Thanks to all the help, my transistor have not been getting hot. I must have hooked something up wrong a few times.
There are thousands of different types of transistors. The chances of a random transistor grabbed out of an old kit being the one you need for your circuit are low. Unless you can conclusively identify the type of transistor by part number you aren’t going to have much luck.
From the circuit, any small current NPN should work.
I do wonder what the point is. You can drive an LED from nine volts with just a resistor.
As stated by fubar.gr and others, this should work fine, and the transistor shouldn’t get hot.
Using your values, the transistor is definitely operating in the saturation region, which is what you want when using the transistor as a switch. Using these values, and assuming my back-of-the-napkin calculations are correct,
I[sub]B[/sub] =800 μA
I[sub]C[/sub] = 6.8 mA (assuming V[sub]D[/sub] = 2 V)
V[sub]BE[/sub] = 0.7 VDC
V[sub]CE[/sub] = 0.2 VDC
V[sub]CB[/sub] = -0.5 VDC
Power dissipated by 1000 Ω base resistor = 0.64 mW (cool to the touch)
Power dissipated by 1000 Ω collector resistor = 46.2 mW (cool to the touch)
Power dissipated by transistor = 1.4 mW (cool to the touch)
Note that the current through the LED is 6.8 mA. I am guessing you can safely crank this up to 20 mA (but check the diode specs first). To do this, change the collector resistor to 340 Ω.
Educational I am guessing. Learning how to use the transistor as a switch (transitioning from cut-off to the saturation region).
In addition, this circuit provides some current regulation to the LED with reference to the 1.5 V on the base (i.e. if the 1.5 V on the base wanders around a lot, the LED will continue to have a constant current trough it).
In undergraduate school, Dr. Pantangia taught us to use 20 mA to run an LED.
Eh, it’s similar in that it will pull the voltage feeding the base down to ground, which will cause no current to flow into the base. The resistor in the base is to limit current. If you wanted to make a true pull down on it, you would connect another resistor from that resistor to ground, and at the same time connect a switch from where the two resistors meet to your supply voltage. Then when the switch is open, the pull down will keep the transistor off, and when you close the switch, the base resistor will prevent too much current from damaging the transistor.
Most transistors don’t leak enough to light an LED. Using an LED should be fine.
Yes, I meant with respect to your common (aka ground).
“Ground” is kind of a funny term in electronics. It usually means your zero volt reference, which is a common connection in your circuit that is often connected to a physical earth ground connection for reasons like noise immunity, safety, etc. This has led to the zero volt common reference being commonly called “ground” even if no earth connection is present.
Voltage is always measured between two points (don’t interrupt me with that mumbo-jumbo about point charges, you physics nerds… 
I’m trying to make a point here). If you refer to a voltage without specifying it’s reference, then the reference is assumed to be your zero volt common.
I’m still wondering about those transistors that lit the LED without a base connection. Why did they read infinite with my ohmmeter?
Could you repost that circuit diagram again please.
It’s possibly that those transistors were FET type, which have gate pins that will read infinite resistance to the drain and source pins, and which can be turned on and off by static charges if the gate isn’t connected to anything. You can’t assume that all of your mystery transistors are BJT. You can’t even assume they’re all transistors for that matter - there are other components that have three pins.
Could be a number of reasons, listed here in terms of Occam’s Razor ranking:
[ol]
[li]Multimeter leads are connected to the wrong transistor legs. The location of base, collector and emitter pins vary between different transistor models.[/li][li]Multimeter is on the resistance range, not the PN junction range (as indicated with the little diode symbol). Often the resistance range won’t bias a PN junction sufficiently to give a reliable reading, hence the need for a dedicated setting for measuring PN junction voltages.[/li][li]The unidentified transistor is a MOSFET, or even a triac or SCR.[/li][/ol]
The datasheet for the 2N3904 may be found at https://www.fairchildsemi.com/datasheets/2N/2N3904.pdf
blood63, there’s nothing wrong with your original circuit (aside from a missing base-emitter resistor, and frankly that’s a woefully common omission). The circuit that fubar.gr posted is called an emitter follower, and while it’s a very useful circuit to know you won’t get any useful results for the application in the OP. Basically, the emitter voltage will be the base voltage minus Vbe (about 0.6 V), so if you’re biasing the base with a 1.5 V cell then you’ll be driving your LED and resistor with about 0.9 V, which might get it glowing very dimly. Don’t forget that you ultimately want to use this circuit with an Arduino board, and the output ports on these are 5 V. Your original circuit will work well with the base resistor being driven from either 1.5 V or 5 V. If you’re wondering how to choose the series base resistor value, then a handy rule of thumb for a transistor you want to drive into saturation is to make the base current about one-tenth of the collector current, e.g. if your LED current is 7 mA then the base current needs to be about 0.7 mA. If you’ve got a 10 kohm resistor from base to emitter then this will shunt away about 60 uA (=0.6V/10kohm) so the current through the series base resistor needs to be 0.7 mA + 60 uA = 0.76 mA. Being driven off a 5 V Arduino port gives a series base resistor value of (5V-0.6V)/0.76mA=5.8kohm, so the 1k resistor in your schematic will turn on that transistor good and hard. It won’t do any harm in this instance, but ensure that you don’t draw more than 20 mA from an Arduino port or it will be damaged.
Another good reason to include a base-emitter resistor is that there will be a small voltage on the Arduino output port even when the output is set low, and this may be enough to partly turn on the transistor if this small current is not shunted away. Also, when a microprocessor boots up, its output ports may briefly be in a floating high-impedance state, so any circuits connected to these ports should have provision for dealing with this (usually just a resistor tied to 0V or the micro’s positive supply, depending on whether you want the line to be forced low or forced high).
The 2N3905 shouldn’t leak more than 50 nA, but MOSFETs can be problematic when driving LEDs as they’ll often leak enough to make the LED glow. The standard workaround for this is to add a shunt resistor across the LED.