I'm going to also add some more explanation. The "load" in the Osgeld schematic is a resistor followed by an LED. To determine the size of the resistor, first determine the max current draw of the LED. Using 20 ma as an example and +5 V as +V, Ohm's law states:
R=E/I
where
E=5 - 0.6 = 4.4 V (the 0.6 is the drop across the diode)
I = 0.020 A
R = 4.4/.020 = 220 ohms.
As suggested you need a resistor in the base circuit. Assuming a beta of 20 (which is ridiculously low) we'll want the transistor Ice current to be 0.020 and the Ibe will be 0.020 / 20 or 0.001 or 1 ma.
We'll assume the voltage drop across the BE junction to be about 0.6 volts. So we want to use Ohm's law again thusly:
R = E / I
R = 4.4 / 0.001
R = 4.4K ohm
Since it won't hurt a thing to pass a bit more current than that, I'd go with a 3.3K ohm resistor since that is a standard size. Practically I suspect anything in the 1K to 10K range would work.
Note beta is a measure of the gain of a transistor and the beta of a 2n3904 is more like 100 so these calculations are very conservative.
YMMV
Jim.