Chapter 6, Problem 6.16TYU

Consider the circuit in Figure 6.70(a). Let $\beta =100$ , $V_{BE}\text{(on)}=0.7\text{V}$ , and $V_A=\infty$ for each transistor. Assume $R_B=10\text{kΩ}$ , $R_C=4\text{kΩ}$ , $I_{Eo}=1\text{mA}$ , $V^{+}=5\text{V}$ , and $V^{-}=-5\text{V}$ . (a) Determine the Q−point values for each transistor. (b) Calculate the small−signal hybrid− $\pi$ parameters for each transistor. (c) Find the overall small−signal voltage gain $A_{\upsilon }=V_o/V_s$ . (d) Find the input resistance $R_i$ . (Ans. (a) $I_{CQ1}=0.0098\text{mA}$ , $V_{CEQ1}=1.7\text{V}$ , $I_{CQ2}=0.990\text{mA}$ , $V_{CEQ2}=2.4\text{V}$ (b) $r_{\pi 1}=265\text{kΩ}$ , $g_{m1}=0.377\text{mA/V}$ , $r_{\pi 2}=2.63\text{kΩ}$ , $g_{m2}=38.1\text{mA/V}$ (c) $A_{\upsilon }=-77.0$ (d) $R_i=531\text{kΩ}$ )

To determine

The value of the Q-point for each transistor.

Answer to Problem 6.16TYU

The Q-point are given as:

  $\begin{array}{c}{I}_{CQ1}=0.0098\text{mA}\\ V_{CEQ1}=1.7\text{V}\\ I_{CQ2}=0.990\text{mA}\\ V_{CEQ2}=2.4\text{V}\end{array}$

Explanation of Solution

Given:

The transistor circuit is provided:

Where,

  $\begin{array}{l}\beta =100\\ V_{BE}=0.7\text{V}\\ V_A=\infty \\ R_B=10\text{k}\Omega \\ R_C=4\text{k}\Omega \\ I_{Eo}=1\text{mA}\\ V^{+}=5\text{V}\\ V^{-}=-5\text{V}\end{array}$

Consider the emitter current of Q2 transistor is same as the quiescent emitter current of transistor 2.

  $I_{Eo}=I_{EQ2}$

The expression for quiescent collector current ( $I_{CQ2}$ ) of transistor 2.

  $I_{CQ2}=\left(\frac{\beta }{1+\beta }\right)\left(I_{EQ2}\right)$

Here ,

  $\beta$ is current gain of transistor and

  $I_{EQ2}$ is quiescent emitter current of transistor 2.

Substitute 1mA for $I_{EQ2}$ and 100 for $\beta$ .

  $\begin{array}{l}{I}_{CQ2}=\left(\frac{\beta }{1+\beta }\right)\left(I_{EQ2}\right)\\ I_{CQ2}=\left(\frac{100}{1+100}\right)\left(\text{1mA}\right)\\ =\left(\frac{100}{101}\right)\left({\text{1×10}}^{\text{-3}}\text{A}\right)\\ =0.990\times {10}^{-3}\text{A}\\ =0.990\text{mA}\end{array}$

The expression for quiescent emitter current $I_{EQ1}$ of transistor 1:

  $I_{EQ1}=\frac{I_{EQ2}}{1+\beta }$

Substituting 1 mA for $I_{EQ2}$ and 100 for $\beta$ .

  $\begin{array}{l}{I}_{EQ1}=\frac{\text{0}\text{.0099mA}}{\text{1+100}}\\ \text{=}\frac{\text{0}{\text{.0099×10}}^{\text{-3}}\text{A}}{\text{101}}\\ \text{=0}{\text{.000098×10}}^{\text{-3}}\text{A}\\ \text{=0}\text{.000098mA}\end{array}$

Now, the expression for quiescent collector current $\left(I_{CQ}\right)$ .

  $I_{CQ1}=\left(\beta \right)I_{BQ1}$

  $I_{CQ1}$ IS quiescent collector current of transistor 1.

