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?

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 $0.7V$.

• So which formula applies for collector current in linear region: this one $$I_C = I_{CO}*(1 +\frac{V_{CE}}{V_a})$$ or this one $$\large I_C =\frac{I_{CO}(V_a+V_{CC})}{I _{CO}R_C + V_a} = I_{CO}\frac{1+\frac{V_{CC}}{V_a}}{1+\frac{R_C}{R_O}}$$ ? Commented Apr 23, 2017 at 18:21
• Shouldn't Ico be replaced with Ic? Like this: $R_O = \frac{V_a}{I_{C}}$ Commented Apr 23, 2017 at 18:47
• @Keno Ico is a collector current without Early effect (perfect transistor). So if you want to find the Ic current and includes the Early effect. We can use this equation Ic = Ico*(1+Vce/Va) (1) but you do not know Vce, hence we need additional equation for Vce. Vce = Vcc-IcRc (2), and we substitute (2) into (1) and solve for Ic we get this: Ic = (Ico (Va + Vcc))/(Ico* Rc + Va) = Ico * (1 + Vcc/Va)/(1 + Rc/Ro) .
– G36
Commented Apr 23, 2017 at 19:29
• @Keno For the bipolar transistor gm (transconductance) is the slope of the collector current to the base-emitter (slope of the Ic=f(Vbe) function). gm=d(Ic)/d(Vbe) = Ic/Vt = Ic/26mV = 40*Ic, But also gm = Ic/Vt = beta/rbe. We also have anther point view (from T-model). We include transresistance re = d(Vbe)/d(Ie) is a dynamic (differential) emitter resistance (often called "little re") re = Vt/Ie and we sometimes use this approximation $r_e = \frac{Vt}{Ie}\approx \frac{Vt}{Ic} \approx \frac{1}{gm}$ but the exact value is gm = beta/((beta +1)*re)
– G36
Commented Apr 24, 2017 at 17:58
• electronics.stackexchange.com/questions/267662/…
– G36
Commented Apr 24, 2017 at 18:02

Have you seen a IV plot for a bipolar

What is the Early voltage for this transistor? Consider that very top line: intercepting the left axis(0 volts) at 4mA, and intercepts the 10_volt line at 7.5mA. [ some very clean numbers, used by the artist ] Our deltas are 7.5mA - 4mA = 2.5mA; 10v - 0v = 10v.

Now extend that to the left, from 0v/4ma point. The intercept with Zero current is 10v * 4mA/2.5mA = 10v * 1.6 = 16 volts to the left. Ve = 16 volts

simulate this circuit – Schematic created using CircuitLab