero each cycle, meaning it would have to ramp up again when the switch turns on.
Dr. Slack wrote ...
In fact even in the limit of a very large inductor, driving the ripple down to a very small value, the peak input and switch current will always be more than the output current.
Does this mean that the power supply being used to power the buck converter also needs to be rated for the peak currents (transformer + bridge rectifier etc)?
Registered Member #72
Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659
There will be an FGC (flippin' great capacitor) between your high voltage low current supply and the buck convertor switch. This is what sources the peak inductor current. If you are worried about HF noise from your buck interfering with other circuits, then you are very careful about where the ground connection to this cap goes, as that terminal is pulsing with the inductor peak current as well. Generally commercial buck converter modules include this capacitor at the input, so that the module is not taking large pulses of current.
The best way to see what's happening is to model a simple buck convertor in one of the free available circuit simulators, and watch the waveforms play out. I suggest one of simetrix - proprietery, free version is crippleware, but totally adequate for this, and a very pretty interface LTSpice - proprietery, full version is free as in beer Qucs - FOSS, still the odd wrinkle to sort, drives like ADS
(actually the term FGC comes from mains power supplies, where you did need 1000s of uF to supply current for the mS between the half cycles. With high speed switcher operation, you only need a few uFs to supply the peak currents to prevent the input supply seeing them).
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Buck converter max switch current
