Determining supply current/voltage to LED

  • Thread starter hl_world
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In summary, in order to supply an LED with a maximum current of 25mA and a maximum voltage of 2.5V, a minimum resistance of 40Ω should be used. The LED will draw less than 3mA of current with a 3V supply, and using a voltage divider with a 1kΩ and 2kΩ resistor, point C will be at 2V above ground. Due to the LED's resistance, some current will still flow through the LED+40Ω branch, but the majority of the current will flow through the 1kΩ and 2kΩ resistors.
  • #1
hl_world
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Bearing in mind safety margins, how much current and voltage should I supply an LED if it's rated as such:

Forward current max: 25mA
Forward voltage max: 2.5V
Reverse voltage max: 5V
Light output min.@ 10mA: 70mcd
Light output typ.@ 10mA: 200mcd
 
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  • #2
Looks like <25mA current and <2.5V voltage.
 
  • #3
hl_world said:
Bearing in mind safety margins, how much current and voltage should I supply an LED if it's rated as such:

Forward current max: 25mA
Forward voltage max: 2.5V
Reverse voltage max: 5V
Light output min.@ 10mA: 70mcd
Light output typ.@ 10mA: 200mcd

In addition to what Mapes said, in order to limit the current to 25mA, you need to know the *minimum* Vf of the LED.

Quiz Question for hl_world -- Why is this so?
 
  • #4
berkeman said:
In addition to what Mapes said, in order to limit the current to 25mA, you need to know the *minimum* Vf of the LED.

Quiz Question for hl_world -- Why is this so?
Hmm.. I think this was in the analogue electronics part of my course but I left after 6 months and that was a few years ago.

So I should try 20mA & 2.0v then? Like this:
http://img526.imageshack.us/img526/631/ledvoltdivcircuitcy4.png
 
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  • #5
It's common practice to do this with just one resistor, in series with the diode:

circuit_LED_basic.gif


EDIT:
"Answer me these questions three..."

Assume the LED voltage is 2 V. What's the voltage across the resistor?
Assume the LED current is 25 mA. What's the current through the resistor?
To have the voltage and current calculated in the first 2 questions, what's the resistance of the resistor?
 
  • #6
I don't understand. I used the voltage divider to ensure 2V between 2 nodes at either side of LED. Does the LED just do that for me by having resistance of its own?

1) 1 volt (3V source - 2V over LED)
2) 25 mA (same throughout circuit)
3) R=V/I so 1/0.025= 40Ω

(To the best of my understanding)
 
  • #7
hl_world said:
1) 1 volt (3V source - 2V over LED)
2) 25 mA (same throughout circuit)
3) R=V/I so 1/0.025= 40Ω

(To the best of my understanding)
Yes, that's right. Once you set up the circuit, you can measure the power supply and LED voltages, and adjust the resistance if needed. But 40Ω is a good safe starting value.

I don't understand. I used the voltage divider to ensure 2V between 2 nodes at either side of LED. Does the LED just do that for me by having resistance of its own?
Yes, the LED is essentially guaranteed to have 2 V (or could be as high as 2.5V) -- as long as it is not connected directly to a fixed voltage source, and it draws some minimal amount of current.

Looking at your voltage divider circuit, here is an observation: because of the 1kΩ resistor, the 3V supply will not produce more than 3 mA of current. So the LED will get less than 3 mA of current.
 
  • #8
http://img4.imageshack.us/img4/7349/ledvoltdivcircuiths5.png

Here, I have labeled the different points in the circuit (except switch & -/+ nodes) for reference and reduced the 100Ω resistor to 40Ω.

Now one of the things I don't get about voltage & current is that between points B & D there is a 2v supply. This will cause the current to flow through the 40Ω resistor a lot more than the 2kΩ.

Using the MAD rule (2000x40)/(2000+40)= 39.216Ω. The current should be 2v/39.216Ω = 51mA which divides between resistors before combining at the negative node. So the current which flows through to 40Ω & LED should be 0.051 x (2000/(40+2000))= 50mA.
I know I've done something wrong here.
 
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  • #9
hl_world said:
& current is that between points B & D there is a 2v supply. This will cause the current to flow through the 40Ω resistor a lot more than the 2kΩ.
That would be true if there were only 40Ω in parallel with the 2kΩ. However, it is an LED+40Ω series combination in parallel with the 2kΩ. We don't know what the effective resistance of the LED+40Ω is, so we can't say that more current flows through that path.

Also, thinking of this as a voltage divider: the resistance of the lower section must be less than 2kΩ, due to the LED+40Ω that is in parallel with the 2kΩ resistor. That would mean VBD is less than 2V.

Another observation: if any appreciable current does flow through the diode, it would have close to 2V, which means close to 2V between C and D. But there is also 2V (or close to it) between B and D. So therefore a very small voltage is between B and C. Just how small we don't really know, but if the circuit were actually built one could measure VBC, and divide it by 40Ω to get the actual current in that path.

One cannot ignore the effect of the LED on the LED+40Ω branch of the circuit.

Hope that helps clear things up :smile:. If not, keep posting. You have a pretty good grasp of the basics, so that helps a lot in composing answers to your questions.
 
  • #10
hl_world said:
http://img4.imageshack.us/img4/7349/ledvoltdivcircuiths5.png

Just thought of a simplified explanation for what's going on in this circuit.

Assume an idealized 2.0V LED:
i=0 for V < 2.0V
V = 2.0V for i > 0​

If ANY current flows through the LED, it will be at 2.0V. That puts point C at 2.0V above ground ( ground is point D or E).

