vedge Posted November 17, 2006 Report Share Posted November 17, 2006 HelloFirst out of many questions.I dont get how to one is supposed to limit the current being fed to a certain pin of "something" without droping voltage getting to it.In a daughter board im making, I want to keep feeding off the main 7805 for +5vdc, but if i use a resistor to limit the current going to my pin, of course the voltage will drop, and i dont want that.For instance, with the current core desing, the +5vdc is fed to pin 32 of the PIC, and pin 25 of the 6581,fine, but both these chips do have spec sheets on which its said that they should only be fed a certain maximum current.250 mA for the PIC, (PIC18FXX2 manual section 22.0), and100 mA for the 6581 (http://stud1.tuwien.ac.at/~e9426444/sidtech3.html)Now call be total newbie in electronics (true), but i just dont get it. is that a power consumption? or a "feeding limit" .. why dont these chips fry if i have 500 mA input?Im working on a few secret projects to drive some other sound chips. Im a very good audio/MIDI/realtime programmer,but a total retard in electronics, and i just have soo much to learn.the CHIP docs say "Supply Voltage (VCC) | MIN 4.5 VDC | TYP 5.0VDC | MAX 5.5VDC""ICC Supply Current | OUtputs Open |Typ 30 mA | Max 50 MaThanks in advance for my stupid questions. Hopefully this one is the stupidest... Quote Link to comment Share on other sites More sharing options...
fluke Posted November 17, 2006 Report Share Posted November 17, 2006 For almost everything except LEDs, the maximum supply current is the most it will draw. If you want to protect the device from drawing too much current (due to a fault such as a short circuit), use a fuse or a polyswitch. You'd usually only use 1 protection device per regulator.LEDs don't limit the current they draw and will overheat and damage themselves if not regulated. Since most LEDs have a voltage drop of less than 5V, you use a resistor to drop the extra voltage and limit the current. If you really want a constant current regulator, use a LM317. Quote Link to comment Share on other sites More sharing options...
doc Posted November 17, 2006 Report Share Posted November 17, 2006 Hi,Now call be total newbie in electronics (true), but i just dont get it.is that a power consumption? or a "feeding limit" .. why dont these chips fry if i have 500 mA input?Sorry, if I don't use the right english terms now (not native english):It's the Ohm law R= U/I that says how much current a device will get.Having a 500mA limit does not mean, that you "blow" 500mA through the device!It's the other way. The device need as much current as the resistor allows.Just calculate it for yourself.It doesn't matter if your power supply delivers 500mA or 10A. It should be enough to drive any device that is connected to it.Back to your resistor-limiting-thing.If you connect a resistor, the overall current gets lower ... look at R=U/I or I=U/R . So: The higher the resistor is, the lower the current. Get it?greetsDoc Quote Link to comment Share on other sites More sharing options...
ZiggY!! Posted November 17, 2006 Report Share Posted November 17, 2006 Hi,Sorry, if I don't use the right english terms now (not native english):It's the Ohm law R= U/I that says how much current a device will get.Having a 500mA limit does not mean, that you "blow" 500mA through the device!It's the other way. The device need as much current as the resistor allows.Just calculate it for yourself.It doesn't matter if your power supply delivers 500mA or 10A. It should be enough to drive any device that is connected to it.Back to your resistor-limiting-thing.If you connect a resistor, the overall current gets lower ... look at R=U/I or I=U/R . So: The higher the resistor is, the lower the current. Get it?greetsDocHey Doc, just thought I'd add something (for the sake of clarity). I guess it depends where you come from, Ohms law is always the same but the symbols for each units will differ. Some equations relating to ohms law include:P=I/V - Power (P) in watts equals current (I) divided by voltage (V). V=IxR - Voltage (V) equals Current (I) multipled by Resistance ®. There is the Ohms Law pie chart that outlines a series of equations for working out voltage, power, current and resistance.Note that depending where you are or where you were taught, some symbols are different. For example voltage can be depicted by the symbol (V) as well as (E). Power can use the symbol (P) or (W) for watts. Resistance is always ® which is given in Ohms. Current can use (I) or (A) for amperes. Quote Link to comment Share on other sites More sharing options...
vedge Posted November 17, 2006 Author Report Share Posted November 17, 2006 If you really want a constant current regulator, use a LM317.But should I?The chip seems to work fine so far, and ive been running it for a few hours and it didnt melt :)I was just curious about that particular Spec sheets and its 50 mA input limit.(thats 50 mA lower than the SID and 150 lower than the PIC specs)I mean, they must write that for a reason other than "dont feed that 10A"!The other thing about the ohm law that confused me was because of my test with a multimeter.If i set my multimeter to check for voltage under this scenario:I naively tought that the voltage was a constant, and figured out, if i wanted5 VCD, and 33 mA, thenR = V/IR = 5/0.033R = 150Trying thisJ2+ ------ 150Ohm_rez ------ + terminal_of_multimeterJ2- -------------------------- - terminal_of_multimeterI STILL get 5vcd, but a new 33 mA of measured current. (and the core PIC didnt reset when mesuring amperage:)Where as when i inserted the 150 ohm res in the actual circuit, powered it up and checked again with the samemultimeter, the voltage was now 2.3 volts.Ok 5VCD is NOT a constant it seems.So all values are really interconnected in the ohm law. and nothing is constantif you change any value. but why the h___ does my multimeter trick me? Quote Link to comment Share on other sites More sharing options...
doc Posted November 17, 2006 Report Share Posted November 17, 2006 .... don't forget:Your multimeter also has an internal resistor!So it produces (a small amount) of load.(and the core PIC didnt reset when mesuring amperage:)No, it won't reset. But yaou put a load of 33mA to measured pin ;)greetsDoc Quote Link to comment Share on other sites More sharing options...
fluke Posted November 17, 2006 Report Share Posted November 17, 2006 But should I?The chip seems to work fine so far, and ive been running it for a few hours and it didnt melt :)I was just curious about that particular Spec sheets and its 50 mA input limit.(thats 50 mA lower than the SID and 150 lower than the PIC specs)I mean, they must write that for a reason other than "dont feed that 10A"!The spec sheet quotes the maximum current for you to design a power supply that can deliver at least that much. ICs don't need external current limiting. It's probably easiest if you think of CMOS working on voltage, not current.Power supplies always supply voltage, not current. An LM317 is a adjustable voltage regulator, but you can configure it to adjust the voltage so the current is constant by having it sense the voltage across a small resistor.When you have resistors in series, they have the same current and "use up" all the voltage. The voltage across each one is proportional to it's resistance, ie V=IR. So when you have 5V to an IC and put a resistor in series, the IC no longer gets 5V.More information can be found if you read up about Kirchhoff's Laws Quote Link to comment Share on other sites More sharing options...
vedge Posted November 17, 2006 Author Report Share Posted November 17, 2006 More information can be found if you read up about Kirchhoff's LawsMany thanks. Question #1 -> Sorted! Quote Link to comment Share on other sites More sharing options...
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