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Picking current limiting resistor for LED circuit.

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alphacat

New Member
I'd like to drive a two anodes, common cathode, bicolor LED.
One of the LED's colors is blue, which according to datasheet, its min. VF equals 3V and max. VF equals 3.8V.
I need to order hunderds of this led circuit (led+driver circuit).
In this case, how can I design a driver circuit that will be appropriate for a range of VF of 3V to 3.8V?
On the one hand, I dont want all 3V blue LEDs to get burned, but on the other hand, I dont want the 3.8V ones to have a weak lighting intensity.

Thanks.

--
More details.
I have two power sources which I can use to drive the LED, 5V and 3.3V.
 
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Boncuk

New Member
Hi,

you will have to use separate current limiting resistors for each colour. Connect the common cathode as usual and connect the current limiting resistors to the anodes.

Boncuk
 

alphacat

New Member
I think that you didnt understand the problem I presented.
First, I wasnt speaking about the other color, only about the blue color.
Second, the problem which I presented was that the blue color's VF ranges from 3V to 3.8V therefore I dont know what current limiting resistor to pick.
As i said in the topic: "On the one hand, I dont want all 3V blue LEDs to get burned, but on the other hand, I dont want the 3.8V ones to have a weak lighting intensity".
 

alphacat

New Member
What value should I write in the Voltage Drop Across LED field?
As I said, my problem is that "On the one hand, I dont want all 3V blue LEDs to get burned, but on the other hand, I dont want the 3.8V ones to have a weak lighting intensity".
And another important detail (which I mentioned) is that I need to order hunderds of this LED circuit (LED + LED driver circuit).
 
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ericgibbs

Well-Known Member
Most Helpful Member
help anyone?

hi,
Assume the Vfwd is the centre voltage of the two limits [3.4V].

Calculate a resistor for 5V operating the 'average' LED at 10% less than the recommended operating current.

What is the recommend mA.???
 
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BrownOut

Banned
You'll probably get satisfactory performance from your led at Vf=3V. Prototype your circuit before ording hundreds of parts.
 

crutschow

Well-Known Member
Most Helpful Member
If the current variation due to LED differences is too large using a resistor to limit the current, you could use a constant current source consisting of a transistor or op amp and associated circuitry (Google "constant current source" for circuit examples).
 
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ericgibbs

Well-Known Member
Most Helpful Member
You'll probably get satisfactory performance from your led at Vf=3V. Prototype your circuit before ording hundreds of parts.

hi,:)

One way would be to select from the total batch, groups that fall into tighter Vfwd spreads and then calc a resistor for each of the groups.
But IMO thats just hard work, unless the LED's are used in a 'matrix' type display or layed out so that the different intensities are required to match.
 

audioguru

Well-Known Member
Most Helpful Member
LEDs are driven from current, not voltage. If you drive them with a current source circuit then their forward voltage does not matter if it is different for each LED. Each LED will have the same current and will have the same brightness.

But a current-limiting resistor is a lot simpler than a current source circuit.
You cannot use a 3.3V supply so calculate the value of the current-limiting resistor using a 5.0V supply:
1) If the LED has a forward voltage of 3.0V and its max allowed current is 30mA then calculate a resistor to supply 25mA. (5.0V - 3.0V)/25mA= 80 ohms. Use 82 ohms which is the closest standard value then the current will be 24.4mA.
2) If the LED is actually 3.8V then its current will be (5.0V - 3.8V)/82 ohms= 14.6mA which will look a little dimmer than the 3.0V LED. Your vision's response to brightness is logarithmic so the LED will not look like half the brightness, just a little less brightness.
 

BrownOut

Banned
hi,:)

One way would be to select from the total batch, groups that fall into tighter Vfwd spreads and then calc a resistor for each of the groups.
But IMO thats just hard work, unless the LED's are used in a 'matrix' type display or layed out so that the different intensities are required to match.

IMO, there probably won't be much difference in brightness from one to another if using an 'nominal' forward current ( see AG above ) Any units that show extreme weak brightness are rejects.
 

MikeMl

Well-Known Member
Most Helpful Member
The missing data in this entire thread is what is the supply voltage?

I think the OP's confusion stems from the fact that the variation in Vf becomes less important as the supply voltage is increased. e.g., if the supply voltage is 12V, then a variation of Vf from 3V to 3.7V causes an insignificant variation in the LED current because the drop across the series resistor goes from 8.3V to 9V; an 8% variation.

OTOH, if the supply voltage is only 4V, then the drop across the series resistor would go from 0.3V to 1V, a 333% variation. This is where an active current limiter (transistor or IC) would make sense.
 
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alphacat

New Member
Thank you guys.

To ericgibbs,
The IF is 20mA.
applying 3.4V on a 3V LED wont burn it?

To BrownOut,
I already have a bicolor led circuit, only with Green and Red colors, and in datasheet, they gave the typical VF of each color, no min. and max. values.
Therefore I didnt have the conflict of picking current limiting resistor, since i used the typical value as the VF.
When giving typical value, it means that still the VF can spread out over 0.8V (for example from 3V to 3.8V)?


To crutschow and audioguru,
Yeah I actually got to current suppliers chapter in the book i read(the course book of analog circuits which deals with amplifiers).
Hopefully I could use what i'd lean there.

MikeML,
Havent though about it like that, thanks :)
 
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BrownOut

Banned
At first I thought that the high and low limits on Vf were specified for min and max forward current. But you're saying that this tolerance is given for a single value of forward current???

I never realized led's had such loose tolerances.
 

audioguru

Well-Known Member
Most Helpful Member
If an LED is 3.0V at 20mA then at 3.4V it might try to draw 67mA which will burn it.
You must limit the current. The LED takes care of the voltage.
 

ericgibbs

Well-Known Member
Most Helpful Member
hi,
With a fixed 5V supply and a max of 20mA/LED and a spread in Vfwd of 3.0V thru 3.8V, using a same value resistor for all LED's of 95.3R.

The graph shows the range of currents and power dissipation for each LED.

It would help if you posted the LED type so that we could check the d/s.
 

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BrownOut

Banned
In the case of loose Vf tolerance, one more spec should be considered, that is luminosity as a function of current. For example, I found a spec for a device which gives these values for relative luminous flux v. forward current:

.9 @ 300ma
.65 @ 250ma
.6 @ 200ma

So, for example, assuming nominal forward voltage drop of 2.5V, and max and min of 3.0V and 2.0V respectively, and assuming 5V supplies, we can easily determint the relative luminosities:

case 1: Vf=2.5V (nominal)

For I=250ma, R=2.5V/.250 = 10 ohms.

relative radiation = .65

Case 2: Vf=2.0V (minimum)

I=3V/10ohms=300ma

relative radiation = .9

case 3: Vf=3.0V (maximum)

I=2V/10ohms=200ma

relative radiation = .6

So, by designing for 'nominal' value of Vf, then using a device at the 'high' limit of Vf gives about 8% less radiation. That might or might not be acceptable. Also, for Vf at the 'low' limit gives a current that might or might not burn the LED.

If any of these limits are unacceptable, then you must either use a different LED, a different voltage or use a current driver.
 
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MikeMl

Well-Known Member
Most Helpful Member

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