Converting from 30v-40v to a higher voltage to decrease power loss in the cable...

Thread Starter

alimash

Joined Oct 12, 2016
67
Powering a load situated more than 200 meters from the power source will result in huge power loss due to the wires resistance....
The load need 1.5A and the power source can deliver up to 3A max.
Designing a boost converter at the output of the source before connecting it to the wire to step up the voltage to for example 300v, then step it back down to 12v before connecting to the load by designing a buck converter.
Would that solve the problem and decrease the power loss in the wire?
I have attached a photo to clarify what i mean.
Thanks
 

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Ramussons

Joined May 3, 2013
1,084
It should work, subject to:
Conversion efficiency of 2 convertors,
Safety measures taken in handling 300 Volts,
Well insulated cable, water proof / Weather proof...

And protection against leakage....
 

Ian0

Joined Aug 7, 2020
3,463
A very simple push-pull circuit (SG3525) working at low frequency (<100Hz) using a mains transformer should work. Send the power as a square-wave, then use a similar transformer to turn it back to DC.
That way, you remove one set of rectifier losses and one set of switching losses.
 

LowQCab

Joined Nov 6, 2012
1,183
That's a ~100-Watt-Supply to a ~25-Watt-Load.

Why not just accept the Wire-Losses and use a Buck-Converter at the Load end ?

With ~16-gauge-Wire, or larger, the losses will probably be acceptable.
That way You don't have to be as concerned about High-Voltage-Insulation,
and You will only have one lossy-conversion,
( this will require a Big-Fat-Capacitor at the end of the run
to compensate for the additional Wire-Resistance ).

This will also result in less radiated Electrical-Noise from the Wiring.

What's the difference between a ~15% Power loss inside a Switching-Converter,
and a ~15% Power loss in the Wiring-Resistance ?

Ian gave a good solution also,
use 2-Transformers,
preferably Toroidal-Transformers.
But a big Capacitor and a single Buck-Converter
at the end of the run is less expensive.

.
.
.
 

Ian0

Joined Aug 7, 2020
3,463
If you use a split-primary transformer, you can connect the primary centretap to earth. That gives 240V down the cable, but with a maximum voltage of 120V on the insulation, like an American 240V split phase supply.

it’s all down to cost vs. Loss.
Power loss for a thicker cable, the loss of two toroids, or the losses of a buck and a boost regulator could all vary from about 5% to 30% depending on implementation.

What is the power source? Is it a power supply? A battery? A generator? Solar panels?
 

Thread Starter

alimash

Joined Oct 12, 2016
67
A very simple push-pull circuit (SG3525) working at low frequency (<100Hz) using a mains transformer should work. Send the power as a square-wave, then use a similar transformer to turn it back to DC.
That way, you remove one set of rectifier losses and one set of switching losses.
At first i thought about AC instead of DC since it is simpler to design and more cost effective , but what about the inductance and capacitance of the cable , we know that at high frequency the capacitance and the inductance increase which cause losses and and that is not the case in DC.
100 hz is low, but the loss will still be there.
 

Thread Starter

alimash

Joined Oct 12, 2016
67
That's a ~100-Watt-Supply to a ~25-Watt-Load.

Why not just accept the Wire-Losses and use a Buck-Converter at the Load end ?

With ~16-gauge-Wire, or larger, the losses will probably be acceptable.
That way You don't have to be as concerned about High-Voltage-Insulation,
and You will only have one lossy-conversion,
( this will require a Big-Fat-Capacitor at the end of the run
to compensate for the additional Wire-Resistance ).

This will also result in less radiated Electrical-Noise from the Wiring.

What's the difference between a ~15% Power loss inside a Switching-Converter,
and a ~15% Power loss in the Wiring-Resistance ?

Ian gave a good solution also,
use 2-Transformers,
preferably Toroidal-Transformers.
But a big Capacitor and a single Buck-Converter
at the end of the run is less expensive.

.
.
.
Actually, my intention is to charge a 12v 100AH lead acid battery which need 10A nominal current :(,i said 2A just for the sake of giving an example, i am know considering Ian's solution.
i think it is simpler since i can connect the output at the end of the cable directly to the battery charger.
 

Ian0

Joined Aug 7, 2020
3,463
At first i thought about AC instead of DC since it is simpler to design and more cost effective , but what about the inductance and capacitance of the cable , we know that at high frequency the capacitance and the inductance increase which cause losses and and that is not the case in DC.
100 hz is low, but the loss will still be there.
I think the principle has been tried before using 50Hz or 60Hz AC to transmit power!
Inductance and capacitance don't introduce loss, only resistance introduces loss. The waveform at the end might be less square and more sine at high enough frequency.
Using 2.5mm twin and earth as an example.
R=13.6mΩ/metre (round trip)
C≈100pF/m
L≈70nH/m
That gives a frequency response that's flat to 100kHz
 

Orson_Cart

Joined Jan 1, 2020
76
A very simple push-pull circuit (SG3525) working at low frequency (<100Hz) using a mains transformer should work. Send the power as a square-wave, then use a similar transformer to turn it back to DC.
That way, you remove one set of rectifier losses and one set of switching losses.
A good idea, look out for voltage spikes from ringing due to the long cable, capacitance in the Tx wdgs - and the square waves you are generating which have high edge frequencies - these will be seen by the rectifiers at the far end and will bump up the no load voltage to possibly 2x - so some sort of reg required.
 

BobTPH

Joined Jun 5, 2013
3,769
What is the power source? If you are running this off the mains, why not wire that to the location of the battery?

Bob
 

Ian0

Joined Aug 7, 2020
3,463
If you hadn't already bought the solar panel, I would have suggested buying several smaller ones (but with the same voltage) and connecting them in series. It would have worked out cheaper and more efficient than the converter.
 
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