Substituting 100 for $\beta$ and 0.000098mA for $I_{BQ1}$ .

  $\begin{array}{l}{I}_{CQ1}=\text{0}\text{.000098mA}\left(\text{100}\right)\\ \text{=0}{\text{.000098×10}}^{\text{-3}}\text{A}\left(\text{100}\right)\\ \text{=0}{\text{.0098×10}}^{\text{-3}}\text{A}\\ \text{=0}\text{.0098mA}\end{array}$

Write the expression for base voltage $\left(V_{B1}\right)$ of transistor 1.

  $V_{B1}=-I_{BQ1}{R}_B$

Here ,

  $R_B$ is base resistance.

Substitute $10k\Omega$ for $R_B$ and 0.000098mAfor $I_{BQ1}$

  $\begin{array}{c}{V}_{B1}=\text{-}\left(\text{0}\text{.000098mA}\right)\left(\text{10kΩ}\right)\\ \text{=-}\left(\text{0}{\text{.000098×10}}^{\text{-3}}\text{A}\right)\left({\text{10×10}}^{\text{3}}\text{Ω}\right)\\ \text{=-0}\text{.000098V}\\ \text{»0V}\end{array}$

Considering the expression for emitter voltage $\left(V_{E1}\right)$ of transistor 1.

  $V_{E1}=V_{B1}-V_{BE\left(on\right)}$

Substituting 0V For $V_{B1}$ , and 0.7V for $V_{BE\left(on\right)}$ .

  $\begin{array}{c}{V}_{E1}=0V-0.7V\\ =-0.7V\end{array}$

Considering the expression for emitter voltage $\left(V_{E2}\right)$ of transistor 2.

  $V_{E2}=V_{B1}-2V_{BE\left(on\right)}$

Substituting 0V FOR $V_{B1}$ , And 0.7V FOR $V_{BE\left(on\right)}$ .

  $\begin{array}{c}{V}_{E2}=\text{0V-2}\left(\text{0}\text{.7V}\right)\\ \text{=-1}\text{.4V}\end{array}$

The expression for current $\left(I_1\right)$ in transistor 1.

  $I_1=I_{CQ1}+I_{CQ2}$

Substitute $\text{0}\text{.0098mA}\text{for}{\text{I}}_{\text{CQ1}}\text{,}\text{and}\text{0}\text{.990mA}\text{For}{\text{I}}_{\text{CQ2}}$

  $\begin{array}{c}{I}_1=\text{0}\text{.0098mA+0}\text{.990mA}\\ \text{=0}{\text{.0098×10}}^{\text{-3}}\text{A+0}{\text{.990×10}}^{\text{-3}}\text{A}\\ {\text{=1×10}}^{\text{-3}}\text{A}\\ \text{=1mA}\end{array}$

The expression for output voltage $\left(V_0\right)$ .

  $V_0=V^{+}-I_1{R}_C$

Substituting 5V for $V^{+}$ , 1mA for $I_1$ and $\text{4kΩ}$ for $R_C$ .

  $\begin{array}{c}{V}_0\text{=5V-1mA}\left(\text{4kΩ}\right)\\ \text{=5V-1}\left(\text{4}\right)\\ \text{=1V}\end{array}$

The expression for collector emitter quiescent voltage $\left(V_{CEQ2}\right)$ of transistor 1.

  $V_{CEQ2}=V_0-V_{E2}$

Substituting 1V for $V_0$ , and -1.4V for $V_{E1}$

  $\begin{array}{c}{V}_{CEQ2}=\text{1V-}\left(\text{-1}\text{.4V}\right)\\ \text{=2}\text{.4V}\end{array}$

The expression for collector emitter quiescent voltage $\left(V_{CEQ1}\right)$ of transistor 1.

  $V_{CEQ1}=V_0-V_{E1}$

Substituting 1V for $V_0$ , and -0.7V for $V_{E1}$ .

