[Ronbo]

You'd have to put a resistor in parallel to simulate the current draw of the lamp...

LED's are typically around 10~20ma at 2v, (you'll have a dropping resistor in the assembly about 1k ohms). To bump the current up, solder a 330 ohm 1/2watt resistor across the LED lamp assembly (including it's dropping resistor if it's external). This will put you closer to 50mA like the lamp would draw.

It's still a lot of leakage though... Even bipolar semiconductors typically only produce a few microamps of leakage.