Global Moderator
UK
Online
Brattain Member
Karma: 137
Posts: 19030
I don't think you connected the grounds, Dave.
|
 |
« Reply #15 on: January 15, 2012, 04:13:34 pm » |
You haven't told us what you're doing, so "full power" is pretty meaningless.
|
|
|
|
|
Logged
|
Pete, it's a fool looks for logic in the chambers of the human heart.
|
|
|
|
Offline
Jr. Member
Karma: 0
Posts: 79
|
 |
« Reply #16 on: January 15, 2012, 04:18:06 pm » |
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:?
|
|
|
|
|
Logged
|
|
|
|
|
UK
Offline
Edison Member
Karma: 42
Posts: 2200
What a host of balls she had seen: gaity, the brass buttons...
|
 |
« Reply #17 on: January 15, 2012, 04:22:41 pm » |
That depends on two factors:
1. The Ah (or mAh) of the battery. 2. 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.
|
|
|
|
|
Logged
|
|
|
|
|
|
|
|
|
Global Moderator
UK
Online
Brattain Member
Karma: 137
Posts: 19030
I don't think you connected the grounds, Dave.
|
 |
« Reply #20 on: January 15, 2012, 04:34:36 pm » |
i am making a robot Can we stop playing "20 Questions" and start putting numbers and specs to this vague hand-waving?
|
|
|
|
|
Logged
|
Pete, it's a fool looks for logic in the chambers of the human heart.
|
|
|
|
Offline
Jr. Member
Karma: 0
Posts: 79
|
 |
« Reply #21 on: January 15, 2012, 04:34:50 pm » |
wait so how do i use a 5v reglater?
|
|
|
|
|
Logged
|
|
|
|
|
UK
Offline
Edison Member
Karma: 42
Posts: 2200
What a host of balls she had seen: gaity, the brass buttons...
|
 |
« Reply #22 on: January 15, 2012, 04:44:13 pm » |
wait so how do i use a 5v reglater? The one I posted a link to - you provide the 9V at one end, and the 5V comes out the other. If you want to roll your own, then the first job is to look for a suitable switching regulator chip, then read the data sheet and build upon the example circuit you will find in there.
|
|
|
|
|
Logged
|
|
|
|
|
Offline
Jr. Member
Karma: 0
Posts: 79
|
 |
« Reply #23 on: January 15, 2012, 04:49:06 pm » |
but what should i do about later i wanna learn and what ones should i get?
|
|
|
|
|
Logged
|
|
|
|
|
Offline
Jr. Member
Karma: 0
Posts: 79
|
 |
« Reply #24 on: January 15, 2012, 04:49:44 pm » |
is this a good one? STMICROELECTRONICS - L4940V5 - IC, LDO VOLT REG, 5V, 1.5A, TO-220
|
|
|
|
|
Logged
|
|
|
|
|
UK
Offline
Edison Member
Karma: 42
Posts: 2200
What a host of balls she had seen: gaity, the brass buttons...
|
 |
« Reply #25 on: January 15, 2012, 04:53:32 pm » |
|
|
|
|
|
Logged
|
|
|
|
|
UK
Offline
Edison Member
Karma: 42
Posts: 2200
What a host of balls she had seen: gaity, the brass buttons...
|
 |
« Reply #26 on: January 15, 2012, 04:55:02 pm » |
Oh, and note that you can't just get a switching regulator and drop it into your supply. You need external components - typically a schottky diode, an inductor or two, some capacitors, resistors, etc....
|
|
|
|
|
Logged
|
|
|
|
|
Offline
Jr. Member
Karma: 0
Posts: 79
|
 |
« Reply #27 on: January 15, 2012, 04:58:32 pm » |
and all of this is just for making the battery last longer? O and what should i put for max, min, vout, lout?
|
|
|
|
|
Logged
|
|
|
|
|
UK
Offline
Tesla Member
Karma: 89
Posts: 6368
-
|
 |
« Reply #28 on: January 15, 2012, 05:02:51 pm » |
You're making a robot? With wheels and motors and arms and wotnot? How are you going to power it? What Voltage and Amperage supplies will it need? It's pretty likely that will use vastly more power than the little Arduino, so it would be sensible to use that as the source for your Arduino's regulator too. Maybe just using the Arduino's existing on-board regulator, if the numbers come out right.
|
|
|
|
|
Logged
|
|
|
|
|
Offline
Jr. Member
Karma: 0
Posts: 79
|
 |
« Reply #29 on: January 15, 2012, 05:10:01 pm » |
I wish there was a easy way like i am just geting more comfuesd.
|
|
|
|
|
Logged
|
|
|
|
|
|