Forums

Pic Currrent

Started by CA Zuke October 15, 2004
Hi,

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.

Thanks in advance
CE AUKE
CA Zuke wrote:

> Hi, > > 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. HTH Ian -- Ian Bell
"CA Zuke" <ceauke@yahoo.com> schreef in bericht
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> Hi, > > Can someone PLEASE tell me what the current (Amperes) should be for > powering a pic?
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.
> 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)
Yes, there are several choices. First find out how much current you need.
> > 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?
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.
> Anyway.. I really want to power my pic off a battery. So please let me > know what the current and voltage should be.
Can't give you an answer here. Depends on your actual circuit. -- Thanks, Frank. (remove 'x' and 'invalid' when replying by email)
On 15 Oct 2004 01:29:28 -0700, ceauke@yahoo.com (CA Zuke) wrote:

>Hi, > >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?
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. -Robert Scott Ypsilanti, Michigan (Reply through this forum, not by direct e-mail to me, as automatic reply address is fake.)
CA Zuke wrote:

> I have 4 1.2V 1800mA batteries but I guess that current is far too > high.
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 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?
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.
> Anyway.. I really want to power my pic off a battery. So please let me > know what the current and voltage should be.
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. Thad
On Fri, 15 Oct 2004 13:28:45 GMT, the renowned no-one@dont-mail-me.com
(Robert Scott) wrote:

>On 15 Oct 2004 01:29:28 -0700, ceauke@yahoo.com (CA Zuke) wrote: > >>Hi, >> >>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? > >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.
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 -- "it's the network..." "The Journey is the reward" speff@interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com
"Spehro Pefhany" <speffSNIP@interlogDOTyou.knowwhat> wrote in message
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> 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.
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. Meindert
> 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. > > HTH > > Ian
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?
"CA Zuke" <ceauke@yahoo.com> wrote in message
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> 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. Meindert
"CA Zuke" <ceauke@yahoo.com> wrote in message

> 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...
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.
> and I give it 10A that everything will work properly if the Voltage is > within the specifications of the pic?
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. Richard [in PE12]