lm3914

[mgeno216]

So i want to set up a led bar that responds to the variance in the voltage supplied to a fan. The fan is controlled by a cpu linearly. As the temp increases the voltage increases. I am using a manual controller and bypasssing the cpus adjustment. I want to use the voltage variance to use the bar as a psuedo temperature display. So here are the specs I have.

My Led Bar (i am going to wire in paralell)
25ma each
2.2v each

My supply
5v-12v
I am going to use resistor to turn 7v into 0v.
That leaves 5v of variance

And I want bar in Bar mode not dot mode

Here is a screen shot because I can't figure out how to wire it. I am using the program posted in the blog.

**broken link removed**

 

[ericgibbs]

hi,
The first point is the Vled supply on the module should be set at 5V, not 2.2V.
To display 7v to 12V directly, the LM3914 must be powered by 15V.
You could then set the RLo voltage at 7V and the RHi voltage at 12V, so that the display would show LED#1 as 7V and LED#10 as 12V.

IF you cannot power the LM3914 with 15V and you power it with say 5V, then you will have to divide the Fans 12V to 0V by a resistor potential divider.

For BAR Mode connect LM3914 pin#9 to +V

Do you follow OK.?

EDIT:
This image shows the setup for 7V to 12V in 0.5V steps/LED ... [Fan voltage for demo is 10V]

 

[mgeno216]

Great answer, but could you explain the 5v to the leds. And why would dividing the 12v to 0v be a good idea, wouldn't that mean that the circut would get no voltage because the max volts (12v) would then be 0v, right?

I googled voltage dividing and found this formula
Vout= (R2/(R1+R2))*Vin

I don't see how I could get the Vout to equal 0. I may not be understanding what you are saying, but am willing to learn

 

[ericgibbs]

Great answer, but could you explain the 5v to the leds. And why would dividing the 12v to 0v be a good idea, wouldn't that mean that the circut would get no voltage because the max volts (12v) would then be 0v, right?

I googled voltage dividing and found this formula
Vout= (R2/(R1+R2))*Vin

I don't see how I could get the Vout to equal 0. I may not be understanding what you are saying, but am willing to learn

hi,
Consider that you connect two 10K resistors in series across the voltage supply to the fan and the fan voltage is 12V.
At the junction of the two resistors you would have +6V measured to 0V/Common.
If the fan voltage was reduced to say 7V, then at the junction of the resistors you would have 3.5V.

So set the LM3914 this way, power the Vled and V+ to 12V.
Set RLo pin4 to approx +3.0V and RHi pin6 & 7 to +6V.
When setting pins 6 & 7 adjust R1 to give about 15mA current thru the LED's.
Look at this image for the settings.

The operation would be when the fan voltage is 12v [ 6v at the divider, all LEDs lit]
when the fan voltage falls to 7V only LED#1 will be lit, each LED will light in 0.3V steps.
If you power the LM3914 Vled with +12V you MUST fit that series resistor, approx 22R at 1Watt.

Do you follow.?

 

[mgeno216]

Sorry I took so long, but I have a few more questions. So why do you in the way have the lm3914 have 12v to the leds. I was under the impression that the 3914 connected the grounds. Could I have a 5v power the leds, that wouldn't change anything else right.

What is the Rlow pin and why is it set to 12v.

And why is Signal set to 6v and then stepped down to 5v.

I was playing around with this and couldn't get a higher current to the leds. My leds I am using run on 25ma and I feel that 15ma is a little low. I am afraid that they will be a little dim.

Sorry for the number of questions, but I am still really confused about this.

Thanks
Mike

 

[ericgibbs]

Sorry I took so long, but I have a few more questions. So why do you in the way have the lm3914 have 12v to the leds. I was under the impression that the 3914 connected the grounds. Could I have a 5v power the leds, that wouldn't change anything else right.
Yes you can use +5V for Vled, I assumed that you had a 12V supply available.

What is the Rlow pin and why is it set to 12v.
Look at the value shown inside of the LM3914 image, its 3V. The 12V is the available source voltage.

And why is Signal set to 6v and then stepped down to 5v.
As I explained the 6V will when the maximum fan voltage is divided down to 6V [resistor divider]
It's set to 5V to show you the reading when the fan voltage is 10V as an example.
I was playing around with this and couldn't get a higher current to the leds. My leds I am using run on 25ma and I feel that 15ma is a little low. I am afraid that they will be a little dim.
You can set the LED mA to 25mA by using R1 and R2 [ make sure that when its 25mA that the Rhi voltage is 6V.

Sorry for the number of questions, but I am still really confused about this.

Thanks
Mike

hi Mike,
Dont worry about asking questions, its not a problem.
If you get really stuck let me know and I will post a full circuit.

EDIT:
Look at this image, its set to show the fan voltage of 7V [3.5v] and approx 25mA thru the LED

EDIT2: I have just noticed you are using the LM3915 module, you must use the LM3914 for linear displays, the LM3915 is for audio use and is logarithmic .!!!

 

[ericgibbs]

hi M216,
Look at these images.
It shows BAR mode, for DOT mode, disconnect pin 9.

 

[audioguru]

The current in R1 and R2 from pin 7 to ground sets the LED current as shown in the datasheet on a chart, "LED Current vs Reference loading". For 25mA from each output the reference must have a load of 3mA.

Then if the supply to the LEDs is 12V, each output of the LM3914 will have a dissipation of (12V - 2.2V) x 25mA= 0.245W and when 10 LEDs are lighted the IC heats with a little more than 2.45W and it will fry.
When properly calculated a resistor is in series with the supply to the LEDs it shares the heat. Then the IC will be close to its max allowed temperature when all 10 LEDs are lighted.

 

[ericgibbs]

The current in R1 and R2 from pin 7 to ground sets the LED current as shown in the datasheet on a chart, "LED Current vs Reference loading". For 25mA from each output the reference must have a load of 3mA.

Then if the supply to the LEDs is 12V, each output of the LM3914 will have a dissipation of (12V - 2.2V) x 25mA= 0.245W and when 10 LEDs are lighted the IC heats with a little more than 2.45W and it will fry.
When properly calculated a resistor is in series with the supply to the LEDs it shares the heat. Then the IC will be close to its max allowed temperature when all 10 LEDs are lighted.

The OP has already stated he will be using +5V for his LED supply and will be operating in the BAR mode.

I am not sure what you are trying to say with your post.??

 

[audioguru]

5V is fine to power the LEDs and I explained that the IC gets too hot if the supply to the LEDs is higher.
I also answered the question about why the LEDs are dim.

 

[ericgibbs]

5V is fine to power the LEDs and I explained that the IC gets too hot if the supply to the LEDs is higher.
I also answered the question about why the LEDs are dim.

hi,
Thanks, I thought I may have overlooked something.