Very cool, THE RB! Thank you very much! (I think I'm beginning to understand what you meant in your earlier post )I modified #12's nice simple circuit from post #12 (spooky) to light the LED when the fan pulse stops;
Hey #12, many thanks for the explanation!1) It is already designed to work with 5 volts. Double the resistor to use 12 volts. 270 becomes 560 ohms.
2)You will probably need to add a current amplifier to get a buzzer to work. The buzzer would be instead of the LED in this circuit, but it probably won't work because it needs more current than an LED. Still, you are welcome to try. You might luck into a high performance transistor that would run a buzzer.
3) The MPSA14 is just the first transistor that I found that is a low power Darlington. Any low power Darlngton transistor should work to run an LED. Much more than that and you'll have to add another stage.
Now, let's see if I'm less scatter brained than yesterday
This is what I'm really thinking about:
All you need to get the LED and buzzer working from the same circuit, is to put the buzzer between the transistor collector and the 12v supply.
Then the LED and it's series resistor also go between the collector and the 12 supply.
Then the transistor can turn both on together.
cool, thank you very much guys! will try it out and report back with the resultsTry it. The worst that can happen is that the buzzer makes no noise or a a silly noise and you add a transistor later.
Following your suggestion, I changed the 10uF cap to a 470uF, but that only made the alarm go off after a relatively long time from the moment the tach signal was cut off.Your fan tach fail detect looks like our original circuit where the tach pulse keeps the darlingon transistor turned off. That looks ok, but I would use a much larger base-emitter cap than your 10uF cap, I would use maybe 470uF.
Checked this, and it looks fine, I think. Stayed constant at 0.65V.Then you check that when the tach pulse is running that the darlington B-E voltage waveform is low, it's highest points should be under 1v.
If that is so, then the darlington is properly OFF when the tach pulse is running.
Swapped places between the PNP and the piezo and indeed the LED stopped blinking when the alarm is off!I don't like your method of making the 555 modulate the piezo. When the 555 output pin goes low, it turns on the PNP transistor and it will flow some small amount of current through the piezo, and through the E-B of the PNP into pin 3 of the 555 timer. That is enough current to make the piezo emit some sound.
If you swap the PNP to the other side of the piezo, then current through PNP E-B will not go through the piezo.
Once the cap is sufficiently large it won't make much difference. But if the cap is too small the voltage there will have ripple, and the high points of the ripple may cause false triggering. The size of the cap depends on the frequency of the tach pulse, which I did not know. You're the one on the front lines, so if 47uF works great then that's done....
After checking out a few other values, I settled for 47uF which is bigger than the original 10uF, but still makes the alarm go off rather quickly (as I'd like it to do).
Is there a specific reason why a bigger cap would be preferable here that I'm not aware of? (from what I saw so far, changing cap values didn't make any other difference apart from the said time difference).
I think you already got it. The 100k resistor will eventually charge the cap, and when >1v will turn the darlington on and make the fault alarm. But the pulse (which is active low) will cause the cap to be discharged through the diode, so the cap can never charge >1v when the pulse is occurring....
Checked this, and it looks fine, I think. Stayed constant at 0.65V.
Btw, I've been trying to figure out how is it possible that the Darlington (which is an NPN) is kept off when the tach pulse is running and "on" when it stops... shouldn't an NPN stay "open" until voltage/current is applied?
... I'd be very grateful if you could clarify how this works ...
Swapped places between the PNP and the piezo and indeed the LED stopped blinking when the alarm is off! ...
I think this one is straightforward. The 555 timer and buzzer need 12v. All the time. Instead of running them from the fan voltage which might vary down to 10v, just run them from the fixed 12v supply inside the PC.... I discovered another hitch...
More specifically, when the fan is powered with the full 12V, the circuit works great.
However, if the voltage that feeds the fan is lowered, at a certain point (from about 10V downward), the buzzer stops emitting sound properly. Instead, it goes down significantly in volume and starts making rapid (and much quitter) clicking sounds.
I wish it was as straight forward as that... but in fact the buzzer and LED are already connected to the 12V of the power supply (not to the fan voltage) exactly as shown in the diagram.I think this one is straightforward. The 555 timer and buzzer need 12v. All the time. Instead of running them from the fan voltage which might vary down to 10v, just run them from the fixed 12v supply inside the PC.
I thought you were already doing that, based on your schematic which shows a "12v" label for their supply.
ok, got it (& thank you very much THE_RB for bearing with me on this )Hmm, ok if you think it's the fan pulse frequency that's going to be easier to diagnose with a 'scope.
Without a scope you have to evaluate the different circuit modules.
If the fan pulse is good, the darlington is OFF, and its base will be well under 1v, and there is zero power to the piezo so it makes NO sound.
Are you saying as the fan voltage is lowered to 10v, it still spins fine but the piezo STARTs making sound? And the sound is "clicking sounds"?
Please provide more info as to why the piezo is running (is the fan stalled at 12? and 10v?) and what all the voltages are.
T1, T3 & T4 VoltagesYou can test that, with the fan voltage low (3v to 10v range), disconnect the FAN_TACH wire, and measure that T1 is turned on >0.6v at its base, that T3 is off (about 0.1v at its base) and T4 darlington shouldbe hard on with about 1v on its base.
Yep, checked the above transistors (in both alarm ON & alarm OFF) while the fan was powered with different voltages (3-12V) and same values as above every timeAnd those voltages shuold also be identical at any time the FAN_TACH wire is disconnected, even if fan PSU voltage is 12v.
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by Luke James
by Luke James
by Luke James