Sunrise Semester # 2 - Ohm's Law and Semiconductors
diodes(semiconductors) don't really have a "resistance" like a resistor.. they kinda do, but it's not really something that you need for calculations.. diodes are like one way valves that let current go one way and not the other.. they do this at a specific voltage.. a typical 4000 series diode like a 1N4007 has a turn on voltage of 0.7V... and that is fixed.. you might get slightly more than .7 but it won't be very noticeable... if you try hooking up a diode like an led or regular diode to more than it's rated voltage, with no resistor in series, you'll blow it up.. the resistor acts like a sponge to soak up the current and burn it off(they actually give off heat, how do you think your stove elements work?).. soooo... what does all this mean? well, by looking at Iceman666's diagram, that 750ohm resistor is actually taking current away(shunting) from the LEDs!! If you were to measure the the voltage across the LEDs with that 750ohm resistor in there, it would be less than the rated voltage of the LEDs.. why? because the 2Kohm resistor is fixed, and the 12V power supply is "fixed"(i know it goes up to 14V, but everything else goes up proportionately).. since these are fixed values, you can use Ohm's Law to verify that the LED voltage goes down.. say for example you have 5V LEDs.. the current through the 750ohm resistor would be 5V/750ohms = 6.67mA.. hold on a second, that doesn't seem right does it? of course it isn't right because if you were to multiply 6.67mA x 2Kohms(the other resistor), you would have 13.34V?? that is not possible because the 12V power supply is 'fixed'.. and Ohm's Law doesn't lie, so something has to give.. the LEDs .. If you pull out that 750ohm resistor your LED's will be brighter, and they will be operating at their normal forward voltage.. That's what you want.. by the way, the 2Kohm resistor seems a bit high.. I would use a 1K or less.. ie, for one LED @ 20mA @ 5V, from a 12V supply, you would need a 350ohm resistor in series(12-5=7, 7/0.02=350).. for 5 LEDs @ 20mA @ 5V, you will need a 70ohm resistor(5 x 20ma = 100ma, 7V/100mA = 70ohms)...V = IR my friends.. Ohm's Law is the Law..
beenana
diodes(semiconductors) don't really have a "resistance" like a resistor.. they kinda do, but it's not really something that you need for calculations.. diodes are like one way valves that let current go one way and not the other.. they do this at a specific voltage.. a typical 4000 series diode like a 1N4007 has a turn on voltage of 0.7V... and that is fixed.. you might get slightly more than .7 but it won't be very noticeable... if you try hooking up a diode like an led or regular diode to more than it's rated voltage, with no resistor in series, you'll blow it up.. the resistor acts like a sponge to soak up the current and burn it off(they actually give off heat, how do you think your stove elements work?).. soooo... what does all this mean? well, by looking at Iceman666's diagram, that 750ohm resistor is actually taking current away(shunting) from the LEDs!! If you were to measure the the voltage across the LEDs with that 750ohm resistor in there, it would be less than the rated voltage of the LEDs.. why? because the 2Kohm resistor is fixed, and the 12V power supply is "fixed"(i know it goes up to 14V, but everything else goes up proportionately).. since these are fixed values, you can use Ohm's Law to verify that the LED voltage goes down.. say for example you have 5V LEDs.. the current through the 750ohm resistor would be 5V/750ohms = 6.67mA.. hold on a second, that doesn't seem right does it? of course it isn't right because if you were to multiply 6.67mA x 2Kohms(the other resistor), you would have 13.34V?? that is not possible because the 12V power supply is 'fixed'.. and Ohm's Law doesn't lie, so something has to give.. the LEDs .. If you pull out that 750ohm resistor your LED's will be brighter, and they will be operating at their normal forward voltage.. That's what you want.. by the way, the 2Kohm resistor seems a bit high.. I would use a 1K or less.. ie, for one LED @ 20mA @ 5V, from a 12V supply, you would need a 350ohm resistor in series(12-5=7, 7/0.02=350).. for 5 LEDs @ 20mA @ 5V, you will need a 70ohm resistor(5 x 20ma = 100ma, 7V/100mA = 70ohms)...V = IR my friends.. Ohm's Law is the Law..
beenana