555 timer circuit with pulse motor

[jasonspinmaster]

i am making a pulse motor and i am trying to use an optic sensor (OPB876) to fire the coil. can some one have a look at my circuit please. as at the it is not firing and i am really really new at this.
the transistor is a mj2955 PNP
optic sensor OPB876

[pese]

i am making a pulse motor and i am trying to use an optic sensor (OPB876) to fire the coil. can some one have a look at my circuit please. as at the it is not firing and i am really really new at this.
the transistor is a mj2955 PNP
optic sensor OPB876

You must ad an rbe 
Resistor between emitter and base.
1 kohm  to 5 kohm

it will TAKE LEACKAGE currents from transistor
2 mak the fall-time faster .
Important this Resistor ,
even if most designer (Naudin enz) forget this or dont know it

Pese

[gyulasun]

i am making a pulse motor and i am trying to use an optic sensor (OPB876) to fire the coil. can some one have a look at my circuit please. as at the it is not firing and i am really really new at this.
the transistor is a mj2955 PNP
optic sensor OPB876

Hi,

You have to use NPN transistor as the simplest remedy to your circuit shown!!!  Because the collector goes to the positive battery pole through the coil and a PNP transistor cannot receive positive voltage at its collector, OK?
So obtain an MJ3055 NPN or 2N3055 or similar NPN transistors, to replace your PNP  (and use heat sink too).

Gyula

EDIT: Here is a useful page on 555: http://www.williamson-labs.com/555-circuits.htm

[jadaro2600]

I don't have a pulse motor, but I have a pulse circuit - and a question:

I measured across the current source and I get a measure of 55ma, regardless of whether or not there's a LED present in the circuitry; can anyone explain this - the LED is bridged across the emitter and the collector in the circuit.

[gyulasun]

If I understand you correctly, you measured 55mA current consumption from a battery (or power supply) so your pulse circuit consumes that current,  right? 
It would be much easier to answer your question if you could show even a free-hand drawing in Windows Paint...

If you mean a real current source then the characteristic of current sources is that they do not let the current taken out change, whatever load you connect to them and in your case this current is set to 55mA...   Really this is a possible guess.

rgds,  Gyula