LM3914N Woes..

[PeterDove]

Hi All,

I am an electronic newbie... there now I said it!

I have managed to build a volt meter using the LM3914N in my software simulator called livewire. It works great...Here is is below

**broken link removed**

The woes I have are due to the fact that I dont really understand WHY it works. And I was hoping someone would help. I have the following questions

1) Given a voltage X for V1 ( which I have set variable from 0 to 2.8V ) how what are the formulaes for adjusting the R1 and R2 to make it work

2) I dont really understand, even after reading the pdf manual, the whole relationship between ADJ pin 8 and the RH pin 6 and OUT pint 7. Is the a relationship there to the signal volage ( ie is there a maximum )

What would be cool if someone could explain in simple language what I have to do and why if I pick a volatage range for V1.

I hope its not too much, the specs just baffle me - I can of course copy from examples but its more of a case of monkey see monkey can do at the moment with not a lot of understanding occurring.

Thanks - some kind soul take pity

Peter

 

[audioguru]

Hi Peter.
Your circuit won't work and your sim program is stupid not to know.
The LM39xx creates a voltage of 1.28V between OUT and ADJ exactly like an LM317 adjustable voltage regulator. The current in the 1.21k resistor also flows in the 12.5k resistor causing 13.2V across it. So the voltage at OUT is 1.28V + 13.2V= 14.5V but the power supply voltage is only 12V so it is impossible.

Use 1.2k for each resistor then the voltage at OUT will be 1.28V + 1.28V= 2.56V. If you connect HI to OUT then the 10th LED will light when the input voltage is 2.56V. The LM3914 separates the voltage to light each LED by 10% of the total so the 9th LED will light when the input voltage is 2.56V - 0.256V= 2.3V, the 8th LED will light when the input voltage is 2.56V - 0.512= 2.05V, etc.

OUT can go no higher than 1.5V less than the supply voltage.
The current in the LEDs is about 10 times the current ftom OUT that flows in the voltage-setting resistors and into HI.

If you want the 10th LED to light when the input voltage is 2.8V then 2.8V - 1.25V= 1.55V will be across R2. So if R1 is 1.2k like before then it has 1.25V across it and has a current of 1.042mA through it so R2= 1.55V/1.042mA= 1.49k.

 

[Nigel Goodwin]

Audioguru is the KING when it comes to LM39xx's

 

[PeterDove]

Hi Peter.
If you want the 10th LED to light when the input voltage is 2.8V then 2.8V - 1.25V= 1.55V will be across R2. So if R1 is 1.2k like before then it has 1.25V across it and has a current of 1.042mA through it so R2= 1.55V/1.042mA= 1.49k.

OMG - hes right, you are the king of LM3914s! Thank you so much - I will reread this a few more times and experiment with the circuits.

Can anyone recommend a good simulation program?

Thanks

Peter

 

[audioguru]

Audioguru is the KING when it comes to LM39xx's

Thanks, Nigel.
I have LED blinky things all over my house and in houses of my family and friends. I have also sold many. Only a single one has an LM3915 in it and I have never seen an LM3914.

 

[PeterDove]

Breadboard not working

Hi,

If I uploaded a breadboard photo would you be able to have a look at it? I think I have it set up on my breadboard right but even with no signal to pin 5 it is lighting up all the LEDs.

I do actually only have pins 1 and 18 ( LEDs 1 and 2 ) attached to LEDs - does this matter? Must all the LEDs be installed and connected to the pins?


Cheers and thanks again

Peter

 

[audioguru]

The input bias current for an LM39xx is positive-going so if the input doesn't have at most a 1M resistor to ground then it will float in a positive direction and the circuit will indicate its voltage.

It works fine without LEDs on all the outputs.

 

[PeterDove]

The breadboard

Hi again,

I have shown the breadboard so you can see if I have made any errors. The board shows the LEDs in an UNLIT state, however as soon as I apply power to the red/black leads you see at the top, both lights come on. However when I apply a 1.2V positive to pin 5 from a + terminal of a 1.2V battery the lights go out - I am putting the negative of that 1.2V battery to the ground of all the other parts of the circuit. I have left the 1.2V battery out of the system for now.

**broken link removed**

A) is the Pin 9 mode attached to the positive rail which brings 10.5V
B) goes to the ground negative rail
c) goes to pin 3 providing 10.5V
D) is the negative from a 10.5V battery pack
E) is the positive from a 10.5V battery pack

Remember the LEDS are actually lit with the 10.5V battery pack - I have dotted the LEDS so you can see when their leads go.

In the green circled area there is a small silver pin joining pin 6 with pin 7. Then from the pin 7 row we have 2 610Ohm resistors connected in series. Pin 8 ADJ connects to the middle of the two resistors through the small orange cable. Then they all get grounded by the big orange cable.

So the lights are on at this point, if I put the + from a 1.2V battery to pin 5 and the negative from the battery to the ground rail, then the lights go out...

