electronics question

[bigegg]

I should know the answer to this, but my brain is like mush at the moment/

If I want to power a 3V led (3.2 forwards voltage, 20ma) with a 5v supply, I use an 82ohm resistor.

If I want to power 2 LEDs from a 10v supply, I can treat each LED+resistor as a discrete component so I put in LED - resistor - LED- resistor.

actual question:

does it matter which order the components are in?
can I do LED - LED - resistor - resistor?

 

[rtbcomp]

If they're all connected in series the order doesn't matter. You could replace the two resistors with single resistor.

Voltage drop across the resistor will be 10 - (2 x 3.2) = 3.6 volts. Current draw = 20mA, resistor = V/I = 3.6/20 = 0.18 K Ohms = 180 Ohms

 

[bigegg]

If they're all connected in series the order doesn't matter. You could replace the two resistors with single resistor.

Voltage drop across the resistor will be 10 - (2 x 3.2) = 3.6 volts. Current draw = 20mA, resistor = V/I = 3.6/20 = 0.18 K Ohms = 180 Ohms

thanks for that.
makes designing the project easier, because I won't have to find room for the resistors


won't the current draw be 40ma for 2 LEDs?
so 90 ohms resistor needed?

 

[Seadog]

Not if they're in series.

 

[bigegg]

just discovered that I can get a better result by chaining four LEDs together across my 10v, without using a resistor -

 

[skotl]

just discovered that I can get a better result by chaining four LEDs together across my 10v, without using a resistor -

But then there is no current limiting. It may well be working now, but it's unlikely to do anything for the life of the LEDs.

Also; chaining the LEDs will mean that if one fails then none of them will light (much more likely without current limiting )

You know you can just solder resistors directly to the LED tails, right, to save some space? Be quick and/or use a croc clip or needle nose pliers as a heat sink close to the LED, to stop the heat damaging them while you solder them.

 

[rtbcomp]

Why do you need current limiting if the total voltage of the LEDs is equal to the supply voltage?

 

[ajlelectronics]

Why do you need current limiting if the total voltage of the LEDs is equal to the supply voltage?

Because LEDs are current devices rather than Voltage.

 

[skotl]

Why do you need current limiting if the total voltage of the LEDs is equal to the supply voltage?

Because without a current limiter, the LED will pull as much current as its internal resistance allows, which is almost certainly going to be more than its rated maximum.
This page gives an excellent explanation of how and why to calculate current limiting resistance for LEDs.

 

[rtbcomp]

^^ that link is pretty much what we've got already. The LED takes the current it needs at the forward voltage, if the supply is greater than the fwd voltage we need a resistor to drop the excess voltage, but when if the supply is 4 x the FWD voltage and we have 4 identical LEDs in series then each will have 1/4 the supply voltage across it, ie the forward voltage, so where's the problem?

Are you saying if we have a 3.2v supply capable of supplying say 5 amps it will blow the LED?

 

[500e]

Here you go @bigegg no brain required

http://led.linear1.org/1led.wiz

 

[dannyp]

Funny things leds. It’s forward voltage will remained pritty much pinned across the actual junction. If you had say a massive perfect led. It would illuminate from micro amps all the way through to 100’s of amps at exavctly the same voltage. In practice that doesn’t happen as mythical component dosent exist. But they are constant current devices treat them as anything else and they will pop way before they should

 

[bigegg]

ta very much.

 

[skotl]

^^ that link is pretty much what we've got already. The LED takes the current it needs at the forward voltage, if the supply is greater than the fwd voltage we need a resistor to drop the excess voltage, but when if the supply is 4 x the FWD voltage and we have 4 identical LEDs in series then each will have 1/4 the supply voltage across it, ie the forward voltage, so where's the problem?

Are you saying if we have a 3.2v supply capable of supplying say 5 amps it will blow the LED?

Yup - LEDs have an internal resistance, but it's very low. The current consumed would be voltage / internal resistance, so it may not pull the full five amps but it's likely to pull more than it is rated for. Cos we can never have enough links () there's a good explanation here: https://electronics.stackexchange.com/questions/76367/accounting-for-led-resistance

Funny things leds. It’s forward voltage will remained pritty much pinned across the actual junction. If you had say a massive perfect led. It would illuminate from micro amps all the way through to 100’s of amps at exavctly the same voltage. In practice that doesn’t happen as mythical component dosent exist. But they are constant current devices treat them as anything else and they will pop way before they should

As Danny says, they are constant current so you should limit that current as a factor of the voltage and max (minus some headroom) current.

---

Edit: I think this is the problem here;
"...The LED takes the current it needs at the forward voltage, if the supply is greater than the fwd voltage we need a resistor to drop the excess voltage..."

You're possible talking about using resistors as a potential divider to get a required voltage, and you've kind of done that by putting the diodes in series, but then forgot about limiting the current. The LED won't take the "current it needs" - it will consume the current that the circuit supplies, which is back to Ic = V/R (where you should supply "R" to be less than or equal to the rated current).

A clumsy example is if we put a nail across the live and neutral pins of a 240V plug it won't take the "current it needs"- it will consume all of the current that the circuit will supply.

 

[bigegg]

ok.
so if I want to power 4 LEDs off the 10v, I'm best off connecting 2 lots of 2 LEDs+2x 82ohm resistors in parallel?

 

[skotl]

ok.
so if I want to power 4 LEDs off the 10v, I'm best off connecting 2 lots of 2 LEDs+2x 82ohm resistors in parallel?

If you want them to draw 20mA each, then that means adding a resistive load of 125ohms or more (4 x 20mA = 80ma, 10V / 0.08). So you could put them in parallel and put a single resistor of 130ohms (or, if you have the 82 ohm resistors handy, put two of them in series to create a total resistance of 164 ohms, which is close enough).

 

[skotl]

Just realised that one other factor we've not been discussing during all of this is wattage for the resistor!

P (current in Watts) = IE (current * voltage)

So in the above diagram, current is 80mW 80mA (0.08), voltage is 10, so the power through the resistor would be 800mW (10 * .08). Most small resistors are 1/4 watt (250mW) so you'd need a 1W 130R resistor for this circuit

Edit: current units updated as per @130

 

[130]

current is 80mW (0.08),

milliamps

 

[ajlelectronics]

Far simpler to just do a parallel set with individual resistors. That will give redundancy too. Exact value of the resistors is unimportant, the brightness does not vary much over a range of currents.

 

[dannyp]

Shouldn’t realy parallel leds to one resistor ether one will allways take more power than the others.

String in series are fine as long as something is there to limit the current.