555 timer circuit with pulse motor

[Thaelin]

   This sounds like the led is acting like the protection diode in this circuit.
It is being lit by the coil collapse  and not by the circuit its self.

Just a thought.

thaelin

[tropes]

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

This thread should be in the Pulse Motor rather than the Magnet Motors section of the forum.
With the advice from Gyula and Hoptoad I have been using a photointerrupter to fire the coils of my Sotropa Motor without any transistor failures provided you remove any reverse current (BEMF,induced current and Flyback EMF) from the coil. Simply add a diode to each end of the coil and collect the reverse current in a capacitor. To demonstrate, I have uploaded this video http://www.youtube.com/watch?v=V6nojNujmoM
Tropes

[jadaro2600]

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

LED is off to the right ...should have flattened out the book a little better.

[gyulasun]

@jadaro2600

Ok, now I understand.  If you study the data sheet of any 555 timer IC, http://www.national.com/ds/LM/LM555.pdf  you find the current consumption of the integrated circuit itself (to make its own working properly) is about 6mA from a 9V supply. 
Now you have to consider R3 1kOhm connected to the output of the 555, it draws about (9-0.65)/1000=8.35mA  [0.65V is the forward voltage of Q1, V_BE.
And finally the current through  (and not across!)  R4, when Q1 is ON: (9-0.2)/270=32.5mA  (The 0.2V is the saturation voltage of Q1,  V_CEsat.)

If you sum up these currents, you get  6 + 8.35 + 32.5=46.85mA. You said 55mA, the difference of roughly 8mA may be explained by the 555 itself (it may draw higher than 6mA due to manufacturing process) and resistor tolerancies.

If you wish to reduce the total consumption to a bare minimum possible (still exluding the LED)  then use the CMOS version of 555 like LMC555CN that has under 1mA self consumption and use a 10 kOhm (or maybe higher value) resistor at R3 to drive the base of Q1 (now you know how to calculate its base current). This way the first two currents will total to about less than 2mA.

The main consumption comes through R4,  270 Ohm when Q1 is ON, about 32.5mA  You can reduce this to any lower current that is still useful or desirable for your LED, ok?  How?
Suppose you wish the LED draw 20mA, right?  (Most LEDs give brilliant and strong enough light at this current.)
You have two choices:

1) You drive the LED directly from the output pin of 555, via a resistor because the 555 is able to drive higher than even 100mA loads, see data sheet.  So to pick the value of the new resistor,   (9V-3.4V)/20mA=280, chose standard 270 Ohm value, ok?  So you omit Q1, R4 and connect a 270 Ohm resistor right to the pin 3 of 555, the anode of the LED goes directly to the other end of this 270 Ohm and the cathode of the LED goes to GND i.e. 9V supply negative.

2)  If you wish to use Q1 to switch ON/OFF the LED, then everything is connected as you showed, except the LED:  you insert the LED in series with the collector of Q1, the LED's anode goes to the lower leg of R4 and the LED's cathode goes to the collector of Q1, ok?  (This way the LED will not be in parallel with the transistor's collector and emitter but in series with the collector;  when Q1 is ON, current will be about:  (9V-3.4V-0.2V)/270=20mA.

Of course if you wish to let less current flow through the LED, then you can choose the value of the resistor (R3 and / or R4) accordingly higher. 

rgds,  Gyula

[X00013]

The easiest way for me at least, was to buy a throw awy camera for 7 bucks and replace the "take picture button" with an if sensor, and if you can produce enuff E to replace battery you have OU? As far as i am concerned, 50 years of kodac engineering will be alot more efficient than what i could do in a lifetime, so why waste your time doing whats already done. Buy the camera for 7 or spend 35 at radio shack. Go figure. Have fun and good luck