Best way to achieve 3.3v and 5v from 12v supply

Thread Starter

Jeremy Ulmer

Joined Jun 18, 2019
9
I have a project that needs a 3.3v regulator to run a esp-12e chip but needs 5v for the logic signal for an LED strip. My supply voltage is 12v. Is it better to drop the voltage to 5v and then wire the 3.3v regulator to that or run them both directly off the 12v supply. Which way would be better as far as dissipating the heat produced?
I have attached a picture of of both ways. Or is there a better way? Maybe a dual output regulator?

Thanks in advance for any advice!

Jeremy

power supply.png

Power Supply 2nd option.png
 

crutschow

Joined Mar 14, 2008
24,372
It makes no difference as far as total heat dissipation is concerned, as that is always the voltage drop times the current.
It does make a difference as to where the heat is dissipated.
Running the 3.3V regulator from the 5V will dissipate relatively more heat in the 5V regulator.

How much current for each voltage?
 

Thread Starter

Jeremy Ulmer

Joined Jun 18, 2019
9
Thanks for the response!
For 3.3v rail:
Looks like the ESP-12e uses about 1 amp
I have an ADC that uses about 10ma
So approximately 1,010ma for 3v3

For the 5v rail:
IR sensor that uses about 50ma
WS2811 LED that uses about 50-60ma
Also a 12v ws2815 LED strip data signal (not sure what current that will use, but I don't think it's too much)

I'm thinking it would be better to just run them separately and dissipate the heat through both of them than to put it all through the 5v regulator since the heat dissipated is the same.
 

AnalogKid

Joined Aug 1, 2013
8,308
With both regulators running directly off the 12 V, the dissipation is:
3.3 V -- 8.787 W
5.0 V -- 0.77 W

Driving the 3.3 V regulator with the 5.0 V output:
3.3 V -- 1.717 W
5.0 V -- 7.84 W

So the heat is more evenly distributed if the 3.3 V regulator input is 5.0 V rather than 12.0 V

ak
 

Thread Starter

Jeremy Ulmer

Joined Jun 18, 2019
9
With both regulators running directly off the 12 V, the dissipation is:
3.3 V -- 8.787 W
5.0 V -- 0.77 W

Driving the 3.3 V regulator with the 5.0 V output:
3.3 V -- 1.717 W
5.0 V -- 7.84 W

So the heat is more evenly distributed if the 3.3 V regulator input is 5.0 V rather than 12.0 V

ak
Wow, Thanks AnalogKid! I should have been able to do that math (P = IV source - P=IV regulator), but for some reason it went right past me! I appreciate you taking the time to work that out for me. Sometimes I tend to overthink things.
 

crutschow

Joined Mar 14, 2008
24,372
You might want to consider a switching regulator for he 3.3V output.
They typically have an efficiency of 85-90%, so the dissipation wouldn't be more than about 1/2W, (requiring no heatsink).
That also reduces the current load on the 12V supply since, for 1A out of the 3.3V switching regulator, the input current will only be about 0.3A from the 12V.

In that case the 5V regulator would only supply the 5V current and likely would also not need a heatsink.
 

Thread Starter

Jeremy Ulmer

Joined Jun 18, 2019
9
You might want to consider a switching regulator for he 3.3V output.
They typically have an efficiency of 85-90%, so the dissipation wouldn't be more than about 1/2W, (requiring no heatsink).
That also reduces the current load on the 12V supply since, for 1A out of the 3.3V switching regulator, the input current will only be about 0.3A from the 12V.

In that case the 5V regulator would only supply the 5V current and likely would also not need a heatsink.
OK. I will look into that. I don't think I have enough room on the circuit board for a switching regulator circuit, bit I will see if I can find something that would work. I would much rather do that if possible. Got some learning to do. I've never done a switching regulator yet.

Thanks again!
 

Wolframore

Joined Jan 21, 2019
1,649
You're going to have to figure something else out. The max current for AMS1117 is 1 A. http://www.advanced-monolithic.com/pdf/ds1117.pdf
It may work but hate designing without head room and safety factor.

8.7 watt dissipation isn't great either It will give you 15C/Watt rise over ambient temp. Approx 131 + 20 to 40... that's over 300F.. .

Is it possible to use a second regulator in parallel?
 

crutschow

Joined Mar 14, 2008
24,372
Here's an example of a fairly simple switching regulator that may fit on your board.
Certainly it should take less room than the heatsink for the linear regulator will require.
 

Thread Starter

Jeremy Ulmer

Joined Jun 18, 2019
9
You're going to have to figure something else out. The max current for AMS1117 is 1 A. http://www.advanced-monolithic.com/pdf/ds1117.pdf
It may work but hate designing without head room and safety factor.

8.7 watt dissipation isn't great either It will give you 15C/Watt rise over ambient temp. Approx 131 + 20 to 40... that's over 300F.. .

Is it possible to use a second regulator in parallel?
I agree. The regulator on the prototype that I designed was getting very hot. That was the reason for my original post. I was trying to see if there was a way to do this with linear regulators, but it's not looking good. I like designing with headroom as well.
 

Wolframore

Joined Jan 21, 2019
1,649
then @crutschow is correct, best solution is buck converter. More efficient just mind the EMI use quality inductor. ti has great resources on their website. Most of the work is done for you.

It is still possible to use parallel LDO regulators but at expense of efficiency and space.
 

Thread Starter

Jeremy Ulmer

Joined Jun 18, 2019
9
OK guys, thanks so much for all of the input! I have a lot to think about. I am VERY limited on space due to all of the things I am trying to pack into a very small enclosure. I'll take all of this into consideration and see what is the smallest footprint buck regulator I can find. I may see if I can stack a couple linear regulators in parallel to see if that cuts the heat down enough.
 

AnalogKid

Joined Aug 1, 2013
8,308
A small, pre-assembled non-isolated buck device is called a POL - Point Of Load - regulator. There are many that are quite small. A couple of companies make them in a faux-TO-220 assembly designed to drop into almost any circuit that uses a 78xx linear regulator.

Also, your output current is low enough that you might find regulator chips with the power MOSFETs built-in. Linear Technology (now part of Analog Devices) is big in this market segment.

ak
 

dendad

Joined Feb 20, 2016
3,322
A couple of the small switch mode boards off the 12V supply, one set for 5V, and the other 3.3V would be even better. They are VERY cheap, and quite small. If the existing regs are on a heat sink, I expect these boards would take up less room than that.
"Dimension: approx.17mm*11mm*3.8mm(L*W*H)"
They are available in fixed and adjustable versions.
 

mmcginty

Joined Feb 4, 2010
37
You're going to have to figure something else out. The max current for AMS1117 is 1 A. http://www.advanced-monolithic.com/pdf/ds1117.pdf
It may work but hate designing without head room and safety factor.

8.7 watt dissipation isn't great either It will give you 15C/Watt rise over ambient temp. Approx 131 + 20 to 40... that's over 300F.. .

Is it possible to use a second regulator in parallel?
The ESP8266-12 isn't going to draw nearly that much, it will hit momentary load peaks of ~700mA, while powering up radios and connecting to an AP, but the vast majority of the time it will only need ~300 mA when using the network and ~170 otherwise, even less if the network radios can be powered down.
 
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