• Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Need help with LM3914 voltage display circuit

Not open for further replies.


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)



Well-Known Member
Most Helpful Member
Do I need to lower the Vled from 5V to something like 2V?
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.
Last edited:
Not open for further replies.

EE World Online Articles