Can someone PLEASE tell me what the current (Amperes) should be for powering a pic?
I use a L7805 voltage regulator but it still seems like I can get a whole range of different levels of current depending on the batteries use. (also you get different Amps for the L7805. Some are 0.1A and others are 1A or even 2A)
I have 4 1.2V 1800mA batteries but I guess that current is far too high. The 9V rechargable I have is 150mAh. How is it possible that the
9V has so little curent compared to he 1.2V??? I aways thought 9V is stronger?
Anyway.. I really want to power my pic off a battery. So please let me know what the current and voltage should be.
The regulators you mention output a given voltage. The amount of current they give depends on the load attached to them e.g. a pic plus peripheral circuits. They regulate the voltage so it stays the same no matter what the load current (within certain limits). The various current you mention for 7805 regulators are the maximum current load you can draw from each one whilst it still regulates the voltage.
Similarly with batteries, the currents given are the maximum load the battery can sustain. (Actually this is not true but it will do for the purposes of this explanantion).
So, you need to pick a power source, either battery or regulator that:
a) provides a voltage suitable for the pic
b) is capable of providing at least the amount of current the pic requires.
AFAIK pics require only a few tens of milliamps current so a 78L05 or your batteries should be fine.
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The current consumption depends on the clock frequency. Look it up in the datasheet (or measure it). It (may) also depend on what you hang on the outputs of the PIC. If the pic sources current to, for instance, leds, the current consumption goes up.
Yes, there are several choices. First find out how much current you need.
Think in terms of power and physical battery size. 1.2 x 1800 -> 2160.
9 x 150 -> 1350. 9V batteries have less energy density, because a lot of space is wasted by the insulation of the individual stacked cells inside the package.
Can't give you an answer here. Depends on your actual circuit.
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Your problem is not current. It is drop-out voltage. Voltage regulators like the 7805 require that the input voltage be a certain amount higher than the output voltage (at least several volts in the case of the 7805). There are special Low Dropout (LDO) Voltage Regulators that can work with difference of only 200 mV or so. For battery powered applications, look into LDO regulators.
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The batteries are rated in milli-ampere-hours, which is the capacity, not the current. The actual current is determined by the circuit that the battery is connected to. If your circuit draws 100 mA and you use 4 cells in series, the life is approximately 1800 mAh/100 mA = 18 hours.
The 9 volt battery has a higher voltage. The energy capacity is 9 x 150 mAh, or
1350 milli-Watt-hours, which is less than the energy density of the AA cell (1.2 x 1800).
A battery approximates a constant voltage source, providing the nominal voltage to the load, as long as the current isn't too high. The resistance of the load determines how much current is drawn
I = V / R
where I is the usual symbol for current. Knowing the current and the battery capacity in ampere hours or milli-ampere-hours, you can determine the approximate life of the battery. The formula is not exact because most batteries decrease the output voltage as they are discharged. At some point the voltage becomes too low to power the circuit properly. Also, the capacity of a battery in mAh depends on current being drawn and temperature. Most batteries lose capacity when they are discharged at low temperatures.
It would help to find a tutorial on electrical units of volts, amperes, watts, ohms.
The voltage requirements of the processor varies with the model, but is in the general range of 3 to 6 volts. Some models can run on slightly less voltage. When run on the lowest voltages, they cannot run as fast. You also need to consider the voltage and current requirements for the peripherals that are connected to the processor.
Exactly. And you probably want the circuit to work down to 4.0V battery voltage or so if there are 4 of them, so a Vdd of 3.3V and an LDO regulator would be about right.
Don't forget to study the data sheet on the LDO regulator. They are much fussier about load capacitance than the old NPN regulators.
Best regards, Spehro Pefhany
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Aren't there any PICS that run on 1.8 to 5.5V? Then, with 4 1.2V cells you don't need a regulator at all.... You could probably get away with 3 or even just 2 cells.
