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### AuthorTopic: Circuit building (long)  (Read 2043 times)

#### (NW)Lynx

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•  ##### Circuit building (long)
« on: April 20, 2003, 01:33:39 PM »

brutal touched on half of this subject about 8 months ago...the thread is http://forums.gideontech.com/viewtopic.php?t=3571&postdays=0&postorder=asc&highlight=dimming+led&start=20

I am basically using an LM317T (voltage regulator) and one of those 100 Ohm 5 watt pots in combination with a switch to bypass the voltage regulator for controlling fans.  Along with this I am trying to get 2v leds to dim and brighten depending on the fan speed. Yes I know it is another rheobus thread but this one is more about electronics and math to bear with me a bit.

I found this formula for working with the LM317T, a fixed resistor in series, and the pot. This is supposed to keep the working range within 6-11v.

Vout = 1.25(1+ [R2 + VR1] / R1) + (Iq * [R2+VR1])

It is from this site:
http://www.cpemma.co.uk/reg.html

The values he had were r2 = 1k, vr1= 1k lin pot, and r1 = 240.  I don't know what Iq means but he had it listed as .00005A so I just plugged that into the formula.  According to him he gets an output of 6.5v to 11.7v.  I created an excel sheet and plugged the numbers in.  I got the same as him so I am assuming the formula is correct.

Now I adjusted it for using a 100 Ohm pot.  I had a bunch of these from back when I built a simple rheobus going off info from Cliff's site.

Unfortunately my adjustment figures look really weird.  To get a range from 4.22v to 12.038v I have r1 = 16 ohms, r2 = 38 ohms.  Resistor value seems low to me but I have nil electronic background.  Is there something I am missing as far as basic theory?

My second question stems from that earlier thread that I mentioned.  (this info came from the same site I got the LM317 formula) I obtained information on how to make an Led dim in conjunction with controlling the fan speed.  He suggests using 5-6 1n4001 diodes and a series resistor of 620 Ohm to give a current forward of .65mA at 6v and a 10.3 at 12v. Basically the 6 diodes bring Vf up about .6v each.  So 6 diodes would raise Vf to 3.6 and then add the Vf of the led (in this case 2.0) to make a total Vf of 5.6. I wrote the formula like this Vf=((numD1*.6)+L1).

To figure the current coming forward with the 100 Ohm pot I used the standard (Vs-Vf)/R.  Now I am probably going to need my formula corrected but this is the way I wrote it down.  Vs=Vout from the above formula.  If = (Vout - [{numD1*.6}+L1]) / R3.  With R3 = 470 my range for the current is 1mA to 17mA.  Does this look correct?

Sorry for the long post.  Thanks in advance for the help. Logged

#### viridius

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• • You know what I mean. ##### Circuit building (long)
« Reply #1 on: April 20, 2003, 02:52:41 PM »

Answer one: It's a voltage divider, so the resistor values are probably correct.  You might just want to spend the dollar and get the 1K pots instead.  I'm not sure if they will affect performance, but you'll be dropping about 100mA through the voltage divider network with the 100 ohm setup compared to six mA for the 1K.
Answer two: I'd use three diodes and a 390 or so ohm resistor. Logged
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#### sonique

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•  ##### Circuit building (long)
« Reply #2 on: April 21, 2003, 07:50:58 AM »

I would recommend using a 1K pot as well And also, you mentioned Iq.
I 'think' this means Quiescent Current, so current at rest/inactive, I think, but don't quote me on that. Logged
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