  $\begin{array}{c}{V}_{CEQ1}=1\text{V-}\left(\text{-0}\text{.7V}\right)\\ \text{=1}\text{.7V}\end{array}$

Therefore, the Q -point of transistor 1:

  $\begin{array}{l}{V}_{CEQ1}=1.7\text{V}\\ I_{CQ1}=0.00987\text{mA}\end{array}$

The Q -point of transistor 2:

. $\begin{array}{l}{V}_{CEQ2}=2.4\text{V}\\ I_{CQ2}=0.990\text{mA}\end{array}$

To determine

The small signal hybrid- $\pi$ parameters for the each transistor.

Answer to Problem 6.16TYU

The results are:

  $\begin{array}{l}{r}_{\pi 1}=265\text{k}\Omega \\ {\text{g}}_{m1}=0.377\text{mA/V}\\ r_{\pi 2}=2.63\text{k}\Omega \\ {\text{g}}_{m2}=38.1\text{mA/V}\end{array}$

Explanation of Solution

Given:

The transistor circuit is provided:

Where,

  $\begin{array}{l}\beta =100\\ V_{BE}=0.7\text{V}\\ V_A=\infty \\ R_B=10\text{k}\Omega \\ R_C=4\text{k}\Omega \\ I_{Eo}=1\text{mA}\\ V^{+}=5\text{V}\\ V^{-}=-5\text{V}\end{array}$

Now, the expression for small signal hybrid $\pi$ parameter ( $r_{\pi 1}$ ) input resistance of transistor 1.

  $r_{\pi 1}=\frac{\beta V_{\tau }}{I_{CQ1}}$

Substituting the 100 for $\beta$ , 0.026V FOR $V_{\tau }$ and 0.0098ma for $I_{CQ1}$ .

  $\begin{array}{l}{r}_{\pi 1}=\frac{\left(100\right)\left(0.026V\right)}{\left(0.0098mA\right)}\\ =\frac{2.6V}{0.0098\times {10}^{-3}A}\\ =265\times {10}^3\Omega \\ =265\text{kΩ}\end{array}$

The expression for transconductance $\left(g_{m1}\right)$ of transistor 1.

  $g_{m1}=\frac{I_{CQ1}}{V_{\tau }}$

Substituting 0.026V for $V_{\tau }$ and 0.0098ma for $I_{CQ1}$ .

  $\begin{array}{c}{g}_{m1}=\frac{\text{0}\text{.0098mA}}{\text{0}\text{.026V}}\\ \text{=}\frac{\text{0}{\text{.0098×10}}^{\text{-3}}\text{A}}{\text{0}\text{.026V}}\\ \text{=0}{\text{.377×10}}^{\text{-3}}\text{A}\\ \text{=0}\text{.377mA}\end{array}$

Consider the values or output resistance $r_{01}=\infty and{r}_{02}=\infty$ .

The expression for small signal hybrid $\pi$ parameter ( $r_{\pi 2}$ ) input resistance of transistor 2.

  $r_{\pi 2}=\frac{\beta V_{\tau }}{I_{CQ1}}$

Substituting 100 for $\beta$ , 0.026V for $V_{\tau }$ and $0.990mA$ for $I_{CQ2}$

  $\begin{array}{c}{r}_{\pi 2}=\frac{\beta V_{\tau }}{I_{CQ2}}\\ r_{\pi 2}=\frac{\left(\text{100}\right)\left(\text{0}\text{.026V}\right)}{\left(\text{0}\text{.990mA}\right)}\\ \text{=}\frac{\text{2}\text{.6V}}{\text{0}{\text{.990×10}}^{\text{-3}}\text{A}}\\ {\text{=263×10}}^{\text{3}}\text{Ω}\\ \text{=2}\text{.63kΩ}\end{array}$

The expression for transconductance $\left(g_{m2}\right)$ of transistor 2.

  $g_{m2}=\frac{I_{CQ2}}{V_{\tau }}$

Substituting 0.026V for $V_{\tau }$ and $0.990mA$ for $I_{CQ2}$ .

