Hello All,

New member here. I just wanted to ask a couple of questions about my project using 2 LM3914's and NTE RSI-ID4-21 Solid State Relays instead of LEDS. I'm using 12 volts to power the lm3914's and also the LED bridge + side.

I have everything hooked up, but my questions are these;

1: When you cascade two lm3914's together, I know you need to put the Vref voltage at the end (which would be the 2nd lm3914. But what about the first lm3914's led current control vs the 2nd one? Do I just put a resistor on pin 7 to ground of the first IC to get the current for the first 10 outputs? How do I make sure the current is the same through all 20 outputs?

2: I hear that sense i'm using 12 volts I would need to use a 7.4 ohm resistor in series with the led + side to keep the power dissipation down on the IC's. yes?

3: As for the calculation on the VRef on the second IC, I have R1=6.5K and R2=1.2K. This would give me about 8 volts of range for a pot of 10K hooked up to the input, which would make it not so sensitive. And 10mA of current for the outputs. Does this sound correct?

Oh BTW, this project is to drive a light meter pole, that a person will use to fake a loudness meter by just moving the pot high or low. The bulbs are 120V 11Watt swirl type bulbs.

Thanks for any help on this. I appreciated it!

Sean

 

TI data sheet shows cascading, & 20 seg meter; might help?

 

Yeah..I've looked at that, but it never explains which chip has Control of the current (or led brightness) when cascading 2. Are they independent or does one current limiting resistor on pin 7 (or 8I can't remember) work for both chips?

 

Both ICs need a current-determining resistor at pin 7.
Simulation shows this resistor set should give you ~10mA per LED and offload some of the IC power dissipation on to Rdrop. You will probably need trimmers to adjust resistor values. There is interaction between the adjustments.