I'm sure your infrared part is either a photodiode or a phototransistor.
Like any load, it has more power if the resistance between the power source and the load is reduced.
A load can be either an input or an output device. It just consumes power from the power source.
So if your circuit has a resistor in series with your infrared part, reduce the resistor, but don't reduce it too much or your infrared part might blow up.
If that doesn't work, try adding an amplifier circuit to the output of the infrared unit.