update....
i built this i believe correctly just minus the transistor mentioned below.... and both of the IR LEDs burned up in like 5 seconds.
what the heck?
why would this happen!?
the only other thing i didn't mention is that there was a wire running to the 555 and it wasn't labeled, so i ran it to pin 7, as it was the only one missing. this was ok right?
BC557 is used here as power switch. You can directly connect an LED to the output of 555 through a resistor without any performance loss because 555 can source as much as 200mA of current which is sufficient to drive 2 IR LEDs.
There is no other replacable device for a transistor except old vacuum tubes :lol: . Why you don't want to buy it? It hardly cost anything. Over here in my country BC557 is available for 0.02 US $
well, i dont see the much use of the transistor since the 555 can suply 200mA.....
but you can use any type of transistor for it, it is quite general purpose, 100mA, 45V, 500mW, hFE>110, .
i suggest some replacements, bc182, 183,184, 337,338,414, and you can find something else.
well, i dont see the much use of the transistor since the 555 can suply 200mA.....
just acts like an inverter, ha.
but you can use any type of transistor for it, it is quite general purpose, 100mA, 45V, 500mW, hFE>110, .
i suggest some replacements, bc182, 183,184, 337,338,414, and you can find something else.
by t
anyone can explain me how this circuit work? I know that for a PNP conduct, the base current must negative (voltage Vbe too), I mean the base current must goes in the Output of the 555 timer... this is what I don't understand....
the pnp base voltage is fix to 7V (approx.) (drop voltage from diode) right?
so if my logic is good the Infrared Leds will be ON when the Output of the 555 (pin 3) will be low?!?! :?: :?:
Yes you are right. For a PNP to go into conduction, the base current should flow out of the transistor and should be grounded some how. So when 555's output is low that means its output is connected to ground internally which in-turn pulls the base of PNP to ground thereby causing a current flow which flows out of its base and turns it on.
You took out the BC557 but what did you do with the 180R resistor and the 1N4148 diodes?
If you just connected everything together in place of the BC557 then all you will do is put 7.6 volts through the leds without the benefit of a current limiting resistor.
you are correct... i just connected everything that was going through the transistor together. so i threw 7v at 2 LEDs.... so why is that bad? shouldn't they have like a 12v max? was that the purpose of the transistor, simply a voltage limiter?