Chapter 3, Problem 3.25QAP

Interpretation Introduction

(a)

Interpretation:

The function of the operational amplifier 1 should be determined.

Concept introduction:

The operational amplifier is a high gain device and it amplifies the voltage. Generally, a voltage amplifier has a single output and differential input. So, basically, an operational amplifier produces an output much higher than the input voltage. The output voltage could be thousands of times higher than the input voltage.

Here, the given electronic circuit is as below:

Explanation of Solution

Given information:

The operational amplifier is a high gain device and it amplifies the voltage. Generally, a voltage amplifier has a single output and differential input. So basically, an operational amplifier produces an output much higher than the input voltage. The output voltage could be thousands of times higher than the input voltage.

Here the given electronic circuit is as below:

It can be seen that the operational amplifier 1 here performing the function of inverting voltage amplifier.

Interpretation Introduction

(b)

Interpretation:

The function of the operational amplifier 2 should be determined.

Concept introduction:

The operational amplifier is a high gain device and it amplifies the voltage. Generally, a voltage amplifier has a single output and differential input. So, basically, an operational amplifier produces an output much higher than the input voltage. The output voltage could be thousands of times higher than the input voltage.

Here the given electronic circuit is as below

Explanation of Solution

The operational amplifier is a high gain device and it amplifies the voltage. Generally, a voltage amplifier has a single output and differential input. So basically, an operational amplifier produces an output much higher than the input voltage. The output voltage could be thousands of times higher than the input voltage.

Here the given electronic circuit is as below:

It can be seen that the operational amplifier 2 here performing the function of integrating circuit and hence the integration.

Interpretation Introduction

(c)

Interpretation:

The output voltage during the interval $\Delta t1$ should be described and plotted. Where $\Delta t1$ is the first period when switches $S_1$ and $S_2$ are closed and switch $S_4$ opens. The function of the operational amplifier 2.

Concept introduction:

The operational amplifier is a high gain device and it amplifies the voltage. Generally, a voltage amplifier has a single output and differential input. So, basically, an operational amplifier produces an output much higher than the input voltage. The output voltage could be thousands of times higher than the input voltage.

Here the given electronic circuit is as below

Explanation of Solution

The operational amplifier is a high gain device and it amplifies the voltage. Generally, a voltage amplifier has a single output and differential input. So basically, an operational amplifier produces an output much higher than the input voltage. The output voltage could be thousands of times higher than the input voltage.

Here the given electronic circuit is as below:

Here the input voltage $v_1$ is linearly increasing voltage. This is given by the equation

$v_1=at+c$

Where $a$ and $c$ are constants and $t$ is time.

As per the problem switches $S_1$ and $S_2$ are closed and switch $S_4$ opens hence the circuit acts as an integrating circuit which amplifies the output.

Here, the output signal can be given by the formula as below

$\begin{array}{c}{v}_o=\frac{1}{RC}{\displaystyle \underset{0}{\overset{t}{\int }}{v}_1}dt\\ =\frac{1}{RC}{\displaystyle \underset{0}{\overset{t}{\int }}(at+c)}dt\\ =\frac{1}{RC}\times \frac{{(at+c)}^2}{2a}\end{array}$

Now assume $RC=\frac{1}{2},a=1,c=0$.

Substitute the values in the above equation and simplify, the result obtained is

$v_o=t^2$

The integrating voltage at different time intervals is given as:

Time(s)	Output Voltage (V)
1	1
4	16
6	36
8	64
11	121
20	40

The plot can be obtained as shown below:

Interpretation Introduction

(d)

Interpretation:

The output voltage during the second interval $\Delta t$ should be described and plotted. Where $\Delta t$ is the second period when switches $S_2$ opens and $S_3$ closes.

Concept introduction:

The operational amplifier is a high gain device and it amplifies the voltage. Generally, a voltage amplifier has a single output and differential input. So, basically, an operational amplifier produces an output much higher than the input voltage. The output voltage could be thousands of times higher than the input voltage.

