From past experience this is right up your alley.Without more information the best you can hope for is a lucky guess at a possible solution. And a lot of unrelated guesses as well.
From past experience this is right up your alley.Without more information the best you can hope for is a lucky guess at a possible solution. And a lot of unrelated guesses as well.


I believe 4NR circuit will work very well with transistors which have threshold voltage about 1 mVCan anyone share some insights or a working schematic specifically on this 4NR circuit diagram approach please?
Piezoelectric source usually has big voltage - volts or even kilovolts (10 kV in lighter's piezoelement)The Voltage source is a Piezoelectric energy harvester or sensor generating low frequencies.
Load current @200 mV and 50 mW is 0.25A, so electrodynamic source can provide such big current.Its supplying 200mV to 400mV with low power about 50mW to 100mW.


Once again this active voltage doubling AC/DC converter is one of the approaches I looked at. So far the circuit again captures the input negative and positive cycles of 200mv alternating input voltage to doubled output at the 100-ohm load. AC input supplying once again ranges from 200mV at min to 400mV at maximum with low power about 50mW to 100mW. I want to add on a boost converter. I want to boost these 2x output voltages at the 100-ohm load to obtain a signal usable DC output forexample 5volts that charges a rechargeable battery. I appreciate the insights Mr Danko and the entire experts in general.I believe 4NR circuit will work very well with transistors which have threshold voltage about 1 mV
and Rds(ON) in few mΩ.
Piezoelectric source usually has big voltage - volts or even kilovolts (10 kV in lighter's piezoelement)
but very small current - microamps.
Load current @200 mV and 50 mW is 0.25A, so electrodynamic source can provide such big current.
Resistance of load is 200mV / 0.25A = 0.8 Ω.
For maximal efficiency series resistance of source's coil should be equal 0.8 Ω too.
Then minimal idle (without load) source voltage is 0.25A * (0.8 Ω + 0.8 Ω) = 400mV.
So, idle source voltage range is from 400 mV to 800 mV, load voltage changes from 200mV to 400 mV
and power in load from 50 mW to 200 mW (not to 100 mW).
Converter on picture below consumes 48.4 mW from electrodynamic source V1 and converts
200 mV AC to 6,53V DC @ 1kΩ load with efficiency 88%.
View attachment 232393
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ADDED:
Your case:
View attachment 232519


It is possible to charge the battery to around 3.5v or 5v if the double 0.400v is boosted with a boost converter first before it outputs to the rechargeable battery.How wil those opamps or comparators be powered?? And charging a 5 volt battery from a 0.400 volt source is going to need quite a special circuit. One more thing is that you can not get more power out than you put in.
Ask him, he's who your talking/posting with.How will the TS assemble this system