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Who's used the display driver LM3914 or LM3915?

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Nora

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What does the difference between dot and bar mode mean?
I'm using it coupled with a 10-LED bar graph in dot mode and my bar graph flickers all the time with no input signal.
Any thoughts?
Thanks!
N_N
 

alec_t

Well-Known Member
Most Helpful Member
Dot mode is one LED at a time; bar mode is a row of LEDs at a time.
my bar graph flickers all the time with no input signal
That could be due to electrical noise pickup if the input is floating.
 

Nora

New Member
Thanks for the explanation.

My input is unfortunately an analog range of about 38mV to 1.2V. Any suggestions for how to scale it to start at 38mV = 0?
Thanks!
N_N
 

audioguru

Well-Known Member
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Connect pin 8 to 0V then pin 7 will be +1.28V. Connect a 470 ohm resistor to 0V from pin 7 to set the LED currents to 22mA each.
Connect the top of the divider pin 6 to pin 7 so that the 10th LED lights when the input is +1.28V.
Connect a resistor from the bottom of the divider pin 4 to 0V so that the current in the divider creates a voltage of 38mV at pin 4 so the 1st LED lights when the input is +38mV. Connect a resistor to 0V at the input pin 5 of 1M or less to keep the input from floating high.

The divider in the LM3914 is about 12k and the divider in the LM3915 is about 28k so adjust the resistor from pin 4 to 0V accordingly.
 
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Nora

New Member
Connect pin 8 to 0V then pin 7 will be +1.28V. Connect a 470 ohm resistor to 0V from pin 7 to set the LED currents to 22mA each.
Connect the top of the divider pin 6 to pin 7 so that the 10th LED lights when the input is +1.28V.
Connect a resistor from the bottom of the divider pin 4 to 0V so that the current in the divider creates a voltage of 38mV at pin 4 so the 1st LED lights when the input is +38mV. Connect a resistor to 0V at the input pin 5 of 1M or less to keep the input from floating high. The divider in the LM3914 is about 12k and the divider in the LM3915 is about 28k so adjust the resistor from pin 4 to 0V accordingly.
As always, Audioguru, you helped me, thank you!

I went to the LM3915 datasheet (Block diagram page) and calculated about 26k (divider R1 is 22.63k), attached this to pin 4 and the other end to ground. I put a 820ohm from pin 5 (signal in) to ground. Still blinky on "no" input. Grounding pin 5 gave the expected no blinky.
I started inching up the resistor on pin 4, all the way up to 117k, which still gives a little bit of blinkiness, but less. Naturally, the bar graph now doesn't work over its entire range.
The other thing is that it is wired in dot mode (Mode pin 9 left open), but gives bar output (lots of bars on at the same time).
My signal in is in the 800-1900MHz band, so presumably there are lots of these floating in the air and I can't control it.
Any thoughts?
My thought is to ditch the bar graph and just use a transistor to one LED. Not as nice-looking, but perhaps easier to control?
Thanks in advance,
N_N
 

audioguru

Well-Known Member
Most Helpful Member
I went to the LM3915 datasheet (Block diagram page) and calculated about 26k (divider R1 is 22.63k), attached this to pin 4 and the other end to ground. I put a 820ohm from pin 5 (signal in) to ground.
Absolutely WRONG!
You did not do what I said.
The block Diagram on National Semi's datasheet is on page 7 and does not have an R1.

The resistor from pin 4 to ground will have a fairly low value, about 820 ohms so that when the input is 38mV then the 1st LED lights. With no input then no LED should light. When the input is 1.32V then the 10th LED should light.

Pin 5 will float high if it does not have a resistor to ground. The resistor can be 1M, 100k or any value that is 1M or less.

The LM391x is designed to show audio signal levels, not radio signal levels. It will show the output of the audio level from a radio.
your circuit is extremely simple so many lEDs will be turned on and will look like a dim blur. All the audio level circuit I built with an LM3915 or two use one of the peak detector circuits shown in the datasheet so that the peak level is clearly shown.
 
