voltage regulator with crazy specs: drop max 50mV, Vout = 2V, startup at 0.7V

[tatus1969]

The output of this then goes to a low voltage boost converter: MCP16251. That's an awesome device that can generate a stable 5V supply from an input voltage as low as a diode drop. Actually, that's what the input practically is. But that input can also rise to 25V, which is way outside the MCP16251's max input voltage of 5.5V. That's why I need some form of pre-regulator (or a totally different approach to turn 0.7V into 5.0V).

The boost converter understandably doesn't allow a large voltage drop between that 0.7V supply and its own input. I checked, the circuit still but barely works with a 3 ohm resistor in series. That drastically reduces its output current, but the final 5V load only consumes a few hundred uA. The input current of the MCP16251 is around 15mA in this situation, resulting in a 75mV maximum drop. In the actual solution, I want to allow 50mV max.

So I am in need for a voltage regulator or limiter that doesn't exist. I thought of using a JFET but there are none with sufficiently low voltage drop at zero gate. My current way to go is to use a PTC fuse, followed by two strong diodes in series to GND. At small voltages, the PTC fuse is conducting and has little resistance, and as the voltage rises the diodes start conducting and eventually cause the fuse to heat up and increase its resistance. That works well but has two drawbacks: a) current consumption peaks at more than 0.5A at a certain input voltage, and b) PTC fuses are not really designed for repeated action.

Do you guys have any other and maybe better idea?

The application circuit will be a high current (50A) ideal diode that works with a minimum input and output voltage of 0V under full load. That allows it to use it to charge large ultracapacitors. Sounds impossible, but isn't 

[tatus1969]

Here is the current circuit of that voltage regulator (0.7V to 30V input, 5V 1mA output):

[MasterT]

There are high current /low Rds MOSFET called "Depletion" - normally ON, same as JFET.

[Alex Nikitin]

Here is the current circuit of that voltage regulator (0.7V to 30V input, 5V 1mA output):

If you only need a few mA current, something like the circuit below should do. The JFET should have a low enough resistance to provide the low drop voltage and the cut-off voltage less than the output voltage required. The CHP3910 available from ON Semi might be OK.

Cheers

Alex

P.S. - If you need about 3 Ohm you could put 5-6 CHP3910 in parallel, these JFETs are not expensive.

[schmitt trigger]

subscribing to interesting thread

[magic]

I suppose a depletion mode MOSFET could replace the JFET. These can have Rds(on) in single digit ohms or less.

Alternatively, simply cascode the boost's Vin with a depletion FET. If you are lucky, an hour of going through datasheets will find you something with sufficiently negative Vgs(off) that it will still conduct your load current with gate tied to ground and therefore Vgs=-0.7. If not, build the regulator posted by Alex or otherwise keep the gate a bit above ground.

[Alex Nikitin]

I suppose a depletion mode MOSFET could replace the JFET. These can have Rds(on) in single digit ohms or less.

Alternatively, simply cascode the boost's Vin with a depletion FET. If you are lucky, an hour of going through datasheets will find you something with sufficiently negative Vgs(off) that it will still conduct your load current with gate tied to ground and therefore Vgs=-0.7. If not, build the regulator posted by Alex or otherwise keep the gate a bit above ground.

Yes, actually the BSP149 from Infineon in place of JFET in the circuit I've posted should do the trick, it has ~3 Ohm resistance at 0V and < -2.1V cut-off voltage.

Cheers

Alex

[tatus1969]

The CHP3910 available from ON Semi might be OK. ... If you need about 3 Ohm you could put 5-6 CHP3910 in parallel, these JFETs are not expensive.

I had checked this one before, but the Vds/Id graph told me that I would have 0.5V of drop at Vgs=0 and Id=15mA. That's why I dropped that option, as I would need to parallel 10 of them to achieve 50mV. Thanks for the simple control circuit, I'll give that a try.

Yes, actually the BSP149 from Infineon in place of JFET in the circuit I've posted should do the trick, it has ~3 Ohm resistance at 0V and < -2.1V cut-off voltage.

I wasn't aware that these are available, that option sounds promising. Did a quick Digikey search and also found the BSS159. Three of these in parallel would yield 1.2 ohms at the same price of a single BSP149. Already put on my shopping list, thanks a lot!

[magic]

IXYS CPC3701 is 1Ω.

Perhaps add a zener/tvs to ground too because those things have capacitance. Or just a filter capacitor, nevermind.

[tatus1969]

IXYS CPC3701 is 1Ω.

Perhaps add a zener/tvs to ground too because those things have capacitance. Or just a filter capacitor, nevermind.

Hmmm, that one would be best but hasn't good availability (Mouser only).

There is 4.7u MLCC at the boost input which should be able to keep input transients away from that delicate chip.

[magic]

I found it at TME.

By the way, has anyone ever seen a P channel depletion MOSFET?

[tatus1969]

I found it at TME.

AFAIK that's not an authorized IXYS distributor, and my EMS won't buy from there either 

[David Hess]

I would consider the LM10 or more modern LT1635 as a control element.

[tatus1969]

I would consider the LM10 or more modern LT1635 as a control element.

Interesting component, but overdone in this case. The preregulator just needs to deliver "enough, but not too much" 

Working on the layout now, I'll post updates once it works.

[tatus1969]

Finished the layout and ordered a few sample PCBs. If (*if*) it works then this will be the first (?) ideal diode module that works with zero output voltage (e.g. a short, or an empty ultracapacitor)  Thanks for your help guys!

[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]

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"

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]

I was waiting for this question, already wondering why it took so long 

Because those who recognize that grey cylinder kinda knew the answer

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]

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.

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)

[magic]

Not bad.

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.

I thought about it and I think it hasn't been done before and I will tell you why: because then another ideal diode is needed to power the boost converter while protecting it from reverse bias. Kinda "you can have an ideal diode if you have an ideal diode" thing
Maybe there is some way to do it but probably not as simple as demanding ground connection.

[capt bullshot]

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.

http://www.ti.com/product/LM74610-Q1

These use an interesting kind of oscillator that starts at 20mV:
https://www.analog.com/media/en/technical-documentation/data-sheets/LTC3108.pdf

[tatus1969]

http://www.ti.com/product/LM74610-Q1

That's why I love this forum, someone always has the answer. That's exactly what I need, and it does everything that my current circuit does in a single chip. Startup at 0.48V is also low enough for my MOSFETs. Available, price okay, demo board ordered :-)

These use an interesting kind of oscillator that starts at 20mV:
https://www.analog.com/media/en/technical-documentation/data-sheets/LTC3108.pdf

I've come across these in the past and had forgot, but the price of these is prohibitively high. Still awesome technology.

[tatus1969]

http://www.ti.com/product/LM74610-Q1

That chip appears to have one caveat for my application, as it needs 'significant' reverse voltage before turning off the MOSFET. It's only 20mV, but this translates to 30A of reverse current. That would make the solution 'a bit' less than 'ideal'. I might still get away with that because that controller naturally duty cycles the MOSFET, and the period can be adjusted to maybe 100ms. This would reduce backfeeding duration, as many power supplies don't like that.