13T3X

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\title{\Huge Bioinstrumentation Workshop 2}
\author{\Large Marko Ruslim \qquad\qquad 841736}
\date{\large March 29, 2019 -- \today}

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Partner: John Kounetas
\newline

\hrule

\section*{Statement of Authorship}
This is a statement of authorship stating that I, Marko Ruslim, am indeed the sole author of this workshop report.

\section*{Aim}

To create an instrumentation amplifier and to investigate its properties. In particular, its differential and common mode gains and bandwidths, its CMRR and how changing the circuitry (differential mode gain) affects these properties.

\section*{Materials and Equipment}
\begin{multicols}{2}
\begin{itemize}
    \item Prototype board
    \item DAQ, function generator and oscilloscope
    \item Resistors and wires
    \item LM348 chip
    \item Wire-cutter and screw-driver
    \item LTspice
\end{itemize}
\end{multicols}

\section*{Method}

The method followed is essentially the same as given in the workshop manual. A few changes have been made: for saturation and slewing calculations, the voltage divider has been changed so that saturation can occur. The specifics is detailed in the report. The oscilloscope has been used for the differential mode bandwidth calculations, since high frequencies are needed. However, we thought we needed high frequencies and the oscilloscope for common mode calculations which is incorrect and wasted a lot of time pursuing this. The questions that we were unable to do were done by simulation on LTspice.

\newpage \lfoot{Marko Ruslim}

\section*{Results and Analysis}

\subsection*{7.\quad Unused Op-Amps within a Package}

\subsubsection*{1. The two circuits in figure \ref{fig:c00} were simulated using LTspice.}

\begin{figure}[H]
     \centering
     \begin{subfigure}{0.49\textwidth}
         \centering
         \includegraphics[width=\textwidth]{q1.jpg}
         \caption{Circuit 1}
         \label{fig:c01}
     \end{subfigure}
     \begin{subfigure}{0.49\textwidth}
         \centering
         \includegraphics[width=\textwidth]{q2.jpg}
         \caption{Circuit 2}
         \label{fig:c02}
     \end{subfigure}
        \caption{Circuits to neutralise unused op amps}
        \label{fig:c00}
\end{figure}

\subsubsection*{2.\quad Which circuit produces a $0 \ \mathrm{V}$ output? Explain why the other one does not.}

\begin{figure}[H]
    \centering
    \includegraphics[width=\textwidth]{q1_2.PNG}
    \caption{The transient voltage output of circuit 1 (green) and circuit 2 (blue)}
    \label{fig:w0}
\end{figure}

Circuit 1 outputs a constant zero voltage. Circuit 2 outputs a non-zero voltage. At the 10ms range it appears constant at around 0.0008 fV = $8 \times 10^{-19}$ V. The first circuit has the two op-amp inputs connected together and to ground. Since these two inputs have the same voltage through it, the output voltage is zero in this simulation. The second circuit has negative feedback, and the two inputs of the op-amp are not connected, so the negative input terminal of the op-amp is still able to vary, which it does. But because of negative feedback, the output will act to make the inputs of the op-amp to be equal. So, what is seen are fluctuations near 0 V.\newline

However, in reality there is noise, which LTspice does not consider. Therefore slight changes in the wires in the first circuit could cause output saturation. In the second circuit there is negative feedback to prevent output saturation, so in real life this would be more ideal.

\subsubsection*{3.\quad Explain the importance of neutralising unused op‐amps.}

Neutralising unused op-amps avoids noise and power problems \cite{maxim}. If not neutralised, the op-amp will consume more power and could contribute to noise in the other circuits. Noise contributions to other circuits could be brought about by fluctuations in current inducing electro-magnetic fields.

\subsubsection*{4.\quad What are the typical and maximum input offset voltages of the LM348?}

For the LM348, its typical input offset voltage is $1 \ \mathrm{mV}$ and its maximum input offset voltage is $6 \ \mathrm{mV}$. These were tested at $25\degree \mathrm{C}$ and with $\mathrm{V_{CC\pm}=\pm15 \ \mathrm{V}}$. With the test condition of full range free-air temperature, the maximum input offset voltage is $7.5 \ \mathrm{mV}$ \cite{lm348}.

