LED's on 10 element display going out after time

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Audioguru, you said it is in the datasheet. I'm looking at it, but there is a lot of information there. I really don't know what I am looking for. That is, will this be some sort of "input voltage" or what?

At the top of page 4, Note 2, it says 3 VDC ≤ V+ ≤ 20 VDC. This is what I am looking for, and the min operating voltage (for the IC to work properly) is 3V, right?

Sorry for being a total noob .
 
darkenreaper57 said:
At the top of page 4, Note 2, it says 3 VDC ≤ V+ ≤ 20 VDC. This is what I am looking for, and the min operating voltage (for the IC to work properly) is 3V, right?
Click to expand...
Nope. That is the supply voltage operating range for the guts of the IC, but the current-regulated outputs can have a separate supply voltage from the LEDs.

The current regulation for the LEDs works by adjusting the conductance of an output transistor to ground. The more voltage across the transistor, the hotter it gets. The transistor "drops out" of regulation when the voltage at the cathode of an LED connected to it is at most, 1.5V. At dropout, the current in the LED is reduced by 10% which you probably wouldn't notice.
 

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So roughly 1.5V is a maximum for "dropout", and thus a minimum for keeping all LEDs properly lit?

This is all new terminology to me, and it is a bit overwhelming :shock: .

Also, the condition set above do not meet mine. Is there a way to find out (yeah, it is probably in the DS, but I don't understand most of the data) what the dropout voltage would be for roughly 10.4 mA per LED and 10.3V across the IC?
 
darkenreaper57 said:
So roughly 1.5V is a maximum for "dropout", and thus a minimum for keeping all LEDs properly lit?
Click to expand...
Correct, for the outputs of the LM3914, plus the voltage of the LEDs to reach the minimum supplyvoltage for the LEDs.
Is there a way to find out (yeah, it is probably in the DS, but I don't understand most of the data) what the dropout voltage would be for roughly 10.4 mA per LED and 10.3V across the IC?
Click to expand...
You don't need to know the actual dropout voltage, it is guaranteed to be less than 1.5V. You do need to know the actual operating voltage of the LEDs to calculate a series "cooling" resistor.
 
Ok, I installed 2 150 ohm resistors in parallel and the 2.2uF cap as shown above. I fired it up, and the resistors got really hot at full 10.2 V. I measured the voltage across them, and it was 9.08V. Using P=V^2/R, I found out the total dissipation was 1.1W (so .55W each resistor).

This is over spec, so I added another 150 ohm resistor for an equivalent resistance of 50 ohms and total 1.5W dissipation. They are much cooler now, and the IC is also cooler. I'll let it run overnight and report back the results. I think it is finally working correctly .
 
Well, it figures I guess. The top LED has started to flicker again.

When this started I quickly shut off the power and swapped in the old LM3914. The same thing happened

Could it be that the LED block is bad, and ceases to work after an extended period of time?
 
The heating of the LM3914 is determined by its current whichis a max of 104mA, and the voltage across it. Your home-made 75 ohm resistor has 7.8V across it and the remainder of the power supply voltage is across trhe LEDs and the LM3914 in series.

What is the voltage across the LM3914's outputs when all the LEDs are lighted? Then you could calculate its power dissipation.
At first the supply was 12V. Later it was 10.4V.
 

Actually, I am using 3 150 ohms resistors in parallel, so the equivalent resistance is 50 ohms and 1.5W max power dissipation.

Anyway, I took some measurements and the voltage across VLo and VHi on the IC is 1.24V. It is 5.75V across the resistors in parallel at max voltage on the fan controller.

Sorry for the confusion about voltage earlier, but here is how it is set up. The sensors are connected to a fan controller with a 12V line. Due to efficiency, the max voltage the fan controller put out is 10.3V or so. This goes to the signal line on the LM3914 circuit. The LM3914 circuit also does have its own 12VC and ground in and outputs. I guess I forgot this, and was hung up on the signal voltage...whoops. This explains why 2 150 ohm resistors in parallel got so darn hot.
 
So with a 12V supply and 1.8V LEDs, 5.75V is dropped across your home-made 50 ohm resistor leaving 4.45V across the outputs of the LM3914.
Therefore with all 10 LEDs lighted, the dissipation of the LM3914 is only 0.51W. That is less than half of its max rating so is fine.

Get a Name-Brand LED array. It won't be intermittent.
 
darkenreaper57 said:
Is there an easy way to desolder a LED array other than using a desoldering braid?
Click to expand...

Personally I rarely ever use desolder braid (perhaps once every five or ten years?), mostly I use a 'solder sucker', a handheld spring loaded device that sucks the solder off the joint. If you have a suitably large (or shaped) soldering iron bit, then you can heat all the pins simultaneously and pull the part out.
 
Hi Nigel,
That aluminum solder sucker is just like mine that has worked well for over 30 years. Most today are big plastic cylinders. :lol:
 
audioguru said:
Hi Nigel,
That aluminum solder sucker is just like mine that has worked well for over 30 years. Most today are big plastic cylinders. :lol:
Click to expand...

I don't have that exact one, mine came from RS Components, and I think is made by AB Electronics? - or something like that?.

I've worn out probably three or four over all the years, and I've thrown all the old nozzles in a draw - I just counted, there's 64 old nozzles in it :lol:
 
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