12v DC supply for 5v and 7v loads on PCB.... direction needed

Thread Starter

johnaustinkaty

Joined Jul 16, 2021
37
Just FYI, I am not an experienced electrical engineer/designer
I am designing a PCB that has several items using varying voltage and current: (supply is 12v DC)
My LED (3up cree high power Luxdrive LED) requires +/-7v @ 450mA
My DVR, Attiny85, LT1215 op amp and electret mic will use 5v (current varies of course but approx 40mA and less)
My board cam will use the 12v supply, so no prob there.

My question is, should I go with two bulky voltage regulators to get the voltage required, then limit current as needed using resistors? (5v & 7v)
I find that limiting 7v @ 450mA burns up (via voltage divider) my 1/4w resistors w/o using the bulky voltage regulator. For a PCB, what is the best approach to get my 7v and 5v supply as well as limiting the 7v to .450A and the 5v to around .04A?
I have looked at voltage and current dividers and the math just doesn't allow me to use 1/4w resistors.
Maybe I need to get BIG resistors? Please help, and thx, and yes, I am willing to pay someone to help
me finalize this (design/consulting). My PCB is designed in Eagle.
Thx in advance
-John
 

Ian0

Joined Aug 7, 2020
3,754
Don't the LEDs require a constant current drive? If so, there are plenty of buck circuits to run LEDs which will run off a 12V supply.
https://www.diodes.com/products/pow.../medium-voltage-dc-dc-led-drivers/#tab-finder
The LEDs will take about 3.5W. Your circuitry which uses 40mA will take about 0.5W if run via a linear regulator, or ~0.25W if run via a separate buck regulator. Is that 0.25W worth saving considering how much power the rest of it is using?
If not a 78L05 will do for the PCB.

You could run the LEDs from the 12V supply simply by using a 12Ω 3W resistor in series, but that wastes 3W (which turns into heat)
 

Tonyr1084

Joined Sep 24, 2015
6,452
limiting 7v @ 450mA burns up (via voltage divider) my 1/4w resistors
First - welcome to AAC.

Now, 7 volts at 450mA is 3.15 watts. A quarter watt resistor will turn into toast rather quickly. Also, using voltage divider circuits to attain the voltages you are looking for is not the way to go about it. First, the load alone will change the divider as adding a parallel resistance to another resistor will lower the total resistance of the two. Then your sought voltage will no longer be what you expected it to be. I used to make the same mistake back in the days when we actually had high school shop classes. I suppose they still do - but back then nobody (that I know of) knew anything about PWM as a voltage regulator.

The buck regulator LesJones recommends
The LEDs will take about 3.5W. Your circuitry which uses 40mA will take about 0.5W if run via a linear regulator, or ~0.25W if run via a separate buck regulator. (underline my own) Is that 0.25W worth saving considering how much power the rest of it is using?
is probably the best approach. [EDIT] This portion of my comment is being removed because the theory is fundamentally incorrect. So as to not cause confusion for others I'm voluntarily removing this portion. [END EDIT]
 
Last edited:

Ian0

Joined Aug 7, 2020
3,754
First - welcome to AAC.

Now, 7 volts at 450mA is 3.15 watts. A quarter watt resistor will turn into toast rather quickly. Also, using voltage divider circuits to attain the voltages you are looking for is not the way to go about it. First, the load alone will change the divider as adding a parallel resistance to another resistor will lower the total resistance of the two. Then your sought voltage will no longer be what you expected it to be. I used to make the same mistake back in the days when we actually had high school shop classes. I suppose they still do - but back then nobody (that I know of) knew anything about PWM as a voltage regulator.

The buck regulator LesJones recommends
is probably the best approach. You can use a LM78L05 for the 5V line, but that's turning 12 volts into 5 volts, which means converting 7 volts into wasted heat energy. Unless he has something in mind I'm not aware of. Maybe - and this is just quick guesswork on my part - an LM78L12 with its ground reference held at 5V with an LM78L05 MIGHT give you the two voltages you're looking for; but for sure I may be talking out my arse on this one. It's nothing I've ever thought to do. But maybe, just may-be, holding the 7812 ground pin at 5 volts might produce the results. But probably not. I'm 100% certain someone here will shoot this theory down rather quickly. But hey! What do you learn if you don't think outside the box every now-and-then.
A 7812 with its 0V pin at 5V would give you 17V - it maintains 12V between output and the 0V reference, much like the LM317 would give an output that is 1.2V more than the reference.
If you want 7V (and can't afford an LM317) then a 7805 with it 0V pin held at 2V would do it. If you don't need an accurate 7V then a green LED in the 0V connection to the 7805 would work.

But regulating the voltage to the LEDs is a non-starter unless there is some other current regulation going on.
 

LesJones

Joined Jan 8, 2017
3,570
An LM317 could be configured as a constant current regulator for 450 mA by connecting an 2.7 ohm resistor between it's output pin and it's adjust pin. the load (LEDs) would be connected between it's adjust pin and the negative supply rail.
There is a few things I am not clear about. 7 volts is an unusual voltage for an LED. Is this a package that contains two LEDs in series ? Also is the 450 mA LED current for each of the three LEDs or is it the total current of the three when connected in parallel ?

Les.
 

BobTPH

Joined Jun 5, 2013
4,032
Lots of conflicting and some incorrect info in this thread.

Listen to Ian0 in post #3. or LesJones in post #4. They are essentially saying the same thing.

You want to drive the LEDs with a constant current of 450mA. Putting them in series does this with just one driver, whereas 3 would be required if you did it in parallel. They will need a supply that can provide constant current or 450mA at 21V.

