chutton01 Well, as promised, I checked my scavanged LEDs in the test circuit.VR ≈ 28.8mvR = 10.0ΩTherefore, current for 1 LED ≈ 28.8/10.0 ≈ 2.9mA, and a thousand LEDs (easy proposition on a large layout) would be 3A, which is not bad at all.For comparison, I also measured an old school late 1980s Red LED. It pulled only 3.9mA, which I found very surprising, I figured on a magnitude higher.Circuit was 10Ω resistor (to measure across), 1KΩ resistor, LED, 9V battery - and an old knife switch I found among my electric part box (for on/off control in style) - all components snuggly plugged into my late 1980s Archer® Experimental Breadboard
Well, as promised, I checked my scavanged LEDs in the test circuit.VR ≈ 28.8mvR = 10.0ΩTherefore, current for 1 LED ≈ 28.8/10.0 ≈ 2.9mA, and a thousand LEDs (easy proposition on a large layout) would be 3A, which is not bad at all.For comparison, I also measured an old school late 1980s Red LED. It pulled only 3.9mA, which I found very surprising, I figured on a magnitude higher.Circuit was 10Ω resistor (to measure across), 1KΩ resistor, LED, 9V battery - and an old knife switch I found among my electric part box (for on/off control in style) - all components snuggly plugged into my late 1980s Archer® Experimental Breadboard
Now you are too low. You aren't calculating or measuring things correctly. When the LED is operating normally in a circuit with a power supply, resistor, and the LED, it's the forward voltage that matters. This is going to be around 3.5 volts for a white LED, unless they are really wierd ones.
Now we need to start witht he absolute basics, before even Ohm's Law. This is Kirchoff's Law. Simplified, what Kirchoff's Laws state is that components in series add voltage. So if you put 12V into the circuit, and the LED 'consumes' 3.5 volts of it, what's left is 'consumed' by the resistor. In this case, 8.5 volts (12 - 3.5). Kirchoff's Laws also state that loads in series share the same current. An LED does not draw a fixed current, but a resistor is a linear device and does. So if we want the LED current to be around 10ma, we also want the resistor current to be around 10ma. Mow we have 2 out of the 3 values for Ohm's Law to come into play - we know the voltage (8.5) across the resistor, and we know the current, 10ma. Solving for R, we get R=E/I, 8.5/.010, 850 ohms. Resistors are not available in every single value, so you pick the closest standard value, or larger. If you use a 1K resistor and work backwards, you get the 8.5ma I mentioned previously.
So hook one up to a 12V power supply with a 1K resistor. Measure the power supply boltage. The use the meter in volts mode and measure the voltage across the LED, it should be somewhere around 3.5 volts. Do the same across the 1K resistor - it should be (power suppy voltage) - (voltage measured across LED). Set the meter to ma mode and insert it in the circuit (anywhere is fine: ps - meter - led - resistor - ps, ps - resistor - meter - led - ps, any combo is all the same per Kirchoff's Law) and it should read 8.5ma or so, depending on the actual ps voltage.
--Randy
and one of these days I will remember to copy this and paste it on my web page so I can just post the link instead of retyping effectively the same thing every time I explain this.
Modeling the Reading Railroad in the 1950's
Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.
rrinkerNow you are too low. You aren't calculating or measuring things correctly. When the LED is operating normally in a circuit with a power supply, resistor, and the LED, it's the forward voltage that matters...
No, he did measure it correctly (I=V/R, current = volts/resistance), although there was no need to add the extra 10 ohm resistor because he could have just measured the voltage drop across the 1k resistor and calculated it from there.. The 10 ohm resistor actually reduced the current draw slightly, but it would have been neglegible in this case. In fact, this is the same thing as inserting a multimeter set to current in the circuit - the meter measures the volatge drop across a low value resistance to determine the current draw.
CSX Robert rrinker Now you are too low. You aren't calculating or measuring things correctly. When the LED is operating normally in a circuit with a power supply, resistor, and the LED, it's the forward voltage that matters... No, he did measure it correctly (I=V/R, current = volts/resistance), although there was no need to add the extra 10 ohm resistor because he could have just measured the voltage drop across the 1k resistor and calculated it from there.. The 10 ohm resistor actually reduced the current draw slightly, but it would have been neglegible in this case. In fact, this is the same thing as inserting a multimeter set to current in the circuit - the meter measures the volatge drop across a low value resistance to determine the current draw.
rrinker Now you are too low. You aren't calculating or measuring things correctly. When the LED is operating normally in a circuit with a power supply, resistor, and the LED, it's the forward voltage that matters...
