Here is a constant-current switch-mode regulator derived from Hero's reference. I show how it handles start-up, turn-off, and variation of input voltage from 12 to 15V. I set the output current to 1.7A. The current is set by the 0.33Ω shunt resistor and the Vbe of the 2N3904.
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No, it won't switch off fast enough without Q3 and D2. The current through R2 may only end up going to the load but the increased switching losses ruin the efficiency, try it yourself.Very nice Hero.
I have a couple of questions;
1. Did you really need Q3 / D2? Was there a problem getting the fet to turn off quick enough? I would have tried just reducing R2 to a couple of hundred ohms or less, and seen if it turned off fast enough. The current used for R2 only ends up going through the load anyway.
Increasing the inductor value doesn't make much difference to the peak current, with a 220µH inductor the peak current is 2.8A and switching frequency drops to just under 4kHz so it will emit a horrible noise.2. The inductor current ripple is much larger than I would like or expect, I think you should change L1 to 100uH or 220uH, there is no benefit to a very high switching speed and it causes more switching losses and more I2R losses on the inductor peak current.
The inductor current is best in a range of +/- 0.3 Iout, so from 1 amp to 2 amp would be much better, 1.2 amp to 1.8 amp better still.
I see why you had it set up the way it was before now.3. did you try putting another cap on the output? Putting the main cap there and reducing C3 should reduce current ripple quite a lot. The way you have it set up now slows the current feedback down a lot (hence larger current ripple).
It will always be a compromise between inductor size, switching frequency and current ripple.PS. If you like I can put your circuit up on the web page after you fix the inductor current.
Mike, your circuit has the advantage of being able to use the full PSU voltage to control the FET gate, with easier FET turn on/off and can operate with less buck headroom.
But all the regulator circuit current is to ground (ie wasted) and the regulator draws power to keep the FET off, so it will be inefficient at low duty cycles (like driving a single 3v LED from 24v) which is the time you need a buck design the most.
But, your circuit is definitely the best when there is only a small headroom, like say a solar 12v -> 3 LEDs where the solar battery might get as low as 11v driving 9v LEDs.
Ok here's my entry in the LED wars, I drafted the schematic in HandNpen 1.0.
I cheated and turned the buck upside down, so it can use a NFET which will be cheaper and have better performance etc...
What did you set the inductor's ESR to?
Look at a typical inductor of the desired value and current rating in a catalogue and set the ESR to the listed value.
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