  $\begin{array}{c}{g}_{m2}=\frac{I_{CQ2}}{V_{\tau }}\\ g_{m2}=\frac{\text{0}\text{.990mA}}{\text{0}\text{.026V}}\\ \text{=}\frac{\text{0}{\text{.990×10}}^{\text{-3}}\text{A}}{\text{0}\text{.026V}}\\ \text{=38}{\text{.1×10}}^{\text{-3}}\text{A}\\ \text{=38}\text{.1}\frac{\text{mA}}{\text{V}}\end{array}$

Thus the small signal parameter values are $r_{\pi 1}$ , $g_{m1}$ , $r_{\pi 2}$ and $g_{m2}$ are $265k\Omega$ , $0.377mA$ , $2.63k\Omega$ and $38.1\frac{mA}{V}$ respectively.

To determine

The small-signal voltage gain.

Answer to Problem 6.16TYU

The results are:

  $A_v=-77$

Explanation of Solution

Given:

The transistor circuit is provided:

Where,

  $\begin{array}{l}\beta =100\\ V_{BE}=0.7\text{V}\\ V_A=\infty \\ R_B=10\text{k}\Omega \\ R_C=4\text{k}\Omega \\ I_{Eo}=1\text{mA}\\ V^{+}=5\text{V}\\ V^{-}=-5\text{V}\end{array}$

The expression for output voltage $\left(V_0\right)$ :

  $V_0=-\left(g_{m1}{V}_{\pi 1}+g_{m2}{V}_{\pi 2}\right)R_C\mathrm{..........}\left(1\right)$

The expression for source voltage $\left(V_s\right)$ :

  $V_s=V_{\pi 1}+V_{\pi 2}\mathrm{...........}\left(2\right)$

Considering the expression for hybrid $\pi$ parameter voltage ( $V_{\pi 2}$ ) of transistor 2.

  $V_{\pi 2}=\left(\frac{V_{\pi 1}}{r_{\pi 1}}+g_{m1}{V}_{\pi 1}\right)r_{\pi 2}\mathrm{..........}\left(3\right)$

Re-arranging the equation (3).

  $\begin{array}{l}{V}_{\pi 2}=\left(\frac{1}{r_{\pi 1}}+g_{m1}\right)V_{\pi 1}{r}_{\pi 2}\\ V_{\pi 2}=\left(\frac{1}{r_{\pi 1}}+g_{m1}\right)V_{\pi 1}{r}_{\pi 2}\mathrm{............}\left(4\right)\end{array}$

Considering the expression for current gain of transistor ( $\beta$ )

  $\beta =g_{m1}{r}_{\pi 1}\mathrm{............}\left(5\right)$

Substituting the equation (5) in (4)

  $V_{\pi 2}=\left(\frac{1+\beta }{r_{\pi 1}}\right)V_{\pi 1}{r}_{\pi 2}\mathrm{.............}\left(6\right)$

Substituting the equation (6) in (1)

  $V_0=-\left(g_{m1}{r}_{\pi 1}+g_{m2}\left(\frac{1+\beta }{r_{\pi 1}}\right)V_{\pi 1}{r}_{\pi 2}\right)R_C\mathrm{..........}\left(7\right)$

Substituting the equation (6) in (2):

  $\begin{array}{l}{V}_s=V_{\pi 1}+\left(\frac{1+\beta }{r_{\pi 1}}\right)V_{\pi 1}{r}_{\pi 2}\\ V_s=V_{\pi 1}\left[1+\left(\frac{1+\beta }{r_{\pi 1}}\right)r_{\pi 2}\right]\mathrm{..........}\left(8\right)\end{array}$

The expression for voltage gain $A_{\nu }$ .

