Voltage divider output voltage

[PNeil]

Hello forum, I need to step down 24 volt DC to 3.2 volt to drive a bicyle LED tail light. Current drawn from light is 19ma which should be 169 ohms at 3.2 volts. Voltage is supplied from two 12v battries. When fully charged voltage is around 27v so i will design for that voltage.

Making R1 5k and R2 670 should get me 3.2 volt at the junction. When factor in the 169 ohm resistance of the load, the output voltage will drop to 709.6mv due to R2 and the load now in parrallel.

My question is should i design the circuit to output 3.2 volt while the load is connected or 3.2 volt with no load connected.

Chaning R1 to 1k should give me 3.2v when the load is connected but 10.8v when no load is connected.

When i experiment with dry cell battries almost no voltage is lost when the load is connected. Now i curious to know why.

 

[alec_t]

One resistor is all that's needed. Taking 3V as the LED forward voltage, the voltage drop is 24-3=21V. At 19mA current the resistor value will be 21/19 kOhms = 1.2k (nearest standard value). It will dissipate 21 x 19 = 400mW, so use a 1W rated resistor.

 

[PNeil]

Thank you very much. I was wondering if i could use 1 resistor in series but thought i didn't know the forward voltage and forward current for the LED to make the calculation. R=(vs - vf) / lf. Thanks for letting me know that the voltage required to operate the light is the forward voltage (vf) and the current drawn by the light is the forward current. (lf). After drawing the circuit everything you said become clear.

I will experiment with 27 volts as Vs because when those batteries are charged the voltage is around 27.0v. In that case R should be 1263, i think the closest value is still 1.2k. Real componet will need adjustment anyway...

 

[alec_t]

Real componet will need adjustment anyway...

I don't see why. If the LED forward voltage is anywhere from 1V to 4V the voltage drop across the resistor is going to be 26V to 23V (assuming a 27V battery). That's only about 12% difference; which won't produce an observable LED brightness change or significant power dissipation change.

 

[audioguru]

An LED does not work from a voltage like a light bulb. A light bulb is a simple resistive piece of white hot wire. But an LED is a diode that sets its own voltage (its forward voltage) and works with the amount of current you feed it. A single series resistor sets the amount of current.

 

[Nigel Goodwin]

Thank you very much. I was wondering if i could use 1 resistor in series but thought i didn't know the forward voltage and forward current for the LED to make the calculation. R=(vs - vf) / lf. Thanks for letting me know that the voltage required to operate the light is the forward voltage (vf) and the current drawn by the light is the forward current. (lf). After drawing the circuit everything you said become clear.

I will experiment with 27 volts as Vs because when those batteries are charged the voltage is around 27.0v. In that case R should be 1263, i think the closest value is still 1.2k. Real componet will need adjustment anyway...

You should design based on 24V, the rated voltage - the difference when fully charged is insignificant.

However, it's a really, really poor way to feed an LED from such a high voltage (very wasteful - only about 10% efficient) - why do you have two 12V batteries in series?.