Pulse bit draws large current


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?


I noticed this as well, the max draw for a pulse bit was close to 30mA (my meter’s max accuracy is 10mA).

Hey 7th, how ya been?

Perhaps they are trying to ensure that even a weak signal will still work. Imagine that the pulse is at the end of a synth and connected to a couple servos and several LEDs. And the battery is getting weak. VCC and the signal line get as high as 3 and a half volts, maybe. This circuit looks like it’s designed to provide a readable pulse under those conditions. Maybe a mosfet would have been better than the bc847bs?

I’m lookin’ at my copy of Mim’sTimer Circuits book, and he uses less ohms for higher current applications. He triggers transistors with as little an no resistor, and up to 10K, for decreasing current loads. The LV321 doesn’t present much of a load, right?

OK, more questions than answers. But this a device with a very wide range of applications, so things like current constraints are limiting. Perhaps the designers got carried away.

No matter the cause - I like your easy fix! Can you detail it out here, or maybe post it in the Inventions area? I think we should encourage activities beyond hooking bits together.