So does that mean that the current of the power source can never break a pic? Because the pic only takes the amount of power it needs? That sounds strange... Does that mean if I have a pic connected to one led, and I give it 10A that everything will work properly if the Voltage is within the specifications of the pic?
I think you don't quite understand how this works. You don't *give* something a current, you apply a voltage. And the devices that gets the voltage determines how much current it draws from the voltage source.
So if you have a power supply of say 5V/1A it means that the output voltage is 5V and it *can* supply a maximum current of 1A. The attached device is the only thing that determines how much current is really needed.
Observe Ohm's Law: U = I * R. You set U by the voltage of the power source and the PIC has a certain resistance (over simplified) that determines how much current actually flows.
I case of the batteries you mentioned, the "current" of 1800mAh is actually the capacity of the battery and it is the product of current and time. To put it simple: a 1800mAh cell can deliver 1800mA during 1 hour, or 180mA over 10 hours and so on.
I would advise you to read a book about basic electronics, to teach yourself a bit about the basics of voltage, current and resistance etc. It will clear up things a bit.
Correct. If you have, say, a huge 4V "car" battery (don't know if you can get them any more), you can happily run a PIC (assuming it'll work at 4V) of it, and it'll run for about a year(?); it will take whatever current it needs. This is true of all electronic circuits.
Yes. You do not "give" it 10A; you "offer" it 10A: it will take whatever it needs.
However, an "electronic circuit" can include a short circuit. That will take
10A (and more). It will probably melt and, in extreme cases, cause a fire. It may or may not damage the PIC, depending on where the short is.
Yes. You cannot 'give it 10A'. If you connect the correct voltage to it then it will take whatever current it needs. The current rating of a regulated voltage power supply for example only tells you how much current it *can* supply if the load demands it. This is not strange but a simple consequence of ohms law which can be expressed as:
I = V/R
Where I is the load current, V is the applied voltage and R the load resistance. The PIC effectively has a certain value of R. When you apply a voltage V it is the above formula gives the current. The PIC will only take more current if you increase the voltage.
Whoa, Silver! Let's stop right here, for a little consideration.
Now, how do I put this politely...: you're in *desparate* need of some very basic education about how electricity works. I strongly advise you stop thinking about microprocessors for a while and go back to lamps, batteries, switches, and maybe some resistors. Learn how they actually behave, learn Ohm's law, Kirchhoff's node and loop rules, and what the symbols in an electronics schematic actually mean.
--
Hans-Bernhard Broeker (broeker@physik.rwth-aachen.de)
Even if all the snow were burnt, ashes would remain.
Or you apply a current, and the device determines how much voltage you have to supply to get that current through it. Devices known as "constant current sources" aren't exactly the typical case, sure, but they do exist. Many households have one these days, known to the laypersons as a (simplistic version of a) battery charger.
--
Hans-Bernhard Broeker (broeker@physik.rwth-aachen.de)
Even if all the snow were burnt, ashes would remain.
A little inaccuracy sometimes saves tons of explanation. --Saki (1870 - 1916)
Best regards, Spehro Pefhany
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Think of electricity as water. If you raise a bucket of water 2 meters, then that's voltage. If you attach a hose to the bottom of the bucket, that the hose is your PIC and the water running through it is the current. As you can probably picture in this (flawed) analogy, if you have a much wider pipe, more "current" can flow. As with the bucket, you "apply" a voltage (the height to which you raise it) and the current (the amount of water running through it) is a function of the size of the pipe (impedence).
You might find this
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interesting and useful. It's a free online textbook about electric circuits and electricity. You'll find that your PIC experiments will be much more sucessful and rewarding if you pause for a bit to get some of the basics. Even if you only read through chapters 1-3, you'll get a lot.
Or accidentally turn the voltage up on a bench supply until it hits the CC limit. Stuff happens.
Best regards, Spehro Pefhany
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"it's the network..." "The Journey is the reward"
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I was quite conscious of the several inaccuracies in my reply but I felt that, given the OPs obvious lack of basic electronics knowledge, further detail would only confuse him. I'm with you and Saki on this one.
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