Base voltage peaks at about 0.8 volts, why?
If you increase the voltage across the base-emitter junction, the base current increases exponentially, according to the Shockley diode equation

$$I_b = I_s\left(\exp(\frac{qV_{be}}{nRT})-1\right)$$

where \$I_s\$ and the other symbols you might not have seen before are parameters of the device.

But in your circuit, the more you increase the base current, the more voltage is dropped across the base resistor. Given a 5 V supply, even if \$V_{be}\$ stayed at 0 V, the most you could have delivered through that resistor is 5 mA. With 5 mA into the base, I'd expect more like 0.6 to 0.7 V \$V_{be}\$ (but I'm not specifically familiar with 2N3904). So it's likely you were actually using something more than 5 V to supply the base, and getting somewhere in the 20 - 100 mA range, where 0.7 or 0.8 V is a reasonable guess at the base-emitter voltage.