Any ideas - I am utterly baffled LOL

Thanks

Peter Dove

 

[audioguru]

You forgot to connect pin 4 to ground.
You also forgot to connect a supply bypass capacitor (10uF is fine) between pins 2 (-) and 3 (+).
Also, the input pin 5 is floating up to some positive voltage because it doesn't have a resistor to ground to hold it at 0V.
I don't know what happens if pin 4 is not connected to anything like you have.

Aren't your resistors 510 ohms each? Then the 10th LED should light when the input is +2.56V. With +1.20V as the input then both #1 and #2 LEDs should light.

 

[qsiguy]

Since we're on the subject Audioguru and you're the expert, I am going to make a similar display when I get some time and I need it to read from 0-1V. It will be in a car so I'll have 13.5 VDC +/- while the engine is running. Guess I'll have a range of 12-14 VDC at any given time to power the circuit but the source I want to read varies from 0-1 volt. What resistors will I need for that?

 

[qsiguy]

Oh, and should I put a voltage regulator circuit on it to keep it at a stable voltage? I'm not sure how stable the 3914 would be with a changing input power especially when I need such small readings between 0 and 1 volt.

 

[audioguru]

Hi Gsiguy,
The LM3914 has a very accurate 1.28V reference voltage circuit that can be amplified to a higher voltage within the IC. Your 1.0V requirement is less so you need to divide it with a pot in series with the built-in resistor ladder.
The resistance of the ladder is from 8k to 17k so if your pot is 10k then you will be able to adjust for a 1.0V full scale. Connect pin 4 to ground and the 10k pot wired as a rheostat between pin 6 and pin 7. Connect a resistor from pin 7 to ground for the LED brightness, try 1k for 10mA in each LED. Connect pin 8 to ground.

In the bar mode with a 13.8V supply then the IC will melt if all LEDs are lighted. Add a resistor in series with the supply to the LEDs to reduce the voltage to them. Then a 10uF capacitor should be added to the LEDs end of the resistor and to ground. 68 ohms will reduce the voltage 6.8V and the resistor will heat nearly 1W instead of the IC heating.

 

[PeterDove]

Finally Working - a last question or two

You also forgot to connect a supply bypass capacitor (10uF is fine) between pins 2 (-) and 3 (+).

Hi,

Its finally working - thanks so much for your patience! It was the floating voltage and incorrect grounding when I added the 1,2V to the system.

I am not sure about the bypass capacitor, not sure how I hook that up - I have had a look at the examples from the spec sheet and couldnt see an example of it ( https://www.electro-tech-online.com/custompdfs/2006/11/LM3914-1.pdf ) - is there a schematic you could point me to? I was also reading the previous posts about the IC overheating.. do you have any schematics of that so I know where to put capactors/resistors?

Anyway, on a final note, thanks so much again - the simulation package was so bad on the LM3914 that it confused the hell out of me, so I have sent an email to the maker hoping they will fix it as even when I put the right resistors in the thing still doesnt behave.

Cheers

Peter

 

[qsiguy]

Very informative audioguru. Thanks for the information. I will have to print this thread out and file it in my project file.

 

[audioguru]

All circuits should have a supply bypass capacitor connected as close to the active parts as possible. For the LM39xx, it is between pin2 (-) and pin 3 (+). The datasheets don't show it because they assume that one is there or that the main filter capacitor of the power supply is very close.

A simple power calculation shows how much heat the LM39xx will produce.
510 ohms between pin 7 and pin 8 creates a reference current of 2.5mA. A graph in the datasheet shows that it causes a current of 22mA in each LED.
You have a 10.5V supply and if 2.0V red LEDs are used then each output of the LM39xx has 22mA and 10.5V - 2.0V= 8.5V across it and dissipates 22mA x 8.5v= 187mW. If all 10 LEDs are on then the IC will dissipate 187mW x 10= 1.87W plus the IC's small operating power. Its absolute max dissipation at 25 degrees C ambient is only 1.365W so it will be too hot and will fail.
Adding a resistor to share some of the heat as explained in the datasheet will cool it.

 

[PeterDove]

All circuits should have a supply bypass capacitor connected as close to the active parts as possible. For the LM39xx, it is between pin2 (-) and pin 3 (+). The datasheets don't show it because they assume that one is there or that the main filter capacitor of the power supply is very close.

Thanks - Found a very nice explanation of it at



I'll remember to add in some resistors to dissipate heat

Peter

 

[Hero999]

Or reduce the current to a more sensible value like 10mA.

 

[audioguru]

I'll remember to add in some resistors to dissipate heat

Only a single resistor is needed. It goes in series with the supply to the LEDs. Then a supply bypass capacitor is needed at the anodes of the LEDs.

 

[PeterDove]

Only a single resistor is needed. It goes in series with the supply to the LEDs. Then a supply bypass capacitor is needed at the anodes of the LEDs.

Yes thanks, got it - read another useful thread you contributed to on power dissipation at

https://www.electro-tech-online.com/threads/leds-on-10-element-display-going-out-after-time.18567/

Extremely helpful! Thanks again, I have written all this up in my electronics notebook fo future reference.

Peter

 

[audioguru]

- read another useful thread you contributed to on power dissipation

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