Differential Amplifier BJT Experimenting
To get a better understanding of how differential BJT amplifier work. 

You must treat the BJT as a Voltage controlled (Vbe) current source (Ic). 

So to be able to test this it try to build this simple circuit: 







simulate this circuit &ndash; Schematic created using CircuitLab

And as a name suggest the differential amplifier "react" on voltage difference.

$$V_{ID} = V_A - V_B$$ 

And if this \$V_{ID}\$ is larger then \$150\textrm{mV}\$ all the \$I_{EE}\$ current will flow through \$Q_1\$. 

And if \$V_A - V_B = 0\textrm{V}\$ (\$R_X = 0\Omega\$) the \$I_{EE}\$  current will split equally between two BJT's. 

So to change \$ V_{ID}\$ voltage all you need to do is to change the\$R_X\$ resistor value between \$0\Omega\$ to \$220\Omega\$

Also notice that \$V_{ID} = V_{BE1} + V_{EB2}\$ (from KVL loop). 
And this means that as \$V_{BE1}\$ increase the \$V_{BE2}\$ must decrease by the same amount.



For example, if \$V_{BE1}\$ changes from 0.60V to 0.61V the \$V_{BE2}\$ will drop by 0.01V from 0.6V to 0.59V. 

And sometimes we can treat the differential amplifier as a cascade connection of the two "basic" BJT amplifiers. The first one is CC (Common Collector) and the second stage is CB (Common base) stage.

You can modify the original circuit by adding two separate voltage divider. 

One for Q1 and the second one for Q2 So you can "set" \$V_A\$ and \$V_B\$ separately.