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Nora

New Member
Absolutely WRONG!
You did not do what I said.
The block Diagram on National Semi's datasheet is on page 7 and does not have an R1.

The resistor from pin 4 to ground will have a fairly low value, about 820 ohms so that when the input is 38mV then the 1st LED lights. With no input then no LED should light. When the input is 1.32V then the 10th LED should light.

Pin 5 will float high if it does not have a resistor to ground. The resistor can be 1M, 100k or any value that is 1M or less.

The LM391x is designed to show audio signal levels, not radio signal levels. It will show the output of the audio level from a radio.
your circuit is extremely simple so many lEDs will be turned on and will look like a dim blur. All the audio level circuit I built with an LM3915 or two use one of the peak detector circuits shown in the datasheet so that the peak level is clearly shown.
No, you misunderstand me. What I meant by R1 was all of the resistors on page 7 (I have TI datasheet, so page 9) added together. R1 = 6.63 + 4.69+3.31+2.34+1.66+1.17+0.83+0.59+).41+1.0. So then R2 came out to be about 26k to make Vout 38mV

Where is my mistake? And where in the datasheet does it say how to program the value at the 10th LED?
Thanks!
N_N
 

audioguru

Well-Known Member
Most Helpful Member
Your arithmatic is wrong.
Pin 7 should be about 1.28V. The typical divider resistance of an LM3915 is said to be 28k (plus or minus 28%) in the datasheet. I don't know if you measured yours. Then the current is almost 1.28V/ 28k= 45.7uA.
The resistor from pin 4 to ground is 38mV/45.7uA= 831 ohms plus or minus 28%.

The 10th LED lights when the input is a little higher than the pin 6 voltage.
 

Nora

New Member
Got it. Thank you. I also had to put a .1uF cap between pins 1 and 2 to discourage flickering.
N_N
 

audioguru

Well-Known Member
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National Semi invented the LM3914, LM3915 and LM3916. I don't look at Texas Instruments copies of them because maybe they copied it wrong.
 

Nora

New Member
Hello again,
Can you make sense of Note 3 for me? It says see tabl for threshold Voltages.
There is a chart with the Threshold Voltage, 5 columns, output, dB, Min, Typ, Max.
My signal input range is around 0 - 1.3V, not up to 10.0V like the chart implies. At 1.3V input signal, the 10th LED lights up.
Thanks in advance,
N_N
 

audioguru

Well-Known Member
Most Helpful Member
You can set the reference voltage to whatever you want. The circuit in figure 1 has two resistors between pin 7 (Ref Out) and pin 8 (Ref Adj) which sets the reference at +10.0V which is applied to pin 6 (Rhi).

You probably have pin 8 connected to 0V so your pin 7 (and pin 6) reference voltage is 1.25V. Then the 10th LED lights when the input is +1.25V.
The threshold voltage is the input that lights a certain LED. In the LM3915 the threshold voltages are 3dB apart (0.707 times or 1.414 times). So if +1.25V lights the 10th LED then 1.25V x 0.707= 0.884V lights the 9th LED and 0.625V lights the 8th LED.
 
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Nora

New Member
OK, thanks for the explanation. I get it now.
I've got something strange going on though. I have only one resistor between pins 7 and 8, 630ohms and a 2.2uF cap between pins 1 and 2. No other passives.
I'm using each output from the LM3914 to trigger a PNP which then triggers an LED panel. The trouble is that when all 10 panels are connected, output 9 from LM3914 causes panel 9 to occasionally blink quickly, out of order, before the rest of the panels.
Is it normal that one output could appear to trigger out of order for a split second? I say appear because I wonder if the switching time of the PNPs could be off from each other ever so slightly.
?
 
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audioguru

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OK, thanks for the explanation. I get it now.
I've got something strange going on though. I have only one resistor between pins 7 and 8, 630ohms and a 2.2uF cap between pins 1 and 2. No other passives.
Since you do not post your schematic then we are just guessing at the details. Is pin 8 connected to 0V? Then the 630 ohm resistor (not a standard resistor value) from pin 7 to 0V causes the current in each lighted LED to be 17.5mA.
You should NEVER connect a capacitor between pin 1 and pin 2.

