I am given a circuit with a transistor and I have to calculate base current. To do that I am using the formula for the linear load:

$$I_c = - \frac{1}{R_c} \cdot U_{CE}+ \frac{U_{CC}}{R_C}$$

So my task:

enter image description here

And the given values:

current boost factor: \$ B = 400\$

\$U_{CC} = 11V\$


\$R_C=3,3k \omega = 3,3 \cdot 10 ^{3} \Omega\$

\$ R_E = 5 k \Omega = 5 \cdot 10^3\$

\$ R_1 = 55 \Omega\$

\$ R_2 = 7 \Omega\$

So I have tried to use the values to get \$U_{CE}\$.

$$I_c = - \frac{1}{R_C} \cdot U_{CE} + \frac{U_{CC}}{R_C}$$ $$0 = - \frac{1}{3,3 \cdot 10^3} \cdot U_{CE} + \frac{11V}{3,3 \cdot 10^3}$$ $$ - \frac{11V}{3,3 \cdot 10^3} = - \frac{1}{3,3 \cdot 10^3} \cdot U_{CE} $$

$$11 V = U_{CE}$$

Question: Does that make any sense? That would mean that \$U_{CE}\$ always equals \$ U_{CC}\$? What is the right way to solve that task ?

  • \$\begingroup\$ It's possible if URC and URE equals 0. but they are no, so you've got something wrong. \$\endgroup\$
    – Pierre
    Mar 6, 2017 at 13:30
  • \$\begingroup\$ The collector current is not zero. \$\endgroup\$
    – Barry
    Mar 6, 2017 at 13:30
  • \$\begingroup\$ You're assuming Ic=0, that's leading your transistor to be in cutoff region, so your Uce will be equal to Ucc since there's no current flowing through Rc and Re. For this exercise you can't assume you're in cutoff region since R1 and R2 are feeding a current in the base of your transistor. \$\endgroup\$ Mar 6, 2017 at 13:33
  • \$\begingroup\$ Already the first equation is wrong. You are adding currents! Instead, you must add three voltages: Vcc=IcRc+Vce+IeRe. However, this equation does not help at all. Start with input voltage loop (as suggested by dannyf). \$\endgroup\$
    – LvW
    Mar 6, 2017 at 15:12
  • \$\begingroup\$ hint: Vb (the voltage at the base) = Ucc * ( R2/(R1+R2)) - now go forth and calculate. \$\endgroup\$ Mar 6, 2017 at 15:42

2 Answers 2

  1. calculate Vb.
  2. from Vb, calculate Ve.
  3. from Ve, calculate Ie.
  4. calculate Vc based Ic ~= Ie.
  5. relax.
  • \$\begingroup\$ Sounds easy. However today I am not even able to do that.. I would apreciate a sample with formulas. \$\endgroup\$
    – jublikon
    Mar 6, 2017 at 13:33
  • \$\begingroup\$ The simplified method dannyf has proposed is sufficiently correct because the base divider has unrealistic low resistor values (55/7 ohms). Really ohms? In this case, the expected base current can be totally neglected - in comparison to the divider current andthe necessary assumption (uncertainty) for Vbe (0.65...0.7 V). Hence, find Ic and recalculate Ib. \$\endgroup\$
    – LvW
    Mar 6, 2017 at 13:48

Maybe this will help. The voltage at the base is Vcc * R2/(R1+R2). This is the Thevinin voltage. Then put R1 and R2 in parallel to get the Thevinin resistance. Redraw the circuit so that Rth is in series with Vth and the base of the transistor. Then write a KVL loop around the base circuit in terms of Ib setting that equation to 0. Then solve for Ib. I hope this suggestion helps.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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