Best way to reduce voltage in higher wattage system?

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Conzo427

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Hello, amateur electronic engineer here. The issue is that I think I am overpowering my Peltier tiles. Here are the components that I have, everything is 12 volts.

4 small fans (connected to radiators for the Peltier tile cooling system): .2 amp each and each need the full 12v

1 water pump: 1.5 amp needing the full 12v

6 Peltier tiles: 5.8 amp, but would like to reduce voltage to the units.

1 200W 12v power supply.

I was planning on using a buck converter, but finding one to those specs has been a challenge, and the ones that I can find that can at least handle the amperage are rather expensive. Someone suggested using a few rectifiers in series. Any ideas on how to effectively reduce the voltage to the Peltier tiles without reducing the voltage to the fan/pump?
 
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6 Peltier tiles: 5.8 amp, but would like to reduce voltage to the units.

1 200W 12v power supply.

6 x 5.8A x 12V = 417W

I can definitely see problems there...

I'd suggest a separate small 12V supply for the fans etc., then modify a 12V 400W to variable voltage, just for the peltier modules.

Indo on modding a 12V PSU to variable here:
 
I think I need to reduce by about 2 volts to the Peltier tiles.

I am in fact running the Peltier tiles in parallel.
 
6 x 5.8A x 12V = 417W

I can definitely see problems there...

I'd suggest a separate small 12V supply for the fans etc., then modify a 12V 400W to variable voltage, just for the peltier modules.

Indo on modding a 12V PSU to variable here:

I considered this as an option, but when I ran the whole system it was only pushing 180w on my bench power supply. I figured with a power reduction to the peltiers I would be well under spec, but according to the math I needed a much larger power supply. Do you know why that would be?

I may just take your advise and call it a day with this project. Do you think that is the best solution with the info above?
 
I think I need to reduce by about 2 volts to the Peltier tiles.

I am in fact running the Peltier tiles in parallel.

Then just use 3 10A diodes in series with each Peltier which will give you ~ 2V
drop. Thats ~ 5W rating each diode. Would need them on heatsinks. You would have
to calc the size of heatsink needed.

Dropping like this is very inefficient but simple to do. Otherwise use a switching power
supply.


Regards, Dana.
 
(6 x 5.8A x 12v)+(4 x .2a x 12v)+(1.5a x 12v) = 444.6W
But the bench power supply was only pulling a total of 180w @12v for the whole system hence the 200 power supply. I think I'm going to order an adjustable 400w power supply for the Peltier tiles and dial it back a bit to 10v, and a seperate 20w for the water pump/fans. Do you agree that this is the best solution? I just feel wierd about dialing back a power supply just to add another power supply.
 
I'm not familiar with that. Those would boost the voltage and reduce the amperage though. Am I reading that right? I need to step it down, so 12v remains to the fans/water pump, and decrease it to the peltier tiles.

Thank you for your help!
 
You would take the main supply, which I gather would be set to 10V, and use this part to boost
back to 12V for the water pump etc.. You have to be careful to pick the boost, make sure it
can do it with 10V in to get 12 out. Just look at specs.

It is typically efficient, 80% or better, so overall power saved. You would divide the
power of all 12V loads by .8 and that would be the power you have to add needed for
the 10V supply for peltiers. So if peltiers consume 100W and 12V loads 20 W then you
would need 120W supply. And you would have to take into account its efficiency in
final calculation.

FYI, there are even, for many applications, buck/boost regulators which will handle
wide range inputs, like batteries discharging, etc.. For this design though you just need
boost.

Regards, Dana.
 
Example




The second example 336W is total energy need from 12 V source. The peltier supply =
200W / .7
 
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