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3-5v to 12v 1A boost converter

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reik

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Hi all,

I have been research for few weeks for this problem, and saw alot of suggestion in this forum or other places.

i know that max668 or mc34063 could do this 3-5v to 12v at 1A, 1 question here:

isit any input current will do this?
my project is solar battery charger, i measure that i could get 4V above but i donno what is the current. Isit any current that go into the component above wil come out 12v at 1A ? normally the voltage is slightly higher or lower?
 

Well, if you want 12V 1A output with 100% efficiency your device will drain 4Amps from 3V battery(?). Are you sure, that you need this sort of device?
 
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    reik

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ok,another question here:

Let say im using LM317 and i have around 18V with 150mA. After the voltage go into the LM317 and i adjust the resistor to become 12V,what would be the output current? is the output current 1.5A as the datasheet stated? or still 150mA?

---------- Post added at 19:02 ---------- Previous post was at 19:01 ----------

ok,another question here:

Let say im using LM317 and i have around 18V with 150mA. After the voltage go into the LM317 and i adjust the resistor to become 12V,what would be the output current? is the output current 1.5A as the datasheet stated? or still 150mA?
 

You are confused!

You don't put the current in at a set value, it is whatever the load takes out at the voltage you have set. Apart from a small current used to power the internal circuit of the LM317, the input current is the same as the load current. The difference in voltage between the input and output in this kind or regulator is converted to heat. The 1.5A rating is the MAXIMUM your load can take subject to the amount of heat being within acceptable limits.

Brian.
 
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    reik

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You are talking for two different kind of regulators,
LM317 is a linear regulator it can only provide a voltage lower than the input and the input/output current is almost the same at any output voltage,
think of it a a series resistor that automatically varies the resistance in order to keep the output voltage steady,
because of that fact the additional voltage will be consumed on that resistor (pass element) as heat.

On the other hand a switching regulator works like a switch that is either on or off, because of that the efficiency is high and the power consumed as heat is low,
there is also some king of energy storage in the output (coil) and depending on the topology it can either give more voltage than the input (step up) or less (step down),
the voltage level is controlled by controlling the time that the switch is on or off (varying duty cycle).

In any case you can never get more energy than the amount you give, so when you have a supply that can give a max of 1v/1A (=1W) to a regulator, you can only get an output that is less than that to about 90% (best case).
When you want 12v/1A (12W) you have to provide at least 13W input and if the input voltage is lower than the output voltage you have to use a switching regulator to step up the voltage.

Alex
 
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    reik

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let say I have a solar panel which will constantly give a input for about 18v 150mA, after i step down it to 12v it is still 150mA right? for my project this current is kinda low, is there any method that could boost the current up to let say 500mA or 1A ?
 

In any case you can never get more energy than the amount you give, so when you have a supply that can give a max of 1v/1A (=1W) to a regulator, you can only get an output that is less than that to about 90% (best case).

You are talking about a 18v*0.15A=2.7W power supply, even with an efficiency of 90% you can not get more than 2.7W*90%=2.43W
2.43W/12v=0.20A

Alex
 
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    reik

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Solar battery? Ummm sounds really good - green energy :)
IMO you should use rechargeable battery to power your device and solar battery to topup accu.
Then question mark - can you make your "black box" more economical?
 
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    reik

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thanks everyone, my project is using solar panel to charge battery using PIC microcontroller. But my solar panel only provide 1W and the maksimum voltage i measure is 10v and the current i have no idea how to measure but according to the datasheet it is around 150mA. In order to charge the 2200maH battery, it might needs 15+ hours ! LOL, what a time, so is there any other method could boost the current up?

i mention above 20v is because i have intended to series up 2 solar panel for this project =)
 

Do some simple math.
You can't create power but you can convert it. The power you have available is the voltage multiplied by the current so in your case you start with 20V and 150mA, that makes 3W.
Some power is lost in the conversion circuit but basically you can convert your 3W into any combination of voltage and current as long as the result of multiplying them is 3.
So you could produce 1V at 3A, 2V at 1.5A, 3V at 1A and so on. When you try to charge a 12V battery, the maximum current you can possibly get is 0.25A (12 x 0.25 = 3).
To do this you need to use a switch mode type of power supply. Bear in mind that 3W is your MAXIMUM power from the solar panel and you only get that when the panel is facing the Sun in a clear sky, at night and in cloudy conditions the power will be much less. Also consider that the circuit converting the power will probably only achieve 80% efficiency so the output will be less than the theoretical figure.

Brian.
 
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I think i confuse here, from the lm317 datasheet, it wrote it can be current regulator by putting a small value resistor, the formula is Iout = 1.25/R
so let say my R is 1.2ohm, my Iout will be 1A. is this correct? but u guys saying if my input current only 150mA, then my output current will be 150mA too,how could this make sense?
 

You are referring to a specific connection of the LM317 which works as a constant current source,
the adjustment of the output current is adjusted using that resistor,
but your source must be able to provide that current.

Please read post #7 again https://www.edaboard.com/threads/203517/#post858955
many others have told you the same thing but for some reason you ignore it.

Alex
 
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    reik

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As Alex explains, if you wire the LM317 as a constant current element, YOU decide what the output current will be by choosing the resistor values. The input current to the LM317 will essentially be the same as the load current you chose except for a small extra amount used internally in the device. In a simple series element like that, you cant put more current in than comes out !
Think of this analogy: You have a hose pipe with water running through it, you step on the pipe and compress it so the flow is restricted and less water comes out. Stepping on the water pipe does the same as the LM317 does to the electrical flow in the output wire. Go back to the pipe, when you step on it, the water leaving the pipe is less and the water entering it must also be less, you can't imagine more going in at one end than leaves the other.

Brian.
 
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    reik

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how about just use bigger solar panel?
 
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ok,i totaly understand. Thanks for everyone =)

Takeshi-san, because the panel is expensive, LOL
 

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