\subsection*{Part A\quad Differential Signal Generator}

\subsection*{Test}

\begin{figure}[H]
    \centering
\begin{circuitikz}
    \draw
    (-1.25, 1.5) node [op amp] (opamp1){}
    (opamp1.out) -- ++ (0,1.5) -- ++ (-2.25,0)
        to[short, -o] (-8,3)
            node[left]{$A_\mathrm{out}$}
    (opamp1.-) -- (-4,2)
        to[R, l_=$10 \ \mathrm{k}\Omega$, *-*] (-4,-2)
    (-4,2) to[R, l=$100 \ \mathrm{k}\Omega$] (-6,2)
        to[short, -*] (-6,3)
    (opamp1.+) -- (-6,1)
        to[R, l_=$1 \ \mathrm{k}\Omega$, *-] (-6,-1)
            node[ground]{}
    (-6,1) to[R, l_=$10 \ \mathrm{k}\Omega$, -o] (-8,1)
        node[left]{input}

    (-1.25,-1.5) node [op amp, yscale=-1] (opamp2){}
    (opamp2.out) -- ++ (0,-1.5) -- ++ (-2.25,0)
    (opamp2.+) -- (-3,-1) node[ground]{}
    (opamp2.-) -- (-4,-2)
        to[R, l=$110 \ \mathrm{k}\Omega$] (-6,-2)
        to[short, -*] (-6,-3)
    (-2.25,-3) to[short, -o] (-8,-3)
        node[left]{$B_\mathrm{out}$}

    (1.25, 1.5) node [op amp, xscale=-1] (opamp3){}
    (opamp3.out) -- ++ (0,1.5) -- ++ (2.25,0)

    (1.25, -1.5) node [op amp, xscale=-1, yscale=-1] (opamp4){}
    (opamp4.out) -- ++ (0,-1.5) -- ++ (2.25,0)

    (-2.25,-3.25) -- (2.25, -3.25) -- (2.25, 3.25) -- (-2.25, 3.25) -- (-2.25,-3.25)
    (-2.25, 0) node[vcc, rotate=90] {+15V}
    (2.25, 0) node[vee, rotate=90] {-15V}
    ;
\end{circuitikz}
    \caption{Differential signal generator schematic, with unused op-amps neutralised}
    \label{fig:c1}
\end{figure}

\subsubsection*{2.\quad Confirm your unused op‐amps output $0 \ \mathrm{V}$ using DMM.}

The unused op-amps were neutralised and they output values close to 0 V using the DMM.

\subsubsection*{4.\quad Apply a $1 \ \mathrm{kHz}$ sine wave with $1.1 \ \mathrm{V}$ amplitude, use oscilloscope to measure voltage.}

This can be seen as the green waveform in figure \ref{fig:o1}.

\subsubsection*{5.\quad Confirm the circuit is functioning by looking at the outputs A and B using an oscilloscope.}

\begin{figure}[H]
    \centering
    \includegraphics[width=\textwidth]{1q5.png}
    \caption{Waveforms of the voltages at $A_\mathrm{out}$ (green) and $B_\mathrm{out}$ (blue); the input is in phase with $A_\mathrm{out}$}
    \label{fig:o1}
\end{figure}

The circuit is functioning, two regular sine waves of the same amplitude and out of phase by 180\degree \ can be seen.

\subsubsection*{6.\quad By what factor is the input voltage reduced when it travels through the voltage divider $\mathrm{R}_4$ and
$\mathrm{R}_5$? Use Ohm’s Law to confirm.}

We are assuming ideal op-amps; there is infinite impedance at the non-inverting input terminal. Ohm's Law is $v = iR$. The current through both resistors is the same.

\begin{align*}
    & V_\mathrm{in} = i \times (10 \ \mathrm{k}\Omega + 1 \ \mathrm{k}\Omega) & V_\mathrm{p} = i \times 1 \ \mathrm{k}\Omega &
\end{align*}

\begin{equation*}
    \frac{V_\mathrm{in}}{V_\mathrm{p}} = \frac{10+1}{1} = 11 \ V/V
\end{equation*}

The input voltage is reduced by a factor of 11 as it passes through the voltage divider.

