# get base current in transistor circuit

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:

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

And the given values:

current boost factor: $B = 400$

$U_{CC} = 11V$

$U_{BE}=0,7V$

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

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

$R_1 = 55 \Omega$

$R_2 = 7 \Omega$

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 ?

• It's possible if URC and URE equals 0. but they are no, so you've got something wrong. Mar 6, 2017 at 13:30
• The collector current is not zero. Mar 6, 2017 at 13:30
• 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. Mar 6, 2017 at 13:33
• 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).
– LvW
Mar 6, 2017 at 15:12
• hint: Vb (the voltage at the base) = Ucc * ( R2/(R1+R2)) - now go forth and calculate. Mar 6, 2017 at 15:42