I checked the power consumption of some of the bits and found out, that the pulse bit draws between 7.5 and almost 28mA, depending on the pulse level. There was no bit connected on the output. So it must be the pulse bit itself. Looking into the schematics you find this 100 Ohm resistor on the base of the NPN transistor ( used as inverter ).
According to my understanding, in case the output of the LM555 goes high, the current flows through the resistor over the base to the emitter to ground. The base-emitter current is defined by the base resistor.
I changed this resistor to 4k7 Ohm and now the power consumption of the bit is between 7.5mA and roughly 3mA, which is still quite high for such a small logic bit, but more reasonable than the 28mA.
Question: is the 100 Ohm resistor used by intention?