transistors - How does early voltage affect collector current?

The transistor in the following circuit is said to feature $I_{C} = 1.0mA$ at $V_{BE} = 0.7V$ and $V_{CE} = 10V$. If $V_{A} = 50V$ and $\beta = 50$, find $V_{B}$ and $V_{C}$.

^{simulate this circuit – Schematic created using CircuitLab}

Attempt:

$V_{B}$ Calculation:

$V_{B} = 10 - 0.7 - 196.5I_{B}$

$I_{B} = \frac {I_{C}} {\beta} $

$V_{B} = 10 - 0.7 - 196.5 *\frac {1mA} {50}\ = 5.3V$

$V_{C}$ Calculation:

$V_{C} = 10 - 1 = 9 V$

How do I incorporate the early voltage $V_{A}$ into the calculations?

Answer

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 – 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 $I_C > 1mA$ so the$V_{BE}$ value will also be slightly larger than $0.7V$.