Chapter 6, Problem 50E

For the circuit of Fig. 6.62, calculate the differential input voltage and the input bias current if the op amp is a(n) (a) μA741; (b) LF411; (c) AD549K.

To determine

Find the differential input voltage and input bias current of op amp.

Answer to Problem 50E

The differential voltage $v_d$ of $\text{μA741}$ is $1.548\times {10}^{-5}\text{V}$ and input bias current $i_d$ is $7.743\times {10}^{-12}\text{A}$.

Explanation of Solution

Given Data:

Differential op amp is $\text{μA741}$.

Calculation:

The redrawn circuit of op amp is shown in Figure 1 as follows:

Refer to the redrawn Figure 1:

The expression for the nodal analysis at the nodal voltage $v_d$ is:

$\frac{-v_d+450\text{mV}}{R_1}+\frac{-v_d+450\text{mV}-v_{\text{out}}}{R_f}+\frac{-v_d}{R_i}=0$ (1)

Here,

$v_d$ is the differential voltage at node terminal of input across op amp,

$v_{\text{in}}$ is the applied input across op amp,

$R_1$ is the input resistance of op amp,

$R_f$ is the feedback resistance of op amp and

$v_{\text{out}}$ is the output voltage of $\text{μA741}$ op amp.

The expression for nodal analysis at nodal voltage $v_{\text{out}}$ is:

$\frac{v_{\text{out}}-\left(-v_d+450\text{mV}\right)}{R_f}+\frac{v_{\text{out}}-A{v}_d}{R_0}=0$ (2)

Here,

$A$ is the open loop gain of op amp and

$R_0$ is the output resistance of op amp.

The expression for the bias input current of differential op amp is:

$v_d=\left(i_d\times R_i\right)$ (3)

Here,

$i_d$ is the bias input current of practical op amp.

Simplify the equation (1) for $v_d$.

$\begin{array}{l}\frac{-v_d+450\text{mV}}{R_1}+\frac{-v_d+450\text{mV}-v_{\text{out}}}{R_f}+\frac{-v_d}{R_i}=0\\ \frac{-v_d}{R_1}-\frac{v_d}{R_f}-\frac{v_d}{R_i}-\frac{v_{\text{out}}}{R_f}+\frac{450\text{mV}}{R_f}=-\frac{450\text{mV}}{R_1}\\ -v_d\left(\frac{1}{R_1}+\frac{1}{R_f}+\frac{1}{R_i}\right)-\frac{v_{out}}{R_f}+\frac{450\text{mV}}{R_f}=-\frac{450\text{mV}}{R_1}\end{array}$

$-v_d\left(\frac{R_i{R}_f+R_1{R}_i+R_1{R}_f}{R_1{R}_f{R}_i}\right)=-\frac{450\text{mV}}{R_1}+\frac{v_{out}}{R_f}-\frac{450\text{mV}}{R_f}$ (4)

Refer to the TABLE 6.3 in the textbook.

Substitute $1.4\text{k}\Omega$ for $R_f$, $250\text{}\Omega$ for $R_1$ and $2\text{M}\Omega$ for $R_i$ in the equation (4).

$\begin{array}{l}-v_d\left(\frac{2\text{M}\Omega \times 1.4\text{k}\Omega +250\Omega \times 2\text{M}\Omega +250\Omega \times 1.4\text{k}\Omega }{2\text{M}\Omega \times 1.4\text{k}\Omega \times 250\Omega }\right)=-\frac{450\text{mV}}{250\Omega }+\frac{v_{out}}{1.4\text{k}\Omega }-\frac{450\text{mV}}{1.4\text{k}\Omega }\\ -v_d\left(\frac{2800\text{M}\Omega \text{+500}\text{}\text{}\text{ M}\Omega \text{+0}\text{.35}\text{M}\Omega }{700000\text{M}\Omega }\right)=-\frac{450\times {10}^{-3}\text{V}}{250\Omega }+\frac{v_{out}}{1.4\times {10}^3\Omega }-3.214\times {10}^{-4}\\ -v_d\left(4.714\times {10}^{-3}\right)=-1.8\times {10}^{-3}+7.143\times {10}^{-4}{v}_{\text{out}}-3.214\times {10}^{-4}\end{array}$

Rearrange for $v_{\text{out}}$.

