Hi Electricman,
Your original IR tx circuit didn't put much current through the IR LED because the output of a 555 is rated for only 200mA, and I don't know if your supply decoupling cap was suitable to provide that much current.
Now you are trying to use this complicated Mosfet circuit that uses the transistor and zener diode solely to convert the 5V output of its 555 to the 9V required by the gate of the Mosfet. If the 555 was powered from the same supply voltage as the LED/Mosfet, then the transistor and zener diode wouldn't be needed, just the 555 driving a P-channel Mosfet instead. Just about like your B circuit except using a 10A LED, no current limiting resistor and an extremely short-duration pulse.
You don't post where you are, so i don't know if you buy European BDxxx power transistors or American TIPxxx ones.
Your photodiode is upside down, it needs to be reverse-biased and the IR radiation causes it to have a leakage current. A diode doesn't have the gain of a transistor so the amplifier would need to have much more gain. Try reducing the value of the amplifier's 1K resistor or shorting it for plenty of gain.
Your original IR tx circuit didn't put much current through the IR LED because the output of a 555 is rated for only 200mA, and I don't know if your supply decoupling cap was suitable to provide that much current.
Now you are trying to use this complicated Mosfet circuit that uses the transistor and zener diode solely to convert the 5V output of its 555 to the 9V required by the gate of the Mosfet. If the 555 was powered from the same supply voltage as the LED/Mosfet, then the transistor and zener diode wouldn't be needed, just the 555 driving a P-channel Mosfet instead. Just about like your B circuit except using a 10A LED, no current limiting resistor and an extremely short-duration pulse.
You don't post where you are, so i don't know if you buy European BDxxx power transistors or American TIPxxx ones.
Your photodiode is upside down, it needs to be reverse-biased and the IR radiation causes it to have a leakage current. A diode doesn't have the gain of a transistor so the amplifier would need to have much more gain. Try reducing the value of the amplifier's 1K resistor or shorting it for plenty of gain.