OK, my turn. What we have here is a total lack of understanding what the terms on the datashhet means. A language barrier/ Let's get past it.
Under electro-optical characteristics we see some Vf values (1.2 to 1.4 V) and a current of 20 mA
This tells us that the LED would LIKE to operated around 20 mA and it will drop 1.2 to 1.4 Volts.
So, if the LED is working, you would measure 1.2 to 1.4 V ACROSS the LED.
So, to light a LED we need to know how our input voltage might vary and make sure the design fits within the ranges and doesn't go above the ABSOLUTE maximums. These let out the magic smoke.
Parallel batteries ARE not a good idea. Maybe we can work on that,
Those specs mean you have to have a supply voltage greater than 1.2V for anything to happen, it could be 3, 4.5, 12 or even 100V. At the low values of V, you get larger variations.
So if we have a 6V source and want 20 mA and our led drops a max of 1.4 V, we can select a resisistor by R=(6V-1.4)/0.020 (I might write that as 20 mA expecting you to use it in AMPS.
Equations like (5-1.4)/20mA might get evaluated wrong by a newby because the units have to be amps. Also watch K ohms/mA for 1K/1mA is the same as 1000/0.001.
So now, our batteries die and go to 4V min. We can check the battery datasheet to figure out what the lifetime and typical numbers are.
If we want to accept a little more voltage drop with better current regulation, we can make a current source.
A wise engineer recomputes the numbers and finds the max power his resistor has to dissipate and adds a safety margin or maybe use what the company bought too many of.
==
So, we just activated a LED that we can't see.
And we that other number of Peak Forward voltage requires some digging. Peak means pulse. So, you have to look at the graphs. If this was a remote control there could be peak pulses of current bigger than the steady state current which won;t destroy the LED.
In your case the LED is pure DC. Always on and not pulsed.
==
The unseen LED. The quick test: What is the voltage across the LED? 6V, then it might be leakage currents. 1.2V - It's probably on. 0V The battery is disconnected or broke; 0.3V probably broken.
But we still can't see it. OK, get out the cell phone camera. The camera sees blue and I see nothing. Going insane? Nah. Just the responses of the camera and the human eye to light.
--
The light works!!
Now for the detector. The easiest way is to use a voltmeter with a diode test mode and maybe one that beeps too.
So, we hastily hook out=r probes up in diode mode. OK. What? Ah!, the transistor has a polarity. Still nothing, Oh, the lights not On.
So, using the diode/beep mode on your meter, if we turn on the led (nothing blocking) the meter beeps and shows a diode drop.
Now, what? Well, I turn on the invisible LED and put a meter on the output and stick something opaque to IF light in the hole.
Insert it, meter beeps. remove it. Meter doesn't beep,
That's all for the lesson for today. Almost! p-n junctions don't like ohmmeters Their job is to fool them. Charlie's meter will act differently than Sam;s etc. and so will the 50K ohm/Volt meter.
Edits: typos
Under electro-optical characteristics we see some Vf values (1.2 to 1.4 V) and a current of 20 mA
This tells us that the LED would LIKE to operated around 20 mA and it will drop 1.2 to 1.4 Volts.
So, if the LED is working, you would measure 1.2 to 1.4 V ACROSS the LED.
So, to light a LED we need to know how our input voltage might vary and make sure the design fits within the ranges and doesn't go above the ABSOLUTE maximums. These let out the magic smoke.
Parallel batteries ARE not a good idea. Maybe we can work on that,
Those specs mean you have to have a supply voltage greater than 1.2V for anything to happen, it could be 3, 4.5, 12 or even 100V. At the low values of V, you get larger variations.
So if we have a 6V source and want 20 mA and our led drops a max of 1.4 V, we can select a resisistor by R=(6V-1.4)/0.020 (I might write that as 20 mA expecting you to use it in AMPS.
Equations like (5-1.4)/20mA might get evaluated wrong by a newby because the units have to be amps. Also watch K ohms/mA for 1K/1mA is the same as 1000/0.001.
So now, our batteries die and go to 4V min. We can check the battery datasheet to figure out what the lifetime and typical numbers are.
If we want to accept a little more voltage drop with better current regulation, we can make a current source.
A wise engineer recomputes the numbers and finds the max power his resistor has to dissipate and adds a safety margin or maybe use what the company bought too many of.
==
So, we just activated a LED that we can't see.
And we that other number of Peak Forward voltage requires some digging. Peak means pulse. So, you have to look at the graphs. If this was a remote control there could be peak pulses of current bigger than the steady state current which won;t destroy the LED.
In your case the LED is pure DC. Always on and not pulsed.
==
The unseen LED. The quick test: What is the voltage across the LED? 6V, then it might be leakage currents. 1.2V - It's probably on. 0V The battery is disconnected or broke; 0.3V probably broken.
But we still can't see it. OK, get out the cell phone camera. The camera sees blue and I see nothing. Going insane? Nah. Just the responses of the camera and the human eye to light.
--
The light works!!
Now for the detector. The easiest way is to use a voltmeter with a diode test mode and maybe one that beeps too.
So, we hastily hook out=r probes up in diode mode. OK. What? Ah!, the transistor has a polarity. Still nothing, Oh, the lights not On.
So, using the diode/beep mode on your meter, if we turn on the led (nothing blocking) the meter beeps and shows a diode drop.
Now, what? Well, I turn on the invisible LED and put a meter on the output and stick something opaque to IF light in the hole.
Insert it, meter beeps. remove it. Meter doesn't beep,
That's all for the lesson for today. Almost! p-n junctions don't like ohmmeters Their job is to fool them. Charlie's meter will act differently than Sam;s etc. and so will the 50K ohm/Volt meter.
Edits: typos
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