\subsubsection*{7.\quad Measure the gain of both op‐amps. Does this agree with the simulated results in from Workshop One?}

When $V_\mathrm{in}$ is at its peak, $V_\mathrm{in} = 0.55 \ V$,\newline

From figure \ref{fig:o1} and since the input is in phase with $A_\mathrm{out}$, when $V_\mathrm{in} = 0.55 \ V$:

\begin{align*}
    & V_\mathrm{A} = 0.55 \ V & V_\mathrm{B} = -0.55 \ V & \\
    & \frac{V_\mathrm{A}}{V_\mathrm{in}} = \frac{0.55}{0.55} = 1 \ V/V & \frac{V_\mathrm{B}}{V_\mathrm{in}} = \frac{-0.55}{0.55} = -1 \ V/V &
\end{align*}

The op-amp circuit in workshop 1 did not involve a voltage divider. If the voltage divider were removed, then a gain of 11 V/V would be seen, which is the results we get in workshop 1.

\subsubsection*{10.\quad At what frequency does slewing occur?}

The voltage divider ($\mathrm{R}_4$ and $\mathrm{R}_5$) has been changed so that saturation occurs. Here, $\mathrm{R}_5$ has changed from $1 \ \mathrm{k}\Omega$ to $10 \ \mathrm{k}\Omega$. Also the input voltage has been increased until saturation can be seen.

\begin{figure}[H]
    \centering
    \includegraphics[width=\textwidth]{1q10a.png}
    \caption{Waveforms of the voltages at $A_\mathrm{out}$ (green) and $B_\mathrm{out}$ (blue) with the voltage divider changed; this demonstrates output saturation}
    \label{fig:o2}
\end{figure}

Next, the frequency of the input has been increased until slewing occurs.

\begin{figure}[H]
    \centering
    \includegraphics[width=\textwidth]{1q10b.png}
    \caption{Waveforms of the voltages at $A_\mathrm{out}$ (green) and $B_\mathrm{out}$ (blue) with the voltage divider changed; this demonstrates slewing}
    \label{fig:o3}
\end{figure}

Slewing can be seen as there is a linear change in voltage over time that represents the maximum rate of change of output voltage that the op-amp can produce. We began to see this at $8 \ \mathrm{kHz}$.

\subsubsection*{11.\quad Slew rate is equal to the positive slope of the sawtooth wave seen. Measure the slope and record it as the slew rate with units $\mathrm{V}/\mu \mathrm{s}$.}

Frequency has been increased so that slewing is more visible.

\begin{figure}[H]
    \centering
    \includegraphics[width=\textwidth]{1q11.png}
    \caption{Waveforms of the voltages at $A_\mathrm{out}$ (green) and $B_\mathrm{out}$ (blue) with the voltage divider changed; slew rate can be calculated from two points in the linear region}
    \label{fig:o4}
\end{figure}

\begin{equation*}
    \mathrm{SR}= \left. \frac{\mathrm{d}v_o}{\mathrm{d}t}\right|_{\mathrm{max}} = \frac{4.05 \ V - (-3.73 \ V)}{15 \ \mu\mathrm{s}} = 0.519 \ V/\mu\mathrm{s}
\end{equation*}

\subsubsection*{12.\quad Compare with the slew rate given in the manufacturer’s datasheet. Explain why your result is different.}

From the data sheet, the slew rate is $0.5 \ V/\mu\mathrm{s}$ \cite{lm348}. This is very close to our measured slew rate. In fact they would be identical if we only looked at one significant figure. Differences could arise due to different test conditions. The data sheet measured the slew rate with an op-amp with unity gain at $\mathrm{T_A} = 25\degree \mathrm{C}$, and also used a load resistor and load capacitor. Whereas our op-amp did not have a load at the outputs and the temperature at the time was not known. Also, some error could be introduced by the oscilloscope.

\subsection*{7\quad Two Op‐Amp Instrumentation Amplifier}

\begin{figure}[H]
    \centering
\begin{circuitikz}[scale = 0.85, transform shape]
    \draw
    (-1.25, 1.5) node [op amp] (opamp1){}
    (opamp1.out) -- ++ (0,1.5) -- ++ (-2.25,0)
        to[short, -o] (-7,3)
            node[left]{$A_\mathrm{out}$}
    (opamp1.-) -- (-4,2)
        to[R, l_=$10 \ \mathrm{k}\Omega$, *-*] (-4,-2)
    (-4,2) to[R, l=$100 \ \mathrm{k}\Omega$] (-6,2)
        to[short, -*] (-6,3)
    (opamp1.+) -- (-6,1)
        to[R, l_=$1 \ \mathrm{k}\Omega$, *-] (-6,-1)
            node[ground]{}
    (-6,1) to[R, l_=$10 \ \mathrm{k}\Omega$, -o] (-8,1)
        node[left]{input}

