I'm running a small microcontroller (12F683) to flash a infrared led.

I now need to step 9Vdc down to 5V for the microcontroller but see the 5V regulator I have can only manage 100mA. (L78L05ACZTR-Datasheet attached, it's a TO92 PACKAGE) My project draws 106mA.

I searched CPC (UK) but cannot find anything in a TO92 package that will do the job. I am limited for room so the TO92 style really helps.

Any suggestions on this would be appreciated, Thanks - JDR04

 

Hello,

When driving a regulator at its limits, the change of the thermal cut-out will be there.

The following will have a current limit of 500 mA.
Would you have room for a SOT223 SMD chip?
http://www2.mouser.com/ProductDetai...=sGAEpiMZZMuLLNXTG1MZam3vXed/oa05NmLTl0zbhpY=

Bertus

 

Just how many gigiwatts is this IR LED taking?

I understand you may need "more amps" but if you want a parts list then you need to quantify your needs.

There is a PQ050ES3MXPQ by Sharp Microelectronics rated for 300mA. Probably the highest you'll find (and I have not checked what conditions that limit applies).

 

If there is no schematic then you project does not exist. Post your schematic.

 

Hi ErnieM, I've got a IR LED thats been flashed at 38Khz with a duty cycle of 20% and 650mA.

The part you suggested looks ideal but the only concern I have is that it has been discontinued.

Any ideas will be appreciated-JDR04

 

Hi spinnaker, I have attached my proposed schematic. As you can see it use a minimum of parts. I also plan to incorporate a diode to prevent any damage through any polarity reversal.

Thanks- JDR04

 

Hi Bertus, thanks for your info.Could I use the same circuit as I have just posted for spinnaker to look at?

Thanks-JDR04

 

Hi spinnaker, I have attached my proposed schematic. As you can see it use a minimum of parts. I also plan to incorporate a diode to prevent any damage through any polarity reversal.

Thanks- JDR04

No the schematic of your whole project so members can detrmine if you are calculating your power needs correctly.


Also have you considered a daughter board for your PSU? While you might not have room on the board itself for a larger regulator, you could have room above the board.

 

This drawing shows 2 ways to steal a few extra ma. Bypass the regulator with a resistor that supplies part of the load or get rid of the 10 ma load in the adjusting circuit. It depends on whether you always have a load. I have some doubts about including the zener diode.

 

Assuming the 5V is fairly well regulated run the LED not from the 5V but from the 9V (with a series larger resistor).

That way the LED current doesn't go thru the regulator so you can use the one you seem to have.

 

I was wondering what sort of battery life I could expect if I wired the IR LED directly to the 9V battery? The reason I chose a 9V battery was to try and extend the battery life. Any comments??

 

I was wondering what sort of battery life I could expect if I wired the IR LED directly to the 9V battery? The reason I chose a 9V battery was to try and extend the battery life. Any comments??

Battery??? Forget a linear regulator then. Consider a buck regulator or just put together a combination of batteries where you won't need a regulator. 2 1.5v in series for example.

 

I can't see the circuit that runs the LED. If it has current limiting, the IR LED is safe and it will not run its current through the regulator, but...A 9V battery is one of the weaker ways to get 5 volts. The higher the voltage, the worse the waste (unless you're using a switching voltage regulator). Extra space is used up in the battery to get to 9 volts when it could be using that space to get you more amp-hours. Plenty of AA batteries will outperform a 9V battery in the current department.

Edit: I just looked it up, even a (AAA) battery (1.2 watt-hours each) can beat a 9V for amp-hours. A 9V battery contains about 4 watt hours. If you use a switcher, you can convert most of that power to your use.
Here's a switcher that might work for you.

 

I know you're running the LED at a 20% duty cycle, but for how long? Forever? Once an hour? Just when someone wants to change the TV station?

That plays into how efficient you want to be.

A 9 volt battery is not an efficient source of power as you need to throw much of it away! The PIC runs off 5V so you burn off 4V for that, and something similar for your LED. Two or three single cells would be a better match. Single cells also cost much less: a quality 9V battery runs me over 5 bucks.

It does get better if you use a switcher but that "feels" like overkill

 

Thanks for the switching regulator suggestion. So what I understand so far, I would be far better off in terms of battery life if I simply used 3 AA batteries.Correct?

If I wanted to be more efficient then i would incorporate a switching regulator? I've never used one before so could somebody please explain in novice terms what it actually does and why its a good thing in my circumstances.

The IR LED Transmitter I'm planning to leave in place permanantly on and would hope to get at least 48 hrs if possible?

Thanks a lot

 

Thanks ErnieM, I am hoping to leave the IR LED Transmitter in place for at least 48 hours if possible. If I could get more I would be well pleased. Its part of a breakbeam system I'm learning about and building.

Thanks for your advice JDR04

 

If I wanted to be more efficient then i would incorporate a switching regulator? I've never used one before so could somebody please explain in novice terms what it actually does and why its a good thing in my circumstances.
Thanks a lot

You will not need a regulator if you uses 3 - 1.5V batteries in series. Just search switching or buck regulator. There is far more out there than can be discussed here.