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Author Topic: Fan control, indicating, proportional  (Read 18508 times)

Roly

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Re: Fan control, indicating, proportional
« Reply #15 on: June 26, 2012, 11:59:10 PM »
Quote from: JC Maillet

yeah, I used 5k instead of 500 because I didn't need 12mA driving my
LED's // that's way too much IMO I was aiming for 1.2mA if I recall ...

plus I'm making my output mirror scalable, so I can overcome the 1/10 factor loss and then go beyond if need be ...
my fans were rated for 120mA full power but I got there on less if I recall

~jc
If you say theory and practice don't agree you haven't applied enough theory.

joecool85

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Re: Fan control, indicating, proportional
« Reply #16 on: June 28, 2012, 11:03:16 AM »
Quote from: JC Maillet

yeah, I used 5k instead of 500 because I didn't need 12mA driving my
LED's // that's way too much IMO I was aiming for 1.2mA if I recall ...

plus I'm making my output mirror scalable, so I can overcome the 1/10 factor loss and then go beyond if need be ...
my fans were rated for 120mA full power but I got there on less if I recall

~jc

Interesting.  If he did 1.2ma I can't imagine his fan did a whole lot seeing as at full tilt it would only receive 12ma total, probably just enough to make it turn.  Of course, that's assuming he turn the pot all the way to 5k, good chance he ran with 1-2k or something giving him something along the lines of 50-60ma total.
Life is what you make it.
Still rockin' the Dean Markley K-20X
"New" amp: Fender Frontman 25 DSP (FM25DSP)
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Roly

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Re: Fan control, indicating, proportional
« Reply #17 on: June 28, 2012, 12:25:26 PM »
If you look at the right side of his circuit you will see that each of the three fans has a BD140 in series with it and that the current drawn by the LM3914 is from their base circuit, so there is considerable current gain available.

Assuming each fan required 120mA for full speed and the BD140's have an Hfe of 100, then the total base current for full fan speed will be;

3 * 120/100 or 3.6mA,

three segments worth with 2* 5k = 10k between Vref and ground.

The current scaling for the LED drivers is determined by the overall value of the two pots, not their setting.  Their settings determine the bottom and top of the LED range, that is the temperature where fan current starts to flow, and the temperature where it reaches its maximum value.

Iref = Vref/Rref = 1.25/10k = 0.125mA
0.125mA x10 internal scaling factor = 1.25mA/segment

I have a video here by JCM of his adaption in operation but it's 6 megs and too big to attach.
If you say theory and practice don't agree you haven't applied enough theory.

joecool85

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Re: Fan control, indicating, proportional
« Reply #18 on: June 29, 2012, 12:31:22 PM »
Roly, that makes much more sense.  I hadn't thought about the BD140.
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Eb7+9

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Re: Fan control, indicating, proportional
« Reply #19 on: October 09, 2012, 03:01:50 AM »
Hi ... thanks for posting my re-working of your circuit Roly

if I may, I'd like to provide a second version of my adaptation as the first one produced a "hung" LED // we suspect on account of an overly high voltage seen at the driver pins ...

a simple solution to bringing that voltage down simply involves including a common-base stage (BD139) to buffer the high voltage coming from the mirror input // in principle this does not change the overall operation of the circuit since the current gain is very close to unity in the new stage ...

I will say that the temperature range adjustments (as drawn in my version) and current gain adjustment in the mirror all worked like a charm ...


// ~jc
« Last Edit: October 09, 2012, 03:16:09 AM by Eb7+9 »

66cccfff

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Re: Fan control, indicating, proportional
« Reply #20 on: October 15, 2018, 09:32:04 AM »
Good idea it might be, but using a transistor which works in linear section to drive a fan is likely to generate lots of heat. As guitar amp geeks, most of us have several Op-Amps and it might be a better idea to build a PWM(pulse width modulation) speed controlling circuit instead of a linear one. As temp probes, LM35 is a great choice for its voltage output exactly mapped to the centigrade temperature. Feeding the signal and another sawtooth or triangle signal - it could be easily generated by using a NE555 timer chip - into the opamp(let it work as a comparator) and use mosfets (try irfz44 or 75nf75) as switching components. Hope it'll work well!

 

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