Im attempting to build the following circuit to learn more about transistors and switching.
http://www.nutsvolts.com/uploads/wygwam/NV_0806_Marston_Figure03.jpg
Can anyone help me identify the Q1 transistor? Im not familiar with that symbol.
The other transistors are 2n3904's.

 

hi Ron,
Check this out.
https://electronics.stackexchange.c...s-collectors-imply-in-a-bjt-schematics-symbol
E

 

Looks like a muliti-emitter transistor. I can't seem to find any discrete options out there. The two emitter act like an AND gate so you may be able to replicate that with several discrete components instead.

 

Thanks guys, I pretty much figured it was an AND configuration from the circuit but I don't know how to locate a single part which is equilivant to the function of the 2n3904.
Also, I don't see a direct way to duplicate this transistor using 2 or more 2n3904.
That's probably where my knowledge is lacking, basically now to translate that type of part into something I can get my hands on.
Thanks again.

 

I've certainly never seen a multi-emitter transistor as a discrete device, but they are core feature of TTL.

You can approximate the function by connecting three signal diodes (1N4148 or similar) together:
- all anodes together to the base of the multi-emitter
- cathode of one becomes the collector of the m-e transistor
- cathodes of the remaining two become emitters (inputs) of the m-e.

When both emitters are open-circuit or connected to the positive supply, no current flows through them, but instead through the "base - collector" diode, turning on Q2. When either of the emitters is connected to circuit common ("ground") the base voltage will be pulled sufficiently low that the collector current will drop to a very low level, turning Q2 off. In the experimental circuit with diodes, adding a resistor of perhaps 22k from the base of Q2 to ground may help assure Q2 does turn off fully.

The two diodes from the inputs to ground aren't required - they are part of the TTL IC structure, whether they are wanted or not.

 

That's a schematic of a 2-input TTL AND gate IC.
You won't find a discrete equivalent for that two emitter transistor.

As the multi-emitter transistor is only acting as a common-base switch with no gain, you can functionally simulate it by eliminating the transistor and using two small-signal diodes in its place, as shown below:

 

@crutschow , seems like a good idea. However, I was looking at the 1N4148 part and it seems a bit over powered for the needs of this circuit. Ill look for a smaller/lower-power switching diode for my testing. This is just for breadboarding anyhow, just want to trace signals and such. Kinda see how it reacts internally to different inputs and look at switching time...

 

However, I was looking at the 1N4148 part and it seems a bit over powered for the needs of this circuit.

It is.
But, unless you are wanting to operate the circuit at high speeds, it will work fine for that circuit.
Even at high speeds, your breadboard circuit operation will likely be more affected by parasitic capacitances then the characteristics of the diode.

The 1N4148 is a very common, small-signal silicon junction diode (the lowest current one I'm familiar with), but if you want to use a smaller one, that's okay too.