Chapter 14, Problem 14.40P

a.

To determine

Effective resistance $R_1^{\text{'}}$ and $R_2^{\text{'}}$ .

Answer to Problem 14.40P

Value of effective resistance $R_1^{\text{'}}=R_2^{\text{'}}=490\Omega$

Explanation of Solution

Given:

Bipolar diff-amp with active load and a pair of offset-null terminal is shown below.

  

Given parameters are:

  $\begin{array}{l}{R}_1=R_2=500\Omega \\ R_X=50\text{k}\Omega \\ V_T=26\text{mV}\end{array}$

Wipe arm of potentiometer is exactly at center.

Calculation:

As wipe arm of potentiometer is exactly at center. So, $R_1\text{and}{R}_2$ will have resistance $\frac{R_X}{2}$ connected in parallel. Therefore,

  $\begin{array}{l}{R}_1^{\text{'}}=R_1\left|\right|\left(\frac{R_X}{2}\right)\\ R_2^{\text{'}}=R_2\left|\right|\left(\frac{R_X}{2}\right)\end{array}$

As $R_1=R_2$

So, $R_1^{\text{'}}=R_2^{\text{'}}=R_1\left|\right|\left(\frac{R_X}{2}\right)$

  $\begin{array}{l}{R}_1^{\text{'}}=R_2^{\text{'}}=500\left|\right|\left(\frac{50\times {10}^3}{2}\right)\\ R_1^{\text{'}}=R_2^{\text{'}}=500\left|\right|25000\\ R_1^{\text{'}}=R_2^{\text{'}}=\frac{500\times 25000}{500+25000}\\ \Rightarrow R_1^{\text{'}}=R_2^{\text{'}}=490\Omega \end{array}$

b.

To determine

Values of $x$ and $\left(1-x\right)$ of potentiometer in order to compensate the mismatches of transistor.

Answer to Problem 14.40P

Value of $x=0.183$ and $\left(1-x\right)=0.817$

Explanation of Solution

Given:

Bipolar diff-amp with active load and a pair of offset-null terminal is shown below.

  

Given parameters are:

  $\begin{array}{l}{R}_1=R_2=500\Omega \\ R_X=50\text{k}\Omega \\ V_T=26\text{mV}\\ I_Q=250\mu \text{A}\\ i_{C1}=i_{C2}=125\mu \text{A}\\ I_{S3}=2\times {10}^{-14}\text{A}\\ I_{S4}=2.2\times {10}^{-14}\text{A}\end{array}$

Calculation:

Now, KVL between terminals of $Q_3,Q_4\text{and}{V}^{-}$

  $v_{BE3}+i_{C1}{R}_1^{\text{'}}=v_{BE4}+i_{C2}{R}_2^{\text{'}}\mathrm{......}\left(1\right)$

Here,

  $R_1^{\text{'}}$ and $R_2^{\text{'}}$ are effective resistances in emitters of $Q_3\text{and}{Q}_4$ .

Now, on including the effect of potentiometer $R_X$ gives,

  $R_1^{\text{'}}=R_1\left|\right|x{R}_X\mathrm{.....}\left(2\right)$

  $R_2^{\text{'}}=R_2\left|\right|\left(1-x\right)R_X\mathrm{......}\left(3\right)$

The base emitter voltages are,

  $v_{BE3}=V_T\ln \left(\frac{i_{C1}}{I_{S3}}\right)\mathrm{.......}\left(4\right)$

And

  $v_{BE4}=V_T\ln \left(\frac{i_{C2}}{I_{S4}}\right)\mathrm{.......}\left(5\right)$

Substituting values from equation (4) and (5) in equation (1)

  $\begin{array}{l}{V}_T\ln \left(\frac{i_{C1}}{I_{S3}}\right)+i_{C1}{R}_1^{\text{'}}=V_T\ln \left(\frac{i_{C2}}{I_{S4}}\right)+i_{C2}{R}_2^{\text{'}}\\ V_T\left[\ln \left(\frac{i_{C1}}{I_{S3}}\right)-\ln \left(\frac{i_{C2}}{I_{S4}}\right)\right]=i_{C2}{R}_2^{\text{'}}-i_{C1}{R}_1^{\text{'}}\\ 26\times {10}^{-3}\left[\ln \left(\frac{125\times {10}^{-6}}{2\times {10}^{-14}}\right)-\ln \left(\frac{125\times {10}^{-6}}{2.2\times {10}^{-14}}\right)\right]=125\times {10}^{-6}{R}_2^{\text{'}}-125\times {10}^{-6}{R}_1^{\text{'}}\\ 26\times {10}^{-3}\left[\ln \left(6.25\times {10}^9\right)-\ln \left(5.682\times {10}^9\right)\right]=125\times {10}^{-6}\left(R_2^{\text{'}}-R_1^{\text{'}}\right)\\ 26\times {10}^{-3}\left[\ln \left(\frac{6.25\times {10}^9}{5.682\times {10}^9}\right)\right]=125\times {10}^{-6}\left(R_2^{\text{'}}-R_1^{\text{'}}\right)\\ R_2^{\text{'}}-R_1^{\text{'}}=\left(\frac{26\times {10}^{-3}}{125\times {10}^{-6}}\right)\times \ln \left(1.1\right)\\ R_2^{\text{'}}-R_1^{\text{'}}=208\times 0.09531\\ \Rightarrow R_2^{\text{'}}-R_1^{\text{'}}=19.8\end{array}$

