Hi, everyone!

Could someone give me a hand on this project? I'd like to make sure that my LED ( 3.0 V - 20mA) can operate in 3V and in 6V, but I'd like that it doesn't lose so much brightness when I just connect to 3V. What can I do?

 

This device (https://www.mouser.com/datasheet/2/268/20005448A-1021782.pdf) is *almost* what you want, but what is does is what you want. It won't work for you because the lowest voltage it can handle is 5V but the idea is a constant current source. As long as you keep, say, 20mA through the LED regardless of supply voltage it will stay the same brightness.

There are many approaches to constant current sources. I expect that people can offer their favorites. The advantage of a purpose built component is simplicity, but there might be a device like this one that has the voltage range you need.

 

What can I do?

Use a current source so you have a current of 20mA regardless of whether you use a 3V or 6V source.

 

If your just connecting it to just 3V or 6V use ohms law: R=E/I. Where E is the voltage you want to drop from the 6V which is 3V @ 20mA. So we get R=3/.02=150 ohms. Add this is series on one leg of the LED. Remove it for 3V.

 

If the forward voltage of your LED is 2.3V or less, you can use this circuit:


If it's higher than 2.3V, but less than about 2.6V, you can replace D2 with a Schottky diode and adjust R2 accordingly.

 

Same circuit basically but shouldn't need to change any parts.

 

use a small 3pin ldo and a resistor to make a constant current source ?

https://www.maximintegrated.com/en/design/technical-documents/app-notes/4/4404.html

 

Same circuit basically but shouldn't need to change any parts.
View attachment 232071

How did you find this resistor values?

 

How did you find this resistor values?

\(I = \frac{V}{R} = \frac{0.7V}{20mA} = 35\Omega\)
The nearest 5% value is 33 ohms which would give about 21mA.

 

\(I = \frac{V}{R} = \frac{0.7V}{20mA} = 35\Omega\)
The nearest 5% value is 33 ohms which would give about 21mA.

What about the 2.2k resistor? Why this value was chosen? The 0.7V is the voltage-drop across the base-emitter of the transistor?

 

If the LED is fairly bright when fed 3V and if the 3V is from a battery (2 AA cells in series?) then it will drop to 2V soon and you can watch the LED get dimmer and dimmer until the voltage is too low for it (2.8V?).

You said your LED is 3V. Did you measure it? Nobody makes a 3V LED, a white or blue LED is rated at 2.8V to 3.6V and yours will be somewhere between those voltages.
If your LED needs 3.6V but you feed it 3.0V then it will not light.
If your LED needs 2.8V and you feed it 3.0V without limiting the current then it will burn out soon.

 

What about the 2.2k resistor? Why this value was chosen? The 0.7V is the voltage-drop across the base-emitter of the transistor?

That resistor needs to provide base current to Q2 and collector current to Q1 while not dropping more than than 1.6V. With a supply voltage of 3V, it would have 1.6V across it giving it a current of 0.73mA. If Q2 needed to saturate, it would require a current of 2mA; so it would appear that that resistor is too large for a 3V supply. It would work with 6V.

In my version of the circuit, the resistor is also a bit too large; 620 ohms would be better.

 

my LED ( 3.0 V - 20mA)

A white LED?

 

The 0.7V is the voltage-drop across the base-emitter of the transistor?

Of Q1.

 

This Current Regulator is available in 20ma, and 25ma versions,
and has only 1-Volt Forward Voltage .
Good to 90- Volts in the D-Pack Package,
available in TO-92 Transistor Package also.
CL520
CL525
.
.
.

 

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This Current Regulator is available in 20ma, and 25ma versions,
and has only 1-Volt Forward Voltage .
Good to 90- Volts in the D-Pack Package,
available in TO-92 Transistor Package also.
CL520
CL525

isn’t the minimum voltage 5V on those?

 

You're Right.
I never tried running one on 3.3V, I forgot that point.
It is possible that it would work anyway.
Same with your Part.
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