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Not getting the ideal 555 output

Yes Colin is right, but I wanted to go through each stage separately. First get the 555 timer working as it should then move backward. Take a step at a time.
Adam
 
Because this output is going to drive a E-mosfet which is connected to an inductance coil. My main objective is charge that coil with a current of 900mA and for 500ms only. So with the amplification of current will it have any effect on what the coil would get from the mosfet?
 
So it will have nothing to do with the output of IC555. And what about the specs of IC555.. I mean will it be able to handle any amplification in current considering worst case scenarios of over current situations?
 
But your driving a MOSFET so should be fine, you might need a small resistor on the gate which depends on the gate capacitance. You need to redraw your circuit and include any changes that Colin mentioned about your regulator.
Adam
 
I implemented this circuit and am getting the output desired but what I observe is that if I dont connect anything across the Zener diode (i.e. if I remove everything after the capacitor connected parallel to zener diode) i'm getting a constant a 12V but if I connect everything I dont get a 12V across the Zener voltage regulator, basically there's a voltage drop.

Could you pls help in identifying where I'm going wrong or maybe I have chosen some wrong value of resistor or capacitor somewhere which is causing this drop, because I need a 12V across the zener voltage regulator when everything is connected.
Circuit Diagram-1.jpg
 
Hello. Colin gave you a clue in his post. It looks like you are drawing too much current from your regulator or the regulator can't supply enough current. See if you can work out what needs to be change. You mentioned it in your last post.
Adam
 
Implemented a capacitive power supply to get the constant 12V.

Colin could you provide an explanation or a link for how the Darlington pair operates in making that short trigger pulse?
 
Implemented a capacitive power supply to get the constant 12V.

Colin could you provide an explanation or a link for how the Darlington pair operates in making that short trigger pulse?

The capacitor initially has the same voltage across it due to the two 10K resistors. The RC at the start of the circuit gives a small settling delay for the circuit. After a short time delay the Darlington switches on and places a low on this side of the plate. The other side of the capacitor follows this and also goes low creating the low pulse. The only slight downside to this circuit is the transistor draws current all the time. There are better ways to do this but the circuit shows the principle of what you need to do.
Adam
 
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