# Extract 5W low voltage power from 3000W high voltage source

I have a solarpanel array that generates 600V, 5A (3kW). I want to create a small electronic unit to measure the DC current and voltage and transmit this wireless. For this I have an electronic board with ADC's and transmitter. This board uses a maximum of 3.3V and 100mA. I want to use part of the power generated by the panels for this board.

I found some components like the LR8 from Supertex, but this is not supplying enough current and has a max of 400V. It seems like whatever I do to convert this voltage down to 3.3V, I loose a lot by heating of powertransistors or mosfets. As the board might be floating I wonder if there are more smart methods to power this board?

Any ideas would be very much appriciated!

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Please be careful with that 600V DC! It can easily fry more than just your ciruit if handled improperly. – JimmyB Jul 3 '14 at 12:44

Super wide-range buck converter based on the VIPer16
ST AN2872 - 500 VAC / 700 VDC input

Low−Cost 100 mA High−Voltage Buck and Buck−Boost Using NCP1052
ONSEMI NCP1052 - AN8098D
<= 700 V in

Pushing the limit - 600 V rated NXP AN11136 SSL2109T/AT/SSL2129AT controller for SSL applications

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Design Considerations for High Step-Down ratio Buck Regulators

You will have some voltage drop between the panel low voltage end and the battery or load. An energy harvesting IC or a custom boost converter running from a fraction of a volt would be able to provide the energy you need.

To produce 3v3 @ 100 mW = 330 mW assume you need 500 mW in.
At 5A, to produce 500 mW you need a potential of 0.5 W/5A = 0.1 V. That would be with all the drop being used by the converter. If you had 0.5V drop available you need 1A. The first essentially requires interrupting the feed cable and the second is still a substantial proportion of the 5A load current.
If you introduced a point voltage drop across eg a MOSFET it would be more controllable.
Say 0.5V at up to 5A available = 2.5W.
Series MOSFET, regulate to 0.5V drop.
Efficiency loss = 0.5V/600V = 0.08%

BUT

You say

• ... I want to use part of the power generated by the panels for this board.

• ... but I do not have the option to use a single panel for this.

• ... As the board might be floating

but you do not say why these conditions apply.
Giving a fuller definition of your problem would help.

You say that your target load is 3V3 x 100 mA = 330 mW.

Adding a small PV panel to power your system would be trivial.
Presumably there is a converter supplying mains and or battery.
Presumably using these is not acceptable. Saying why may help.

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Thanks for your ideas! I also had this idea of using a voltage drop in the wire, but how to do that was for me too difficult. The reason I want to use the panel power for the board is price, iT is a university project and I need it be as cheap as possible. Adding a separate solar panel is too expensive as the housing needs to be small as well. I can only access the cables that go to the inverter, too much text to explain.....sorry. The floating idea was what I meant by interrupting one line and put something in series with that. Hope this helps? – Enrico Jul 3 '14 at 19:36
If your transmitter send occasionally power drain is even lower that the 330 mW you state. A \$1 solar panel charging an eg portable phone battery (3 x NimH ~ AA shrink wrapped) should suffice. A few dollars. Cheaper would be hard. – Russell McMahon Jul 4 '14 at 15:10
I did not think of that solution, seems actually quite simple. Problem is only to find a 600V solar panel charger. I think I will go for this solution and try to work it out. Thank you so much!! – Enrico Jul 7 '14 at 6:59

There are parts that will easily withstand a 600VDC input. TI has a reference design of a converter that will handle 800V input and produce a 25V/0.25A (6W) output that could easily be used to create your 3.3V required power.

I do suggest that you consider shopping for such a converter on the market though- making the transformer will require some technique for safety as well as functionality and will not be very cost-effective for a one-off (sourcing custom magnetics in small quantity seldom is).

Note that the ratio of 600V:3.3V or even 24V is way beyond what you can reasonably expect to achieve without a transformer or tapped inductor (about 10:1 as a rule of thumb), and that probably means you need a custom magnetic component unless you want to do it in a bunch of stages.

