Please check if I did a correct circuit for my idea.

This is for my motorcycle' tail light.

The tail light of a motorcycle (or any vehicle for that matter) operates in two modes. 1; used during night time, the tail light is dim and becomes bright whent he brake is used. 2; used at day time, the light is off until brake is used.

My idea is to use LED and have the current regulated. The current will also be used to make the LEDs shine brighter by providing more current when the brake is used.

Attached here is the diagram...

The current alteration is done by reducing the resistance of the resistor between pin 1 and 2 of LM317. This is done by connecting another resistor in parallel to the existing resistor. This is done by using an NPN transistor as a switch.

So, everytime the brake is used, the transistor is activated. At night time, the circuit still works even if the brake is not used.

My question is, did i do it correctly? It seems to be working but I am open to other idea.

So, nothing's wrong with my circuit, eh?

WildSwan said:

So, nothing's wrong with my circuit, eh?

Click to expand...

No reply doesn't necessarily mean nothing is wrong with your circuit.

The output voltage has to be taken off the adjust pin.

Also think about using a miniature relay to short one resistor. It's more accurate.

Please compare.

Currents have been calculated using Iout=Vref/Rref.

Iout(1)=1.25V/62.5Ω - Iout(1)=0.02A (20mA)
Iout(2)=1.25V/1,062.5Ω - Iout(2)=0.0011764A (1.17mA)

Boncuk

Why go to the expense of using an LM 317?

Just use resistors and a transistor.

colin55 said:

Why go to the expense of using an LM 317?

Just use resistors and a transistor.

Click to expand...

Show me then. I have few LM317 in my parts bin and it's the simplest circuit.

The first problem i see is that you need a diode to block back EMF into the LM317. The 1n4001 should be in parallel with the LM with the anode on the output.
secondly, you can probably get away with using an 78l12 or 7812 regulator, depending on how much current you plan on pulling.

Here's all you need:


**broken link removed**

Boncuk said:

No reply doesn't necessarily mean nothing is wrong with your circuit.

The output voltage has to be taken off the adjust pin.

Also think about using a miniature relay to short one resistor. It's more accurate.

Please compare.

Currents have been calculated using Iout=Vref/Rref.

Iout(1)=1.25V/62.5Ω - Iout(1)=0.02A (20mA)
Iout(2)=1.25V/1,062.5Ω - Iout(2)=0.0011764A (1.17mA)

Boncuk

Click to expand...

I appreciate the circuit you've posted but it will not work with the tail light of my motor cycle.

Maybe you can help me better if I explain it again.

There are three lines going to the tail light;

Line 1, is a positive line which get switched on everytime you hit the brake.
Line 2, is a positive line which get switched on when you turn on the head lamp, which is usually when it is dark.
Line 3, is a common ground.

The bulb has two filaments inside. One is rated at 10 watts, this is for braking. The other is rated at 5 watts, the tail light. The low rated filament is there so that you can be seen when it is dark. Of course, this line stays on as long as the head lamp is operating.

Now, I am replacing the bulb with LED. The current will help me emulate the two filaments of the bulb. Lower current will simulate the tail light and when I hit the brake, the current should be at maximum settings for maximum LED glow. This is why on my diagram, I have the L+ (tail light), B+ (brake light) and G (ground) connection. My idea is that everytime the brake (B+) gets activated, I would lessen the resistance accross Adj and Out pins of LM317, Making it produce more current to drive my LEDs.

My question is, am I using the transistor effectively as a switch to parallel a resistor to an existing one to increase the current?

Thanks.

You simply connect Line 1 and line 2 together and take them to the two resistors in the circuit above. The other wire is "common" or "earth" or "negative" or "Chassis" or "0v" and the cathode of the LED goes to this line.
The LED in the diagram represents about 20 LEDs connected in series/parallel and the resistors represent the value of resistance for each "string."