Electronics Gurus needed

[richamies]

I would like to put a couple of LED's on the front of my pc, to let me know the cooling fans for my CPU are working. I was thinking that I can possibly run 1 LED in series with each fan, would this work ok?

I understand that running the LED in parallel with the fan would over-volt it, but it seems a good idea to run it in series, that if the fan dies so will the circuit and therefore the LED will die also.

Any suggestions/advice? I plan to try it out tommorrow(27th)

Many thanks in anticipation of your greatly valued answers (does sucking up still work on here? http://www.sysopt.com/forum/smile.gif )

[Nathan G]

...

[This message has been edited by Nathan G (edited 01-07-2000).]

[jad1097]

Nathen G is correct. The only thing the LED you want to put in would tell you have power to the fan. Good idea though.

[BBA]

You asked for it!!!

Ok..., to power a LED you need about 2.5 volts, with a current in the LED of 25mA or so.

A resistor in parallel with the LED which is in series with the fan will develop a voltage drop dependant on the current in the total circuit and its resistance.

How to figure it out? Well Well.....

First...basic Ohms law..."Voltage (E) = Current (I) times Resistance (R)

"V=IxR"

Say we have a normal fan that uses 12v at 0.14Amps. the resistance of the fan is going to be R=E/I or about 85 Ohms. This is the first peice of data we need...85 Ohms.

So to get the needed voltage drop across the resistor, we need to calculate the new current of the fan in the new circuit so we can calculate the resistors needed value.

Do this by saying the 85 Ohm fan will now have 9.5V across it (which is the 12V source voltage minus the 2.5V needed for the LED in series with it)... so since I=E/R, I=9.5V / 85 Ohms = .111 Amps, this is the next piece of data needed.

So, to get the value of resistance needed by the LED and resistor combination, we take the voltage drop across the resistor and LED, which is 2.5 and divide it by the overall new current value of .111 Amps. so R=2.5V / .111 Amps = 22.52 Ohms total of both devices...

since the LED has a resistance equal to 2.5v / .025 Amps (thats the 25 mAmps LED current) which equals 100 Ohms, we can calculate the resistors actual value by calculating for parallel resistance of the 2 devices (LED and Resistor) The formula is R(total) = 1/[1/R(led)+1/R(resistor)]

Or restated algebraically...

R(resistor)= 1/[1/R(total)-1/R(led)] = 1/[1/22-1/100] = 28.2 Ohms

So... Whew... a 28 Ohm resistor in series with the fan, with a common LED across the resistor will give you the indication you want! All this means is that if the fan shorts out and draws more current than you want it too, the LED will get very bright! And if the fan opens its circuit, the LED will go out!

This is all based on estimated values...the actual resistance of the fan in operation (not the DC coil resistance...that is very different) will be a little off...but this will work in general. You may have to recalculate for your particular fan.

This lower voltage to the fan will also slow the fan down a little bit. It will also increase its life span.

It's up to you if you think the indication is needed...but I had a power supply fan quit on me once and had no indication of it until I started to smell hot melting insulation...

You could go better than this and build the same circuit with a .4 volt drop across the resistor and a transistor biased on by this voltage to indicate too high or too low a current on the fan and thereforee give either a audible warning or a visual warning if the current is too high or too low...course I'd tell you to take a vocational ed course in electronics first... http://www.sysopt.com/forum/smile.gif

[This message has been edited by BBA (edited 12-26-1999).]

[Roy]

There surely is a way to indicate the fan is drawing current. My electronics guru and I will put our heads together Tuesday and try to conjure up the circuit. Hang in there, we'll be back.

If simply knowing there is power available to the fan, a 1000 ohm resistor in series with the LED, wired in parallel with the fan will protect the LED from excessive current.

Since there is virtually no power being drawn, a 1/8W resistor is OK, however, if the more common 1/2W variety is cheaper, use that. Also, don't pay extra for 5% precision. 20% is good enough.

LEDs can handle a maximum of 20mA. 1000 ohms limits the current to 12mA across 12V.

Another note ~ the shorter lead on the LED is the negative end. There also usually is a "flat" on the LED body on the side of the negative lead.

EDIT ~ BBA's method will work, but the fan will be somewhat slowed by the series resistor. I'll try to present another way that doesn't rob the fan on Tuesday.

[This message has been edited by Roy (edited 12-26-1999).]

[BBA]

Roy...I guess you posted too closely to the same time I did...but you are a little off in your assumption...

If you put a 1K Ohm resistor in series with the fan and LED you will stop the fan from running! And one in parallel with the fan will only indicate the voltager for the fan is there..nothing about the fan itself

a led in series with the fan with a shunt resistor is all that is needed

[This message has been edited by BBA (edited 12-26-1999).]

[Nathan G]

Monitoring tacho gives you the ability to detect failures BEFORE the blade actually stops spinning(and coil melts). Wouldnt you like to know the fan is going bad PRIOR to your precious gear over-heating???
ya know that smell... when the enamel coating on motor windings are baking away. LOL http://www.sysopt.com/forum/wink.gif

[richamies]

In all honesty it was just an idea, I dont want to spend any omney on it at this time, but I had the LED's lying around here, got a whole bunch of them actually.

[Roy]

BBA ~ I thought that is what I said. If simply knowing there is power available to the fan... Nothing about putting it in series with the fan.

richamies ~ Once I figure out the circuit, I'll send you the parts.

[Roy]

Update ~ Easier said than done, it seems. However, it seems like a useful circuit/device. If one person wants it, many others might also. I'll continue the research.