    (-1.25,-1.5) node [op amp, yscale=-1] (opamp2){}
    (opamp2.out) -- ++ (0,-1.5) -- ++ (-2.25,0)
    (opamp2.+) -- (-3,-1) node[ground]{}
    (opamp2.-) -- (-4,-2)
        to[R, l=$110 \ \mathrm{k}\Omega$] (-6,-2)
        to[short, -*] (-6,-3)
    (-2.25,-3) to[short, -o] (-7,-3)
        node[left]{$B_\mathrm{out}$}

    (1.25, 1.5) node [op amp, xscale=-1] (opamp3){}
    (opamp3.out) -- ++ (0,1.5) -- ++ (2.25,0)
    (-3,3) to[short, *-] (-3,4)
        -- (3,4) -- (3,1) -- (opamp3.+)
    (opamp3.-) -- (4,2)
        to [R, l=$100 \ \mathrm{k}\Omega$] (6,2) to [short, -*] (6,3)
    (2.25,3) to [short, -o] (7,3) node[right] {output}

    (1.25, -1.5) node [op amp, xscale=-1, yscale=-1] (opamp4){}
    (opamp4.out) -- ++ (0,-1.5) -- ++ (2.25,0)
    (-3,-3) to [short, *-] (-3,-4)
        -- (3,-4) -- (3,-1) -- (opamp4.+)
    (2.25,-3) -- (4,-3)
        to [R, l=$10 \ \mathrm{k}\Omega$] (6,-3)
        to [short, -*] (6,-2)
    (opamp4.-) -- (6,-2)
        to [R, l=$100 \ \mathrm{k}\Omega$] (8,-2)
    (8,-2) node[ground] {}
    (4,-3) to[R, l_=$10 \ \mathrm{k}\Omega$, *-*] (4,2)

    (-2.25,-3.25) -- (2.25, -3.25) -- (2.25, 3.25) -- (-2.25, 3.25) -- (-2.25,-3.25)
    (-2.25, 0) node[vcc, rotate=90] {+15V}
    (2.25, 0) node[vee, rotate=90] {-15V}
    ;
\end{circuitikz}
    \caption{Two op‐amp instrumentation amplifier schematic}
    \label{fig:c2}
\end{figure}

\subsubsection*{4.\quad Observe the signal at the output of the instrumentation amplifier. Is it what you expected?}

\begin{figure}[H]
    \centering
    \includegraphics[width=\textwidth]{2q4.png}
    \caption{Waveforms of the voltages at $A_\mathrm{in}$ (green) and output (blue)}
    \label{fig:o5}
\end{figure}

In this figure, the green line is $A_\mathrm{in}$. Introducing both plots onto the figure introduced some offset / $y$-translation. However, if the output graph is centred, then an amplitude of 1.1 V is seen, which is what is expected. Also the output is in phase with the voltage at $A$.

\subsubsection*{5.\quad Switch the $\mathrm{A_{in}}$ and $\mathrm{B_{in}}$ inputs of the instrumentation amplifier, and again observe the signal at the output.}

\begin{figure}[H]
    \centering
    \includegraphics[width=\textwidth]{2q5.png}
    \caption{Waveforms of the voltages at $A_\mathrm{in}$ (green) and output (blue); $A_\mathrm{in}$ and $B_\mathrm{in}$ have been swapped}
    \label{fig:o6}
\end{figure}

The shape of the output is the same as in the previous question: a sine wave with 1.1 V amplitude at frequency 1 kHz. However, the output is now out of phase by 180\degree with $A$. This is what is expected.

