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Like I said in the title, I am using the LM3914 IC to drive a 10 segment LED Bar-graph. The actual subsystem has the signal pin on the IC connected to a capacitor, thus displays the amount of charge the capacitor is holding on the Bar-graph.

I've done a test of my entire circuit on a breadboard and found that every subsystem - including the bar-graph circuit - worked a treat.

However, the LED segments on the Bar-graph are so dim that they are barely visible under bright light. Is there a way of making the 3914 increase the current through the LED segments to make them brighter?

Here is a rough schematic of the circuit I am using: enter image description here

(Supply and resistor values are actual)

Any pointers would be extremely helpful thanks!

Here is the specific bar-graph if it is of use: https://www.kingbrightusa.com/images/catalog/spec/DC10GWA.pdf

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From the datasheet:

The current drawn out of the reference voltage pin (pin 7) determines LED current. Approximately 10 times this current will be drawn through each lighted LED ...

Since the voltage is 1.25V, the current in the schematic will be:

\$ {12.5\text{V} \over {1\text{k}\Omega+3.9\text{k}\Omega}} \approx 2.55mA\$

Reduce the resistances in the divider to about 1/8 of their current values.

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  • \$\begingroup\$ Oh wow, I didn't spot that in the data sheet at all XD \$\endgroup\$ – Tony G. Dec 23 '16 at 1:14
  • \$\begingroup\$ Thanks mate, I replaced it with a 100 ohm and a 390 ohm and it worked superbly! \$\endgroup\$ – Tony G. Dec 23 '16 at 1:14
  • \$\begingroup\$ @tonyg. At 1/10 the divider, your putting 25.5 mA through each led which is above the normal 20mA limit for normal life of the led. Also, your putting another 25 mA through the divider, which may be bad for battery life, if battery powered. \$\endgroup\$ – Passerby Dec 23 '16 at 6:40
  • \$\begingroup\$ Well no, the divider gets 1/10 of the LED current, so 2.55mA. \$\endgroup\$ – Ignacio Vazquez-Abrams Dec 23 '16 at 8:06

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