Will this work as a battery clip? For the MEGA 2560:http://www.amazon.com/Shipping-Camera-Battery-Adaptor-5-5mm/dp/B005EPSGY8/ref=sr_1_3?s=electronics&ie=UTF8&qid=1326654800&sr=1-3
Or can you link me to one for Arduino? On amazon.com?
I almost hesitate to ask, but why would you want to use a 9volt battery with a Mega?
Couldn't you ask your butler to find a clip?
What you mean i that i need one do you know one i can use thoD: all i want is an anser.
What you mean i that i need one do you know one i can use thoD: all i want is an anser
Again, with punctuation and English, please?
Ok. You can stop being a **** and just say something!
You could post a link to what you want.
And mind your language.
o wow all i seid is "****". And all i want to know is if this is for Arduino: http://www.amazon.com/Shipping-Camera-Battery-Adaptor-5-5mm/dp/B005EPSGY8/ref=sr_1_3?ie=UTF8&qid=1326658143&sr=8-3
Yes, it probably is.
But the question remains: can you really afford to waste all that expensive 9V battery heat?
What do you mean? how would i make my MEGA 2560 on the go then?
If you mean, "how would I power my Mega when away from a mains supply?" I would say "not with a 9V battery, unless you have deep pockets" (which is why I assumed you had a butler)
But, if you have deep pockets, sure go ahead.
If you don't, examine other options, like AA batteries.
o so sorry i that when you seid that that was a insalt
A 9v alkaline battery will only power a Mega for around 9 or 10 hours. Add other devices (LEDs, relays etc.) and the battery life will drop further. A 9v lithium battery will last about twice as long, but costs a lot more. I do use 9v batteries sometimes, but only for custom-built projects that run at low clock speeds and use other measures to reduce power requirements.
You want to build (or acquire) yourself a 5V switching regulator. These are far far more efficient than the linear regulator on the board.
A typical car cigarette lighter USB adapter would work well - rip the board out of that, power it from the 9V, and connect the USB out of the board to the USB in of the Mega. You'll find the battery lasts somewhat longer than connecting it direct to the DC in of the Mega.
how long will the 9 volt battery last if i am using full power?
You haven't told us what you're doing, so "full power" is pretty meaningless.
i am making a robot and for just testing to. But how do i make a 5V switching regulator? or like what should i doD:?
That depends on two factors:
- The Ah (or mAh) of the battery.
- The current draw of your device.
The battery will be able to supply a limited amount of current. This is measured in Ah or mAh - Ampere-hours or Milliampere-hours. If you draw this amount of current, the battery will be flat in an hour.
If you draw half the Ah rating of current, the battery will last about twice as long.
For example, a battery rated at 2200mAh (that's 2.2Ah) can supply 2.2A and it will be flat in an hour. If you draw just 1.1 amps from it you can expect about 2 hours from it. 0.55A (550mA - roughly what USB can supply) would last you about 4 hours.
If you use the on-board regulator, things get even worse. This is a "linear" regulator. The current you pull from it is the current it draws from the battery, but the voltage drops. That dropped voltage is converted into heat, so it gets hot.
Say you draw 0.5A from the battery. It's a 9V battery, with a 5V regulator. So, we are getting 5V times 0.5A, which is 2.5W of power. But, the regulator is drawing the same 0.5A but at 9V. That is a whopping 4.5W. Two whole watts of power go to waste. That's almost 50% of your power. So, you can expect your battery to last about half as long!
Use a switching regulator, and the efficiency goes up from around the 15% of a linear regulator, to say 85% or 90%. A switching regulator converts the excess voltage into current, so if you ask for 0.5A at 5V, the 9V gets stepped down (kind of like with a transformer) and the current gets stepped up. So, if we're drawing 2.5W, with a 90% efficient switching regulator, that would mean the regulator would be drawing 2.78W - considerably less. That would equate to 2.78W/9V = 0.3A.
That's quite a saving.
OK> thank you alot:D here i wanna buy this: Amazon.com
So it takes the 9 volts and it truns it into 5 volts? and saves the uther 4 volts?
No. That's a linear regulator.
Switching regulators are somewhat more complex and require external components.
You are best off (unless you are making a professional product) butchering a car USB adaptor like this: http://www.amazon.com/Cigarette-Lighter-Charger-Adapter-Sony-PSP/dp/B002LTWF8S/ref=sr_1_27?ie=UTF8&qid=1326663025&sr=8-27