RAVPower 10400mAh Li-on Battery Power Bank Test 10. By Matt on July 6, 2013 Hardware, Power. This week I tested a RAVPower 10400mAh Power Bank to see how long it could power a Pi running my standard battery test. RAVPower Power Bank (RP-PB07) The RP-PB07 (also known as Dynamo On-the-Go Power Bank) box contained the following items.
I'm thinking of installing my Pi down at my wife's stables which have a 40W solar charged 12V DC lighting circuit.
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Since the Pi runs at 5V I'm thinking that a quick and easy way to step down the voltage safely would be to use a microUSB mobile phone car charger such as this
Also I figure that the Pi should only be drawing about 2.5W (5V x 500mA) so the demand on the available power should be fine & the phone charger can cope with up to 850mA current so should be able to cope with a Pi.
I'll then use one of these for the display (12V so should be simple).
So the questions:
DaveGDaveG
6 Answers
Go to a hobby store (or DX.com) and buy a BEC (battery eliminator circuit) intended for model aircraft. $5 gets you a very efficient switch mode voltage regulator, good for around 3A. I use these with all my Pis.
Here is an example: http://dx.com/p/hobbywing-5v-6v-3a-switch-mode-ultimate-bec-ubec-15149
Christopher BiggsChristopher Biggs
The question seems to be inadequately answered. The best alternative would be a switching mode power supply with simple circuitry such as this:
Although 7805s work well, it's not scalable in the long run. Say you want to add a Camera Module or some other module. It'll be able to drive the circuit but cannot withstand high loads for a long time.
The data sheet of LM2596 says it can suply upto 3A. So, you'll be able to drive a few servos in addition to supplying the required power to the pi and that too at efficiences close to 80%!
You can find more info here
Community♦
user2155601user2155601
One thing to be aware of is that those chargers are not regulated, so any spikes, surges, etc will be passed on to the Pi. I would think that regulator circuit like the one below would be a safer long term solution than a car charger. For prototyping, testing etc, your car charger would be fine.
ButtersButters
They have a DC to DC converter here which is supposedly very efficient - You can find it here
user25704user25704
This is an old post I just stumbled on however. Lets get something straight,Switch mode power supplies are fine until you use them with an audio device (Radio, MP3 player etc.) Switch mode power supplies generate noise, lots of noise. In fact LF and HF radio are nearly impossible now because of them. It is easy to clean the Power supply output with a Zener, inductors and capacitors; also needed is possible ferrite cores on in and out leads and a screened case. Too much trouble?
Using a regulator (voltage dropper) is a better idea but the difference between input voltage (DC) and output voltage multiplied by the drawn input current will give you the power demanded. This power has to be dissipated as heat through a heat sink.
In the case of a Raspberry Pi being powered by some kind of regulator (switch mode or dropper), with up to 13V in and 5V out at up to 2.5A gives you an idea of the power to dissipate (up to 20 Watts). That is going to be a sizable heat sink. This would be fine in non-summer months in Northern Europe, I live 70Km south of the Sahara Desert and it gets down to 25'c in Winter - the rest of the year it destroys (Chinese) electronics.
The point I am making is...... Don't care about electrical noise? Use a switch mode with good ventilation.
Need Low noise.... (Good audio, HF radio, SDR etc.) Use a regulated voltage dropper and a BIG heat sink (The car chassis is good). Use this with decent capacitors on in and out legs and you will be getting somewhere.Just a note for the Hi-Fi enthusiasts.... Quality used to measured by weight... that is a huge, heavy torroidal transformer and well designed regulated circuit, will give you as near DC as you can get to supply the circuits. Use a (light weight) Switch mode power supply and you will spend more on cleaning the DC and screening the noise than on the rest of the circuit.
So back to your post, cheap Chinese 12V to 5v Cigar lighter power converters may only supply a maximum of 1A and the electrical noise could be a problem.Right, another point, Cable size. Rule of thumb I use: at 20'c, maximum 7A per mm squared. Run this on long distances and you will see a volt drop and heating of the cable. Very short runs (say, less than 1 metre) you could use 9A per mm squared with open cable (not with others... heat dissipation!). Beyond that you are looking at tri-rated cable and that would be absolute maximum of 13A per mm squared and that's emergencies only. So no you know why lighting circuits in houses have 1 mm squared cable and a 5A fuse (some use 6A now) and 2.5 mm squared for 13A (15Abreaker).
Cheap nasty 'hook-up' cable (0.5 mm square) and bad connections are not going to help with high current circuits especially in cars. Find your highest peak current and multiply it by 2 for you general cabling. This is also the same with capacitor rated voltages. Brand new electrolytic capacitors rated at 16V and connected across 12V at 20'c will probably just about do it. Raise the ambient and expect your capacitor to smooth big pulses will destroy it. Try using 25 or 35V - if you don't believe me ask Bose sound systems - they had major problems with capacitors with low ratings and bad dielectrics in all their In Car Entertainment systems (Audi comes to mind) a few years back.
