LM3914 question

Discussion in 'Homework Help' started by Acce, Mar 31, 2010.

1. Acce Thread Starter New Member

Mar 4, 2010
4
0
Hello,

I created a 20-segment ledbar using 2 LM3914s. It is designed for a 0-5V signal, and uses 9V voltage source. It works all fine now, but one thing left me wondering:

The 600 ohm resistor was originally 1,2k one, but it didnt work. According to the datasheet, The LED current is about 10*(1,25V / R ), also the current drawn from REFout. With 1,2k it should give about 10mA as LED current, and with 600 double that. However, with the 600 one in use, the LEDs are not brighter than the ones behind 1,2k resistor in the second LM3914.

Could someone explain me why?

Attached a picture of the circuit, left out the mode, V+ and V- pins. Mode, V+ and LEDv are also 9V, and V- is connected to ground.

Also video of it if you want to see that the LEDs really aren't brighter:

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2. hondabones Senior Member

Sep 29, 2009
123
1
My Theory:

If you look at IC2 the set up is basically as recommended in the datasheet, where:

Vout = Vref(1 + R2/R1) + IadjR2

for math purposes let:

VREF = 1 V
R2 = 1.2 kΩ
R1 = 600 Ω

so...

Vout = 1 V(1 + 1.2 kΩ/600 Ω) + (1 mA)(1.2 kΩ)
Vout = 4.2 V

If you look at it from IC1:

R2 essentially is 0 Ω (based on how you have it wired and the datasheet recommendations)

so...

Vout = 1 V(1 + 0 Ω/600 Ω) + (1 mA)(0 Ω)
Vout = 1 V

Less voltage means less current.

I think this may be the reason. I have never seen this chip tied together like how you have it.

What did you do to get it functioning properly?

3. hondabones Senior Member

Sep 29, 2009
123
1
You may want to try something more like this:

4. Acce Thread Starter New Member

Mar 4, 2010
4
0
Well I was measuring the voltages over the leds and noticed that the other IC was driving too small current through the leds, and then I guessed that since the LED current is decided by the current drawn from REFout, I had to increase that current somehow, so I halved the 1,2k resistor I had there before, and then it worked.

Also your calculations seem right, but they can't be, since the measured Vout(REFout pin voltage) at IC2 was around 2,5 V, not 4,2V.

5. Ron H AAC Fanatic!

Apr 14, 2005
7,018
682
Of the 20mA flowing through the 600Ω resistor, 10mA comes from the lower 3914 and sums with the 10mA from the upper 3914, providing the required 20mA through the 600Ω. This is why your LEDs have equal brightness.
EDIT: Your voltage measurements are correct. You have simply stacked the two 1.25V reference supplies.

Last edited: Apr 5, 2010