From 24V to 5V DC voltage regulator

Hello,

I have a power supply of 24V, I use this to power my lights. I would like to add a breadboard Arduino UNO so I can switch the lights on or off using a light sensor, PIR sensor,...etc

I know that feeding the 24V into a voltage regulator of 5V will blow the thing up because I need a proper head sink, also to much power will be dissipated.

Now because the 5V will only be used to for powering the UNO and some sensors, can't I use a voltage divider with resistors and feed the resulting (for example) 9V in the voltage regulator?

Thanks!

In a word NO. Voltage dividers are totally useless as power supply systems - they are only suitable for providing minimal current into sense ports. You need a regulator device, linear chip such as a 7805 if you don't care about lack of efficiency and can make allowances for heat dissipation or one of the switching regulators that appear in most posts , cost very little and are extremely efficient (circa 94% or so)

probably best to get a 24v capable UBEC for efficient high current voltage reduction.

Another one...

i'm using a 7805 in one of my circuits and im getting interference issues i havent exactly been able to put my finger on it yet but could it be an advantage changing my regulator for something like this?

not cheap tho

You should use something like this

Have read here that current Uno regulators don't like being reverse biased. No first hand knowledge of that myself.
Perhaps get a 7.5V output, 9V, somewhere in there, and supply the barrel jack or the Vin header pin.

Reading the Uno's regulator datasheet, page 10:

It would seem to me that a 1N4001 diode from 5V to Vin would prevent any damage should Vin be unpowered and 5V driven directly from an external source:

"Protection Diodes
The NCP1117 family has two internal low impedance
diode paths that normally do not require protection when
used in the typical regulator applications. The first path
connects between Vout and Vin, and it can withstand a peak
surge current of about 15 A. Normal cycling of Vin cannot
generate a current surge of this magnitude. Only when Vin
is shorted or crowbarred to ground and Cout is greater than
50 F, it becomes possible for device damage to occur.
Under these conditions, diode D1 is required to protect the
device. The second path connects between Cadj and Vout, and
it can withstand a peak surge current of about 150 mA.
Protection diode D2 is required if the output is shorted or
crowbarred to ground and Cadj is greater than 1.0 F.

A combination of protection diodes D1 and D2 may be
required in the event that Vin is shorted to ground and Cadj
is greater than 50 F. The peak current capability stated for
the internal diodes are for a time of 100 s with a junction
temperature of 25°C. These values may vary and are to be
used as a general guide."

... Use a 1N5822 from 5V to ground anode to ground and cathode to 5V . There is already an input protector in the DCin jack, A series diode pointing to the regulator.
Here is My $0.02 worth on Switchers @ $1.70 Ea...
http://www.electrodragon.com/?product=better-than-lm2596-dc-dc-step-down-adjustable-power-supply-module

Always better Prices.

Doc

Docedison:
... Use a 1N5822 from 5V to ground anode to ground and cathode to 5V . There is already an input protector in the DCin jack, A series diode pointing to the regulator.
Here is My $0.02 worth on Switchers @ $1.70 Ea...
http://www.electrodragon.com/?product=better-than-lm2596-dc-dc-step-down-adjustable-power-supply-module

Always better Prices.

Doc

Yes, great price. But they do need to work on their usage of engineering units:

Output Current: 2V continuously, and peak value at 4V

Lefty

Docedison:
Here is My $0.02 worth on Switchers @ $1.70 Ea...
http://www.electrodragon.com/?product=better-than-lm2596-dc-dc-step-down-adjustable-power-supply-module
Always better Prices.

That $1.70 has magically turned into $3
I'm using an LM2596 module.
Only $0.99 on eBay.

maybe this schematics will help you i tried all of them and working. but best is MC34063

See schematic on PAGE-7 of datasheet

EVAL BOARD

EVAL BOARD SCHEMATIC

Has anyone ever used a 4 or 8 sensor board?

I have 8 PNP 24vdc sensors I need to step down.

Thanks

-Angel(Newbie)

READ THE DATASHEET.
The Max input voltage for an LM7805 is 18V, not 24V so the heatsink isn't going to do squat.
You could use a power dropping resistor to dissipate 24-7.5=16.5 V at 1.5A = 25W. (24.75)

Rdropping = (Vin - VLoad)/ ILoad = 11 ohms
PRdropping = IRdropping x VRdropping = 24.75 W

Would it be better to use a switching dc-dc converter (like the LM2596) from ebay ? (probably)
Do you have time to wait for that to arrive ? (I don't know)
Can you use the resistor until it arrives ? (yes)

My Arduino draws less than 100mA. So a 7809 voltage regulator can easily do the job. A small heat sink is not a bad thing, even if the 7809 has built in self protection.
Just make sure you use the plug, where the power supply would be attached (8-12 Volts input)

We Germans like the most efficient solution.

Lots of people simply want to show how smart they are.

We Germans like the most efficient solution.

I have a newsflash for you . The LM7809 has a MAX input voltage of 18V, NOT 24v.
The OP clearly stated that his input voltage is 24V. If your German you know that's > 18V.
And just for the record, the LM7809 is not the most efficient because the arduino doesn't need anything higher than 7V. if the OP had a lower input voltage , (like 18V) I would NOT have recommended the LM7809. I would have recommended the LM7808. If there was an LM7807 (like this one, I would recommend that. The arduino onboard 5V regulator is an LDO so 7V would be ideal, not 9V.

So there.

The max input voltage is 35 Volts - according to the Fairchild datasheet ( same for a lot of other manufacturers) for LM7805 thru 7018 and 40 Volts for LM7824.

And when you need less than 100mA - who cares if you waste a little bit of energy in an external voltage regulator or the regulator integrated in the Arduino.

OK, if you prefer a LM7808 ........ (what is the difference ?) - 100mW more load on the integrated regulator.

I am impressed and will change my way of thinking.

I found a datasheet by some other manufacturer.
It's best to post the datasheet when you make the statement to eliminate time spent trying to verify the statement.
I concede you are correct:

LM7805 FAIRCHILD

http://www.mouser.com/ds/2/149/LM7809A-186029.pdf

page 2

Vo = output voltage / the column "value" states the max. input voltage

i know, sometimes datasheets are not easy to read

If you are apply Vout + >=(5V to 7V) to Vin then the regulator is going to get really hot - better plan on having a good heatsink, that tab will get hot enough to blister your skin if very much current is drawn.
For example: Power dissipated as heat = (Vin - Vout) x current. (35V - 5V) x .1A = 3W. Hot!