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I built the following circuit but am getting a result I wasn't expecting.

LED module circuit (image)

Here is a link to the CircuitLab model (I don't know how to upload a saved circuit to StackExchange):

LED_module_circuit_diagram

The circuit design was supposed to work as follows:

The bank of LEDs labelled FWD are three white LEDS, the bank of LEDs labelled REV are two red LEDs. These represent the lights at one one end of a model locomotive which may move forward or reverse and the lights respond accordingly. The lights on the rear of the locomotive would behave the opposite way.

The four inputs on the left are logic high (+5V) or low (0V) (from a MCU), except Vss which is tied to the positive supply (+5V). For the purposes of this question the input nINTEN may be ignored.

A logic low on input nEXTEN is supposed to enable use of the FWD and REV banks of LEDs. A logic high should switch off both banks. While nEXTEN is logic low the input FWDnREV should dictate which bank of LEDs is lit. With logic high the white LEDs should light (red LEDs off), with logic LOW the red LEDs should light (white LEDs off).

The circuit works as desired except in one case. With nEXTEN low, and FWDnREV low, the red LEDs light as desired, but the white LEDs are dimly lit as oposed to fully extinguished.

What is the cause of this and what should solve it?

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  • \$\begingroup\$ I don't know how to upload a saved circuit to StackExchange .... edit your post ... click on the circuit button ... design circuit ... click save and insert ... done \$\endgroup\$
    – jsotola
    Commented Jan 26 at 23:13

3 Answers 3

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To turn on Q4, there must be a very small current through white LEDs to base of Q4. The current is small but it is enough to light up the LEDs so that they are visible.

You could try to put e.g. a 1k resistor over the white LEDs to keep the voltage over LEDs low enough while the 1k resistor passes current instead of the LEDs.

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  • \$\begingroup\$ Would it then be the case that Rb4 is redundant and the biasing of Q4 could be achieved solely by the resistor that you have proposed? \$\endgroup\$
    – RickyBoy
    Commented Jan 27 at 12:40
  • \$\begingroup\$ @RickyBoy It depends what would be the exact circuit you mean. If you omit the Rb4 between LED cathodes and base of Q4, then when Q3 is off, there will be even more current through white LEDs to base of Q4, and white LEDs are almost fully on. \$\endgroup\$
    – Justme
    Commented Jan 27 at 13:01
  • \$\begingroup\$ To clarify ... if we include the 1k resistor (though possibly modify its value as necessary) as you suggest - let us call it RJ - and remove Rb4, then doesn't RJ achieve both the biasing of Q4 and the "bypass" of the LEDs when Q3 is on? \$\endgroup\$
    – RickyBoy
    Commented Jan 27 at 13:16
  • \$\begingroup\$ @RickyBoy If I am not mistaken what you mean, the current from 5V through 47 ohms and 1k to Q4 base is around 4.1mA without the white LEDs. Which means, there will be 4.1V over the 1k resistor. But white LEDs likely have less drop than 4.1V so some current will flow through white LEDs. The 1k resistor should be lower in value, but then respectively more current flows, so even halving it to 500 ohms will cause 3.9 volts over 500 ohms, the white LEDs could still be visibly on. \$\endgroup\$
    – Justme
    Commented Jan 27 at 15:00
  • \$\begingroup\$ You're right. The white LEDs have drop approx 3.5V per datasheet. So my suggestion doesn't work. \$\endgroup\$
    – RickyBoy
    Commented Jan 27 at 15:06
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You could try something like the following. If the red LEDs don't shut off completely, you could add a diode in series with the emitter of Q1.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Will try it out \$\endgroup\$
    – RickyBoy
    Commented Jan 26 at 19:37
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Rb4 should be connected between Vss and the base of Q4. The collector of Q3 should be connected to the base of Q4 and your “fwd” node.

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