\$V_{BE}\$ increasing from saturation to active mode of BJT with \$I_{B}\$ constant
The collector-base junction is in parallel with the base-emitter junction and is forward-biased when the collector is at less than Vbe. 

As it turns out the voltage drop of the collector-base junction is typically somewhat less than that of the emitter-base junction for the same current as well, so having the collector voltage at 0V means that some of the base current flows out of the collector to ground, so Vbe is less than with the collector open. This is a consequence of the asymmetry in the construction of most modern transistors.  

Above maybe 100mV there should be little change in Vbe regardless of Vce (before breakdown). 

If you run the below simulation, you will find that more than 95% of the base current flows out through the collector, reducing the Vbe by maybe 5-10mV.  







simulate this circuit &ndash; Schematic created using CircuitLab