Now, substituting value from equation (2) and (3) in above equation,

  $\begin{array}{l}\left[R_2\left|\right|\left(1-x\right)R_X\right]-\left[R_1\left|\right|x{R}_X\right]=19.8\\ \left[\frac{R_2\times \left(1-x\right)R_X}{R_2+\left(1-x\right)R_X}\right]-\left[\frac{R_1\times x{R}_X}{R_1+x{R}_X}\right]=19.8\\ \left[\frac{\left(1-x\right)R_X{R}_2\left(R_1+x{R}_X\right)-x{R}_X{R}_1\left(R_2+\left(1-x\right)R_X\right)}{\left(R_2+\left(1-x\right)R_X\right)\left(R_1+x{R}_X\right)}\right]=19.8\\ \left[\frac{\left(1-x\right)R_X{R}_2{R}_1+x\left(1-x\right)R_X^2{R}_2=x{R}_X{R}_1{R}_2-x\left(1-x\right)R_X^2{R}_1}{\left(R_2+\left(1-x\right)R_X\right)\left(R_1+x{R}_X\right)}\right]=19.8\\ \left[\frac{\left(1-2x\right)R_X{R}_1{R}_2}{R_1{R}_2+x{R}_X{R}_2+\left(1-x\right)R_X{R}_1+x\left(1-x\right)R_X^2}\right]=19.8\end{array}$

Substituting the values

  $\left[\frac{\left(1-2x\right)\left(50\times {10}^3\right)\left(500\right)\left(500\right)}{\left(500\right)\left(500\right)+x\left(50\times {10}^3\right)\left(500\right)+\left(1-x\right)\left(50\times {10}^3\right)\left(500\right)+x\left(1-x\right){\left(50\times {10}^3\right)}^2}\right]=19.8$

  $\left[\frac{\left(1-2x\right)\left(1.25\times {10}^{10}\right)}{\left(25\times {10}^4\right)+\left(25\times {10}^6\right)+x\left(25\times {10}^8\right)-x^2\left(25\times {10}^8\right)}\right]=19.8$

  $\left[\frac{1.25\times {10}^{10}-2\times 1.25\times {10}^{10}x}{25.25\times {10}^6+x\left(25\times {10}^8\right)-x^2\left(25\times {10}^8\right)}\right]=19.8$

  $1.25\times {10}^{10}-2.5\times {10}^{10}x=19.8\left[25.25\times {10}^6+x\left(25\times {10}^8\right)-x^2\left(25\times {10}^8\right)\right]$

  $1.25\times {10}^{10}-2.5\times {10}^{10}x=\left(499.95\times {10}^6\right)+x\left(495\times {10}^8\right)-x^2\left(495\times {10}^8\right)$

  $1.25\times {10}^{10}-\left(499.95\times {10}^6\right)=2.5\times {10}^{10}x+\left(495\times {10}^8\right)x-x^2\left(495\times {10}^8\right)$

  $1.2\times {10}^{10}=\left(7.45\times {10}^{10}\right)x-\left(4.95\times {10}^{10}\right)x^2$

  $4.95x^2-7.45x+1.2=0$

So,

  $\begin{array}{l}x=\frac{-\left(-7.45\right)\pm \sqrt{{\left(-7.45\right)}^2-\left(4\times 4.95\times 1.2\right)}}{2\times 4.95}\\ x=\frac{7.45\pm \sqrt{31.7425}}{9.9}\\ x=\frac{7.45\pm 5.634}{9.9}\\ x=\frac{7.45+5.634}{9.9}\text{or}\frac{7.45-5.634}{9.9}\\ x=\frac{13.084}{9.9}\text{or}\frac{1.816}{9.9}\\ \Rightarrow x=1.322\text{or}0.183\end{array}$

Since, $x$ should be less than 1. Therefore, $x=0.813$

So, to compensate the mismatches of transistor the value of $x=0.183$ and

  $\begin{array}{l}1-x=1-0.183\\ 1-x=0.817\end{array}$