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Remote Jammer

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daviddoria

New Member
If i just hook an IR LED up to an astable 555, will that mess with my tv? is there a certain rate of blink or how does all that work?

also, how can i tell the voltage out of a 555 before i explode more LEDs? if i want to find R using R=I/V, then i have to know I and V.... so how do i know this?

can i turn on the circuit (9v) and use a multimeter without there being an LED there?

still not quite sure why in the diagram linked below
http://www.bandtank.com/~calculusap/jammer.gif
there are diodes and that transistor.

why cant it just be like a normal 555 flasher?

david
 

bogdanfirst

New Member
first R=V/I and the voltege out of the 555 is almost the voltage in so you can condider that. also the voltage of an IR led is most of the time below 2V, usually like 1.4-1.8V so be carefull.
as for the frequency you will have to use the same frequency that the remote uses i think it is around 40KHz but i am not sure. somebody might know it. also it depends on what model you have.
also, why it uses the trensistor. at any 555 you have the on time>off time , but by putting that transistor you get that the on time of the leds is smaller than the off one.. otherwise i dont see why you need it.
also i think that you should add a capacitor in paralel with the batt something like 50-100uF and increase the value of the 180R resistor, maybe up to 470R or less.
 

mechie

New Member
That pesky Ohm guy again !

Carry on! There is no need for an LED - the 555 will work without it.

For your calculations...
The standard 555 timer is capable of sourcing or sinking 200mA from its output. A typical IR LED seems to be rated at upto 100mA and will drop about 1.3v when doing this. The 555 can happily drive the LEDs then, you just need that resistor calculating. :?

I assume you have a single LED and a 9v power supply.
If you connect directly to the 555 output you need a series resistor ...

R = V/I , the resistor will drop 7.7volts, the LED dropping the other 1.3v, I guess you want 100mA (the max the LED can stand) for maximum 'brightness'.

The resistor will therefore be...
7.7v/0.1A = 77 Ohms

For two LEDs in series (dropping 1.3v each, 2.6v total)
The resistor will be 6.4v/0.1A = 64 Ohms

The circuit you linked to shows an output transistor (PNP) with two diodes from the base to the positive line. This configuration is a constant current supply --- but I calculate it to be providing only 3.5mA :!:
Don't copy this if you want your circuit to work ... I reckon the circuit as shown should have the 180 Ohm emitter resistor changed for a 7 Ohm one to supply 100mA.
 

daviddoria

New Member
haha i apologize i got the formula backwards.

anyone know what the diodes were for?

also, i know this:
High Time = 0.693*(R1+R2)*C1/1000
Low Time = 0.693*R2*C1/1000

but if i want F=40KHz, how do i find R1, R2, AND C1?
 

Gene

New Member
What do you mean by "mess with my TV"?

What is this schematic designed to do?

I agree with mechie, the twin diodes appear to keep the transistor turned 'on' constantly and, therefore, keep the LED 'on' constantly.
 

daviddoria

New Member
haha i dont know what i mean by "mess with" the tv. i just figured electronic equiptment that takes in an IR signal can be confused by transmitting a garbage IR signal (this device we are making).

2 remaining questions.

1) why use 2 diodes instead of just a wire going from + to the transistor?
2) if you put +9v to the 555, doesn't any of it get taken up by the timing resistors? is it actually 9v coming out of pin 3?
2b) what is the amperage coming out of a 9v? is it completely dependant on the resistance?? that seems like you could then get infinite amperage by making 0 resistance.... thats too wierd.

david
 

bogdanfirst

New Member
you are right about the infinite current.
but this will only be possible(i wish it was) it the bettery wouldnt have an internal resistance, but it has, so there is a limited current that can pass through a battery. also if the current is too high it will destroy the paths and wires.
ablou replacing the diodes with a piece of wire it will not have the same effect, because you have to remember that a silicon diode has a voltage drop of about 0.6-0.7V across it.
about what amperage you can get out of a 9V battery i think that you cant get more than 1A at short circuit, maybe if you use a rechargable one...
but the 555 suports a maximum output of 200mA wich is more than the leds need....so back to the same problem ....why is the transistor needed?
 

mechie

New Member
Re: That pesky Ohm guy again !

mechie said:
The circuit you linked to shows an output transistor (PNP) with two diodes from the base to the positive line. This configuration is a constant current supply --- but I calculate it to be providing only 3.5mA :!:
 

daviddoria

New Member
so i think i figured everything out... the only thing left is how fast to make the light blink to be in any kind of range of normal electronic equiptment IR speeds.... any thoughts?

david
 
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