Bob
 

Thread Starter

johnaustinkaty

Joined Jul 16, 2021
37
Just to be clear, here's what I'm WONDERING if it's a viable solution:
View attachment 243811
All I know is you have a 12 volt DC source is from a battery, which is probably the case, but I don't know that for certain. You might have a 12 volt transformer rectified, which will typically produce, when filtered with the 10µF caps, will give you about 15.5 volts peak (after diode forward voltage drops). More information on your power source will prove beneficial in finding help for what you want to do. The circuit above is CLEARLY something I just made up in my dusty head. It might be totally bogus. I'm not an engineer either. Just an average joe who, from many years ago, learned electronics - and not deeply at that. I can manage the basics. So my advice needs to be looked at with a very critical eye towards scrutiny.
Your schematic shows what I was hoping to be able to do. I am using a 12v battery for Vcc. My LED is supplied by a PWM signal (fade on/fade off) from the attiny85 triggering a mosfet (7V regulated to drain and source to GND), which seems to be working very well.
If I go with your scenario, it appears that I won't be wasting any power as heat, or am I wrong about that?
I'm hoping others will chime in with confirmation and opinions.
Thx to all for the great input!!
P.s. I did find that the voltage divider scenario wouldn't work once a load was added.... lots of frustration there
 

Thread Starter

johnaustinkaty

Joined Jul 16, 2021
37
If I am using pwm and the LM7812 voltage regulator, will I still need to use this constant current regulator? Or, maybe I could use
an LM317 current regulator?
1626555369193.png



const curr regulator.PNG
 

Thread Starter

johnaustinkaty

Joined Jul 16, 2021
37
Lots of conflicting and some incorrect info in this thread.

Listen to Ian0 in post #3. or LesJones in post #4. They are essentially saying the same thing.

You want to drive the LEDs with a constant current of 450mA. Putting them in series does this with just one driver, whereas 3 would be required if you did it in parallel. They will need a supply that can provide constant current or 450mA at 21V.

Bob
Here is the LED I am using (3 LEDs in series on one board) Forward voltage is min 6.6v to 8.4v max
1626555965907.png
 

ElectricSpidey

Joined Dec 2, 2017
1,961
You don't see anything wrong with the circuit in post #6 ?

If I were you I would simply use the 12 volts to supply the LED with the proper dropping resistor, and just use a 7805 or a DC to DC converter for the micro.

If you feel you really need constant current then you could use the circuit you show in post #11 to drive the LED.

Really same difference power wise.
 

Thread Starter

johnaustinkaty

Joined Jul 16, 2021
37
You don't see anything wrong with the circuit in post #6 ?

If I were you I would simply use the 12 volts to supply the LED with the proper dropping resistor, and just use a 7805 or a DC to DC converter for the micro.

If you feel you really need constant current then you could use the circuit you show in post #11 to drive the LED.

Really same difference power wise.
Thanks for your reply. I wish I knew enough to see what is wrong in post #6. I would like to know though so I can avoid it in the future. I'm assuming it has to do with grounding?
I'll plan on using the LM7805 for my 5v needs. I will use 12v for the LEDs and regulate the current using an LM317 circuit.
Thx again!!!
 

ElectricSpidey

Joined Dec 2, 2017
1,961
Well be sure to test your circuit before you make a final build, PWM and constant current don't always get along.

Every time the MOSFET switches the regulator will try to adjust the output.

What is wrong with the circuit in post #6 is the fact that the farther you take the adjust/ground pin from ground the higher the output, not the other way around.
 

Thread Starter

johnaustinkaty

Joined Jul 16, 2021
37
Well be sure to test your circuit before you make a final build, PWM and constant current don't always get along.

Every time the MOSFET switches the regulator will try to adjust the output.

What is wrong with the circuit in post #6 is the fact that the farther you take the adjust/ground pin from ground the higher the output, not the other way around.
I will definitely do thorough testing (breadboard and oscilloscope).
It looks like I will need to use a 2.8 ohm resistor in my LM317 circuit to get 450mA at 12v. It looks like I will need a resistor that will handle .56w. Can I use a 1w, 2.8ohm(+/-) resistor? My amperage just needs to be around 450mA but can be as high as 900mA according to the LED's datasheet. I just don't need to go that high (bright).
 

ElectricSpidey

Joined Dec 2, 2017
1,961
You can use a 1 watt resistor, but I would go higher.

And be sure to properly heatsink that LED...all the constant current in the world can't protect an LED from poor thermal management.
 

Thread Starter

johnaustinkaty

Joined Jul 16, 2021
37
You can use a 1 watt resistor, but I would go higher.

And be sure to properly heatsink that LED...all the constant current in the world can't protect an LED from poor thermal management.
Yes, the LED is mounted in an aluminum holder that is also a heat sink. Thx for your help. I am routing my nets in Eagle now. I will post the ugly results when done... this PCB will only be 2.5"x1.75". Great results and much smaller than I had expected.

I appreciate everyone's input. Thank you.

-John
 

Thread Starter

johnaustinkaty

Joined Jul 16, 2021
37
I need someone who is very knowledgeable to analyze/check/optimize my circuit. As I have stated, I am not an electrical engineer or designer. I would like to have someone go through my design (Eagle schematic & board) to make sure it is correct.
The PCB will have several separate devices connected to it (mini DVR module, board camera, electret mic, PIR sensor, 12v jack and LED. I will need advice on connector part types and numbers also. I have datasheets on everything of course.
I would insist on paying for hours spent on this. I don't want to order a PCB until I verify that it is ready
Anyone interested?.
 
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