\subsection*{7.1\quad Differential Gain and Bandwidth}

\subsubsection*{1.\quad Measure the gain of the instrumentation amplifier $\frac{\mathrm{V_o}}{\mathrm{V_{A_{in}}} - \mathrm{V_{B_{in}}}}$. Since it is receiving a differential signal input, we call this the differential gain $\mathrm{A_d}$.}

\begin{figure}[H]
    \centering
    \includegraphics[width=\textwidth]{3q1.png}
    \caption{Waveforms of the voltages of $A_\mathrm{in}-B_\mathrm{in}$ (green) and output (blue)}
    \label{fig:o7}
\end{figure}

We chose to measure peak to peak voltage (because that is what the oscilloscope measures). The peak to peak voltage of $A_\mathrm{in}-B_\mathrm{in}$ is $201.12 \ \mathrm{mV}$, the voltage at output is $2.209 \ \mathrm{V}$. (Overall it does not matter as the waveforms are symmetric and so its peak is half of these values). Therefore, the differential gain is:

\begin{equation*}
    \mathrm{A_d} = \frac{\mathrm{V_o}}{\mathrm{V_{A_{in}}} - \mathrm{V_{B_{in}}}} = \frac{2.209}{0.20112} = 10.98 \ V/V
\end{equation*}

\subsubsection*{2.\quad Compare the measured differential gain to that predicted theoretically.}

\begin{figure}[H]
    \centering
\begin{circuitikz}[scale = 1, transform shape]
    \draw
    (0,0) node[ground]{}
        to[R=$\mathrm{R_6}$, -*] ++ (2,0) node[below]{$\mathrm{V_{B}}$}
        to[R=$\mathrm{R_7}$, -*] ++ (2,0)
        to[R=$\mathrm{R_8}$, -*] ++ (2,0) node[below]{$\mathrm{V_{A}}$}
        to[R=$\mathrm{R_9}$, -*] ++ (2,0)
        to [short, f<=$i$, -o] ++(1,0) node[right]{output}
    ;
\end{circuitikz}
    \caption{Over-simplified circuit, under the assumption of ideal op-amps}
    \label{fig:c2.5}
\end{figure}

Under the assumptions of ideal op-amps, there is infinite input impedance, so current does not flow into the inputs of the op-amps. Also, there is negative feedback occurring due to the connection between negative input terminals to output terminals of each op-amp. This means that the voltages at the input terminals of the op-amp are equal (virtual short circuit), and this has also been labelled in the circuit diagram. Applying nodal voltage calculations:

\begin{equation*}
\begin{aligned}[c]
\frac{\mathrm{V_B}}{\mathrm{R_6}} &= i\\
\Rightarrow\mathrm{V_B}&=100i
\end{aligned}
\qquad\qquad
\begin{aligned}[c]
\frac{\mathrm{V_A}}{\mathrm{R_6+R_7+R_8}} &= i\\
\Rightarrow\mathrm{V_A}&=120i
\end{aligned}
\qquad\qquad
\begin{aligned}[c]
\frac{\mathrm{V_o}}{\mathrm{R_6+R_7+R_8+R_9}} &= i\\
\Rightarrow\mathrm{V_o}&=220i
\end{aligned}
\end{equation*}

\begin{equation*}
    \therefore\frac{\mathrm{V_o}}{\mathrm{V_A-V_B}} = \frac{220i}{120i-100i} = 11 \ V/V
\end{equation*}

This is very similar to what is obtained experimentally.

\subsubsection*{3.\quad Slowly reduce the input frequency and confirm the differential gain does not decrease even at very low frequencies (i.e. the low break‐frequency, $f_\mathrm{L} = 0\mathrm{Hz}$).}

\begin{figure}[H]
    \centering
    \includegraphics[width=\textwidth]{3q3.png}
    \caption{Waveforms of the voltages of $A_\mathrm{in}-B_\mathrm{in}$ (green) and output (blue) with input at a low frequency}
    \label{fig:o8}
\end{figure}

Here at $f = 500 \ \mathrm{mHz}$, the differential gain is $\mathrm{A_d} = 11.0$. The differential gain does not decrease at very low frequencies.

\subsubsection*{6.\quad Determine $\frac{\mathrm{A_d}}{\sqrt{2}}$ ($\mathrm{A_d}=$ differential gain measured in step 1) Note: $\frac{1}{\sqrt{2}} = 70\%$}

\begin{equation*}
    \frac{\mathrm{A_d}}{\sqrt{2}} = 7.75 \ V/V
\end{equation*}

\subsubsection*{7.\quad Increase input frequency until a gain of $\frac{\mathrm{A_d}}{\sqrt{2}}$ is found. Mark this frequency as the high break
frequency $f_\mathrm{H}$.}

\begin{figure}[H]
    \centering
    \includegraphics[width=0.8\textwidth]{3q7.jpg}
    \caption{Waveforms of output voltage (CH1) and $A_\mathrm{in}$ (CH2) at 30 kHz}
    \label{fig:o9}
\end{figure}