The voltage divider will tend to put point B at 2.0V above ground also. That would put 0V across the 40Ω resistor (points B and C at the same potential), hence zero current through the 40Ω and LED (path BCD).

With zero current going through path BCD, all current must flow straight through the 1kΩ and 2kΩ resistors. This current is
i = 3V / (1+2)kΩ = 1 mA​

So we have:
A at 3V
B & C at 2V
D & E at 0V

0 mA through path BCD
1 mA through path AE

Hope that helps. In reality there will be a small fraction of the current taking path BCD through the LED, and the LED voltage will be a little less than 2.0V.

Regards,

Mark
 
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  • #11
hl_world -- I haven't read the last couple posts in detail, but do not put a voltage divider around an LED. If you do that in a job interview, you will be shown the door.

You put a resistor in series with the LED to determine the LED current, based on the supply voltage and the expected LED forward voltage drop. Nothing else.
 
  • #12
berkeman said:
... do not put a voltage divider around an LED. If you do that in a job interview, you will be shown the door.
I would add, "Never use a voltage divider to power something, use them only for making a voltage reference". Would you agree?
 
  • #13
Redbelly98 said:
I would add, "Never use a voltage divider to power something, use them only for making a voltage reference". Would you agree?

Absolutely. Otherwise, you're just wasting power for no reason.
 
  • #14
Thanks, Redbelly & berkeman. It's been a while since college but recently I've been building LED circuits on a breadboard. Just 1 more question for now though:- The specs of one of the LEDs I ordered says it should be supplied at max forward current/voltage of 100mA/4V. Why does it specify both; wouldn't current be the only relevant factor?
 
  • #15
hl_world said:
Thanks, Redbelly & berkeman. It's been a while since college but recently I've been building LED circuits on a breadboard. Just 1 more question for now though:- The specs of one of the LEDs I ordered says it should be supplied at max forward current/voltage of 100mA/4V. Why does it specify both; wouldn't current be the only relevant factor?

That's probably just the diode voltage at 100 mA. They supply the (nominal) operating voltage along with the current so you can easily figure out whether or not the power supply you have is appropriate for turning it on.

EDIT: For example, if you had a 3 V, 1 A (max) supply available, would you spec out LEDs with a 4.5V voltage?
 
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  • #16
Well, I figured it was a simple case of taking the supply voltage, dividing it by the ideal forward current of the LED and applying that resistance to the series circuit.
 
  • #17
hl_world said:
Well, I figured it was a simple case of taking the supply voltage, dividing it by the ideal forward current of the LED and applying that resistance to the series circuit.

Well, this would give higher current than you want. It's better to use the following:
[tex]R_{LED}=\frac{V_{Supply}-V_{LED}}{I_{LED}}[/tex]

EDIT: More info:
http://alan-parekh.com/led_resistor_calculator.html
 
  • #18
hl_world said:
Well, I figured it was a simple case of taking the supply voltage, dividing it by the ideal forward current of the LED and applying that resistance to the series circuit.

Not sure what you mean by that. To bias an LED, you subtract the LED Vf from the supply voltage (and any other voltage drops, like the Vol of a drive gate, or Vsat of a driving transistor), and divide that resistor voltave Vr by the current you want to have passing through the LED. That determines the value of the series resistor.
 
  • #19
I mean I thought if you connected an LED that needs 4.5v / 100mA in a series circuit, and it was powered by a 3v supply, you would add a 30Ω resistor, it will get a 100mA flow and that would be it (ignoring voltage requirements).
 
  • #20
If the LED needs 4.5 V, then a 3V supply will never be able to power it, no matter what resistor you use.

As berkeman said, you subtract the LED voltage (4.5) from the supply voltage, and then divide by current to get the resistor.

The supply voltage must be greater than the LED voltage for this to work.
 

1. How do you determine the supply current for an LED?

To determine the supply current for an LED, you will need to know the forward voltage, or Vf, of the LED and the maximum current rating of the LED. The supply current can be calculated by dividing the supply voltage by the forward voltage and then multiplying by the maximum current rating. For example, if the supply voltage is 5V and the LED has a Vf of 2V and a maximum current rating of 20mA, the supply current would be (5V/2V) * 20mA = 50mA.

2. How do you determine the supply voltage for an LED?

The supply voltage for an LED can be determined by multiplying the forward voltage of the LED by the desired supply current and then adding any additional voltage drops from resistors or other components in the circuit. For example, if the LED has a Vf of 3V and a desired supply current of 30mA, and there is a 100 ohm resistor in the circuit with a voltage drop of 1V, the supply voltage would be (3V * 30mA) + 1V = 10V.

3. What is the importance of determining the supply current and voltage for an LED?

Determining the supply current and voltage for an LED is important because it ensures that the LED is operating within its safe operating parameters. Supplying too much current or voltage to an LED can cause it to overheat and potentially fail. Additionally, knowing the supply current and voltage allows for proper selection of components in the circuit to ensure the LED is receiving the necessary power for optimal performance.

4. How do you measure the forward voltage of an LED?

The forward voltage of an LED can be measured using a multimeter. Set the multimeter to the diode test function and place the positive lead on the anode of the LED and the negative lead on the cathode. The voltage reading on the multimeter will be the forward voltage of the LED.

5. Can the supply current and voltage for an LED change over time?

Yes, the supply current and voltage for an LED can change over time due to factors such as temperature, aging, and fluctuations in the power supply. It is important to regularly check and adjust the supply current and voltage for an LED to ensure it is operating within safe parameters.

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