$\begin{array}{l}-v_d\left(4.714\times {10}^{-3}\right)=-1.8\times {10}^{-3}+7.143\times {10}^{-4}{v}_{\text{out}}-3.214\times {10}^{-4}\\ -v_d\left(4.714\times {10}^{-3}\right)=7.143\times {10}^{-4}{v}_{\text{out}}-2.121\times {10}^{-3}\\ 7.143\times {10}^{-4}{v}_{\text{out}}=2.121\times {10}^{-3}-v_d\left(4.714\times {10}^{-3}\right)\\ v_{\text{out}}=\frac{2.121\times {10}^{-3}-v_d\left(4.714\times {10}^{-3}\right)}{7.143\times {10}^{-4}}\end{array}$

$v_{\text{out}}=2.969-6.599v_d$ (5)

Substitute $2.969-6.599v_d$ for $v_{\text{out}}$ from equation (5) in equation (2).

$\begin{array}{l}\frac{\left(2.969-6.599v_d\right)-\left(-v_d+450\text{mV}\right)}{1.4\text{}\text{k}\Omega }+\frac{\left(2.969-6.599v_d\right)-A\times v_d}{75\text{}\Omega }=0\\ \frac{-6.599v_d}{1.4\text{}\text{k}\Omega }+\frac{v_d}{1.4\text{}\text{k}\Omega }-\frac{6.599v_d}{75\text{}\Omega }-\frac{A\times v_d}{75\text{}\text{}\Omega }+\frac{2.969}{1.4\times {10}^3\Omega }-\frac{450\times {10}^{-3}\text{V}}{1.4\times {10}^3\text{}\Omega }+\frac{2.969}{75\text{}\Omega }=0\\ v_d\left(\frac{-6.599}{1.4\text{}\text{k}\Omega }+\frac{1}{1.4\text{}\text{k}\Omega }-\frac{6.599}{75\text{}\Omega }-\frac{A}{75\text{}\text{}\Omega }\right)+2.121\times {10}^{-3}-3.214\times {10}^{-4}+0.0395=0\end{array}$

$v_d\left(-4.713\times {10}^{-3}+7.14\times {10}^{-4}-0.0879-\frac{A}{75\text{}\text{}\Omega }\right)+0.0413=0$ (6)

Substitute $2\times {10}^5\text{V/V}$ for $A$ in the equation (6).

$\begin{array}{l}{v}_d\left(-4.713\times {10}^{-3}+7.14\times {10}^{-4}-0.0879-\frac{2\times {10}^5\text{V/V}}{75\text{}\text{}\Omega }\right)+0.0413=0\\ v_d\left(-2666.758\right)+0.0413=0\\ v_d=\frac{0.0413}{2666.758}\text{V}\\ v_d=1.548\times {10}^{-5}\text{V}\end{array}$

Substitute $1.548\times {10}^{-5}\text{V}$ for $v_d$ in equation (3).

$\begin{array}{l}1.548\times {10}^{-5}\text{V}=\left(i_d\times 2\text{M}\Omega \right)\\ 1.548\times {10}^{-5}\text{V}=\left(i_d\times 2\times {\text{10}}^6\Omega \right)\left\{\because 1\text{M}\Omega =\text{1}\times {\text{10}}^6\Omega \right\}\end{array}$

Rearrange for $i_d$.

$i_d=7.743\times {10}^{-12}\text{A}$

Conclusion:

Thus, the differential voltage $v_d$ of $\text{μA741}$ is $1.548\times {10}^{-5}\text{V}$ and input bias current $i_d$ is $7.743\times {10}^{-12}\text{A}$.

To determine

Find the differential input voltage and input bias current of op amp.

Answer to Problem 50E

The differential voltage $v_d$ of $\text{LF411}$ is $1.485\times {10}^{-5}\text{V}$ and input bias current $i_d$ is $1.485\times {10}^{-17}\text{A}$.

Explanation of Solution

Given Data:

Differential op amp is $\text{LF411}$.

Calculation:

Refer to the TABLE 6.3 in the textbook.

Substitute $1.4\text{k}\Omega$ for $R_f$, $250\text{}\Omega$ for $R_1$ and $1\text{T}\Omega$ for $R_i$ in the equation (4).

