a. Using the characteristics of Fig. 3.13 determine ICEO at VCE = 10 V.b. Determine βdc at...

From Figure 1, the Collector emitter voltage is,

$$ V_{C E}=10 \mathrm{~V} $$

The collector to emitter current when the base leg is open \(I_{C x o}\) is \(0.3 \mathrm{~mA}\)

Therefore, the collector to emitter current when the base leg is open \(I_{c: o}\) is \(0.3 \mathrm{~mA}\).

(b)

From Figure 1,

The Base current is,

$$ I_{n}=10 \mu \mathrm{A} $$

The Collector emitter voltage is,

$$ V_{C E}=10 \mathrm{~V} $$

The corresponding value of Collector current \(I_{C}\) is \(1.35 \mathrm{~mA}\) Calculate the value of dc beta \(\beta_{\mathrm{c}}\) factor as follows:

$$ \begin{aligned} \beta_{d c} &=\frac{I_{C}}{I_{n}} \\ &=\frac{1.35 \mathrm{~mA}}{10 \mu \mathrm{A}} \\ &=135 \end{aligned} $$

(c)

The factor \(\alpha\) is,

$$ \begin{aligned} \alpha &=\frac{\beta}{\beta+1} \\ &=\frac{135}{135+1} \\ &=\frac{135}{136} \\ &=0.9926 \end{aligned} $$

Calculate Collector to Base Current when the emitter leg is open \(\left(I_{C B o}\right)\).

$$ \begin{aligned} I_{C B O} &=(1-\alpha) I_{C E O} \\ &=(1-0.9926)(0.3 \mathrm{~mA}) \\ &=7.4 \times 10^{-3}\left(0.3 \times 10^{-3} \mathrm{~A}\right) \\ &=2.2 \mu \mathrm{A} \end{aligned} $$

Calculate the value of collector to emitter current when the base leg is open \(\left(I_{C E O}\right)\).

$$ \begin{aligned} I_{C D O} & \cong \beta I_{C B O} \\ &=135(2.2 \mu \mathrm{A}) \\ &=135\left(2.2 \times 10^{-6} \mathrm{~A}\right) \\ & \approx 0.3 \mathrm{~mA} \end{aligned} $$