0
\$\begingroup\$

First of all here's the datasheet for the IC, a dot/bar display driver.

What I'm confused about are the following pins: 4,6,7,8

pinout from datasheet

The datasheet mentions a 1.25V reference source. Is that built into the IC so that no matter how I configure it there is a 1.25V source there? From what I can see, this involves pin 7 and 8 (REF OUT and REF ADJ). Assuming pin 8 is connected directly to ground, what does this do? If I increased/decreased the voltage at the REF OUT pin 7 what changes? I understand that the LOAD at pin 7 determines the brightness (through current draw) but I don't understand what the voltage at pin 7 does.

The next two pins I'm confused about are 4 and 6 (RHI and RLO). They seem to be connected in series with the 10 1k resistors. Assuming RLO is connected to ground, as I decrease/increase the voltage at RHI, what happens? From looking at the block diagram, it looks as if the "LED Steps" are directly proportional to the voltage at RHI. Does this mean for instance, if RHI = 1V then every every 100mV from the signal source will turn on another LED? So 200mV = 2 LEDs, 400mv = 4 LEDs etc..? And if I put 3V at RHI would it behave similarly?

EDIT:

So in the schematic example they're connecting REF OUT to RHI which is why they say in that example each comparator would activate every 125mV? Does that mean I can connect whatever voltage I want there or do I have to use the reference voltage? Regarding resistors R1 and R2 connected to pin 7 and 8, I see that equation gives you the VOUT. But what's the point of VOUT? Why am I bothering to use a resistor divider between those two pins? Is it just to control LED brightness? If so why would they have two resistors instead of just 1?

What also confuses me is in the figure above they call it a "0V - 5V Bar Graph Meter." With a 0-5V signal source. If there were connecting REF_OUT directly to RHI as they do in that diagram, wouldn't the max voltage display be 1.25V? At that point wouldnt all the LEDs be on? So why do they call it a 0-5V bar meter? I notice they connect the signal source to RLO as well. What's up with that? Wouldn't you normally connect it to ground? Wouldnt your voltage across the "resistor ladder" constantly be changing if you connected RLO to the signal source? What's going on here?

It also looks like I connect the LEDs directly to the V+ source from the diagrams as well? This seems strange to me. For instance, what type of voltage to I need to hook up as VLED?

\$\endgroup\$
3
\$\begingroup\$

The datasheet mentions a 1.25V reference source. Is that built into the IC so that no matter how I configure it there is a 1.25V source there?

From that same datasheet:

block diag

So yes, that's built in.


Assuming pin 8 is connected directly to ground, what does this do?

You might simply want to actually read the datasheet :) it says:

The reference is designed to be adjustable and develops a nominal 1.25V between the REF OUT (pin 7) and REF ADJ (pin 8) terminals. The reference voltage is impressed across program resistor R1 and, since the voltage is constant, a constant current I1 then flows through the output set resistor R2 giving an output voltage of:

formula schematic

you can assume Iadj to be very small.


The next two pins I'm confused about are 4 and 6 (RHI and RLO). They seem to be connected in series with the 10 1k resistors. Assuming RLO is connected to ground, as I decrease/increase the voltage at RHI, what happens?

Look at the block diagram at the top of my answer. That's actually the resistive "ladder", ie. what your device does. It's hard to explain it better in words than with the schematic show: You divide the Voltage between pin 6 and 4 into ten equal voltages, and then have a comparator the "partial sums" of these voltages to the input signal.

\$\endgroup\$
  • \$\begingroup\$ So in the schematic example they're connecting REF OUT to RHI which is why they say in that example each comparator would activate every 125mV? Does that mean I can connect whatever voltage I want there or do I have to use the reference voltage? Regarding resistors R1 and R2 connected to pin 7 and 8, I see that equation gives you the VOUT. But what's the point of VOUT? Why am I bothering to use a resistor divider between those two pins? Is it just to control LED brightness? If so why would they have two resistors instead of just 1? \$\endgroup\$ – Zearia Mar 2 '17 at 23:04
  • 1
    \$\begingroup\$ The point of the voltage reference is to give you a moderately precise controlable voltage that you can pass to Rhi, which you can adjust by changing the ratio of the values R1 and R2 (in accordance with the published formula). The SUM of R1 and R2 allow you to program the LED current. With two pins and two resistances, you get two different tweakable parameters (output voltage for the comparator chain, and LED current control) \$\endgroup\$ – Lawrence NK1G Mar 3 '17 at 2:23
  • \$\begingroup\$ So then this means that if you choose the correct resistor values for R1 and R2, you and use the internal voltage to pass a voltage greater than 1.25V to RHI? And that equation for VOUT (in this answer) becomes the voltage being passed to RHI? If that's the case if Iadj is very small, is it possible to approximate VOUT by ignoring it since I don't know the value of Iadj? Are you saying that the ratio of R2 to R1 controls the VOUT voltage while the resistance of R1 and R2 controlls the LED brightness? \$\endgroup\$ – Zearia Mar 3 '17 at 2:47
0
\$\begingroup\$

The reference out voltage at pin 7 of the LM3914 is the same as the OUT voltage pin of an LM317 adjustable voltage regulator and the reference adjustment pin 8 is the same as the ADJ pin of the LM317. There is always 1.25V between the pins and Ohms Law calculates the resistor values. The reference output voltage at pin 7 can be set from 1.25V up to 1V less than the supply voltage.

The datasheet clearly says exactly how to do it.

The DIL through holes package of the LM3914 is obsolete now. The tiny surface mounted version made now might need a robot and an oven to solder it.

\$\endgroup\$
  • \$\begingroup\$ Is it really obsolete? The TI product page says it's still active. \$\endgroup\$ – Richard the Spacecat Jul 7 at 23:19
  • \$\begingroup\$ It also comes in a PLCC variant (which I'd expect to be obsolete, but it somehow is not). You really don't need a robot for that one either; 1.27mm pitch is still very much manageable by hand. \$\endgroup\$ – Richard the Spacecat Jul 7 at 23:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.