$\begin{array}{l}-v_d\left(\frac{1\text{T}\Omega \times 1.4\text{k}\Omega +250\Omega \times 1\text{T}\Omega +250\Omega \times 1.4\text{k}\Omega }{1\text{T}\Omega \times 1.4\text{k}\Omega \times 250\Omega }\right)=-\frac{450\text{mV}}{250\Omega }+\frac{v_{\text{out}}}{1.4\text{k}\Omega }-\frac{450\text{mV}}{1.4\text{k}\Omega }\\ -v_d\left(\frac{1400\text{T}\Omega \text{+250}\text{}\text{}\text{ T}\Omega \text{+0}\text{.00035}\text{T}\Omega }{350000\text{T}\Omega }\right)=-\frac{450\times {10}^{-3}\text{V}}{250\Omega }+\frac{v_{\text{out}}}{1.4\times {10}^3\Omega }-3.214\times {10}^{-4}\\ -v_d\left(4.714\times {10}^{-3}\right)=-1.8\times {10}^{-3}+7.143\times {10}^{-4}{v}_{\text{out}}-3.214\times {10}^{-4}\end{array}$

Rearrange for $v_{\text{out}}$.

$\begin{array}{l}-v_d\left(4.714\times {10}^{-3}\right)=7.143\times {10}^{-4}{v}_{\text{out}}-2.121\times {10}^{-3}\\ 7.143\times {10}^{-4}{v}_{\text{out}}=2.121\times {10}^{-3}-v_d\left(4.714\times {10}^{-3}\right)\\ v_{\text{out}}=\frac{2.121\times {10}^{-3}-v_d\left(4.714\times {10}^{-3}\right)}{7.143\times {10}^{-4}}\end{array}$

$v_{\text{out}}=2.969-6.599v_d$

Substitute $2.969-6.599v_d$ for $v_{\text{out}}$ in equation (2).

$\begin{array}{l}\frac{\left(2.969-6.599v_d\right)-\left(-v_d+450\text{mV}\right)}{1.4\text{}\text{k}\Omega }+\frac{\left(2.969-6.599v_d\right)-A\times v_d}{1\text{}\Omega }=0\\ \frac{-6.599v_d}{1.4\text{}\text{k}\Omega }+\frac{v_d}{1.4\text{}\text{k}\Omega }-\frac{6.599v_d}{1\text{}\Omega }-\frac{A\times v_d}{\text{1 }\Omega }+\frac{2.969}{1.4\times {10}^3\Omega }-\frac{450\times {10}^{-3}\text{V}}{1.4\times {10}^3\text{}\Omega }+\frac{2.969}{1\text{}\Omega }=0\\ v_d\left(\frac{-6.599}{1.4\text{}\text{k}\Omega }+\frac{1}{1.4\text{}\text{k}\Omega }-\frac{6.599}{\text{1 }\Omega }-\frac{A}{\text{1 }\Omega }\right)+2.121\times {10}^{-3}-3.214\times {10}^{-4}+2.969=0\end{array}$

$v_d\left(-4.713\times {10}^{-3}+7.14\times {10}^{-4}-6.599-A\right)+2.970=0$ (7)

Substitute $2\times {10}^5\text{V/V}$ for $A$ in the equation (7).

$\begin{array}{l}{v}_d\left(-4.713\times {10}^{-3}+7.14\times {10}^{-4}-6.599-2\times {10}^5\right)+2.970=0\\ v_d\left(-200006.603\right)+0.0349=0\\ v_d=\frac{2.970}{200006.603}\text{V}\\ v_d=1.485\times {10}^{-5}\text{V}\end{array}$

Substitute $1.485\times {10}^{-5}\text{V}$ for $v_d$ in equation (3).

$\begin{array}{l}1.485\times {10}^{-5}\text{V}=\left(i_d\times 1\text{T}\Omega \right)\\ 1.485\times {10}^{-5}\text{V}=\left(i_d\times 1\times {\text{10}}^{12}\Omega \right)\left\{\because 1\text{T}\Omega =\text{1}\times {\text{10}}^{12}\Omega \right\}\end{array}$

Rearrange for $i_d$.

$i_d=1.485\times {10}^{-17}\text{A}$

Conclusion:

Thus, the differential voltage $v_d$ of $\text{LF411}$ is $1.485\times {10}^{-5}\text{V}$ and input bias current $i_d$ is $1.485\times {10}^{-17}\text{A}$.

To determine

Find the differential input voltage and input bias current of $\text{AD549K}$ op amp.

Answer to Problem 50E

The differential voltage $v_d$ of $\text{AD549K}$ is $1.99\times {10}^{-7}\text{V}$ and input bias current $i_d$ is $1.996\times {10}^{-20}\text{A}$.

Explanation of Solution

Given Data:

Differential op amp is $\text{AD549K}$.

Calculation:

Refer to the TABLE 6.3 in the textbook.

Substitute $1.4\text{k}\Omega$ for $R_f$, $250\text{}\Omega$ for $R_1$ and $10\text{T}\Omega$ for $R_i$ in the equation (4).

