Here's the scheme I tried to simulate with Proteus.
The working principle is next: the voltage rating I'm trying to detect is 8.4V --6V (fully charged and almost fully discharged li-ion battery pack) witch gives a 2.4V of difference.
Since the lowes possible value of the voltage difference between Rhi and Rlo is 1.2V, and the Vdif is 2.4 then logically that the voltage for Rlo1 for the first LM-circuit must be:
8.4-1.2=7.2V (Rhi1 = 8.4V).
The Rlo2 voltage for the second LM-circuit is:
7.2-1.2=6V (Rhi2 = 7.2V)
So virtually it's working pretty fine.
But some problems appear in the real world.
The Rlo voltages which I set with R2 and R4 are less than I expected. So I thought that it must be trouble with the reference voltage of the IC.
I tried to measure that 1.25V between VRO (pin 7) and ADJ (pin 8) but for me, it's equal to 200mV. I tried 4 different IC but the values are the same for each.
Moreover the REf. voltage source in the assembled scheme is in depending of the power voltage, provided to the IC while it should be stable for the Rlo reference.
Could you guys tell me what I'm doing wrong here and why there isn't a so much desired stable 1,25 at the "naked" LM3914 IC.
Thanks!
upd.
Forgot to mention where the value of 440 R appeared from.
The LED current I want to have is 30mA. For that by the formula from the datasheet, I get approx. 416R for R1. Then I take into account leakage current from the ADJ pin and a bunch of resistors inside of the IC's comparators divider, so I end up with 430R (pretty close to 440R I have).
I'm not familiar with the LM3914, however if I were in your situation I would troubleshoot it like this:
I would connect one of the IC's exactly like the TI datasheet. Same voltages connections etc. A quick look at the spec shows it's not 100% clear (to me) the limits of the input signal for a workable range.
Apart from the issues that you face, I think you need to reconsider the resistive divider that sets your input range (6 to 8.4V). The effect of the precision voltage regulator is that negative feedback forces 1.25V between Ref out (VRO) and Ref adj, which you have jumpered to VHI and VLO. The desired range covers 8.4-6V = 2.4V. That is not the same. So you need an additional resistor in the upper leg of the voltage divider to increase the range. VRO would connect to the junction. The thing to use when you design it, at least one way, is to realize that the current of all the resistors in the divider is set by 1.25/R1.
I'm too lazy to do the math.
Another thing I missed, what is the actual supply voltage applied to the LM3914 V+ pin? It's doubtful, although possible, that the part can measure right up to its own supply voltage.
ICs are supplied with 8.4V
Sorry, I can't grasp what do you mean.
The 2.4V range is the range for both ICs, hence for each one the reference will be 1.2V
I can't cover the 1.2V range for two IC's because the minimum voltage they can pick up is 1V.
And I also can't change the R1 resistance, it will decrease the brightness of the LEDs.
Could you describe a bit more detail, please?
upd.
I've found that the current across R1 is changing by the R2 value, so its like impossible to set the proper voltage, idk why is it so
