OK, attached is the schematic for the project I'm working on, I've circled the power on reset and the transistor bits, because those are the areas I have concerns about
Namely, will the power on reset work, will the transistors work as wired, and what value of current limiting resistors do i need for the transistors??
the pair of two because I've never used a transistor for that type of application before, and one because it's a 6 volt EL wire unit, and I'm planning on using a 6 volt power supply for all of this.
(BC547B & BC327, if you can't read it)

Thanks guys

 

The power on reset circuit needs to pull down the reset inputs after the initial HIGH pulse was applied. This would be done via the 100k resistor.... if there wasn't all these diodes.

With Q4 of IC4 LOW and D2 not conducting the base of the transistor at pin 4 of IC9 will be in an undefined state.

You need a resistor from the cathode of D2 to Gnd. (or the base of the mentioned transistor)

The BC547B has a minimum DC current gain of 200. At 10k collector resistor you can easily calculate the maximum base resistor.

 

The power on reset circuit needs to pull down the reset inputs after the initial HIGH pulse was applied. This would be done via the 100k resistor.... if there wasn't all these diodes.

With Q4 of IC4 LOW and D2 not conducting the base of the transistor at pin 4 of IC9 will be in an undefined state.

You need a resistor from the cathode of D2 to Gnd. (or the base of the mentioned transistor)

I thought the diodes might mess something up
I think I'll just make 2 resets and eliminate those two diodes
(also, could someone please explain how that reset works? how do you know what the voltage is going to be on the low side of the capacitor?

The BC547B has a minimum DC current gain of 200. At 10k collector resistor you can easily calculate the maximum base resistor.

how do I find how much CE current I want though? How can I be sure that the voltage will drop enough to trigger the 555?

 

If you guys can tell me how to calculate the resistor, I can do it, I'm just not sure if the process is different since voltage at the collector is the main goal here, rather than total current draw

 

If you guys can tell me how to calculate the resistor, I can do it, I'm just not sure if the process is different since voltage at the collector is the main goal here, rather than total current draw

(also, could someone please explain how that reset works? how do you know what the voltage is going to be on the low side of the capacitor?

How can I be sure that the voltage will drop enough to trigger the 555?

Take what you have at hand, it will work with various combinations. Take for example the 10k, at 200 DC gain that would be 2M base resistor (just ignoring voltage drops on the transistor ). It will work. You will get a steeper edge if you increase base current, say if you use 100k.

Reset: When the circuit is powered up the capacitor acts as a short circuit because it's discharged. +6V gets therefore to the reset inputs. The capacitor charges via the 100k resistor, the voltage at the capacitor negative pole goes to 0V, the diodes cease to conduct.

555: As far as I remember the trigger input is compared to a voltage level at 1/3 of the power supply voltage, i.e. 2V. The VCE saturation voltage of your reset switch at the trigger input needs therefore to be lower than 2V.
What exactly is this 555 doing there?

 

Take what you have at hand, it will work with various combinations. Take for example the 10k, at 200 DC gain that would be 2M base resistor (just ignoring voltage drops on the transistor ). It will work. You will get a steeper edge if you increase base current, say if you use 100k.

Reset: When the circuit is powered up the capacitor acts as a short circuit because it's discharged. +6V gets therefore to the reset inputs. The capacitor charges via the 100k resistor, the voltage at the capacitor negative pole goes to 0V, the diodes cease to conduct.

OK, that all makes sense, thankyou

555: As far as I remember the trigger input is compared to a voltage level at 1/3 of the power supply voltage, i.e. 2V. The VCE saturation voltage of your reset switch at the trigger input needs therefore to be lower than 2V.
What exactly is this 555 doing there?

Acting as a bistable latch to keep the output high until the reset is triggered via Q3

[EDIT]
wait, Q0 is automatically high on reset, isnt it... I'll offset all the outputs by one, reset is now Q4