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voltage regulator with crazy specs: drop max 50mV, Vout = 2V, startup at 0.7V
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magic:
Even disregarding that diode is a 2 terminal device, how exactly is it supposed to keep the FETs driven for long when both Vin+ and Vout+ are at Vin-?
Or is it just ideal "for long enough for everyone" :)
tatus1969:

--- Quote from: magic on August 16, 2019, 06:24:07 am ---Even disregarding that diode is a 2 terminal device, how exactly is it supposed to keep the FETs driven for long when both Vin+ and Vout+ are at Vin-?
Or is it just ideal "for long enough for everyone" :)

--- End quote ---
I was waiting for this question, already wondering why it took so long  ;) I use a bootstrapping technique here. And that's why I need to be able to generate 5V from such a low voltage. The circuit draws its supply from the diode drop across the MOSFETs while they are off (output shorted -> input raises to 0.7V). Once the regulator has built up enough energy in the large electrolytic cap in the picture, it turns on the MOSFET and then lives off that charge as long as it can. The duty cycle is ~ 98%, so "long enough for the MOSFETs to not overheat".

Actually your side cut regarding the additional terminals gave me another idea. The current circuit stops bootstrapping as soon as the input voltage is high enough for permanent operation, but it could actually be a good idea to drop that function. When modifying it such that the voltage regulator is connected across  the MOSFET, instead of being connected to the input terminals, then this could actually become a real two-terminal (almost) ideal diode. I wonder if that has been done before.
magic:

--- Quote from: tatus1969 on August 16, 2019, 09:41:11 am ---I was waiting for this question, already wondering why it took so long  ;)
--- End quote ---
Because those who recognize that grey cylinder kinda knew the answer :D

A bit surprised that duty cycle is so low, I expected that one charge would suffice to maintain a FET for seconds if not longer. I guess the biggest power hog is whatever watchdog circuit that turns it off to prevent slow and painful transition through linear region. Or is it some protected gate rubbish? ;)
tatus1969:

--- Quote from: magic on August 16, 2019, 08:27:38 pm ---A bit surprised that duty cycle is so low, I expected that one charge would suffice to maintain a FET for seconds if not longer. I guess the biggest power hog is whatever watchdog circuit that turns it off to prevent slow and painful transition through linear region.

--- End quote ---
Exactly. I'm using an LTC4352 ideal diode controller here. That's the only one that I could find that works down to 0V. It uses a charge pump to drive a floating NMOS in the positive rail. Because of this it consumes a lot of current, around 1.5mA. And the boost converter at the same time operates at its absolute minimum input voltage, where it cannot deliver much current anymore for recharging. As soon as the intended ultracap at the output starts to charge, the situation quickly gets better. And at 0.35V at the output, the MCP16251 is already able to maintain regulation and the MOSFETs can stay conducting. Pretty amazing if you ask me  :)
tatus1969:
Got the prototype up and running - almost... It turns out that the massively paralleleled MOSFETs have a very low diode drop of only 0.65V, which is just below the startup voltage of the MCP16251 boost converter. But I've already found a better chip: TPS61202. This one is guaranteed to start into full load at 0.5V if I should believe the datasheet. Eval board ordered.

The good part is that the depletion MOSFETs work very well - I measure only 50mV of drop across them. Thanks again guys, this solution is so much better than my attempt with TVS/PTC.

Here are a few first impressions. The continuous output current is 50A without heatsink and 70A with, that was exactly my design target. The halogen bulbs are 20 x 50W but didn't run at full brightness during the test, roughly 550W total (8.3V and 65A).








(without heatsink at 65A)


(with heatsink at 65A)
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