Question

Question: Volumes have been written on transistor biasing, but Figure 10.45 gets at the main idea. Suppose that and that the "input" produces its own voltage . The total resistance is in the input loop, which goes clockwise from the emitter through the various components to the base, then back to the emitter through the base-emitter diode. this diode is forward biased with the base at all times 0.7 V higher than the emitter. Suppose also that V_cc = 12 V and that the "out- put" is$350K\Omega$ . Now. given that for every 201 electrons entering the emitter, I passes out the base and 200 out the collector, calculate the maximum and minimum in the sinusoidally varying

(a) Current in the base emitter circuit.

(b) Power delivered by the input.

(c) Power delivered to the output.

(d) Power delivered byV_ce.

(e) what does most of the work.

Short answer

Answer

a) The highest current is $90\mu A$ and the lowest current is$70\mu A$ .

b) Maximum power is $\left|\begin{array}{c}9\mu W\end{array}\right|$, minimum power is $\left|7\mu W\right|$.

c) Maximum output power is 113mW , minimum output power is 69mW .

d) The maximum power is 216mW, minimum power is 168mW .

e) Most work is done by the $v_{\alpha }$.

Step-by-step solution

Given data

$V_{\text{input}}=0.1\sin \omega t$

Resistance$=10\text{K}\Omega$.

The highest current is $90\mu A$ and the lowest current is $70\mu A$.

The maximum power done by $V_{\alpha }$ is 216mW and the minimum power done by $V_{\alpha }$is168mW .

Ohm’s law

$$\begin{gathered} \mathbf{\text{Ohm'slaw:}}{\mathit{I}}_{\mathbf{h}}\mathbf{=}\frac{{\mathbf{V}}_{\dot{\mathbf{h}}}}{\mathbf{R}} \\ \mathbf{\text{PowerdeliveredP=VI}} \\ \mathbf{\text{Powerdelivered}}\mathit{P}\mathbf{=}{\mathit{I}}^{\mathbf{2}}\mathit{R} \end{gathered}$$

Calculate the Voltage

(a)

The voltage across the resistor in the input circuit is expressed as,

.$V=V_{he}-0.7V-0.1V\sin \left(\omega t\right)$

Thus the highest voltage across the resistor would be:

$$\begin{gathered} V_h=V_{he}-0.7\text{V}-0.1\text{V}\sin \left(\omega t\right) \\ =1.5\text{V}-0.7\text{V}+0.1\text{V} \\ =0.9\text{V} \end{gathered}$$

And the lowest voltage across the resistor would be:

$$\begin{gathered} V_I=V_{he}-0.7 \text{V}-0.1V\sin \left(\omega t\right) \\ =1.5\text{V}-0.7 \text{V}-0.1 \text{V} \\ =0.7V \end{gathered}$$

Calculate the Current

The resistor follows Ohm's Law, so the highest current is shown below.

$$\begin{gathered} I_h=\frac{V_h}{R} \\ =\frac{0.9V}{10K\Omega } \\ =9\times {10}^{-5} \text{A} \\ =90\mu A \end{gathered}$$

And the lowest current is given below.

$$\begin{gathered} I_I=\frac{V_I}{R} \\ =7\times {10}^{-5} \text{A}\left[\frac{1\mu A}{{10}^{-6}A}\right] \\ \text{=7μA} \end{gathered}$$

The highest current is $90\mu A$ and the lowest current is $70\mu A$.

Calculate the Maximum Power 

(b)

The maximum power delivered by the input is shown below.

$$\begin{gathered} P_h=V{I}_h \\ =\left(0.1V\right)\left(90\mu A\right) \\ =9\mu A \end{gathered}$$

The minimum power delivered by the input is:

$$\begin{gathered} P_I=V{I}_I \\ =\left(0.1V\right)\left(70\mu A\right) \\ =7\mu A \end{gathered}$$

The maximum power delivered by the input is $\left|9\mu W\right|$ and the minimum power delivered by the P is$\left|7\mu W\right|$ .

Calculate the Current in Case 

(c)

The highest current in the output circuit is:

$$\begin{gathered} I_{n,h}=I_h\times 200 \\ =\left(90\mu A\right)\left(200\right) \\ =18mA \end{gathered}$$

So the lowest current in the output circuit is shown below.

$$\begin{gathered} I_{n,j}=I_h\times 200 \\ =\left(70\mu A\right)\left(200\right) \\ =14mA \end{gathered}$$

Calculate the Maximum and Minimum Output Power

The maximum output power is shown below.

$$\begin{gathered} P=I_{n,h}^2{R}_n \\ =\left[18mA\left(\frac{{10}^{-3}A}{1mA}\right)\right]350\Omega \\ =113mW \end{gathered}$$

The minimum output power is:

$$\begin{gathered} P=I_{n,J}^2{R}_n \\ =\left[14mA\left(\frac{{10}^{-3}A}{1mA}\right)\right]350\Omega \\ =69mW \end{gathered}$$

The maximum output power is $113mW$ and the minimum output power is 69mW .

Calculate the Maximum and Minimum Output Power in Case (d)

(d)

The maximum power done by $V_{\alpha }$ is given below.

role="math" localid="1658398499816" $$\begin{gathered} P_{\alpha ,h}=V_{\alpha }{I}_{\alpha ,h} \\ =\left(12v\right)\left(18mA\right) \\ =216mW \end{gathered}$$

The minimum power done by $V_{\alpha }$ is shown below.

$$\begin{gathered} P_{\alpha ,h}=V_{\alpha }{I}_{\alpha ,J} \\ =\left(12v\right)\left(14mA\right) \\ =168mW \end{gathered}$$

The maximum power done by$V_{\alpha }$is 216mW and the minimum power done by$V_a$is 168mW .

Determine the Most Work done by using the Power Delivered relation 

(e)

Power delivered$P=VI$.

Thus, greater the power greater will be work done.

The maximum power done by$V_{\alpha }$ is 216 mW and the minimum power done by$V_a$is 168mW .

Hence most work is done by the $V_a$.