You're absolutely right. Mother nature is willing to trade but gives nothing free.
The secret lies in storing energy in a field. You connect a 5V source to a coil, and the current will increase and build up a field which basically stores the energy (you can't make or destroy energy; it has to come from or go somewhere - Google 2nd law of thermodynamics. But I digress).
Then, through some fancy switches, you connect the coil to the output circuitry and the field is turned back into current again that stores voltage into a capacitor.
So, since energy cannot be destroyed or created, whatever the Watts you put into the field has to come out (minus some losses in ohmic wires etc; say you recover 80%, a realistic figure) in that output capacitor.
If the load on the output is less than the load represented by the circuit when you connected the 5V to the coil, the output cap will charge to a higher voltage, albeit with less current capacity, such that the output I * V will be around 80% of the input I * V.
So if your output is say 15V at 150mA, that's 2.25W. This is 80% of the input power which therefor was 2.7W, so the circuit draws from the 5V supply a current of about 560mA to provide 15V at 150mA. The losses (2.7W-2.25W = 0.45W) turn up as heat in the circuit.
Jan
PS Yes I know I switched from energy to power but I think the concept is clear;-)
The secret lies in storing energy in a field. You connect a 5V source to a coil, and the current will increase and build up a field which basically stores the energy (you can't make or destroy energy; it has to come from or go somewhere - Google 2nd law of thermodynamics. But I digress).
Then, through some fancy switches, you connect the coil to the output circuitry and the field is turned back into current again that stores voltage into a capacitor.
So, since energy cannot be destroyed or created, whatever the Watts you put into the field has to come out (minus some losses in ohmic wires etc; say you recover 80%, a realistic figure) in that output capacitor.
If the load on the output is less than the load represented by the circuit when you connected the 5V to the coil, the output cap will charge to a higher voltage, albeit with less current capacity, such that the output I * V will be around 80% of the input I * V.
So if your output is say 15V at 150mA, that's 2.25W. This is 80% of the input power which therefor was 2.7W, so the circuit draws from the 5V supply a current of about 560mA to provide 15V at 150mA. The losses (2.7W-2.25W = 0.45W) turn up as heat in the circuit.
Jan
PS Yes I know I switched from energy to power but I think the concept is clear;-)
Yes, but you need to feed it a bit more than 5V I think. I don't know the exact numbers, but the limit is the internal FET switch current that puts the energy into the field. IIRC that switch current is limited at 1A. So with 5V, the max you can get is say 80% of 5W which is 4W. Then at 15V that 4W will provide for max. 267mA - borderline.
Edit - the switch limit is 1.3A so with 5V in, 15V at 250mA seems OK.
But if you feed it with say 9V, the max input power is 9V * 1A * 80% is 7.2W which means you can extract like 480mA from the 15V output!
These are all ballpark figures but you get the idea.
Jan
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off to some kickstarting!
It's on Kickstarter: https://www.kickstarter.com/project...her-mains-free-low-noise-15v-and-5v?ref=email
Jan
Jan
