Nobody makes an LED with a voltage rating like you say. For 400mA the forward voltage of your LED Might be 3.1V (typical) but it also might be 2.8V (minimum) or 3.4V (maximum) or anywhere in between. The forward voltage of an LED also changes when its temperature changes.
Please post the part number or datasheet of the LED so we can show ya.
As AudioGuru implies putting LEDs in parallel is a BAD idea unless they are perfectly matched.
But I would disagree with him, that the factory matches the Vf exactly. Usually they rely on the batch or lot having the same Vf. {so they are not always having exactly the same Vf}
Yours worked because the small batteries limited the current.
They will probably start to fail with a more substantial power supply,
If you insist, use a 7805 wired as constant current [ https://www.eleccircuit.com/7805-current-constant-for-battery-charger/ ] and vary the resistor R1 in the link between appropriate max current and min current.
Max current will likely by 23 * 20mA
Min current [low enough so the LEDs are barely on] could be 23 * 1 mA
To change the brightness of an LED you are supposed to change the current, not change the voltage fed to it.
The forward voltage of an LED decreases as it heats up. Then if you feed it a steady voltage its forward voltage drops which increases its current which makes it hotter which makes the current increase more which makes it hotter which makes the current increase more which makes it hotter which makes the current increase more which makes it hotter .... This is called "thermal runaway" and can destroy the LEDs.
Instead you feed it a variable current to change its brightness. A current source can simply be a resistor in series with the LED. When the resistor has some voltage across it then the current does not change much when the LED heats and its forward voltage drops. You want your paralleled 23 LEDs to have a maximum total current of 400mA so each one is about 400mA/23= 17.4mA (20mA is normal for a bright 5mm diameter LED). Then use a resistor that is about 1V/400mA= 2.5 ohms (use 2.7 ohms) rated at (400mA squared x 2.7 ohms x 2=) 1W.
Feed the series resistor and LEDs a variable voltage from 2V or 3V to 4.2V.
You do not want an LM317 current limiter because a simple resistor in series with the LEDs limits the current and converts changing the voltage into changing the current.
The circuit you found uses a 2k ohms trimpot. Then when the trimpot is set to halfway the output voltage from the circuit is 6.9V. Turn it down so the voltage is about 4.2V for maximum brightness when the series resistor is 2.7 ohms.
The slow turn on circuit will not immediately light the LEDs dimly but instead the LEDs will not turn on until the voltage has risen to about 2.4V, a long delay time. You probably want the LEDs to immediately light dimly then slowly get brighter. Then the transistor in the circuit must be part of a voltage divider so that the output voltage starts at about 2.4V. Add a 220 ohms or 270 ohms resistor in series with its collector.
You do not want an LM317 current limiter because a simple resistor in series with the LEDs limits the current and converts changing the voltage into changing the current.
The circuit you found uses a 2k ohms trimpot. Then when the trimpot is set to halfway the output voltage from the circuit is 6.9V. Turn it down so the voltage is about 4.2V for maximum brightness when the series resistor is 2.7 ohms.
The slow turn on circuit will not immediately light the LEDs dimly but instead the LEDs will not turn on until the voltage has risen to about 2.4V, a long delay time. You probably want the LEDs to immediately light dimly then slowly get brighter. Then the transistor in the circuit must be part of a voltage divider so that the output voltage starts at about 2.4V. Add a 220 ohms or 270 ohms resistor in series with its collector.
Your resistor value and voltage are so high that your circuit wastes a lot of power making heat.
The resistor heats with (0.44A squared x 20 ohms=) 3.9W. But if you use 2.7 ohms then it heats with only (0.44A squared x 2.7 ohms=) 0.52W.
You are not applying 12V to the LEDs. Instead you are applying 12V to the 20 ohm resistor in series with the LEDs so that the maximum current is 440mA.
If you use a 2.7 ohm resistor in series with the LEDs and feed it enough voltage that the current is 440mA then the LEDs will be exactly as bright as when you used 20 ohms making a lot of heat and 12V.
The voltage across the 2.7 ohm resistor is calculated with Ohm's Law: V= 440mA x 2.7 ohms= 1.188V. The maximum (brightest LEDs) output of the LM317 will be the LED voltage (3.1V) plus this 1.188V= 4.288V. For the dimmest LEDs you want the output of the LM317 to be about 2.4V so the resistor in series with the voltage-control pot should be 200 ohms. For the brightest LEDs the output of the LM317 should be 4.288V so the pot should have across it (4.288V - 1.25V=) 3.04V. If you use a pot with a total resistance of (1070 ohms - 200 ohms=) 870 ohms (use a 1k ohm pot) then you will have plenty of adjustment range.
Does "headed up" mean heating up?if I do it like you say and feed the regulator with a 12v input, the regulator will also be headed up too much wouldn't it?
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