I have to simply get 38 volts 16 MA down to about 2 volts 16 MA (to control a LED). What would be the best way of doing it?

 

A 2.2k resistor in series with the LED will limit current to slightly less than 16ma depending on the voltage drop across the LED.

 

Led's are really current operated devices. They take a certain voltage to place them in conduction (that 2 volts you mention), but it's all current after that. Your simplest way out is to use a 2.2K resistor in series with the voltage source to limit the current. Here's a link into our Ebook that describes LED's - http://www.allaboutcircuits.com/vol_3/chpt_3/12.html

 

Hello,

Is the incoming voltage AC or DC?
You will loose a lot of power in the resistor.
If the voltage is DC the resistor will be (38 - 2) / 0.016 = 2250 Ohms.
The nearest higher value is a 2700 Ohms resistor.
The current will be than 36 / 2700 = 0.013333 = 13.3 mA.
The power dissipated by the resistor is 0.013333 * 0.013333 * 2700 = 0.4799 Watt.
A 1 Watt resistor of 2700 Ohms will be needed for your project.

Bertus

 

The voltage coming in is DC. Sorry I didn't specify that. Thanks for the quick replies. I appreciate also showing how you come to the conclusions that you guys came to. I absolutely love this forum. I am in the process of reading allaboutcircuits book which is awesome. I will then try to find a resistor that matches these specs. Again, I very much appreciate the quick replies.

 

That should be easy - 2.2K 1/4 watt is a very standard value. You can even use 1/2 watt if that's what the store has.

 

That should be easy - 2.2K 1/4 watt is a very standard value. You can even use 1/2 watt if that's what the store has.

I would never have a 1/4W or 1/2W resistor dissipate 0.59W. It would be very hot, maybe "incandescent".

 

I put a 1 watt 2.7k ohm resistor in this circuit and the resistor is still getting very hot. Would this be normal for a 1 watt resistor? If it is normal, would there be anything wrong with going with a 2 watt resistor--this would keep temperatures down right?

 

Hello,

A resistor of 1 Watt dissipating about 0.5 Watt will get rather warm.
If you want to have less heat produced you can take a 2 or 5 Watt resistor.
The higher the power of the resistor the more the heat is spread.
The total amount of heat will be the same.

Bertus

 

Thanks Bertus for first of all your reply a few days ago. I appreciate your explanation. And thanks for your recent reply too.

 

Hello,

No thanks. We are here to help.
For some background information on resistors you could take a look at the following page:
http://hobby_elec.piclist.com/e_resistor.htm

This is a link from the EDUCYPEDIA page on resistors:
http://www.educypedia.be/electronics/resistors.htm

There are pages like that for other components too:

Passive semiconductors technology: Batteries Diodes Photovoltaic systems Switches Buzzers Diode types Quartz crystals Thermoelectric Capacitors General overview Relays Transformers Capacitor types Heat Sinks Resistors Transformers-RF Coils Photo diode Resistors-Nonlinear Transformer types

Greetings,
Bertus

 

Other ideas:

* Use a buck converter, to convert the voltage to 3.3V or a similar voltage that is closer to the LED's forward voltage, and then a smaller resistor will do the final current limiting. It depends on the priorities. This is more costly but will waste much less power and not make much heat. Here is an example of one buck converter that could do it. You need several other components like capacitors and an inductor.

* If converting power to heat is okay for you, then you could also put several resistors in series so each one only dissipates part of the power. They should add up to about 2.7K Ohms.

 

Have you read this?

LEDs, 555s, Flashers, and Light Chasers