Here the given electronic circuit is as below

Explanation of Solution

The operational amplifier is a high gain device and it amplifies the voltage. Generally, a voltage amplifier has a single output and differential input. So basically, an operational amplifier produces an output much higher than the input voltage. The output voltage could be thousands of times higher than the input voltage.

Here the given electronic circuit is as below

Here the input voltage $v_1$ is linearly increasing voltage. This is given by the equation

$v_1=at+c$

Where $a$ and $c$ are constants and $t$ is time.

As per the problem switches $S_2$ opens and $S_3$ closes hence the circuit acts as an combination of the inverting voltage amplifier and integrating circuit.

Here the output signal can be given by the formula as below

$v_o=\frac{R_f}{R_i}{v}_i$

Where $R_f$ and $R_i$ are resistance. Also assume that $R_f=R_i$.

Therefore, it can be said that

$v_o=-v_i$

The output of the second operational amplifier which is actually the final output can be given by the equation as below:

$v_o=-\frac{1}{RC}\times \frac{{\left(at+c\right)}^2}{2a}$

Where $a$ and $c$ are constants.

Now assume $RC=\frac{1}{2},a=1,c=0$.

Substitute the values in the above equation and simplify, the result obtained is

$v_o=-t^2$

The integrating voltage at different time intervals is given as:

Time(s)	Output Voltage (V)
1	-1
4	-16
6	-36
8	-64
11	-121
20	-40

The plot can be obtained as shown below:

Interpretation Introduction

(e)

Interpretation:

The output voltage $V_o$ at the end of the measurement cycle is given by $vo=k\times$ input rate should be shown.

Concept introduction:

The operational amplifier is a high gain device and it amplifies the voltage. Generally, a voltage amplifier has a single output and differential input. So, basically, an operational amplifier produces an output much higher than the input voltage. The output voltage could be thousands of times higher than the input voltage.

Here the given electronic circuit is as below

Explanation of Solution

The operational amplifier is a high gain device and it amplifies the voltage. Generally, a voltage amplifier has a single output and differential input. So basically, an operational amplifier produces an output much higher than the input voltage. The output voltage could be thousands of times higher than the input voltage.

Here the given electronic circuit is as below

Since there is no current so capacitor will discharge. And circuit will act as differentiator.

$\frac{v_o}{R_f}=C\times \frac{d{v}_i}{dt}$

Further, it can be re-written as

$v_o=-R_{f}C\times \frac{d{v}_i}{dt}$

The output can be given as

$v_o=k\times \frac{d{v}_i}{dt}$

Interpretation Introduction

(f)

Interpretation:

The advantages and disadvantages of the given circuit over the normal amplifier should be described.

Concept introduction:

The operational amplifier is a high gain device and it amplifies the voltage. Generally, a voltage amplifier has a single output and differential input. So, basically, an operational amplifier produces an output much higher than the input voltage. The output voltage could be thousands of times higher than the input voltage.

Here the given electronic circuit is as below

Explanation of Solution

The operational amplifier is a high gain device and it amplifies the voltage. Generally, a voltage amplifier has a single output and differential input. So basically, an operational amplifier produces an output much higher than the input voltage. The output voltage could be thousands of times higher than the input voltage.

Here the given electronic circuit is as below

Since there is no current so capacitor will discharge. And circuit will act as differentiator.

In general, the normal operational amplifier circuit can be used as a differentiator circuit only. But the given integrating circuit can be used as differentiator as well as integrator. Just one disadvantage is that the given circuit has higher noise level as compared to the normal circuit.

Interpretation Introduction

(g)

Interpretation:

The scenario if the input signal gets changes with the slope during the measurement cycle should be determined.

Concept introduction:

The operational amplifier is a high gain device and it amplifies the voltage. Generally, a voltage amplifier has a single output and differential input. So, basically, an operational amplifier produces an output much higher than the input voltage. The output voltage could be thousands of times higher than the input voltage.

Here the given electronic circuit is as below

Explanation of Solution

The operational amplifier is a high gain device and it amplifies the voltage. Generally, a voltage amplifier has a single output and differential input. So basically, an operational amplifier produces an output much higher than the input voltage. The output voltage could be thousands of times higher than the input voltage.

