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Joe Dunfee
11-09-2006, 08:33 PM
A project to control a Roomba vacuum with a basic stamp, will be powered by the Roomba's 14 volt battery. However, the Stamp 2e Motherboard manual states that it can be powered by 6 to 9 volts...obviously out of range for my battery.

I started to investigate replacing the regulator on the I/O daughterboard.··But, when I·looked up the existing regulator which was a LM2940, SOT-223 package, the datasheet on that seems to allow up to 26 volts. It has been many years since my electronics schooling, so forgive my ignorance.· I know it is more complicated than just getting the voltage in range.·

A voltage regulator of this type (if I recall correctly) will act as a variable resistor and all of the voltage drop is experienced at the regulator... so 14 volt input - 5 volt output = 9 volt. If I allow for power consumption of 0.1 Amps, then this works out to 0.1 watts. This seems to be well within the rated capacity.

So, the big question is if I can safely ignore the 9 volt limit to to the unregulated input·on both the motherboard or daughterboard (I don't know my total power consumption yet, but even if it is 2 or 3 times greater, I seem to be in range)?

If not, what would you recommend?· Perhaps a DC/DC converter?

Joe Dunfee

Lee Harker
11-09-2006, 10:02 PM
Joe,
One thing to keep in mind is the power dissipated by the regulator. Each package type allows a finite amount of heat to be dissipated safely. You could use a larger pre-regulator to get the voltage in the proper range. Maybe something in a TO220 package would do.

Lee

allanlane5
11-09-2006, 10:18 PM
Wow, you know enough to check the power dissappation. Well done. Yes, this should work.

The limit is so that people won't put a '26 volt, unregulated wall-wart' on the part -- because most people don't understand the power dissapation limit issue, AND most people don't understand a 26-volt unregulated power supply actually puts out more than 30 volts into a small load. Plus, people have a tendency to try to drive external loads with the on-module regulator, which raises the current and necessary power dissapation. If you're not doing any of these things (and it sounds like you're not) then you should be ok with 14 volts.

Most BS2 installations use an external regulator, because most installations are driving servo's, or relays, or some other power hungry devices. But it sounds like you've done the appropriate analysis for your voltage regulator, so it should be ok.

In theory, the part should go into thermal shut-down before any permanent damage is done anyway. Though I have read a post from somebody where the regulator got so hot the solder melted, and it 'slid' off the module. Sounds impossible to me.

Edit:· Assuming you use 10 mA for the BS2e, the power dissapation needed on the regulator is 9 volts * 10 mA (P == I*V) == 90 mWatts.· I'm not sure WHAT the power is that the BS2e needs just by itself, but that IS cruising kind of close to the limit, no?· Should work, but might run the regulator 'hot'.


Post Edited (allanlane5) : 11/9/2006 3:23:24 PM GMT

Joe Dunfee
11-10-2006, 12:26 AM
allanlane5, Thank you for the detailed reply.

The thing I am confused about from your post,·is the power limits on the regulator.

allanlane5 said...
Assuming you use 10 mA for the BS2e, the power dissapation needed on the regulator is 9 volts * 10 mA (P == I*V) == 90 mWatts. I'm not sure WHAT the power is that the BS2e needs just by itself, but that IS cruising kind of close to the limit, no?
My calculation for my project was 100 mA, requiring the regulator to dissapate 900 mWatts. In the quote above, you calculated power for 10 mA which was 90 mWatts... so we seem to be in agreement with the calculations.· But, you called that 10 mA "crusing close to the limit".· Perhaps my 100mA is close to the limit.· Here is a link to the data sheet.

http://www.national.com/ds/LM/LM2940.pdf

Still, it wouldn't be to hard to just put in an external regulator.· One person on the roomba forum said

I'd suggest a switching regulator instead of a linear one to avoid wasting current. TI Power Solutions (available through DigiKey) makes a nice one that is pin-for-pin compatable with the old '7805 but with eighty-something percent efficiency
But, I can't find the one he suggested or even one similar.

Any more ideas?

Joe Dunfee

allanlane5
11-10-2006, 01:06 AM
Very good point. I hadn't looked at the actual PDF -- thanks for the link. The Power Dissapation graph on page 11 does seem to indicate that with 1 square inch of copper for heat dissappation, the tiny package on the BS2e could disappate 1000 down to 500 mWatts of heat (depending on ambient temperature). I doubt the BS2e HAS 1 square inch of copper for that purpose, though.

I used 10 mA, because Parallax recommends a limit of 50 mA (in addition to the module's current requirements) and I just wanted a 'rule-of-thumb' estimate. 100 mA is kind of high, and then when you factor in the 14 volt supply (9-volt drop) that's on or over the bleeding edge. About 1 watt of heat is a LOT for that small device to dissappate, especially in the 'tight' space it has on the BS2e module.

An external, TO-220 device will give you a design with a lot more 'room for growth'. On the good side, if you DO try to disappate "too much" heat, the device will go into thermal shutdown until it cools. So you can try it without permanently destroying anything, and see how hot the regulator gets.

metron9
11-10-2006, 03:42 AM
Another thing when using unregulated wall warts to pick up on what Allanlane5 said about Voltage.
Of course I lost the source of the information again but as I remember it

The higher the mA rateing on an unregulated supply the higher the voltage will be this is because the voltage is rated with a load the size of the Ma Rating.

So If you have a 2Amp 6Volt supply, some cheap ones that I have measure 9Volts with no load but you have to draw 2 amps for the voltage to drop to the rated 6V. A 500ma 6V wall wart would have much lower voltages for a load of much less. So , the capacity of the power supply in an unregulated unit has a direct influence on voltage output at low mA loads.

Some wall warts, typically the ones that are battery rechargers put out only a half wave.

So it would seem if you are using an unregulated wall supply to supply a regulator and you are dropping to a lower voltage, a supply with a lower mA rating would require less voltage drop and produce less heat than a higher rated supply.

I realize the OP is using batteries but the above is something I just ran across yesterday. I thought I read it on the MPJA.com site but I can't find it there.

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