I'm trying to solve a transistor problem, but I'm stuck on the I2. Is I2 equal to I0 ? The question is: The transistor circuit in Fig. 3.45 has β = 80 and VBE = 0.7 Find V0 and I0 . Answers : 12V, 600 mA.
My equations:
For \$I_1\$,
\$-1+120 \mathrm{k} \cdot I_1 + V_{BE} = 0\$.
Therefore, \$I_1=2.5μA\$.
For \$I_3\$,
\$10\mathrm{k} \cdot I_3+20+10\mathrm{k} \cdot I_3 – 10\mathrm{k} \cdot I_2=0\$.
\$2\left(I_3\right )-I_2=-2\mathrm{mA}\$
If \$I_C=I_B \cdot \beta \$, \$I_B\$ is equal to \$I_1\$, isn't it ?
However, I get \$I_2=\$ 0.8μA
Best Answer
The answer given is clearly wrong.
20 volts across a 10K resistor will give 2 mA, so \$I_0\$ can't possibly be 600 mA.
If we omit the transistor, the two 10K resistors form a voltage divider, dividing the 20 volt supply in half, making \$V_0\$ 10 volts.
The transistor will draw some current in parallel with the lower 10K, so \$V_0\$ must be less than 10 volts.