  $A_{\nu }=\frac{V_0}{V_s}\mathrm{........}\left(9\right)$

Substituting the equation (7) and (8) in (9):

  $\begin{array}{c}{A}_{\nu }=\frac{-\left(g_{m1}{r}_{\pi 1}+g_{m2}\left(\frac{1+\beta }{r_{\pi 1}}\right)V_{\pi 1}{r}_{\pi 2}\right)R_C}{V_{\pi 1}\left[1+\left(\frac{1+\beta }{r_{\pi 1}}\right)r_{\pi 2}\right]}\\ =\frac{-V_{\pi 1}\left(g_{m1}+g_{m2}\left(\frac{1+\beta }{r_{\pi 1}}\right)r_{\pi 2}\right)R_C}{V_{\pi 1}\left[1+\left(\frac{1+\beta }{r_{\pi 1}}\right)r_{\pi 2}\right]}\\ =\frac{-\left(g_{m1}+g_{m2}\left(\frac{1+\beta }{r_{\pi 1}}\right)r_{\pi 2}\right)R_C}{\left[1+\left(\frac{1+\beta }{r_{\pi 1}}\right)r_{\pi 2}\right]}\mathrm{...........}\left(10\right)\\ \end{array}$

Substituting 0.377mA for $g_{m1}$ , $38.1\frac{mA}{V}$ for $g_{m2}$ , 100 for $\beta$ , $4k\Omega$ for $R_C$ , $2.63k\Omega$ for $r_{\pi 2}$ , and $265k\Omega$ for $r_{\pi 1}$ in equation (10).

  $\begin{array}{c}{A}_{\nu }=\frac{\text{-}\left(\text{0}\text{.3777mA+38}\text{.1}\frac{\text{mA}}{\text{V}}\left(\frac{\text{1+100}}{\text{265kΩ}}\right)\text{2}\text{.63kΩ}\right)\text{4kΩ}}{\left[\text{1+}\left(\frac{\text{1+10}}{\text{265kΩ}}\right)\text{2}\text{.63kΩ}\right]}\\ \text{=}\frac{\text{-}\left(\text{0}{\text{.3777×10}}^{\text{-3}}\text{A+38}{\text{.1×10}}^{\text{-3}}\frac{\text{A}}{\text{V}}\left(\frac{\text{101}}{{\text{265×10}}^{\text{3}}\text{kΩ}}\right)\text{2}{\text{.63×10}}^{\text{3}}\text{Ω}\right){\text{4×10}}^{\text{3}}\text{Ω}}{\left[\text{1+}\left(\frac{\text{101}}{\text{265kΩ}}\right)\text{2}\text{.63kΩ}\right]}\\ \text{=-77}\text{.01}\end{array}$

Thus , the voltage gain is -77.01

To determine

The input resistance R_i .

Answer to Problem 6.16TYU

The input resistance are:

  $R_i=531\text{k}\Omega$

Explanation of Solution

Given:

The transistor circuit is provided:

Where,

  $\begin{array}{l}\beta =100\\ V_{BE}=0.7\text{V}\\ V_A=\infty \\ R_B=10\text{k}\Omega \\ R_C=4\text{k}\Omega \\ I_{Eo}=1\text{mA}\\ V^{+}=5\text{V}\\ V^{-}=-5\text{V}\end{array}$

The expression for input resistance $\left(R_i\right)$ is given as:

  $R_i=r_{\pi 1}+\left(1+\beta \right)r_{\pi 2}$

Substituting the 100 for $\beta$ , $2.63k\Omega$ for $r_{\pi 2}$ and $265k\Omega$ for $r_{\pi 1}$ .

  $\begin{array}{c}{R}_i=\text{265kΩ+}\left(\text{1+100}\right)\text{2}\text{.63kΩ}\\ {\text{=265×10}}^{\text{3}}\text{Ω+}\left(\text{1+100}\right)\text{2}{\text{.63kΩ×10}}^{\text{3}}\text{Ω}\\ {\text{=265×10}}^{\text{3}}\text{Ω+}\left(\text{265}{\text{.63×10}}^{\text{3}}\right)\\ \text{=530}{\text{.63×10}}^{\text{3}}\text{Ω}\\ \text{=531kΩ}\end{array}$

Therefore, the input resistance value is $531\text{kΩ}$ .