THANKS MAN! U MADE THIS PROJECT SO CLEAR AND EASY , I'M GONNA ACE THS PRESENTATION

 

OH AND COULD U EXPLAIN TO ME HOW INSERTNG THE 2 RESSTORS BETWEEN THE PINS WOULD MAKE IT 500R? AND WHAT WATTAGE SHOULD I IUSE FOR THE TWO RESISTORS? 1/2 WATT OR 1/4 WATT?

 

1/4 will be ok. Its not easy to explain with words. Imagine the wiper is half way of the full rotation it can do. Thus, the resistance between the wiper pin and one of the two other pins will be 500 R. If you put the 500R resistors I told you they will be in parallel with the two 500R resistors the pot makes (the wiper half way). Thus 500R in parallel with 500R equals 250R. The same for the other half of the pot. Thus total resistance across the pot is 500R. Make the same calculations for other wiper positions if you want or simulate it.

 

Make the two resistors I told you to add 1 K, not 500R.

 

MIKE............U WOULDNT BELIEVE IT!!! THE STORE GAVE ME A 5K LINEAR POT!!!!!!!!! NOW HOW IN THE WORLD AM I GOING TO MAKE THAT A 500R POT?!!!!!! SMT!!!!!!

 

Will you PLEASE stop typing in all capital letters? It's VERY annoying.

Take the 5k pot back to the store, and get the correct value.

Sadly, most people in electronics stores nowadays don't know any more about electronics than how to run the cash register and credit card machine.

 

MIK3 when u get online hit me up

 

The best solution is, as SGTwookie said, go back and get a 500R pot. But if you want a quick solution just connect two 500R resistors on the pot as before. It won't work exactly as a 500R pot but you will make your job.

 

the thing is..........the store doesnt hav a 500R. so if i put the 500R across the pins it would hav the same behavior as the 500R linear pot?

 

Well, not exactly but it will work.

 

what will that do exactly?

 

what will that do exactly?

Its better to simulate it to see how it behaves. The resistance won't vary linearly with the wiper position.

 

but logrithmically huh?

 

then i might as well use a log pot?

 

do u hav PSpice?

 

then i might as well use a log pot?

Yes, you can use one but it will be harder for you to determine the current limit. Better a log one than the 5K combined with resistors.

 

do u hav PSpice?

Yes, I have the demo version. It works good but you can't build very large circuits. You can download it from here:

http://www.cadence.com/products/orcad/Pages/default.aspx

 

I think he meant R5 was physically large (that is, high wattage). It's an area where electronics jargon is confusing to novices. Because the most important resistor specification is generally the value (that is, the ohms), we use "large" to mean a high value, such as 1 meg-ohm (1,000,000), or "small" for a low value, such as 1 ohm. (And you have to get used to the range of commonly available values to know what's large or small.)

R5 is there to generate a voltage proportional to the output current, and the entire output current passes through it. Therefore it has to be a small value and a high wattage. That is, it has to be able to dissipate lots of heat.