It might work, but - when the 555 output is high, current id flowing through *both* resistors. The diode and transistor effectively "short out" the red LED to keep it from illuminating, but they do not open the current path through its resistor; in fact, they increase its current above what it is when the red LED is on. This will shorten battery life significantly.

There is a better 1-transistor way, but it is buried in my old posts. I'll try to find it.

ak

 

That circuit is inefficient because current flows even when the LED is off. You could simplify it and eliminate the diode if all you want to do is switch a load on and off.

 

That circuit is inefficient because current flows even when the LED is off. You could simplify it and eliminate the diode if all you want to do is switch a load on and off.

Post #1, the two LEDs toggle, and both currents exceed a 555 output capability.

ak

 

Post #1, the two LEDs toggle, and both currents exceed a 555 output capability.

I thought we were now discussing #19 with just one load on a single transistor.

 

That circuit has two loads effectively. One is switched directly by the transistor, and one is shorted out by the transistor and the diode. See post #21.

ak

 

Here is a basic concept for switching two LEDs with one transistor. If the Vf of the two LEDs are approximately equal, then the extra Vf of the signal diode will prevent the green LED from coming on when the red LED is grounded through Q1. Any MOSFET appropriate for the current will work.

ak

 

Again, I think the TS is now switching one LED, not two.

 

Here is a basic concept for switching two LEDs with one transistor. If the Vf of the two LEDs are approximately equal, then the extra Vf of the signal diode will prevent the green LED from coming on when the red LED is grounded through Q1. Any MOSFET appropriate for the current will work.

ak
View attachment 118107

Cool circuit AnalogKid!