In the circuit below, when the switch is closed the LED won't turn on because R1 takes most of the current, am I right? But there's a voltage drop across D1/R1 about 1.4V. I'm just studying some behaviors in basic circuits and got a bit confused about this one.

 

Hello,

Have a look at the datasheet.
The green led has a voltage drop of 2.4 Volts (typ) @ 20 mA.
The 1.4 Volt will not light the led.

Calculator - https://www.allaboutcircuits.com/tools/parallel-resistance-calculator/

Bertus

 

hi blue,
Look at this LED datasheet,note the Green operating voltage.
E

 

Ok, that's because R1 forces that voltage drop on D1. If the switch is open, then the LED takes what it needs.

 

LED won't turn on because R1 takes most of the current, am I right?

Nope, Its not about current but voltage drop in this case, when you close the switch, the voltage at the junction of R1 and R2 became 11.8V, thus 13.25 - 11.8 = 1.45V, not enough Vf.

 

Ok, that's because R1 forces that voltage drop on D1.

hi,
Calculate the current flowing thru those 3 resistors, then calculate the voltage drop across the 750R.????
The voltage across the 750R is lower than the turn On voltage of the Green LED.
E

 

Ok, that's because R1 forces that voltage drop on D1.

The manufacturer doesn't give characteristics for forward voltage less than about 1.6V:

If the switch is open, then the LED takes what it needs.

In this situation, it would be more accurate to say it takes what it can get.

 

I'm struggling to understand why you would want to turn off an LED like that... I mean other than an exercise.

 

I'm struggling to understand why you would want to turn off an LED like that... I mean other than an exercise.

Actually, I saw that in an old design. That's why it caught my attention. But if I were the designer, I wouldn't do that.

 

Actually, I saw that in an old design. That's why it caught my attention. But if I were the designer, I wouldn't do that.

If they were going to do something that goofy, they could have just omitted the resistor and saved a penny.

 

LOL... good point! but it costs you slightly more in power which they don't seem to care about.

 

In the circuit below, when the switch is closed the LED won't turn on because R1 takes most of the current, am I right? But there's a voltage drop across D1/R1 about 1.4V. I'm just studying some behaviors in basic circuits and got a bit confused about this one.

@Blue_Electronx
Or perhaps it's about both voltage and current. So long as the voltage across the LED remains low enough that (essentially) no current flows, no light is emitted. Once the voltage is high enough to produce current flow, the LED lights. Once lighted, the light intensity is directly proportional to LED current, not voltage.

 

@TeeKay6 when you described it like above it reminded me of an analog peak light... maybe we are missing the big picture and it's not a DC source as shown and the signal goes somewhere after this... but I still fail to see why it has to be switchable... again missing the big picture.

 

Hello,

Have a look at this post:
https://forum.allaboutcircuits.com/threads/leds-555s-flashers-and-light-chasers.19075/#post-117639
Fig 3.2 shows a resistor/led only voltmeter.

Bertus