----- Original Message ----- From: "Kieren Johnstone" To: Sent: Friday, July 19, 2002 3:23 AM Subject: Re: [PIC]: Flashing LED Episode II (small attachment) > Thanks, this seems most helpful. > Am I to understand that the resistor is there only to blow the circuit up if > there's a short circuit / something drawing too much current? ****** The LED is a "light emitting diode". So it is a diode and diodes act as an infinite resistance till a certain voltage is reached across it, say 1 volt. Then it becomes an short circuit to any voltage above that. So without the "limiting resister" too much current would flow through the led and destroy it. I think I use a 150 ohm resister in series with a led. 4v/150 ohms = 27 ma of current. ***** >I mean, > obviously it adds to the global resistance of the circuit, so if I wanted to > use a higher-current device, I'd have to draw it's power off in parallel > from the battery and use a transistor? (that's a question :)) The only > problem in my head with that could be there's a 5V supply connected to the > base of a transistor that's controlling a 6V "load"; would the internals of > the transistor stop current flowing back down the base, or would I need to > add a diode, or..? ************** You are thinking too deep. just be sure the power supply can supply enuff current at the required voltage. *********** > Other strange question: > One other thing that's got my head spinning; assuming I've got a pure 5v > power supply.. now, the PIC datasheet says the chip consumes 20uA typically > (at 3V but lets just assume its the same). Even without using Ohm's Law > it's clear to see that I=V/R is going to be really quite large. Is my > assumption correct therefore when I assume that the chip is basically just a > resistor? I mean, using crazy logic I could add a massive resistor to the > circuit, and the PIC would still work OK because it only needs 20uA, but > then my head melts further when I think when the circuit has no resistor, > wouldnt the chip be overloaded with current and burn (5V / 0.00verysmall R = > veryhigh A). > > Thanks :/ > Kieren **************** The Pic is a "semiconductor" device. Which means that it is part conductor, part resistor. Just think of it as a variable resistor. And what it is doing will determine it's effective resistance. Yes, I=V/R is correct,, but R = V/I. and 5V/0.000020A or 250 thousand ohms. But when the led turns on, then the chip would draw 20ma of current. That would be 5V/0.020A or 250 ohms. A thousand times less. So you have to insure that your power supply can supply the maximum amount of current you may need while keeping the voltage supply constant at 5V. Bill ************** > > ----- Original Message ----- > From: "Mircea Chiriciuc" > To: > Sent: Friday, July 19, 2002 8:46 AM > Subject: Re: [PIC]: Flashing LED Episode II (small attachment) > > > > Use this. With a 1V voltage drop across the resistor the current is also > > limmited at about 30mA. > > The voltage drop parammeter for a zenner diode states for the voltage drop > > across the diode when reversed biased. So in the drawing attached the > diode > > is "keeping" the voltage at a maximum of 5.1V. If the voltage tends to > > increase the zenned will permit current to flow trough it so the voltage > > drop will remain aprox. constant untill the current limit trough the diode > > is passed and the diode will becone a wire. If the diode becomes a wire, > the > > resistor will get the full current of your batteries I=U/R=6/33=181mA. The > > power the resistor will try to excange in heat will be > P=U*I=6*0.181=1.09W. > > If you use a 1/4W resistor it will provide also protection for your > circuit > > as it will burn and open the circuit if the current will exceed 50mA. > > > > Good Luck, > > Mircea Chiriciuc. > > > > > > -- > http://www.piclist.com#nomail Going offline? Don't AutoReply us! > email listserv@mitvma.mit.edu with SET PICList DIGEST in the body > > -- http://www.piclist.com#nomail Going offline? Don't AutoReply us! email listserv@mitvma.mit.edu with SET PICList DIGEST in the body