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Thanks Spehro. I see that in this schematics the LM5022 is fed by directly by the Q100, is this possible only because the LM5022 usage very little power?\ – Enrico Jul 3 '14 at 14:14
It's fed for start-up only by Q100 (a 1kV MOSFET) which forms a kind of linear regulator as a source follower from the zener. Once the chip gets going, it feeds itself via D105 from the transformer. Otherwise Q100 would get very hot. If you tried to make a linear regulator to supply 3.3V at 100mA it would burn 60W. – Spehro Pefhany Jul 3 '14 at 14:20

I would use 2 power diodes in series , generating 1.2V, enough to run a small boost converter chip to say 3.3V to power the rest. max dissipation in the diodes is 5x1.2V=6W but that is nothing compared to the 3kW. this may work as soon as you draw more than 100mA off the solar array (60W)

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Note that the OP will need to draw at least as much current from the panels as the boost converter requires. Still, I like the solution for its safety and simplicity. Also note that diodes don't need to be rated at 600V, since reverse voltage would never be applied to them. – Dmitry Grigoryev Oct 26 '15 at 11:07

What you want is a buck converter. That is a type of switching power supply that makes a lower voltage out than in.

Since you only want to drive a microcontroller and a little electronics around it, you don't need super high effeciency. 100 mA at 3.3 V (330 mW) would be a lot for your circuit. Even if this overall draws 1 W from the solar panel, that shouldn't cause any trouble.

I'd keep this power supply simple, probably a pule on demand system. The tricky part will be controlling the high side switch accross 600 V. The high voltage will also greatly restrict the parts available for that switch. Arranging for something that produces a 500 ns pulse, for example, whenever the low side power voltage is below a threshold would be good enough. I wouldn't try to regulat the 3.3 V directly. Make something like 4 V minimum, then have a linear regulator make 3.3 V. That 4 V minimum might go up to 6 V at its peak right after a pulse, but again, your power levels are so low that efficiency really isn't a big issue here.

A 100 mΩ sense resistor can be used to measure the current, and a voltage divider to measure the voltage. The output of these would be measured by the micro, which would send on the information digitally, perhaps over a opto-isolator.

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Thanks Olin, I see many of these buck converter circuits, but they all need some lower (<100V) input voltage to control the oscillator circuit for buck conversion. I do not have this lower voltage available, only the very high voltage. Do you have any idea of how to make a solution for this? – Enrico Jul 3 '14 at 13:00
there are some linear regulators that work for really low powers, but 600VDC is quite the upper limit if I remember it right – Vladimir Cravero Jul 3 '14 at 13:37
@Vlad: I know I said effeciency isn't a big issue, but that was within the context of a buck converter. A linear regulator from 600 V would be so hugely inefficient as to be a problem. Even if the micro and surrounding circuitry only draws 10 mA, that would be 6 W dissipated by the regulator. That would definitely need a heat sink. – Olin Lathrop Jul 3 '14 at 13:58
@OlinLathrop yeah I know, that was about the "how to power the oscillator" question. – Vladimir Cravero Jul 3 '14 at 13:59

For powering your circuit you may look into the possibility to tap into the series-connected array. Usually individual panels have voltages much below the final system voltage, e.g. 12 or 36V. If you connect your circuit to only one panel you may have the power supply problem solved:

simulate this circuit – Schematic created using CircuitLab

Where "CIRCUIT" would be your circuit/power supply while the dashed box represents the whole array of panels (four in this case).

Measuring 5A should not be hard to do either, but someone else will have to hint you at viable solutions to safely measure those 600V.

Disclaimer: Your project is potentially dangerous to you or your environment. What I said above is my personal opinion and may be completely wrong. Proceed at your own risk, but do so very carefully.

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Thanks Hanno, sorry but I do not have the option to use a single panel for this. But I agree that would be a simple safe solution – Enrico Jul 3 '14 at 12:57
Another note: Have a look at the specs of the panel array. It may well have a nominal (MPP) voltage of 600V but a low-/no-load/idle voltage that's significantly higher. A 12V-nominal panel may well yield 18V with no load, which, extrapolated, would be 900V max on your array, and your circuit would have to deal with that too. – JimmyB Jul 3 '14 at 15:22
Yes you are right, the 600V is based on the open circuit voltages of the panels, so the absolute maximum without any load. I just have no access in this situation to separate modules, it is no option. Otherwise it would be simple... – Enrico Jul 3 '14 at 20:13

Would it be possible to generate 5W through induction off the main feedline?

Otherwise, why not go the simpler approach and add a separate, smaller panel, to generate the 5W needed for the electronics?

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How would you generate power from induction when there is only DC current in the wire? The smaller panel is simple but for this project not an option....sorry – Enrico Jul 3 '14 at 20:14