I'm using each output from the LM3914 to trigger a PNP which then triggers an LED panel. The trouble is that when all 10 panels are connected, output 9 from LM3914 causes panel 9 to occasionally blink quickly, out of order, before the rest of the panels.
Is it normal that one output could appear to trigger out of order for a split second? I say appear because I wonder if the switching time of the PNPs could be off from each other ever so slightly.
Since you do not post your schematic then maybe the supply voltage is collapsing when more than a few PNP transistors have 17.5mA of base current. You don't even say what is the output current of each PNP transistor. You don't even say what are the supply voltages and currenty capability.

Instead of talking about some of the circuit, please post the entire detailed schematic.
 

Nora

New Member
View attachment 64022

Pin 5 is signal into the LM3914, sorry the cap is between supply and ground, not pins 1 and 2.
Supply to the transistors is now 16V because that seems to work the best. Since there is one supply, all 12V are now 16V.
I am using a benchtop adjustable power supply rated for 30A.
TIP117 (Darlington) saturates (measured) at 0.77V.
Each panel is on a progressively longer wire. Panel 10/LM3914 output 10 is on the longest wire (about 20 ft.).
Panel 10 draws about 0.44A while Panel 1/LM3914 output 1 draws 1.08A.

The black boxes connected between collector and ground represent the LED panels. Each panel without long wires draws about 1A and is designed for 12V.

Can you explain further what you mean about supply voltage collapsing? Why? How can I design this out?

The wire is 18AWG and drops 1.27V per 100 feet.
 
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audioguru

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Most Helpful Member
What limits the LED current? The darlington transistors are slammed on hard with a base current of 17.5mA so their max collector current can be many many Amps. For a very good output of 1A the datasheet shows that they need a max base current of only 4mA.

Why do you have many 1.2k base resistors for the darlingtons? They are not needed because the output of the LM391x is adjusted with the value of the resistor from pin 7 to 0V but the value of your resistor is too low so it might cause smoke.

Your circuit is missing an extremely important local supply bypass capacitor.
 

Nora

New Member
The LED current is limited by the panels. They are made from LED reels which are configured like so: many sets of 3 red LEDs (1.8Vdrop) in series with 330 ohm resistor all in parallel with each other. About 50 sets of these totaling about 1 Amp draw (measured).

I chose the value of resistor to give close to 20mA for current, originally I thought the panels might draw more than 1A and wanted to make sure it would be enough.

Where is the circuit missing the bypass cap? Do you mean on the 12V? I have a capacitor there.

How accurate is the LM3914? It looks pretty good right now, just has that one panel (number9/output 9) that flashes for a split second before the others.....occasionally.

Thank you for your help!
 
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audioguru

Well-Known Member
Most Helpful Member
You do not understand that the TIP117 turns on very well when its base current is 1/250th its collector current. Then since the max collector current is only 1A then its base current should be only only 4mA. Yours is much higher. Why?

Maybe you built the circuit on a breadboard that has high capacitance between the wires and between the rows of contacts. Real working circuits are not built like that.
Maybe your input cable is not shielded audio cable so it picks up mains hum.
 
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Nora

New Member
Original transistor had gain of 50. Left old circuit w/ old transistor but swapped transistor for the high gain one, lazy.

Circuit is not built on breadboard, it is soldered on etched board. I rarely breadboard circuits.
The input wires are not shielded, will change to shielded wire.
Thanks again for your help.
 

audioguru

Well-Known Member
Most Helpful Member
A transistor that is used as a linear amplifier with plenty of collector to emitter voltage has a gain of 50. But a saturated transistor used as a switch has a gain of only 10 and sometimes less.

A TIP117 darlington has a minimum current gain of 100 when used as an amplifier. its minimum gain is 250 when used as a switch.

Do you understand that the 1.2k base resistors are not needed because you can set the amount of output current from the LM3914 by the value of the resistor from pin 7 to 0V.
 
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