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« Reply #15 on: January 04, 2013, 08:31:24 am » |
The gsm shield required is 1.5A. A nice heater element.
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« Reply #16 on: January 04, 2013, 08:43:26 am » |
Here a NPN setup 
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« Reply #17 on: January 04, 2013, 09:15:55 am » |
Here a NPN setup The output would fluctuate with the load. You need 1 small signal pnp, + 2 resistors.
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« Reply #18 on: January 04, 2013, 03:31:28 pm » |
The GSM shield requires 5v @ 1.5A? Really?
I would use an unregulated PSU for the solenoids, because there's not likely to be any good reason to worry about the exact voltage going through a coil. Same for the relays. As long as it's rectified and filtered to remove any serious ripple, they should be perfectly happy. A simple transistor (and protection diodes) to fire them would suffice.
That leaves the control circuitry and the GSM module. At a load of 1.5A, you really need to move to switching regulators. That's way, way too much current to stick with linear. You'll be generating a TON of heat. That means, you're wasting quite a bit of power, and you need sufficient means to remove that heat. Those LM78xx chips say they'll drive 1.5A, but don't even think about asking them to do it IRL.
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« Reply #19 on: January 04, 2013, 06:47:12 pm » |
Yes, because i will use calls too. http://tronixstuff.wordpress.com/2011/01/19/tutorial-arduino-and-gsm-cellular-part-one/The GSM shield can often require up to 2A of current in short bursts – especially when turned on, reset, or initiating a call. However your Arduino board can only supply up to just under 1A. It is highly recommended that you use an external 5V power supply capable of delivering 2A of current – from an AC adaptor, large battery with power regulator, etc. Otherwise there is a very strong probability of damaging your shield and Arduino. Ignore this at your own risk. When connecting this supply DO NOT use the DC socket on the Arduino. Instead, connect the 5V (positive) from the supply to the 5V pin on the GSM shield, and the negative to the GND pin. I'm using linear regulator because it is easy to make and it is only a school project. So i will only use it once. I already have a big heatsink and thermal paste so generating heat is not a problem.
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« Reply #20 on: January 04, 2013, 09:05:08 pm » |
Follow the circuit posted that uses a PNP 'wrapped' around the regulator. It has short circuit protection, is a reprint from the National Semi Linear app notes book (1 of 3) and I would LOVE to get copies of those in digital format $$$... It works very well as I've built several dozen copies of it in the past 20 or so years... and I can tell you from personal experience that although later an that same app note AN??? it (the app note) shows a version that used an NPN transistor as the pass transistor. The NPN requires either a variable regulator or a diode in the ground lead of the regulator to correct for the .7 V loss in the resultant emitter follower that results from adding the NPN transistor. The NPN version doesn't work as well and requires more parts. The single one I made was real problematical.. the current limiter was an issue... Use the PNP.
Bob
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“The solution of every problem is another problem.” -Johann Wolfgang von Goethe
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« Reply #21 on: January 04, 2013, 10:59:45 pm » |
I'm using linear regulator because it is easy to make and it is only a school project. So i will only use it once. I already have a big heatsink and thermal paste so generating heat is not a problem.
I think you under-estimate how hot a regulator will get passing 1.5A of current. I'm too lazy to do the math, but if it's dropping any voltage whatsoever, it may be impractical to keep the die temperature cool enough to survive. I built a 28v-to-15v regulator to supply 100mA or so of current, and the little 1"x2"x0.5" heatsinks got hot enough that I couldn't touch them comfortably. You're asking for a lot more current than that. Trust me -- that "1.5A" spec on the regulator is nothing short of wishful thinking. There's no way you'll keep the case cool enough that the die won't damage itself at that kind of load. That is unless the 1.5A spec is for a very brief, worst-case, transient load... and average current is really *significantly* less.
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« Reply #22 on: January 04, 2013, 11:37:37 pm » |
Do you really think you'll use that much power. The link the OP posted said bust up to 2 amps The Lm78 can handle 2.2 Amps in short burst.
These burst of the SM5100B cellular module only last 4.6 mS
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« Reply #23 on: January 05, 2013, 02:23:37 pm » |
You don't want to do that there not going to do what you think. One will try to put out more the it should and basically cutoff the second one. See there not going to supply the same voltage.
I would do this use a PNP to get you more current
Here's a circuit that I use for more regulated power... the small resistor values cause the 78XX part to provide current limit (around 10 amps with these values) and since the regulator and pass transistor are on the same heatsink, the thermal protection of the 78XX is also provided. Note the way the output is connected... this provides remote current sensing which compensates for drop along long wires if they are used. Only disadvantage to this circuit is that a minimum load of a few milliamps is required because of the design of the 78XX regulator. If you use an LM-317 instead, then this is not a problem. (edit to add): If you use a 79XX regulator and an NPN 2N3055 and reverse the plus and minus terminals, you get the same circuit.
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« Last Edit: January 05, 2013, 02:27:21 pm by Krupski »
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« Reply #24 on: January 05, 2013, 03:07:37 pm » |
thank god we don't live in a simulator.
You don't really know that. (Sorry, off topic, but couldn't resist.  )
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« Reply #25 on: January 05, 2013, 04:08:17 pm » |
Here's a circuit The 0.15ohm ressitor needs to be 100x of its current value. You can buffer that pnp with a power npn / n-ch mosfet so you can use a smaller pnp device instead.
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« Reply #26 on: January 05, 2013, 04:09:11 pm » |
thank god we don't live in a simulator.
You don't really know that. (Sorry, off topic, but couldn't resist.  ) The latest on that is that they have proposed tests to see if we are real or not.
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The last thing you did is where you should start looking.
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« Reply #27 on: January 05, 2013, 05:50:06 pm » |
As long as there is a ground connection between the power supplies you don't have to connect the Positive outputs together. Therefore you can have 1,2,3,4,5... regulators connected together (as long as there is a common ground)!
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« Reply #28 on: January 06, 2013, 08:35:57 am » |
I'm using linear regulator because it is easy to make and it is only a school project. So i will only use it once. I already have a big heatsink and thermal paste so generating heat is not a problem.
I think you under-estimate how hot a regulator will get passing 1.5A of current. I'm too lazy to do the math, but if it's dropping any voltage whatsoever, it may be impractical to keep the die temperature cool enough to survive. I built a 28v-to-15v regulator to supply 100mA or so of current, and the little 1"x2"x0.5" heatsinks got hot enough that I couldn't touch them comfortably. You're asking for a lot more current than that. Trust me -- that "1.5A" spec on the regulator is nothing short of wishful thinking. There's no way you'll keep the case cool enough that the die won't damage itself at that kind of load. That is unless the 1.5A spec is for a very brief, worst-case, transient load... and average current is really *significantly* less. I see, but based on the replies in my topic i can draw a lot of current from pnp transistor not the regulator. My heatsink is 80mmx41mmx15mm. Thank you for the warning. Now i will focus on adding a pnp transistor.
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« Reply #29 on: January 06, 2013, 08:39:16 am » |
I tried this circuit a while ago in a breadboard and it is not a working. I tried to draw 1 ampere by using a load of 10 ohm 5 watt resistor and the regulator gets very hot the pnp transistor is very cold and it seems it is not the one supplying it. I'm sure my connections are correct. http://www.zen22142.zen.co.uk/Circuits/Power/boosti.htm
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