$\begin{array}{l}-v_d\left(\frac{10\text{T}\Omega \times 1.4\text{k}\Omega +250\Omega \times 10\text{T}\Omega +250\Omega \times 1.4\text{k}\Omega }{10\text{T}\Omega \times 1.4\text{k}\Omega \times 250\Omega }\right)=-\frac{450\text{mV}}{250\Omega }+\frac{v_{\text{out}}}{1.4\text{k}\Omega }-\frac{450\text{mV}}{1.4\text{k}\Omega }\\ -v_d\left(\frac{14000\text{T}\Omega \text{+2500}\text{}\text{}\text{ T}\Omega \text{+0}\text{.00035}\text{T}\Omega }{3500000\text{T}\Omega }\right)=-\frac{450\times {10}^{-3}\text{V}}{250\Omega }+\frac{v_{\text{out}}}{1.4\times {10}^3\Omega }-3.214\times {10}^{-4}\\ -v_d\left(4.714\times {10}^{-3}\right)=-1.8\times {10}^{-3}+7.143\times {10}^{-4}{v}_{\text{out}}-3.214\times {10}^{-4}\end{array}$

Rearrange for $v_{\text{out}}$.

$\begin{array}{l}-v_d\left(4.714\times {10}^{-3}\right)=7.143\times {10}^{-4}{v}_{\text{out}}-2.121\times {10}^{-3}\\ 7.143\times {10}^{-4}{v}_{\text{out}}=2.121\times {10}^{-3}-v_d\left(4.714\times {10}^{-3}\right)\\ v_{\text{out}}=\frac{2.121\times {10}^{-3}-v_d\left(4.714\times {10}^{-3}\right)}{7.143\times {10}^{-4}}\end{array}$

$v_{\text{out}}=2.969-6.599v_d$

Substitute $2.969-6.599v_d$ for $v_{\text{out}}$ in equation (2).

$\begin{array}{l}\frac{\left(2.969-6.599v_d\right)-\left(-v_d+450\text{mV}\right)}{1.4\text{}\text{k}\Omega }+\frac{\left(2.969-6.599v_d\text{V}\right)-A\times v_d}{15\text{}\Omega }=0\\ \frac{-6.599v_d}{1.4\text{}\text{k}\Omega }+\frac{v_d}{1.4\text{}\text{k}\Omega }-\frac{6.599v_d}{15\text{}\Omega }-\frac{A\times v_d}{\text{15 }\Omega }+\frac{2.969}{1.4\times {10}^3\Omega }-\frac{450\times {10}^{-3}\text{V}}{1.4\times {10}^3\text{}\Omega }+\frac{2.969}{15\text{}\Omega }=0\\ v_d\left(\frac{-6.599}{1.4\text{}\text{k}\Omega }+\frac{1}{1.4\text{}\text{k}\Omega }-\frac{6.599}{\text{15 }\Omega }-\frac{A}{\text{15 }\Omega }\right)+2.121\times {10}^{-3}-3.214\times {10}^{-4}+0.198=0\end{array}$

$v_d\left(-4.713\times {10}^{-3}+7.14\times {10}^{-4}-0.439-A\right)+0.1997=0$ (8)

Substitute ${10}^6\text{V/V}$ for $A$ in equation (8).

$\begin{array}{l}{v}_d\left(-4.713\times {10}^{-3}+7.14\times {10}^{-4}-0.439-{10}^6\right)+0.1997=0\\ v_d\left(-1000000.443\right)+0.1997=0\\ v_d=\frac{0.1997}{1000000.443}\text{V}\\ v_d=1.99\times {10}^{-7}\text{V}\end{array}$

Substitute $1.99\times {10}^{-7}\text{V}$ for $v_d$ in equation (3).

$\begin{array}{l}1.99\times {10}^{-7}\text{V}=\left(i_d\times 10\text{T}\Omega \right)\\ 1.99\times {10}^{-7}\text{V}=\left(i_d\times 10\times {\text{10}}^{12}\Omega \right)\left\{\because 1\text{T}\Omega =\text{1}\times {\text{10}}^{12}\Omega \right\}\end{array}$

Rearrange for $i_d$,

$i_d=1.996\times {10}^{-20}\text{A}$

Conclusion:

Thus, the differential voltage $v_d$ of $\text{AD549K}$ is $1.99\times {10}^{-7}\text{V}$ and input bias current $i_d$ is $1.996\times {10}^{-20}\text{A}$.