How does early voltage affect collector current?
Early effect (base-width modulation) means that \$I_C\$ current will change his value as \$V_{CE}\$ change, even if \$V_{BE}\$ and \$(I_B)\$ is kept constant. 

So we have another source of a nonlinearity.

For your example circuit we have:

\$\beta = 50\$,\$V_{CC} = 10V\$,\$R_C=1k\Omega\$,\$R_B=196.5k\Omega\$; and the Early Voltage is \$V_a=50\$

Without Early effect the DC operation point is: 

$$I_B=\frac{V_{CC} - V_{BE}}{R_B} = \frac{10V - 0.7V}{196.5k\Omega} = 47.328\mu A $$ 

And \$V_B = V_{BE}\$

Hence the collector current (without Early effect) is equal to: 

\$I_{CO} =\beta*I_B = 47.328\mu A * 50 = 2.366mA\$  

and the \$V_{CEO}=V_C\$ voltage (without Early effect).  

\$V_{CEO} = V_{CC} - I_{CO}*R_C = 10V - 2.366mA*1k\Omega = 7.6335V\$.

But if we include Early effect \$I_C\$ current will change. 

We have

$$I_C = I_{CO}*(1 +\frac{V_{CE}}{V_a}) $$ 

$$V_{CE} = V_{CC} - I_{C}*R_C$$

And if we solve this for \$I_C\$ current we will end up with this: 

$$\large I_C =\frac{I_{CO}(V_a+V_{CC})}{I_{CO}R_C + V_a} = I_{CO}\frac{1+\frac{V_{CC}}{V_a}}{1+\frac{R_C}{R_O}}$$

$$I_C =  2.366mA\frac{1+\frac{10V}{50V}}{1+\frac{1k\Omega}{21.129k\Omega}} = 2.36641mA * 1.14577 = 2.71137mA $$

where \$R_O = \frac{V_a}{I_{CO}} = \frac{50V}{2.36641mA} = 21.129k\Omega \$ 

All this mean that Early effect can be model as a resistor \$R_O\$ connected from the collector to the emitter of an “perfect” transistor.







simulate this circuit &ndash; Schematic created using CircuitLab

Also, I deliberately skip the fact that the \$V_{BE}\$ value was given for \$I_C=1mA\$. And here we have \$I_C &gt; 1mA\$ so the\$V_{BE}\$ value will also be slightly larger than \$0.7V\$.