Mark R. BroadwayLion [snip] ....The LONG lead goes to the NEGATIVE side of the circuit, (I use GROUND for this), the shorter lead goes to the + dc voltage. Watch out for LIONS. ROAR The long lead is positive, not negative .... Mark.
BroadwayLion [snip] ....The LONG lead goes to the NEGATIVE side of the circuit, (I use GROUND for this), the shorter lead goes to the + dc voltage. Watch out for LIONS. ROAR
[snip] ....The LONG lead goes to the NEGATIVE side of the circuit, (I use GROUND for this), the shorter lead goes to the + dc voltage.
Watch out for LIONS.
ROAR
The long lead is positive, not negative ....
Mark.
Correct. LIONS are dyslexic and ALWAYS mix up any binary problem.
The Negative pole has the BIG flag (inside the LED) and the short leg outside)
The Positive pole has the LITTLE flag and the long leg.
LION always looks at the flags rather than the legs. BIG = Negative. I think I have that written on the wall in the train room, but not in the computer room.
The Route of the Broadway Lion The Largest Subway Layout in North Dakota.
Here there be cats. LIONS with CAMERAS
BroadwayLionThe Negative pole has the BIG flag (inside the LED) and the short leg outside) The Positive pole has the LITTLE flag and the long leg. LION always looks at the flags rather than the legs. BIG = Negative. I think I have that written on the wall in the train room, but not in the computer room.
One lead longer is common but by no means universal. ANother common indication is a flat in the case. The problem of course with lead length is what if you are reusing an LED that you previous wired and clipped all the wires nice and neat? If you can see through the LED by holding it up to the light, this is the most foolproof method, I have yet to see an LED manufactured 'upside down' internally, probbaly because the small leg doesn;t have enough mass to heat sink the actual diode, plus trying to make a wire bonder that could do that would be an interesting task.
If you measured across the 10 ohm resistor, all you got was the voltage dropped by the 10 ohm resistor, not the LED. And with a pure DC source like a battery, where are you getting a Vr value from? There is 0 volts across the LED backwards if it's running off a battery.
rrinkerOne lead longer is common but by no means universal. ANother common indication is a flat in the case. The problem of course with lead length is what if you are reusing an LED that you previous wired and clipped all the wires nice and neat? If you can see through the LED by holding it up to the light, this is the most foolproof method
I have yet to see an LED manufactured 'upside down' internally, probbaly because the small leg doesn;t have enough mass to heat sink the actual diode, plus trying to make a wire bonder that could do that would be an interesting task.
If you measured across the 10 ohm resistor, all you got was the voltage dropped by the 10 ohm resistor, not the LED.
And with a pure DC source like a battery, where are you getting a Vr value from? There is 0 volts across the LED backwards if it's running off a battery.
No, when someone talks about a diode and Vr without posting a diagram of the circuit under test, I assume reverse voltage of the diode, not voltage across a resistor. Hey, I was up til 1:30am on a conference call and I'm working on about 3 hours of sleep.
I probably would have just measured the drop across the 1K, and also measured the actual resistence of the 1K especially if it was a standard 10% one.
I pretty much dount anyone would make an LED with the flags reversed. Mainly for the heat, like I said. We tend to think of LEDs as not getting hot, but at the junction they can be quite warm. But it's such a miniscule spot of heat, it's easily carried away by the substrate and large metal lead.
What I haven't seen is a 2 lead bicolor LED, all the ones I've ever had were opaque so you couldn;t see the leads once they get inside the case.Is it two large flags, each with a diode, and a whisker from each flag to the opposite diode junction? I'd suspect this is more likely than there being the usual large flag at the bottom with one diode, and the other one upside-down on the small flag
Another reason the actual dide junction wouldn;t be inside the case upside down - the light would all try to get out the lead side, not the top. The light comes from the diode junction, and you'd want that to point up and out, not down towards whatever board the LED was installed in.