Here the given electronic circuit is as below

Since there is no current so capacitor will discharge. And circuit will act as differentiator.

The output of the second operational amplifier which is actually the final output can be given by the equation as below:

$v_o=-\frac{1}{RC}\times \frac{{\left(at+c\right)}^2}{2a}$

Where $a$ and $c$ are constants.

Now assume $RC=\frac{1}{2},a=1,c=0$.

Substitute the values in the above equation and simplify, the result obtained is

$v_o=-a{t}^2$

Further it can be written that

$\frac{v_{o1}}{v_{o2}}=-\frac{a_1}{a_2}{t}^2$

Interpretation Introduction

(h)

Interpretation:

The result obtained if the two-time intervals are separated by a time delay $\Delta t_3$ should be determined.

Concept introduction:

The operational amplifier is a high gain device and it amplifies the voltage. Generally, a voltage amplifier has a single output and differential input. So, basically, an operational amplifier produces an output much higher than the input voltage. The output voltage could be thousands of times higher than the input voltage.

Here the given electronic circuit is as below

Explanation of Solution

The operational amplifier is a high gain device and it amplifies the voltage. Generally, a voltage amplifier has a single output and differential input. So basically, an operational amplifier produces an output much higher than the input voltage. The output voltage could be thousands of times higher than the input voltage.

Here the given electronic circuit is as below

For the given circuit the during the first time interval $\Delta t_1$, the output voltage is an integration voltage over a period of $\Delta t_1$. Also, in the same way the output voltage is integrating voltage during the second interval $\Delta t_2$ but in this case the phase is opposite of the previous phase.

Now if the two-time intervals are separated by a time delay $\Delta t_3$ then there will be a sudden change in the output voltage. The change is output voltage can be from $225v$ to $-1.$

Interpretation Introduction

(i)

Interpretation:

The result if the two-time intervals were of different duration should be determined.

Concept introduction:

The operational amplifier is a high gain device and it amplifies the voltage. Generally, a voltage amplifier has a single output and differential input. So, basically, an operational amplifier produces an output much higher than the input voltage. The output voltage could be thousands of times higher than the input voltage.

Here the given electronic circuit is as below

Explanation of Solution

The operational amplifier is a high gain device and it amplifies the voltage. Generally, a voltage amplifier has a single output and differential input. So basically, an operational amplifier produces an output much higher than the input voltage. The output voltage could be thousands of times higher than the input voltage.

Here the given electronic circuit is as below

As its known that, if the two-time intervals are same then the plot of the output voltage is same in magnitude and opposite in sign for both the interval. But if the two-time intervals are of different duration then the magnitude of output signal will be different for both the time periods.

Interpretation Introduction

(j)

Interpretation:

The reason should be discussed for why the time interval is desirable to be as large as possible in measuring enzyme kinematics.

Concept introduction:

The operational amplifier is a high gain device and it amplifies the voltage. Generally, a voltage amplifier has a single output and differential input. So, basically, an operational amplifier produces an output much higher than the input voltage. The output voltage could be thousands of times higher than the input voltage.

Here the given electronic circuit is as below:

Explanation of Solution

The operational amplifier is a high gain device and it amplifies the voltage. Generally, a voltage amplifier has a single output and differential input. So basically, an operational amplifier produces an output much higher than the input voltage. The output voltage could be thousands of times higher than the input voltage.

Here the given electronic circuit is as below

Since there is no current so capacitor will discharge. And circuit will act as differentiator.

The output of the second operational amplifier which is actually the final output can be given by the equation as below:

$v_o=-\frac{1}{RC}\times \frac{{\left(at+c\right)}^2}{2a}$

Where $a$ and $c$ are constants.

Now, assume $RC=\frac{1}{2},a=1,c=0$.

Substitute the values in the above equation and simplify, the result obtained is

$v_o=-a{t}^2$

If the time gets larger then the output voltage also gets larger.

Enzyme kinematics are generally time-consuming process. So, its desirable to have larger time.