This one?
1Aerosemi Technology Co., Ltd
MT3608
AEROSEMI
High Efficiency 1.2MHz
2A Step Up Converter
FEATURES
• Integrated 80mΩ Power MOSFET
• 2V to 24V Input Voltage
• 1.2MHz Fixed Switching Frequency • Internal 4A Switch Current Limit
Why? Where does 1A come from?
E = 0.5 C V^2 = 4J in 2 sec = 2W
So not 20W?
The 2 * 10mF caps with a 20ohm series res has a time constant of 0.4 sec and will be fully charged (from a zero impedance 20V supply) in 2 seconds.
With a 1k resistor the circuit could not proivide the 40mA needed by the load.
Does anyone do the sums?
Its already been said. A bridge rectifier on the generator, smooth it with a capacitor, and a series resistor ( maybe 20 ohm 2W) to limit the inrush current to the caps.
It's already been said. That doesn't work for some reason linked to the unique nature of the mystery power source used.
What's also been said:
But that would have been too obvious, wouldn't it?
This is one of those cases where a real solution might be straightforward, but the option is kept out of reach by withholding application information. It's one of those problems that doesn't "want" to be solved. So I think what we should do in this case is offer emotional support instead. "There, there, OP. We feel for you. Yes, it's a challenge. Life is hard sometimes!"
I was thinking in something like this:
The idea would be that the load gets always the needed power, and the remaining small current charges slowly the caps.
It doesn't matter if the caps charge slowly, as far as the load gets enough current meanwhile.
And what about the voltage doubler? I think that it could work in this case. The big capacitors charge control would be the same, tough.
Well, I said above, adjusting the resistor to have a current small enough to keep the load working. I don't know which concrete resistor value, but I suppose it can be set high enough. Actually the margin seems to be very small.
I don't know the source total current, I don't know the load max or average current... this something to check and test by the OP.
I understand, but I guess as soon as there is not power, energy will come from the capacitors through the resistor and I would like to calculate the energy loss.
I just realized the diode direction now
It will certainly be a brilliant solution. It's a pity I can't apply it, the negative cycle is being used by the generator itself for its own stuff and I can't “steal” single mW so I have to keep only the positive part.
Please take a look at the edit, I have calculated about 215mW but I am not sure if this test is accurate enough or I am missing something, let me know what do you think.
The load will be permanet with a maximum requirement of 132mW (including a safety margin for regulator inefficiency etc..)
Yes, exactly
I'm sorry but I must have made a mistake somewhere, In order to fix it, where have you been interpreted that there are 20w available?
I saw you coming in your first post and I wasn't wrong.
My concern was always to limit the output of the booster, I also indicated in my original post that there might be other better solutions but I had to stick to the current one as much as possible.
If you have come to the conclusion that the solution is to change a wonky 200mW AC mystery power source for a 20A stable DC power supply.... Congratulations
EDIT: sorry, I didn't realize of the mW. I correct the calculations:
If you measurements are correct and based on this arrangement:
to charge the caps you have maximum left of: 215mW - 132mW = 83mW , right?
At 20V => 83mW / 20V = 4.15mA
In the worst case, when the caps are empty, you will have a voltage drop of 20V across the resistor. To limit the current to 4.15mA you need:
20V / 4.15mA = 4.8KΩ.
With that resistor, the approximate time to charge the caps will be:
~5 times R*C = 5 * 4.8KΩ * 20mF = 480 seconds (from zero to 20V)
This is quite a lot, you margin is small.
Maybe it will also work with an smaller resistor, because the caps will be empty (worst case) only the first time. Usually the voltage drop will be less and they will take less current.
And you mentioned that the load survives about 30 seconds with the caps, so probably you can also charge them in about 60 seconds.
Maybe you should just move the large caps before the boost, not after...
Maybe with a small resistor to prevent other things on the same transformer get into trouble...
Als 20 mF is a bit much (certainly after the boost). 20 microF is a bit too little...
Thank you so much for the detailed reply, I greatly appreciate it, I'm going to test it and update any news
The sole purpose of the boost is to maximize the energy that the capacitors will store.
If I put even more capacity before the boost than after, then it makes no sense and it is better to remove the boost instead.
A 1000 uF capacitor before the boost would guarantee that the boost can run continuously (under normal operation)....
The doubler already proposed may be more robust than a boost on half wave...
Maybe you should think of a supercapacitor...
The big caps after the boost will act as a dead short during startup (as explained before). You are lucky that the boost has a built in protection circuit. Otherwise it would be dead by now...
While waiting to receive the necessary components, I have done some preliminary testing and it seems to be working properly
Indeed, the capacitors take ages to be filled and it is a damn shame because the 5mA limitation means that most of the time they are being charged well below the available power.
I wonder if I could improve this with some sort of “variable resistor”. I have seen this video which I don't know if it could be a good alternative. Not sure about the power waste of this approach.
You could do something with a servo and a small signal MOSFET. Use a low-voltage opamp in combination with a shunt resistor to measure the cap charge current and then use that to control a PMOS that's in the charge path. It'd take some playing around in Spice to get something that works and that doesn't defeat the purpose (the latter in particular might be a challenge).
Actually that shouldn't be a problem. It will take a couple of minutes, but finally the caps will fully charge to 20V and stay there until the power source fails.
Anyway try a smaller resistor, probably it will still work but charge a bit faster.
Or maybe a constant current limiter with a TL431? Something like this, not tested:
(The return diode is not included)
I'm not sure if it would work, too hard for my last neuron to run the simulation
But maybe is bit overkill, if it could work just adjusting the single resistor.
Yes, that would be simpler and I imagine it should work. The drawback is that the TL431 needs 2.5V + overhead, while the servo approach does not necessarily need much overhead at all. The error amplifier of the servo could be fed off of a small secondary power supply powered by the mystery generator, but would draw so very little current that it shouldn't present a problem in that sense. Decent startup behavior would require some thought.
But if I understand it right, in this arrangement the 2.5V is the drop accross R2, to sense the current. That's why I choose R2 = 600Ω, to have constant 4.15mA all the way.
If we want a maximum of 4.15 mA, then:
2.5V / 4.14mA = 602 Ω
If the current goes up, the transistor will reduce it. If it goes down, the transistor will conduct more.
At the end the transistor will be like a variable resistor to keep the current constant at around 4.15 mA. It means that it will charge faster than with the fix resistor.
So what will the max voltage be that the two caps at the bottom will charge up to, assuming 20V in from the top? Also assuming that it'll start in the first place.
