Power regulation (trying to avoid smoke)

[do_i_dare]

Concept: Regulate 48vdc input to 13.5vdc output using 1 or 2 LM2596HVS step-down modules (no heat-sinks) to provide at least 2.0A constant current out @12v.

Problem: Probability of smoke -- Unknown to this novice.

LM2596HVS module specs: 5-55vdc in, 3-48v out @2.4A (3A max with HS) ~ probably overstated
Output voltage adjustable but current adjment is unknown (has 2 multi turn pots but 2nd one may only be for charging threshold LED)
Can be connected in series for better efficiency - Is this true?, does it matter?
No external components required.

Use: Power source and battery charger.

Constraints: Limited space for components, no room for heat-sinks, limited ventilation.

Other: Minimum wire gauge?
Input power switch min. rating?

Info and recommendations are highly appreciated,

Thanks

 

[Pommie]

The TI datasheet states they are rated at 1A. Another datasheet states they are rated at 3A but max 37V input. I'v used the 2A modules of Ebay without problems but again, they state 37V max input.

Mike.

 

[AnalogKid]

Can be connected in series for better efficiency - Is this true?

No.

For peak efficiency, use one regulator running close to its max rated output current. However, this is also when it will run the hottest. I recommend the tradeoff - for lowest operating temperature, use a larger regulator rated for 2x your peak current requirement. Running at "half power" the efficiency will be less than the max possible, but the temperature also will be lower than the max rating. #1 killer of electronics - heat.

ak

 

[Pommie]

Concept: Regulate 48vdc input to 13.5vdc output using 1 or 2 LM2596HVS step-down modules (no heat-sinks) to provide at least 2.0A constant current out @12v

Do you want 13.5V or a constant 12V or a constant 2A?

Mike.

 

[do_i_dare]

Thanks,
Looks like TI made 3 versions
I think the HV and HVS versions are rated at higher input Voltages.
I found this datasheet:
https://cms.nacsemi.com/content/AuthDatasheets/HTC/LM2596HV_R1.0.pdf
Supply Voltage
Vin 4.5 - 60 V
Load Current
LOAD - 3.0 A

If I'm wrong, what would you suggest for a better regulator module? Size is also a factor - there is only about 2.5 x 2.5 x 1 inches of space available.

Jim

 

[do_i_dare]

No.

For peak efficiency, use one regulator running close to its max rated output current. However, this is also when it will run the hottest. I recommend the tradeoff - for lowest operating temperature, use a larger regulator rated for 2x your peak current requirement. Running at "half power" the efficiency will be less than the max possible, but the temperature also will be lower than the max rating. #1 killer of electronics - heat.

ak

I see. I had thought running two modules in series at about 75% current each would help keep the heat down because I have a physical space limit and poor ventilation. If that won't work I'll have to find a different module per your suggestion. Any ideas?

 

[do_i_dare]

Do you want 13.5V or a constant 12V or a constant 2A?

Mike.

12V is the min. 2A is desired but as close as possible without overheating would have to do.

Thanks

 

[crutschow]

For peak efficiency, use one regulator running close to its max rated output current.

That's not generally true for two identical switching regulators running together.
The increase in efficiency with load at smaller loads is due to the fixed circuit losses (IC power, etc.) becoming a smaller proportion of the output load.
At some point this efficiency increase stops and then starts to decrease with load, mainly due to the increase in I²R losses.
This means running one regulator at near its maximum won't affect the fixed loses of the two t0gether and thus the overall efficiency due to these loses.

Actually running one at a higher load can reduce the overall efficiency of the two because its I²R losses in the MOSFET and inductor resistances will be higher than if the two ran at the same current, which would be the minimum loss and maximum overall efficiency point.

 

[do_i_dare]

Actually running one at a higher load can reduce the overall efficiency of the two because its I²R losses in the MOSFET and inductor resistances will be higher than if the two ran at the same current, which would be the minimum loss and maximum overall efficiency point.

Thanks for that, sorry I know so little about this stuff.
So, if I understand you, using two modules to regulate 55VIN (12V lead acid batt x 4) with module 1 vin/2=27.5VOUT >> mod. 2 vin/2=13.5VOUT should be close to max efficiency (& less heat) for the set? Or is using just one regulator with 4A+ max. current @13.5VOUT a better choice (per AK)?
All that, bearing in mind that the parts (+ wiring and a switch) will be in a confined space with no heat-sink(s) and little ventilation.

 

[tomizett]

That's not generally true for two identical switching regulators running together.

I'd imagine efficiency characteristics will vary between different device types - best check the datasheet for this part and see how to best approach peak efficiency.

I think limited ventilation could be your biggest problem - you'll have to do your best to get a conductive path to free air.

 

[do_i_dare]

I think limited ventilation could be your biggest problem

I thought that may be inviting smoke. Those modules (which I have) show a 40C rise @max current so I hoped running them at half that would reduce the heat to something manageable. Doesn't look promising. If I can't find a way to cool them I may have to re-think this project. Thanks.

 

[tomizett]

40 degC doesn't sound catastrophic... but that's in free air, right?
You could try to make an estimate of the thermal resistance of your enclosure by fitting a resistor and a thermometer (like a thermocouple) where the module will go. If you apply a known power to the resistor and measure the temperature rise, you can work out the thermal resistance. Then, of course, if you know your expected power throughput and efficiency at that power, you can calculate what the temperature rise of the converter module should be. All a bit approximate, but better than nothing.

If you provide some more details of the application, maybe a sketch, then maybe people more experienced than I will be able to make some suggestions.

Of course, with no ventilation you will never actually see the smoke, so why worry?

 

[do_i_dare]

Of course, with no ventilation you will never actually see the smoke, so why worry?

LOL! Why didn't I think of that? -- Prob. solved!

You did give me an idea though. I could breadboard the circuit, find or make a similarly sized means to enclose it and measure the heat rise under test conditions. Not precise, but maybe close enough.

I still might not see smoke, but I'd know if I started a fire!

Thank you!