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Really basic led question ?

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crazyjohn

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Why does the led not burn out in this configuration if the forward maximum voltage of the led is 1.8v and this has the full 5v on the anode ?

5v->anode[led]cathode->resistor(330ohm)->0v
 
Because there is 3.2 V across the resistor.

That gives just under 10 mA through the LED, which is fine.

The absolute voltage is not important for any component. It is the voltage difference from one terminal to the other that matters.

The total voltage in your circuit is 5 V, and in a series circuit, that is equal to the sum of the voltages across each component. The voltage across the LED is 1.8 V, so that leaves 3.2 V across the resistor.
 
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Where's the schematic?

I am new to this forum and do not see any diagram of the posters LED circuit. Are you guys telepathic:) or am I missing the circuit?
Bruce in the Peg
 
You need to see that
5v->anode[led]cathode->resistor(330ohm)->0v
is the same as
5v->resistor(330ohm)->anode[led]cathode->0v
(series circuit)

then it will be obvious :)
 
Cartoon provided. :)

Ron
 

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NO not really. The idea is the LED and the resistor both have a voltage drop and those drops are additive in nature. So if I drop 1.8 volts (LED) and 3.2 volts (Resistor) I have the 5 volts I started with. We can't change the value of the LED short of a different LED but if we change the value of the series resistor all bets are off. :)

Ron
 
To expand somewhat on Ron's answer, the LED and resistor would have the same voltage drops, regardless of resistor value, but the current through both would change. This is because LEDs (like other diodes) have a (roughly) constant voltage drop throughout their current range. So if the LED in question is a 1.8v LED, it will always drop 1.8v, and the other components in series with it will have to take the rest of the voltage.
The downside is, of course, you need another component to handle the extra voltage as well as to limit the current. Diodes (once their knee voltage is reached) have a very low resistance. Not only that, they can only handle a small amount of current (a 1.8v standard red LED is somewhere around 22mA). Connecting the diode to exactly 1.8v will not work, since the low package resistance (less than 20Ω when it's 'on') would lead to overcurrent and a blown diode.
To remedy this, you add a resistor. To find the right value, use the following formula:
Resistor = (power supply voltage - diode voltage)/Spec current (this is on the diode datasheet, if you don't know and it's a typical looking diode, go for 20mA).
So, for a 5V supply and a 20mA LED, it would look like this:
R = (5v-1.8v)/20mA = 3.2/.02 = 160Ω
If we change that to a 330Ω resistor, the current would drop to about 10mA (3.2v/330Ω = 9.69mA) but the voltages would stay the same.

It's like paying your monthly bills, the phone company doesn't get all your money because it's the first bill. you pay what you owe them, then move on to the next bill. Same thing here. The LED takes what it needs, then sends the rest of the voltage on to the other components.
 
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HI
As JohnBruse replied, diodes have to response to the voltage after their cut in voltage.
Once it reaches that voltage, the voltage across the diode remains same but current increases exponentially.
So the diode doesnt burn since the current through it is limitted by the 330ohm resistor.
 
You may know what you want to say, but you are not conveying it properly.

With a 5V source and no resistor, theoretically the diode would drop 1.8V and the remaining 3.2V would be dissipated by the source. If the source has a small resistance, they current of this circuit will be very high (this is why we need the resistor).

The only phenomena which is "exponential" in nature is the power dissipated in the resistor or source.
 
"the diode doesnt burn since the current through it is limitted by the 330ohm resistor. "
This is the same thing mentioned in the response.
And about the exponential rise in current, what i meant was
even if the current rises to a high value , the voltage remains the same.
 
The forward voltage does not remain the same. When the current is increased then the forward voltage rises a little like any diode does.
 

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Ok, so the voltage does go up, but not by much.

From the graph, 1.5 V to 1.8 V as the current goes from 1 to 20 mA. That is about 15 Ohms. For a first approximation, which is probably good enough for a lot of applications, that 15 Ohms can be ignored when there is a 330 Ohm resistor in series.
 
The current is a function of the voltage, and the voltage is a function of the current.

That is because the voltage - current curve has always got a positive slope, and it doesn't have hysterisis.

The graphs happen to be drawn with the voltage on the x axis, but they could just as well be drawn with the current on the x axis. I don't see that either is necessarily the dependent variable.

The slope can be measured in ohms. I reacon that it is about 15 ohms, give or take, in the 1 - 20 mA range. It will be very different at other currents.That is 5% ish of the 330 ohm resistor.

15 ohms is only 5 % of the 330 ohm resistor so it is only going to make about 5% difference to the current, and for an LED that is a fairly small variation for most applications.
 
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