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Topic: ATTiny Power (Read 8964 times) previous topic - next topic

STDummy

Hi Guys

I have a question regarding powering an ATTiny85 that is being used to control a single white led.  The power source I have availabe for use is pair of CR2032 3v batteries connected in series supplying 6v.
Now, the maximum voltage rating of an ATTiny85 is listed as 5.5v, so my question is; do I need to worry about the extra 0.5v, ie will it cause the ATTiny85 to overheat and/or fry itself, or is it forgiving enough to take it?
If I do have to regulate the voltage down, what's the best method?  I could use a L78L05C, which I estimate would deliver 4.3v out for 6v input but it seems a bit it seems a bit "over the top" for dropping just a half volt.  What's the best (ie read cheaper) alternative do you think?

Bernie
If you learn by your mistakes then I am the model pupil

lemming

#1
Jan 16, 2013, 10:53 am Last Edit: Jan 16, 2013, 11:31 am by lemming Reason: 1
Run it on one cell rather than two (or for longer life two connected in parallel). The ATTiny85 will run down to 2.7 volts or 1.8 for the "V" series.

The clock may have to be lowered to something like 1 Mhz. (I assume you would be using the internal clock).
Scrap that; the standard version will run at 10Mhz at 2.7volts. You will be fine with the factory default of 8Mhz.

Have a look at this post for low power usage:

http://www.sparkfun.com/tutorials/309

STDummy

Thats a good idea Lemming, and thanks for the useful link - some useful stuff there. :)

The problem I have though, is that the ATTiny to be installed into an existing installation that already derives its power from the 6v. I don't have access to the batteries myself - I've programmed the chip and attached the led that it controls (without, as yet its resistor - its value will depend on the input voltage to the ATTiny) and will be sending it to the end user, but he just wants to attach it to the existing battery power supply and be up and running without modifying battery carriers etc.

Bernie
If you learn by your mistakes then I am the model pupil

lemming

You could try a switch mode regulator like this:

http://www.digikey.com.au/product-detail/en/R-783.3-0.5/945-1035-ND/2256215

rather than a linear one that would burn off the excess voltage as heat. Less efficient for those small batteries.

STDummy

Thanks again Lemming :)

I can only see one fly in the ointment using a DC-DC converter of that type (picky ain't I? :D).  Thats the input range of the R-781.0-0.5.  Thats the 5v converter I'd need, but its input voltage range is listed as 6.5v - 32v and I've only got 6v to work with. :~

Bernie
If you learn by your mistakes then I am the model pupil

Erni

Maybe you could use a diode (cheap) or two. That would give you a voltage drop of 0,7V per diode

fungus


The clock may have to be lowered to something like 1 Mhz. (I assume you would be using the internal clock).
Scrap that; the standard version will run at 10Mhz at 2.7volts. You will be fine with the factory default of 8Mhz.


It's a good idea to run at 1MHz anyway if it's running on batteries, it will use less power. Also turn off the parts of the chip you're not using, see the "PRR" (Power Reduction Register) in the power management section of the datasheet.

No, I don't answer questions sent in private messages (but I do accept thank-you notes...)

fungus


Maybe you could use a diode (cheap) or two. That would give you a voltage drop of 0,7V per diode


Yep, this is probably the cheapest/easiest way for a job like this. Voltage drops by 0.7V every time it passes through a normal silicon diode.
No, I don't answer questions sent in private messages (but I do accept thank-you notes...)

lemming

#8
Jan 16, 2013, 01:13 pm Last Edit: Jan 16, 2013, 01:17 pm by lemming Reason: 1
Quote
Thats the 5v converter converter I'd need..


I gave a link to a 3.3v one which would give you more than enough voltage to run the 85 while giving you more headroom for a lower input voltage.


Quote
Voltage drops by 0.7V every time it passes through a normal silicon diode.


Does the energy of that .7V drop get expended as heat like in a linear regulator? i.e. low efficiency?

Quote
It's a good idea to run at 1MHz anyway if it's running on batteries...


It looks like the factory default fuse settings are 8Mhz internal with divide by 8 enabled. You will be running at 1 Mhz without having to make any changes to the fuses.  




fungus

#9
Jan 16, 2013, 01:45 pm Last Edit: Jan 16, 2013, 01:50 pm by fungus Reason: 1

Quote
Voltage drops by 0.7V every time it passes through a normal silicon diode.


Does the energy of that .7V drop get expended as heat like in a linear regulator? i.e. low efficiency?


Yes, but I don't think there's any way around that. The DC converters mentioned above don't look as if they'll be efficient at these voltages/currents either.

Besides, a properly configured Tiny85 can run on less than 1mA. The power hog is the LED. Dropping the LED current by a couple of mA will make more difference than using one of those DC converters instead of a diode.
No, I don't answer questions sent in private messages (but I do accept thank-you notes...)

STDummy

All good advise guys - thanks.

I think I'll be going with a 1N4001 or two in the power line.  That'll give me 4.6 - 5.3v available for the chip and led/resistor with minimum of components.

Bernie
If you learn by your mistakes then I am the model pupil

Docedison

There is no heat generated in the diode except the normal resistive losses and the current must be fairly large to see any heat generated at all.

Bob
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fungus


There is no heat generated in the diode except the normal resistive losses...


The normal resistive losses are heat.
No, I don't answer questions sent in private messages (but I do accept thank-you notes...)

dc42

I would use a micropower regulator such as http://uk.farnell.com/microchip/mcp1702-3302e-to/ic-v-reg-ldo-250ma-to-92-3/dp/1331485. But a diode may be adequate if you don't mind the voltage dropping as the batteries run down.
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SirNickity

I can't seem to find any concrete reference specifying the typical diode loss as a resistive loss at all.  I was under the impression it's a carrier loss and is not represented by Ohm's Law, and therefore does not have a heat byproduct (other than natural resistive losses in any real-world conductive material).  Pedantic as it may be, anyone know for sure?

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