0
\$\begingroup\$

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}\$.

schematic

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?

\$\endgroup\$
1
\$\begingroup\$

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.

schematic

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\$.

\$\endgroup\$
  • \$\begingroup\$ 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 $$\large I_C =\frac{I_{CO}(V_a+V_{CC})}{I _{CO}R_C + V_a} = I_{CO}\frac{1+\frac{V_{CC}}{V_a}}{1+\frac{R_C}{R_O}}$$ ? \$\endgroup\$ – Keno Apr 23 '17 at 18:21
  • \$\begingroup\$ Shouldn't Ico be replaced with Ic? Like this: \$R_O = \frac{V_a}{I_{C}} \$ \$\endgroup\$ – Keno Apr 23 '17 at 18:47
  • 1
    \$\begingroup\$ @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) . \$\endgroup\$ – G36 Apr 23 '17 at 19:29
  • 1
    \$\begingroup\$ @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) \$\endgroup\$ – G36 Apr 24 '17 at 17:58
  • 1
    \$\begingroup\$ electronics.stackexchange.com/questions/267662/… \$\endgroup\$ – G36 Apr 24 '17 at 18:02
2
\$\begingroup\$

Have you seen a IV plot for a bipolar

enter image description here

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

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.