ULN2003 driving Relays with status indicator
simulate this circuit &ndash; Schematic created using CircuitLab

Figure 1. On and off indications using one Darlington buffer output.


R1 drives the green LED as in your schematic.
When Q1 turns off current flows through RLY1 to light the LED. If, for example, the coil resistance was 300 Ω then about 5.5 mA would flow through that circuit. It is very unlikely that the relay will remain on with 5 mA through the coil.


To give yourself some confidence check the minimum hold-on voltage for your relay and measure the voltage across the coil when R2 and the red LED are in series.


  [@Sparky256's comment:] If Q1 is OFF, won't there be current flow through R1 and R2 and both LED's? Or enough to keep the Green LED dimly lit.




Figure 2. A Finder 12 V relay with a 300 Ω coil.

At 5.5 mA calculated above the voltage across the coil of this relay would be \$ V = IR = 5.5m \times 300 \approx 1.7 V \$. Let's be generous and say that the LED forward voltage will only be 1.5 V at small current. That leaves 0.2 V across the 1.8k series resistor so the current R1 and the green LED would be \$ I = \frac {V}{R} = \frac {0.2}{300} \approx 0.7~mA \$.



Figure 3. LED forward current and voltage relationship.

Depending on the efficiency of the LED this might be a problem. It will be very faint in relationship to the red LED which would be on at this time.


  @Spehro's comment: I would suggest adding an NPN voltage follower (eg. 2N4401) to the lower side of the coil (collector to +12, base to ULN2003 output Q1, emitter to R2 (1.5K). That way the current through the coil when off will certainly be negligible.








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Figure 4. Schematic for Spehro to edit.