Turn ON LED that behaves like OR gate with the same intensity

Question

Suppose that A and B in the image are connected to the two output pins of LED driver, which can be independently switched on and off a constant current of 30 mA.
D1, D2, and D3 are all the same type of LED and I want D3 to act like the output of an OR gate, i.e., to turn ON when either or both D1 or D2 is ON.
However, in my circuit, D3 will have 60mA and will not have the same brightness when both D1 and D2 are ON.

Is it possible to make D3 have the same current as D1 and D2, i.e. 30mA, and still behave like an OR gate??
Of note, other pins of the LED driver are not available because they are occupied for different purposes.

Edits: These are LEDs and a LED driver I am going to use.

Answer 1

This is a way to limit the current through D3 to approximately 30mA, which is not quite what you asked for:

^{simulate this circuit – Schematic created using CircuitLab}

The TL431 circuit is fairly accurate and stable with temperature but responds slowly compared to PWM. The transistor is less accurate and stable with temperature but responds fast. The effect of "accuracy" is that the brightness may change noticeably when D1+D2 are on vs. just D1 or D2. To reduce that effect, you could limit the current to something like 25mA (increase the resistor values by about 30/25) so that it is always limiting.

Answer 2

Here I used a single PNP bipolar transistor to act as a 30mA current source:

^{simulate this circuit – Schematic created using CircuitLab}

_{Ignore Da and Db, which are only present to constrain voltage across sources I1 and I2 when their switches are open.}

When either R1 or R2 passes 30mA, they develop 3V. Diodes D4 and D5 wire-or that voltage to the base of Q1, which places about 1.8V across R3, setting current through it to:

$$ I_{R3} \approx \frac{1.8V}{60\Omega} = 30mA $$

All of that current passes via D3, so if either D1 or D2 is lit (or both), so is D3.

There will be 19V across Q1, so it will dissipate \$19V\times 30mA=0.6W\$. That's quite a lot, so consider reducing the supply voltage to 12V. Your driver IC will be dissipating the same power per channel, so you'll be helping that too.

Answer 3

No.

You have defined a hypothetical question with some current-mode logic using diodes combined into one so-called "diode-OR" voltage logic. That would violate Kirchoff Laws to do what you asked.

Maybe. You could limit diode current with a precision current sensor, reference voltage and switch on an active shunt. But no simple RTL logic solution as you might be thinking.

A simple CMOS OR/NOR gate could replace this attempted question but use only 3.3V to 5V logic rather than waste 24V.

Answer 4

Here is a way to limit the current in D3 to 30mA, while using the same current that flows in the other LEDs - ie: D3 does not consume more current from the power supply than what is already being used by D1 and D2.

^{simulate this circuit – Schematic created using CircuitLab}

Circuit Description
When D1 or D2 has 30mA of forward current, the same current is passed through D3, as set by the current source formed by Q1, Q2, R1, R2. However, when both D1 and D2 are on, the extra 30mA bypasses D3 via D4 (zener) and D5.

The components R1, R2, Q1 & Q2 form a constant current source of about 30mA, which is only energised when either of the current sinks for D1 or D2 (or both) are turned on. If both D1 and D2 are off, then it remains in the off state.

You could introduce another LED in series with D4 to indicate the AND function, ie: when both D1 and D2 are on (refer D5).

The voltage required for zener D4 should be selected based on testing of your circuit. The value shown (5V) here is merely a starting point. Note that D4 will dissipate more power as its break-over voltage increases. Select the breakover voltage of D4 to be as low as possible while still providing voltage margin for the current source and LED D3 to work correctly. The anode of D3 only needs to be about 1V below Vcc_24V for the current source to work properly.

R1 is selected to give 30mA of current into the emitter of Q1. R2 is selected to provide sufficient base current for Q1, and needs to be adjusted if supply voltage (24V) is varied significantly.