EEVblog #204 – LM3914 - Getting strange results

[John Wilson]

Hello everyone!

This is my first post here and hopefully I've placed this in the correct spot.

I've been watching Dave's videos and learning quite a bit. The #204 on designing a battery gauge with the LM3914 was particularly interesting.

I modified the circuit (or attempted to) to give me a range of 4.00v - 5.25v, keeping the range of 1.25v for simplicity sake. I calculated R1 as 13.04K and R2 as 14.55K. Since I'm using a low voltage, I'm not using a voltage divider. I've attached my circuit in a quickly-drawn mspaint image.

With a voltage of 5.09V and pin 9 floating, the LED on pin 10 lit. As I decrease the voltage, LED pin 11 lights, until I get down to around 3.4 volts and LEDs on pins 14,13,12,11,10 are lit. With pin 9 floating, it's supposed to be in DOT mode.

R1 and R2 are variable resistors and I've adjusted each to try to change the behaviour of the circuit, but haven't had any luck.

I've replaced every component, tried different LM3914s (I have LM3914N-1), different batteries. I'm missing something basic I think. Is it how I have removed the voltage divider?

Any help you can give would be most appreciated. I'm scratching my head and trying to get this to work.

Thanks!

john

[John Wilson]

I was hoping that someone would have some suggestions.

I tried the example on page 11 of the datasheet (Expanded Scale Meter, Dot or Bar), which measures from 4.46 to 5.54 V. I still get the exact same results. The LED on pin 10 lights, and that's it. If I change the voltage, then several LEDs come on, even with the circuit set to DOT mode.

I can't understand how the unit can measure and scale the voltage when the voltage itself constantly changes. I must be missing something here.

I tried using 5V to run the chip and a lower voltage to measure (with the grounds connected), with the same results. I thought that perhaps the chip needs to be powered by a voltage higher than the largest measurable one (due to loss in the circuit), so I powered the unit from 9V and connected the variable voltage to pin 5.

Adjusting R4 as per the datasheet instructions does nothing - the LED on pin 10 stays lit and nothing changes.

What am I missing?

[IanB]

Perhaps you should try building the 0-5 V bar graph circuit on page 2 of the data sheet and work up from there?

The best way to diagnose problems is often to start with the simplest working design and then modify it one thing at a time and confirm that each change leads to the outcome you expect.

With the basic bar graph circuit you could power the chip from a 9 V battery and then apply the 0-5 V measured signal using a potentiometer.

[John Wilson]

Thanks, Ian. Good advice.

I switched the breadboard, thinking that might be the problem, too. Here's what I have.

Building the circuit on page 2, using a constant 5V input and varying the voltage on pin 5, I can get the full range in bar- and dot-modes and everything works as it should. I know that everything is ok, I haven't blown the chip, etc. etc.

My problem seems to stem from trying to measure the voltage of the cell (or cells, or power supply) that is actually powering the circuit, and this is what I'm not understanding. As I power the entire circuit (and pin 5) from lower voltage, then I always have Pin 10 LED lit indicating that the voltage is at the maximum. This makes sense to me, in that the voltage I'm measuring is exactly the same as the voltage that is supplying the circuit - therefore I have "full" voltage. If I use a voltage divider, I could see how I could get different results, because I'm dividing the measured voltage from the working voltage of the circuit.

I guess I'm at a loss to understand how I can measure the power supply voltage when that same power supply voltage is powering everything - won't it always see pin 5 input as maximum? I suppose I need to go back to Dave's video again, too, and refresh.

Any thoughts? Hopefully I'm missing something very basic here, but I just don't see how you can measure a voltage level and show it as a percentage of full when the entire circuit sees the same voltage.

john

[IanB]

According to the data sheet the maximum voltage you can sense (V HI) is 1.5 V below the supply voltage (V+). So if you wanted to measure the actual power supply that is also powering the IC you would have to divide down the measured voltage before applying it to the sense input of the IC. You could, for example, use a precision divider to divide the measured voltage by two. The input bias current on the sense pin has a typical value of 25 nA so you would have to allow for that when choosing resistors for your divider so that the sense current does not perturb the voltage too much.