LM3914 Question, Help!

[TheSwede73]

Thanks!
Yes i have the datasheet already, but to a newbie its all confusing, but with this i will learn a little more..

I dont have the resistors shown, so that's the next purchase then... Without the resistors it wont work at all? or will i fry it?

Thanks

 

[ericgibbs]

Thanks!
Yes i have the datasheet already, but to a newbie its all confusing, but with this i will learn a little more..

I dont have the resistors shown, so that's the next purchase then... Without the resistors it wont work at all? or will i fry it?

Thanks

Dont use it without resistors.

Do you have a 2K2 resistor in place of the 1K21 and

a 4K7 in place of the 3K83..?

This will give you a working LM3914 to try out.

 

[TheSwede73]

I just have resistors for Led's they are higher in values i think.. well, its a visit to the electronics shop tomorrow then!

Then comes the big challenge taking this from the breadboard and making a real circuit board..
But that should'nt be so hard, the connections and circuits are easy tp place...

 

[ericgibbs]

I just have resistors for Led's they are higher in values i think.. well, its a visit to the electronics shop tomorrow then!

Then comes the big challenge taking this from the breadboard and making a real circuit board..
But that should'nt be so hard, the connections and circuits are easy tp place...

Remember when you make the pcb/wiring, is to allow for pin 9 [MODE] wire to be repositioned to give a 'bar' or 'dot' graph.

 

[TheSwede73]

Yes!, thanks for the reminder, i will put two pins there which can be connected by a black "cap" sort of, well you know... so its easy to change if necessary..

Thanks for all the help!

 

[audioguru]

Remember when you make the pcb/wiring, is to allow for pin 9 [MODE] wire to be repositioned to give a 'bar' or 'dot' graph.

The LM3914 will be extremely hot if it is set for BAR MODE, the supply for the LEDs is 12V, the LEDs are 2V red ones and all 10 LEDs are lighted:
1) LED current= 12.5/1.21k= 10.3mA.
2) Dissipation in each output transistor= (12V - 2V) x 10.3mA= 103mW.
3) Total dissipation for the LM3914= 103mW x 10= 1.03W.

If a 47 ohm/1W resistor is added in series with the LEDs then it will dissipate 499mW and the LM3914 will dissipate only 531mW. Then the 2.2uF capacitor is used as shown in the datasheet.

 

[TheSwede73]

Ah ok, that wont change the leds light in any way i understand, so they still light up with full strenght?

I just add the resistor directly after the battery going to the leds then..

The capacitor as i understands it will take care of the "powerloss" and make them go as bright as they would with out the resistor and the capacitor....?

 

[TheSwede73]

And the resistors from pin 7 is Kohm not ohm i understand, 3.83 kohm is 3830 ohm or am i wrong...

I can only get resistors with 3.9kohm and 1.2 kohm, but thats almost the same....that will work?

 

[audioguru]

The resistor values of the circuit give an LED current of only 10.3mA so they will not be "full strength". Most LEDs have a max allowed continuous current of 30mA, some 40mA.

The 47 ohm resistor reduces the heat in the LM3914, not the brightness of the LEDs.

I have a Sound Level Indicator project that operates its LM3915 with 32mA in its LEDs. My supply voltage is 9V, my LEDs are 3.6V each and a 10 ohm/1W resistor is in series with all the LEDs to share the heat. It is very bright.

 

[TheSwede73]

So to get the leds in full brightness i should skip the resistors or one of them?

 

[audioguru]

So to get the leds in full brightness i should skip the resistors or one of them?

You need to re-calculate the values for the two resistors from pin 7 to 0V. Their ratio should remain the same.

Then you need to reduce the supply voltage or limit the voltage to the LEDs with a pretty big resistor.

Instead of making a lot of heat, why not make a lot of light?
Connect LEDs in series at each output of the LM3914. If the LEDs are 2V red ones then up to 5 can be connected in series at each output since the supply is so high at 12V. Then the LM3914 will not get hot.

 

[ericgibbs]

So to get the leds in full brightness i should skip the resistors or one of them?

The LED current is determined by the 1.21K resistor to about 10mA, this is only approximate.

The point thats been raised is the heat generated within the LM3914 when all the LEDs are lit when using the 'bar' mode.

A solution to reduce the heat within the LM3914 is to fit a series resistor in series with each led. The led current will be the same.

What colour of led to you plan to use.?

Another project I am helping with has two blue leds in series, a blue led can have a voltage drop of 3 to 4 volts, so when the two are in series the voltage drop is say 7v, so the other 5v at say 10mA is dissapated as heat in the LM3914. Thats approx 5 * 0.01 50mW/led output.

 

[audioguru]

A solution to reduce the heat within the LM3914 is to fit a series resistor in series with each led. The led current will be the same.

Only a single resistor is needed to share the total heat. The LED current will also be the same.

 

[TheSwede73]

Im using a bargraph, the average kind, 7 green, 3 red led's

 

[TheSwede73]

A single resistor for the led's or do i have to put a resistor at every led? on the + or - ? they are joined togheter with the kathode so there or from every "numbered" pin from the lm3914

Thanks!

 

[audioguru]

A single resistor for the led's or do i have to put a resistor at every led? on the + or - ? they are joined togheter with the kathode so there or from every "numbered" pin from the lm3914

Thanks!

Connect together the anodes of the LEDs and feed them with a single power resistor. Add a capacitor.
It is in the datasheet:

 

[TheSwede73]

that was no sheet, just text....

 

[TheSwede73]

I know, all of you are pulling your hair thinking, god...why cant he understand!!

Sorry, im so new to this, and you are so helpful, i told some friends and they couldnt belive it.. that i got so much help..

So thankyou for being there!,, i will be needing you....

next step : purchasing the right resistors.. after that, experiment again

 

[audioguru]

Tell your friends that we are glad to help you.

 

[TheSwede73]

Thanks for all help!

I finally did it...

I bought a double output labpowerunit, very nice.

Got the resistors needed and started mounting and pretty soon i got it to work!

Now it's next challange, trying to make a circuitboard out of it..

Thanks again!