12v or 5v - whats safer

I want to install small LED 'lamps' to each room to provide minimum lighting in the event of a power failure ( a regular thing in our country ). My 12vDC system runs off a UPS, so was thinking that using LEDs the best option due to the little current they draw.

I have found a local source of small 4 LED bulb lights ( typically used in cupboards, etc ) which supply enough light for me to see my way around in the dark. They are powered with 3 small AAA batteries ( 1.5v each ).

My thinking is to replace the batteries with an on/off switch connected to my 12v supply and a 7805 regulator to drop the power to 5v.

Which of the following would you consider the safer option ( in terms of wiring heat / potential damage / danger, etc ), considering the wiring will be running from my supply to each room through the ceiling :

A. run the 12v from supply to the room, then have a wall mount box containing the 7805 and on / off switch ( one box in each room ), or
B. have a single 7805 at the supply source location, and run 5v to each room with only the on / off switch the wall mount box ?

Also, for the 7805, I have read conflicting information :

  1. that the 7805 can be used as is to drop the 12v to 5v,
  2. that one should add 2 capacitors to the step-down circuit.

Which is the correct method ?

  1. that the 7805 can be used as is to drop the 12v to 5v,

Wrong

  1. that one should add 2 capacitors to the step-down circuit.

Right but not for the reason you quote, the capacitors are for circuit stability, they stop it oscillating.

There is no difference in safety, 12V and 5V are equally safe. Remember a child's train set has 12V exposed rails.

Local regulation however is better to minimise line volts drops.

Pesonally, I would connect the LEDs in string of 4 or 5 with a current limit resistor and run them from 12V if that is an option.

Thanks for the reply Grumpy_Mike.

Grumpy_Mike:

  1. that the 7805 can be used as is to drop the 12v to 5v,

Wrong

Are you saying the 7805 should not be used to drop 12v to 5v, or only when used with 2 capacitors in the circuit to stabilise it ?

Grumpy_Mike:
There is no difference in safety, 12V and 5V are equally safe. Remember a child's train set has 12V exposed rails.

Local regulation however is better to minimise line volts drops.

Very true - I have a 12v slot car track in the lounge as we speak.

"better to minimise line volts drops" - are you saying better to run the 12v and avoid the drop to 5v until the last possible point, or have only 1 7805 at the source ?

Sorry to ask, but your reply confused me a little.

CrossRoads:
Pesonally, I would connect the LEDs in string of 4 or 5 with a current limit resistor and run them from 12V if that is an option.

Hi CrossRoads

Thank You for the idea. I had considered that, but a few factors ruled that out :

  1. the LED units, although not too expensive, would become so if I wanted to add 3 or 4 per room.
  2. the additional light provided would not be required - 1 unit is enough.
  3. they would drain my UPS faster.
  4. would be more of an eye-sore in the room than a single 4 bulb LED unit.

Regards
Dave

DaveO: if you are using a UPS, try not to waste current while doing a voltage conversion. Depending on the current used, the 7805 will waste energy producing heat.

Are you saying the 7805 should not be used to drop 12v to 5v, or only when used with 2 capacitors in the circuit to stabilise it ?

only when used with 2 capacitors in the circuit to stabilise it.

are you saying better to run the 12v and avoid the drop to 5v until the last possible point,

Yes, that way any volts drop getting to the regulator is eliminated. If you regulate before your long run then you might not get 5V at the end.

If you only want 1 per room, I'd find the specs for the LEDs (foward voltage, recommended safe current) and feed it with 12volts through a series resistor of suitable value. LEDs aren't rated by voltage, using a suitable resistor you could run them off 200 volts if you chose to (not recommended). Personally I'd wire 3 in series with 12 volts and a smaller resistor (it would use the same amount of electricity for 3* the light . Buying small (5mm) high brightness LEDs is cheap in comparison to the effort involved in positioning and wiring them. They can be had for pennies apiece on ebay. Running 1 LED from 12V is no more wasteful than using a 7805 to step down the voltage.

Carefully drill a 5mm hole (so as not to damage the surface) where you want the LED and stick a single LED in the hole from the back and hold in place with blutack/tape/hot melt/whatever and its barely noticable unless its on and dark in the room.

pluggy:
If you only want 1 per room, I'd find the specs for the LEDs (foward voltage, recommended safe current) and feed it with 12volts through a series resistor of suitable value. LEDs aren't rated by voltage, using a suitable resistor you could run them off 200 volts if you chose to (not recommended). Personally I'd wire 3 in series with 12 volts and a smaller resistor (it would use the same amount of electricity for 3* the light . Buying small (5mm) high brightness LEDs is cheap in comparison to the effort involved in positioning and wiring them. They can be had for pennies apiece on ebay. Running 1 LED from 12V is no more wasteful than using a 7805 to step down the voltage.

Hi Pluggy

So you're saying it may use less ( or very similar ) power to have, say 5 LEDs 2v 20mA in series with a single 100 ohm resistor connected to a 12v supply, than to use a 7805 to drop the 12v to 5v and connect a single LED with a 150 ohm resistor connected to the 5v ?

Grumpy_Mike:

Are you saying the 7805 should not be used to drop 12v to 5v, or only when used with 2 capacitors in the circuit to stabilise it ?

only when used with 2 capacitors in the circuit to stabilise it.

are you saying better to run the 12v and avoid the drop to 5v until the last possible point,

Yes, that way any volts drop getting to the regulator is eliminated. If you regulate before your long run then you might not get 5V at the end.

Thanks again Grumpy_Mike.

Any links you can point me to for the correct diagram and capacitor selection ?

I hadn't thought about the loss over the long run. Good point there.

White LED's typically have a forward voltage of 3.5 volts so you can''t use more than 3 in series on 12V. If they are using the same current using 3 with use the same amount of juice as 1 using a series resistor, the higher value resistor to use 1 will waste more as heat. The 7805 will waste the stepdown voltage as heat.

http://cgi.ebay.co.uk/50-x-White-LED-5mm-Ultra-Bright-13000mcd-New-UK-/300536909231?pt=UK_BOI_Electrical_Components_Supplies_ET&hash=item45f9654daf

Using these (which are pretty typical) I'd use a single 70 ohm resistor with 3 in series or a 470 ohm if just using 1 with 12V. Using a 7805 is a waste of money.

Any links you can point me to for the correct diagram and capacitor selection

It depends on the exact type and manufacturer of the regulator. See the data sheet for the device you are using and that will tell you the minimum capacitor values to use.

Got several thoughts going on above. I'll only comment on one.

If you are looking to conserve energy, than 2-3 ultrabright white LEDS, capable of 15,000mCD on just 20mA, are the way to go.
Heck, just use 1 even, with a diffuser over it. I put 3 in a scoring machine in series from 12V, had to dial the current down to just 2-3 mA because it was so darn bright!
Someone in Australia posted about getting these recently for very little money, less than the 43 cents (USD) each that I paid for them.

CrossRoads:
If you are looking to conserve energy, than 2-3 ultrabright white LEDS, capable of 15,000mCD on just 20mA, are the way to go.

Hi CrossRoads

Are you talking about something like this :

So I don't waste the energy on the 7805, but instead use that energy on 3 of these connected in series, with a 56 ohm resistor ( 0.5 W ) onto the 12vDC supply ?

I assume that if I need to tone them down, I can just increase the resistor value ?

Yes and Yes.
Keep the current under 20mA tho - go with 82 ohm to start.