Author Topic: 10k POT Question  (Read 3650 times)

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Offline Ian

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10k POT Question
« on: May 04, 2011, 06:04:47 pm »
So i am working on a project with some leds. My Variable psu can only do down to about 1.4v and my leds are 1.28v. So i was wondering if i use a 10k pot could i use it to control the voltage?

Offline Zero999

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Re: 10k POT Question
« Reply #1 on: May 04, 2011, 06:09:35 pm »
The normal way to dim LED is to use PWM.

A pot will work but you should keep the series resistor in place to protect the LED from the pot being set to full voltage.

Offline Ian

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Re: 10k POT Question
« Reply #2 on: May 04, 2011, 06:12:41 pm »
should i use some capacitors to despike the voltage 

Offline tekfan

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Re: 10k POT Question
« Reply #3 on: May 04, 2011, 06:56:29 pm »
You could use a big wirewound pot Unless you have a small carbon pot, it is is a very bad solution because it will cause the track to melt inside if you have a bright LED connected to it. I'd put several power diodes in series  with the power supply output so you get the voltage drop. You can then have an almost 0 output voltage out of your PSU. Still remember to have a resistor to limit the current going to the LED.
One can never have enough oscilloscopes.

Offline jimmc

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Re: 10k POT Question
« Reply #4 on: May 04, 2011, 07:13:39 pm »
Simple answer is that it is not wise to.

It is the current through the LED that should be controlled, the forward voltage drop varies from device to device and has a negative temperature coefficient (ie it reduces as the LED warms up).
Unless you have good reason not to, it would be better to set the PSU to a higher voltage (say 5v) and use a fixed resistor to define the current [R = (5-1.28)/I] eg for I=20ma R=186R.
There is a simple calculator here (note it chooses the nearest preferred value above the calculated value).

Unless your pot is something quite special, it will have a maximum power dissipation of 1Watt or less.
This is for when all the track is in use so the maximum current allowed will be 10mA (from I = sqrt(P/R)).

No capacitors required.


Offline tecman

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Re: 10k POT Question
« Reply #5 on: May 04, 2011, 07:18:57 pm »
I agree with Jim.  A simple series resistor is the right choice.  Your 10K pot will likely not last since the current will be too high, and control will be next to impossible since yoy need about 1% of the rotation.

I would differ, however in the calculation.  20 mA is the usual max current for a typical LED.  Most will be bright enough at 5 mA or 10 mA.  Use his formula but cut the current for most apps.  Also higher supply voltage is better, as Jim indicated, since it will reducte temperature effects.


Offline Jon Chandler

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Re: 10k POT Question
« Reply #6 on: May 04, 2011, 07:28:34 pm »
Ummm....forgive me for saying all the replies answered the question you asked but really missed the point.

An LED is typically used in series with a resistor, and connected to whatever voltage you want to apply.  Connecting an LED without a series resistor is almost never done.

The resistor value is calculated to allow the desired current to flow through the LED.  An LED has a voltage drop dependent on the current through it.  This is the 1.8 volts specified on the data sheet.

Vsource = Vled + Vr where

Vsource = supply voltage
Vled = forward voltage of the LED
Vr = voltage drop needed across resistor.

By Ohm's Law

Vr = I * R where

I = current in amps
R = resistance in Ohms

Rather than go into all the details here, please look at two articles I've published:

LED Calculations

LED Calculations - The Lab Section

Offline sacherjj

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Re: 10k POT Question
« Reply #7 on: May 04, 2011, 07:31:23 pm »
I'll just echo everyone's advice.  This is a good place to learn or reinforce how to implement Ohm's Law.

V = IR    

Voltage across the resistor is equal to Vsource (voltage of however you drive it) minus Vled (Usually listed a Vf in LED specs).  You will also need If for the LED, the current required to turn it on.

Using Ohm's:
R = (Vsource - Vled) / Iled

Now, it will be important to also make sure the resistor you want to use is rated properly for the power.  

P = IV = (Vsource - Vled) * Iled

If this comes out 0.5W and you try to put a 1/4 W resistor in there, you will make a nice charred resistor.

Edit: Jon must have pushed post just after I started typing... :)
« Last Edit: May 04, 2011, 07:32:59 pm by sacherjj »

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