Darth Vader♦
TiberiusJTiberiusJ
Remember! That wattage draw will be by the USB charger. Unfortunately they are very inefficient.
So if the USB charger says a MAX of 2A @ 5V it could use the full 10Watts regardless if the Pi is on or off. Pretty stupid but that is the trade off for being cheap and cheerful.
Switching power supplies can be up to 90% efficient. That means it will only use the power it needs. On low power the minimum wattage used when the Pi is off will be up to 9 times less than that of those USB chargers. These power supplies work very well with high wattage equipment or equipment that changes wattage very often.
Remember that it needs to be a 'switching' power supply and should show the efficiency somewhere but on small power supplies the efficiency almost doesn't matter. These are also regulated, so they protect your Pi from power surges, short circuits and brownouts.
Wikipeadia
Unlike a linear power supply, the pass transistor of a switching-mode supply continually switches between low-dissipation, full-on and full-off states, and spends very little time in the high dissipation transitions, which minimizes wasted energy.
Piotr KulaPiotr Kula
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Logarithms: Simplifying with
'The Relationship' (page 2 of 3)
Sections: Introduction to logs, Simplifying log expressions, Common and natural logs
This log is equal to some number, which I'll call y. This naming gives me the equation log2(8) = y. Then the Relationship says:
2 y = 8
That is, log2(8), also known as y, is the power that, when put on 2, will turn 2 into 8. The power that does this is 3:
23 = 8
Since 2 y = 8 = 23, then it must be true that y = 3, and I get:
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log2(8) = 3
The Relationship says that, since log5(25) = y, then 5 y = 25. This means that the given log log5(25) is equal to the power y that, when put on 5, turns 5 into 25. The required power is 2, because 52 = 25. Then 52 = 5 y = 25, so:
log5(25) = 2
The Relationship says that this log represents the power y that, when put on 64, turns it into 4. Remembering that 43 is 64, and remembering that fractional exponents correspond to roots, this means that the cube root of 64 is 4, so 64(1/3) = 4. Then:
log64(4) = 1/3
This last example highlights the fact that, to be able to work intelligently with logs, you need to be pretty good with your exponents. So take the time to review them, if you're feeling a little shaky.
The Relationship says that, since log6(6) = y, then 6 y = 6. But 6 = 61, so 6 y = 61, and y = 1. That is:
log6(6) = 1
This is always true: logb(b) = 1 for any base b, not just for b = 6.
The Relationship says that, since log3(1) = y, then 3 y = 1. But 1 = 30, so 3 y = 30, and y = 0. That is:
log3(1) = 0
This is always true: logb(1) = 0 for any base b, not just for b = 3.
The Relationship says that, since log4(–16) = y, then 4 y = –16. But wait! What power y could possibly turn a positive4 into a negative16? This just isn't possible, so the answer is:
no solution
This is always true: logb(a) is undefined for any negative argumenta, regardless of what the base is.
The Relationship says that, since log2(0) = y, then 2 y = 0. But wait! What power y could possibly turn a 2 into a zero? This just isn't possible, so the answer is:
no solution
This is always true: logb(0) is undefined for any base b, not just for b = 2.
The Relationship says that 'logb(b3) = y' means 'b y = b3'. Then clearly y = 3, so:
logb(b3) = 3
This is always true: logb(bn) = nfor any base b.
Some students like to think of the above simplification as meaning that the b and the log-base-b 'cancel out'. This is not technically correct, but it can be a useful way of thinking of things. Just don't say it out loud in front of your instructor.
Remember that a logarithm is just a power; it's a lumpy and long way of writing the power, but it's just a power, nonetheless. The expression 'log2(9)' means 'the power which, when put on 2, turns 2 into 9.' And they've put that power onto 2, which means that the 2 has been turned into 9!
Looking at it another way, '2log2(9) = y' means 'log2(y) = log2(9)' (which is the equivalent logarithmic statement), so y = 9. But y = 2log2(9), so 2log2(9) = 9.
While the second way is technically correct, I find the first way to be more intuitive and understandable. Either way, though, I get an answer of:
2log2(9) = 9
This last example probably looks very complicated, and, in the technical explanation, it is. Look instead at the intuitive explanation (in the first paragraph). Some students even view the above problem as the 2 and the log-base-2 as 'cancelling out', which is not technically correct, but can be a useful way of remembering how this type of problem works.
To synopsize, these are the things you should know from this lesson so far:
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