Chapter 14, Problem 14.55P

a.

To determine

Worst case output voltage for given circuit.

Answer to Problem 14.55P

The worst case output voltage is

  $v_o=9\text{mV}$

Explanation of Solution

Given:

The given circuit is shown below.

  

Input offset voltage $V_{0S}=2\text{mV}$ at $T=25°\text{C}$

Average input bias current $I_B=500\text{nA}$

Input offset current $I_{OS}=100\text{nA}$

  $V_{0S}=2\text{mV}$ at $T=25°\text{C}$

Calculation:

On applying $V_{0S}$ for both op-amp, the output voltage $v_0$ due to input offset voltage effect as shown below.

  

For given circuit $V_Y=V_{OS}$

For an op-amp $V_Y=V_X$

Then, $V_X=V_{OS}$

Now, KCL at input nodes:

  $\begin{array}{l}\frac{V_X}{R_1}+\frac{V_X-v_o}{R_2}=0\\ \frac{R_2{V}_X+R_1\left(V_X-v_o\right)}{R_1{R}_2}=0\\ R_2{V}_X+R_1\left(V_X-v_o\right)=0\\ \left(R_1+R_2\right)V_X=R_1{v}_o\\ v_o=\left(\frac{R_1+R_2}{R_1}\right)V_X\\ v_o=\left(1+\frac{R_2}{R_1}\right)V_X\\ v_o=\left(1+\frac{R_2}{R_1}\right)V_{OS}\mathrm{.....}\left(1\right)\end{array}$

For op-amp circuit

  $\begin{array}{l}{R}_2=R_1=50\text{k}\Omega \\ R_3=25\text{k}\Omega \end{array}$

Now, output voltage due to input bias current and input offset current is,

  $\begin{array}{l}{v}_o=R_2{I}_{OS}\\ v_o=50\times {10}^3\times 100\times {10}^{-9}\\ v_o=5\text{mV}\end{array}$

Due to offset voltage,

From equation (1)

  $\begin{array}{l}{v}_o=\left(1+\frac{50\text{k}}{50\text{k}}\right)\times 2\times {10}^{-3}\\ v_o=2\times 2\times {10}^{-3}\\ \Rightarrow v_o=4\text{mV}\end{array}$

So, the worst case output voltage for the circuit will be,

  $\begin{array}{l}{v}_o=4\text{mV}+5\text{mV}\\ \Rightarrow v_o=9\text{mV}\end{array}$

b.

To determine

Worst case output voltage for given circuit.

Answer to Problem 14.55P

The worst case output voltage is

  $v_o=-1.0815\text{V}$

Explanation of Solution

Given:

The given circuit is shown below.

  

Input offset voltage $V_{0S}=2\text{mV}$ at $T=25°\text{C}$

Average input bias current $I_B=500\text{nA}$

Input offset current $I_{OS}=100\text{nA}$

  $V_{0S}=2\text{mV}$ at $T=25°\text{C}$

Calculation:

Consider the circuit shwon below.

  

Apply KCL at non-inverting terminal

  $\begin{array}{l}\frac{V_Y}{R_3}+I_{B2}=0\\ V_Y=-I_{B2}{R}_3\end{array}$

For op-amp $V_X=V_Y$

So, $V_X=-I_{B2}{R}_3\mathrm{.....}\left(2\right)$

Now, KCL at inverting terminal,

  $\begin{array}{l}\frac{V_X}{R_1}+\frac{V_X-v_o}{R_2}+I_{B1}=0\\ V_X\left(\frac{1}{R_1}+\frac{1}{R_3}\right)-\frac{v_o}{R_2}+I_{B1}=0\\ \frac{v_o}{R_2}=V_X\left(\frac{1}{R_1}+\frac{1}{R_3}\right)+I_{B1}\\ v_o=V_X\left(\frac{R_2}{R_1}+\frac{R_2}{R_3}\right)+I_{B1}{R}_2\end{array}$

Putting $V_X$ from (1)$\begin{array}{l}{v}_o=-\left(I_{B2}{R}_3\right)\left(\frac{R_2}{R_1}+\frac{R_2}{R_3}\right)+I_{B1}{R}_2\\ v_o=I_{B1}{R}_2-I_{B2}{R}_3\left(\frac{R_2}{R_1}+1\right)\mathrm{.....}\left(3\right)\end{array}$

As,

  $\begin{array}{l}{I}_B=\frac{I_{B1}+I_{B2}}{2}\\ I_{B1}+I_{B2}=2I_B\\ I_{B1}+I_{B2}=2\times 500\times {10}^{-9}\\ I_{B1}+I_{B2}=1000\times {10}^{-9}\mathrm{.....}\left(4\right)\end{array}$

And

  $\begin{array}{l}{I}_{OS}=|I_{B1}-I_{B2}|\\ I_{B1}-I_{B2}=\pm I_{OS}\\ I_{B1}-I_{B2}=\pm 100\times {10}^{-9}\mathrm{....}\left(5\right)\end{array}$

Adding (4) and (5)

  $\begin{array}{l}{I}_{B1}+I_{B2}+I_{B1}-I_{B2}=1000\times {10}^{-9}\pm 100\times {10}^{-9}\\ 2I_{B1}=1000\times {10}^{-9}\pm 100\times {10}^{-9}\\ \therefore I_{B1}=\frac{1100\times {10}^{-9}}{2}\text{or}\frac{900\times {10}^{-9}}{2}\\ \Rightarrow I_{B1}=550\times {10}^{-9}\text{or}\text{450}\times {10}^{-9}\end{array}$

From (4)

  $I_{B2}=1000\times {10}^{-9}-I_{B1}$

For $I_{B1}=550\text{nA}$

  $\begin{array}{l}{I}_{B2}=1000\times {10}^{-9}-550\text{n}\\ I_{B2}=450\text{nA}\end{array}$

For $I_{B1}=450\text{nA}$

  $\begin{array}{l}{I}_{B2}=1000\times {10}^{-9}-450\text{n}\\ I_{B2}=550\text{nA}\end{array}$

For the circuit

  $\begin{array}{l}{R}_2=R_1=50\text{k}\Omega \\ R_3=1\text{M}\Omega \end{array}$

If offset voltage is negative, then from (1)

  $\begin{array}{l}{v}_o=\left(1+\frac{50\text{k}}{50\text{k}}\right)\times (-2\times {10}^{-3})\\ v_o=-2\times 2\times {10}^{-3}\\ \Rightarrow v_o=-4\text{mV}\end{array}$

Let $I_{B1}=450\text{nA}$ and $I_{B2}=550\text{nA}$ due to input bias current and input offset current

Putting $I_{B1},I_{B2}$ and resistance values in (3)$\begin{array}{l}{v}_o=\left(450\times {10}^{-9}\times 50\times {10}^3\right)-550\times {10}^{-9}\times 1\times {10}^6\left(\frac{50\text{k}}{50\text{k}}+1\right)\\ v_o=\left(0.0225\right)-0.55\left(2\right)\\ v_o=-1.0775\text{V}\end{array}$

So, the worst case output voltage for circuit is

  $\begin{array}{l}{v}_o=-4\times {10}^{-3}-1.0775\\ \therefore v_o=-1.0815\text{V}\end{array}$