I am currently attempting to build a circuit for a TV Remote Jammer. I have all the parts, but i am having a little trouble reading the schematic. I am not familiar with the battery layout. The schematic in question: http://http://www.instructables.com/id/SG61H3YF82EZB3L/ The reason I am confused is because the circuits I am used to building have the ground and the power going to the battery. In the above schematic, the power, a 9v battery, only has a positive lead coming out of it, and the ground is at the bottom of the schematic. If anyone could help me out, I would appreciate it.

 

Your link doesn't work.
Why don't you save the schematic then attach it to your reply here??

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Uncle $crooge

 

The link is http://www.instructables.com/id/SG61H3YF82EZB3L/

The ground sign means common or 0v you connect this to the - of the battery

 

The 180 ohm resistor limits the LED current to less than 4mA which is much lower than a signal from a remote control. I would change it to 27 ohms for 26mA.

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Uncle $crooge

 

thanks thats what was confusing me the most.

 

Would the variable resistor have anything to do with lowering the resistance of that 180 ohm resistor?

 

i just re-looked at the schematics and realized the VR wouldn't have much of anything to do with the 180 ohm resistor.

 

would a 30ohm resistor be suitable. It is the closest thing to 27ohm i have.

 

The value of the 180 ohm resistor determines the LED current and its brightness.
The value of the pot determines the frequency.

With 27 ohms, the current in the LEDs is 0.7V/27= 26mA.
With 30 ohms the current is 0.7V/30= 23.3mA. Not much difference but much more than the original circuit.

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Uncle $crooge

 

Hi
yes Uncle $crooge write. out put frequency(f out) will effect by pot value
fout=1/ {ln(2).(470+Rpot+2000)10e-9} Hz
so by turning the pot u can figure out exact signal jam point

 

Like tazman2087, I'm new to electronics. (The ground = the negative terminal of the battery was a suspicion with me, too, not a fact.)

So, if the schematic starts with 9V, how did audioguru end up with 0.7V so quickly?

 

The two diodes, the transistor and the 180 ohm resistor make a constant current source.
The two diodes in series have a voltage drop of 2 x 0.7V= 1.4V.
The base-emitter of the transistor has a voltage drop of 0.7V. Then the 180 ohm resistor has 1.4v - 0.7V= 0.7V across it. Ohm's Law calculates the current at 0.7V/180 ohms= 3.9ma.

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Uncle $crooge

 

Uncle $crooge,

I'm beginning to understand. You are concentrating 4 of the components, ignoring the other 9 in the schematic.

The datasheet for 1N4148 has a max Forward Voltage, V(FM), or 0.715V. The datasheet for BC557 has a max Base-Emitter On Voltage of -750 mV.
Is that where your 0.7V came from?

On the NTE30001 datasheet, the High Output is at I(F) = 20 mA. Is that how you chose the 27 ohms?

If so, how did you (how would I) know to ignore the rest of the schematic, to see if the resistor should be 180 ohm or 30 ohm? At this stage in my understanding, I would have thought I needed to account for all 9 volts.

Thanks a lot.

 

The current in the two 1N4148 diodes is pretty high at about 13mA so their total voltage is typically 1.5V.
the base-emitter voltage of a BC557 is typically o.65V when its collector current is 4.7mA.
Then the voltage across the 180 ohm resistor is 1.5V - 0.65V= 0.85V and the constant current is 0.85V/180 ohms= 4.7mA.

The 4.7mA is not affected much by the supply voltage since it is regulated by the voltage of the two diodes minus the base-emitter diode of the transistor.

Now that the datasheets show the voltage across the current-setting resistor is 0.85V and not 0.7V then a 27 ohm resistor would produce 0.85V/27 ohms= 31.5ma.
Your IR diode has a max allowed continuous current of 50mA so 31.5mA is fine.

Only the 555 IC has a supply voltage of 9V in this circuit.

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Uncle $crooge

 

Uncle $crooge,

Okay, this is a public forum. And I am trying to come to grips with my personal ignorance. Public vs. private. So I'm prepared to be told this is not the place (especially by tazman).

But, if it's okay to continue, I'd like to concentrate on that diode, and figure out why you're quoting 13mA and 0.7V.

Attached please find a small circuit. Please assume the diode is a 1N4148, and that V(dc) is 9V. If I (or you, more carefully) measured amps and volts across the diode, what would I get? And where do I look on the datasheet for confirmation of that?

Thanks.

Attached Images

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