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Need help with LM3914 voltage display circuit

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Egz

New Member
Hello, first time poster, so please be gentle.

I want to build a circuit to display the value of a signal on a LED bar graph. I've gotten my breadboard prototype working, but not perfect. The more and more I research, the more and more confused I become.

The circuit closely follows this design:


Except its in DOT mode, the optional capacitor isn't present, and I have a voltage regulator dropping the source voltage from 12v to 5v, as to keep the chip from working overtime dissipating heat.

The voltage range I would like for it to operate within is 0.08v to 3.87v. The LED bar graph I'm using is a HDSP-4836, and used a current of 0.030A in my calculations. My initial design is using the following resistors: R1 = 360Ω, R2 = 755Ω

It works, but my ranges aren't perfect I could have my resistors off a little since I ordered the wrong ones initially, so I had to combine some to get close to what I wanted.

So after some more research, I stumbled upon the calculator hosted here. Playing with it at home, it was yelling at me about the LED current. So I tried using a lower current of 0.020A, with a R1 = 570Ω and R2 = 1190Ω, and its still blinking at me.

So, this is making me wonder:
Do I need to lower the Vled from 5V to something like 2V?
Am I using the wrong value for iLED?
What exactly are those colored boxes in the calculator and how do they differ from the sliders?
How do I alter RLo to allow my range to operate from 0.08 to 3.87? (I figured 0 was close enough, but now I'm not sure)

Thanks!
 

alec_t

Well-Known Member
Most Helpful Member
Do I need to lower the Vled from 5V to something like 2V?
No.
Am I using the wrong value for iLED?
Iled = 12.5/570 = 21.9mA.
I figured 0 was close enough
Agreed; but if you really want .08V as the lower part of the range you could add a third resistor, R3, between the lower end of R2 and ground. Connect pin 4 to the junction of R2/R3 instead of to ground. Choose R1 = 12.5/Iled and R2 and R3 such that the voltage drop across R3 = 0.08V and the drop across R2 = 3.87 - 1.25 - 0.08 = 2.54.
 
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