LM3914N Woes..

[PeterDove]

Yeah, we have a yearly alarm that makes us talk about these same things year after year after year after ...

If you have a few mins, would you mind sketching in on my breadboardwhere you would put the capacitors? I am completely happy with the resistors for power dissipation but not quite sure about when you meant when you said 'between pin 2 and 3' for the supply bypass. Do you mean putting the capacitor legs across the track of both pin 2 and 3?

---|--- (track of pin 2 with capacitor pin in it )
....O ( capacitor body)
---|--- (track of pin 3 with capacitor pin in it )

Like the above?

Thanks

Peter

 

[audioguru]

If you have a few mins, would you mind sketching in on my breadboardwhere you would put the capacitors? I am completely happy with the resistors for power dissipation but not quite sure about when you meant when you said 'between pin 2 and 3' for the supply bypass. Do you mean putting the capacitor legs across the track of both pin 2 and 3?

---|--- (track of pin 2 with capacitor pin in it )
....O ( capacitor body)
---|--- (track of pin 3 with capacitor pin in it )

Like the above?

Your capacitor across pin 2 and pin 3 of the IC is fine. The other capacitor and single cooling resistor (re-calculate its value for your circuit's conditions) are shown here:

 

[TheSwede73]

A big newbie here also...
I just want to make a led bargraph, that simulates power increase by turning a pot. that should be extremely simple with the LM3914N ?

Could anyone post a diagram for this?

Thanks from Sweden!

 

[audioguru]

A big newbie here also...
I just want to make a led bargraph, that simulates power increase by turning a pot. that should be extremely simple with the LM3914N ?

Could anyone post a diagram for this?

Thanks from Sweden!

The LM3914 is a voltmeter. Simply connect a pot to its reference voltage and ground, then connect the slider of the pot to the input.
As you turn the pot the LM3914 shows the voltage changing at its slider.

 

[TheSwede73]

Thanks!
But its "label" is "Led-driver" isn't it? or is it mostly used as a voltmeter, or as i understand it the voltmeter function can be used for many Led-projects

Thanks!!

 

[audioguru]

The LM3914 displays voltage in 10 steps. Its display is LEDs.
It can also display current, resistance. With a suitable sensor it can display pressure or temperature.
It is linear so it doesn't display audio level or light brightness properly, an LM3915 is logarithmic and does.

 

[TheSwede73]

I forgot the questionmarks it's a led-driver? or a voltmeter?, that is used for led purposes? excuse my lack of knowledge with these things, i'm learning....

 

[MrAl]

Hi,


In addition to the previous posts here are some design formulas that make
the resistor selections take about 1 minute or less...

We will use two simple formulas to get the two resistor values:
R1=3750/(380*iLED-1)
R2=(4/5*Vsignal-1)*R1

Note we must calculate R1 first, and that the formula for R1 is valid for
LED currents from 3ma up to 25ma.

First, we know what current we want in each LED, and we'll call this iLED.
Resistor R1 is then dependent SOLELY on this LED current:

R1=3750/(380*iLED-1)

Next, now we know what R1 is so we can easily calculate R2 knowing what
our max signal voltage (Vsignal) is:

R2=(4/5*Vsignal-1)*R1

Now we know both R1 and R2 so we are done, unless we want to check our
results:

Vref=1.25*(1+R2/R1), (should equal Vsignal)
I=12.5/R1, (approximate just for checking)



Design Example 1:

Lets say we want iLED (LED current) to equal 10ma, this means:
iLED=0.010

Lets say we also want our max input voltage to be 5v, this means:
Vsignal=5

Plugging iLED into the formula:
R1=3750/(380*0.010-1)
so
R1=1339.29 ohms

Next, plug that R1 and Vsignal into the other formula:
R2=(4/5*Vsignal-1)*R1
R2=(4/5*5.00-1)*1339.29
and we get:
R2=4018 ohms

We might check our answers next just to make sure we did everything right:

Vref=1.25*(1+R2/R1)
I=12.5/R1

and we get Vref=5v and I=9.33ma, and that is close so we are ok here.


Design Example 2:

We have LED current of 10ma again, but this time Vsignal=3v.
After doing the calculations, we get:

R1=1339.29 ohms
R2=1875 ohms

Note R1 is the same as last time because the LED current was the same.
R2 changed because the max input signal voltage changed.

 

[TheSwede73]

I get it now! thankyou so much, just ordered the proper things from my supplier..
I will take it slow, just want to make a functioning "power" thing, with the led bargraph increasing with the turn of the pot.

Thanks!
I will get back to you with other questions, so i'll hope i'm welcome!

 

[The Question]

I was wondering if there is a way to use multiple LM3914N in series or something similar.
Would it be as simple as just putting a resistor on pin 5 on the latter LM3914N?

I want to make a 20 or 30 LED bar display.

Also the heating issue has me a little concerned. I want to use 12v 2000 mcd LED, pretty sure they draw more mah than your typical LEDs, and am afraid the the outputs from the LM3914N might not be enough to light it.

I was thinking of setting up everything the same as in the post that Peter had (only referring to a single LM3914N for now) and having the LEDS draw power straight from the source and having transistors go inline with them and having its signal come from the LM3914N. Have any suggestions on which transistors I should use or if there is a better way of doing this?

 

[audioguru]

The datasheet of the LM3914 shows how to have a 20 segment meter with two LM3914 ICs. A third one can be added the same.

Each output can sink up to 30mA.
LEDs can be connected in series if the supply voltage is high enough.

 

[The Question]

The datasheet of the LM3914 shows how to have a 20 segment meter with two LM3914 ICs. A third one can be added the same.

Each output can sink up to 30mA.
LEDs can be connected in series if the supply voltage is high enough.

Thanks!

So in series? Like putting it on with the cathode connected to the anode with all the LEDs and then each joined "connection" to the output of the IC?

Will this still work in bar mode (pin 9) without any drop in brightness? Also the datasheet did mention putting on more LEDS, but from my understanding from reading it, is that it will only work in "dot" mode.

 

[audioguru]

If you have a 12V supply and 2V red LEDs then each output of the LM3914 can drive five of the LEDs in series and all will have the current that you set the outputs to have (up to 30mA). Bar or Dot mode doesn't make any difference.

The datasheet shows a very odd method of using a very high voltage supply with all the LEDs in series and only for the Dot mode but you don't want it like that.