I simulated the schematic diagram, how does it look?

after the schematic image they are all the same simulation of the schematic but from different angles

********************REVISED***********************









*******************REVISED**********************

after the schematic image they are all the same simulation of the schematic but from different angles

 

I can't see the transistor's part number. If it is an NPN like in your schematic then the zener diode is connected backwards.

The transistor also might be connected backwards.
The power supply also might be connected backwards.

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Uncle $crooge

 

2n3904
w67
I am surprised it doesnt say A B C or D anywhere, but my friend told me that it wouldnt matter which transistor I use because the amperage consumed by the transmitter is so little. However, it does matter if is npn or pnp because....?

 

You still don't know how much current the transmitter uses. It might be more than the 200mA max of the little 2N3904 which would melt with such a high power dissipation.

The 1k resistor feeding it and the zener diode has a value that is way too high because the minimum current gain at a collector current of 100mA is only 30 and is much less at higher currents. So at 200mA the base current might be 13mA and the base resistor would need to be about 200 ohms.

Your zener diode is definitely backwards. Its end with the black bar should connect to the base of the transistor.

__________________
Uncle $crooge

 

Why doesn't the schematic match with the connection on the breadboard?
As shown the yellow wire is the collector, so the red wire (assumed as Vcc) should be connected to the collector, and the other end of the resistor as well. The emitter should be grounded.
The pins of 2N3904 are EBC, the flat surface is facing you and the pins are pointing down.

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Superman returns..

 

Okay... Replying to the original post,

I believe that the emitter of a (NPN) transistor puts up the same voltage as that what is supplied to the base (with a small drop). Consider it as a PN Junction diode.

The current that is available at the emitter is that that is coming from the collector as the base current is very small - few microAmpers.

The zener and the resistor forms a potential divider to give a constant stable voltage to the transistor base irrespective of the change in input voltage..

__________________
Bharath Bhushan Lohray.
M.Sc. Electronics.

 

Either this is some strange kind of breadboard, or it's not hooked up right.

I see:
Red, Yellow, E, and C are connected together.
Base isn't connected to anything.
1.2K resistor isn't connected to anything.
Diode cathode is connected to Black.
Diode anode isn't connected to anything.

Summary: output equals input. Circuit has no effect.

Needs to be rewired after an understanding of the breadboard's interconnect is gained.

 

YUP YOU ARE RIGHT i was doing this too late at night, thanks for pointing it out. Ill revised it

 

yes i conected most of it wrong, sorry should have revised it before. Where can i find the specs for my 2n3904 transistor? I'll google it see what i can find

 

I get datasheets from www.datasheetarchive.com . You can even select the manufacturer.

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Uncle $crooge