We were not able to measure $A_\mathrm{in}-B_\mathrm{in}$ in CH2 on the oscilloscope, we assume the peak to peak voltage is the same as previously measured.\newline

At this frequency of 30 kHz:

\begin{equation*}
    \mathrm{A_d} = \frac{1.52}{0.20359} = 7.56 \ V/V
\end{equation*}

This was the closest value we had to $\frac{\mathrm{A_d}}{\sqrt{2}}$. The high break frequency is therefore (slightly less than) 30 kHz. In hindsight, we should have recorded all the voltage data with each frequency we tested to create a Bode plot. (We did this in the other circuits.)

\subsubsection*{8.\quad As a check, measure also the phase shift between the input and output waveforms.}

The phase shift should be 45\degree. However, in the oscilloscope, we did not observe this phase shift. Especially since the input voltage was very noisy. We simulated this on LTspice and observed a 45\degree \ phase shift.

\subsubsection*{9.\quad Determine the differential mode bandwidth of the amplifier: $\mathrm{BW} = f_\mathrm{H} - f_\mathrm{L}$}

\begin{align*}
    f_\mathrm{H} & = 30 \ \mathrm{kHz}, \qquad f_\mathrm{L} = 0 \ \mathrm{kHz} \\
    \therefore \mathrm{BW} & = 30 \ \mathrm{kHz}
\end{align*}

\subsection*{7.2\quad Common Mode Gain, Bandwidth and CMRR}

\begin{figure}[H]
    \centering
\begin{circuitikz}[scale = 1, transform shape]
    \draw
    (-1.25, 1.5) node [op amp] (opamp1){}
    (opamp1.out) -- ++ (0,1.5) -- ++ (-2.25,0)

    (-1.25,-1.5) node [op amp, yscale=-1] (opamp2){}
    (opamp2.out) -- ++ (0,-1.5) -- ++ (-2.25,0)

    (1.25, 1.5) node [op amp, xscale=-1] (opamp3){}
    (opamp3.out) -- ++ (0,1.5) -- ++ (2.25,0)
    (opamp3.-) -- (4,2)
        to [R, l=$100 \ \mathrm{k}\Omega$] (6,2) to [short, -*] (6,3)
    (2.25,3) to [short, -o] (7,3) node[right] {output}

    (1.25, -1.5) node [op amp, xscale=-1, yscale=-1] (opamp4){}
    (opamp4.out) -- ++ (0,-1.5) -- ++ (2.25,0)
    (2.25,-3) -- (4,-3)
        to [R, l=$10 \ \mathrm{k}\Omega$] (6,-3)
        to [short, -*] (6,-2)
    (opamp4.-) -- (6,-2)
        to [R, l=$100 \ \mathrm{k}\Omega$] (8,-2)
    (8,-2) node[ground] {}
    (4,-3) to[short, *-] (4,-2)
        to[R, l_=$10 \ \mathrm{k}\Omega$, -*] (4,2)

    (opamp3.+) -- (6,1) -- (6,-1) -- (opamp4.+)
    (6,0) to[short, *-o] (7,0) node[right]{input}

    (-2.25,-3.25) -- (2.25, -3.25) -- (2.25, 3.25) -- (-2.25, 3.25) -- (-2.25,-3.25)
    (-2.25, 0) node[vcc, rotate=90] {+15V}
    (2.25, 0) node[vee, rotate=90] {-15V}
    ;
\end{circuitikz}
    \caption{Two op‐amp instrumentation amplifier schematic}
    \label{fig:c3}
\end{figure}

\subsubsection*{3.\quad Measure the common mode gain}

\begin{figure}[H]
    \centering
    \includegraphics[width=\textwidth]{4q3.png}
    \caption{Waveforms of output voltage (green) and input (blue)}
    \label{fig:o10}
\end{figure}

The common mode gain, $\mathrm{A_c}$ is:

\begin{equation*}
    \mathrm{A_c} = \frac{\mathrm{V_o}}{\frac{1}{2}\left(\mathrm{V_{A_{in}}} + \mathrm{V_{B_{in}}}\right)} = \frac{\mathrm{V_o}}{\mathrm{V_{in}}} = \frac{0.10127}{2.497} = 0.041 \ V/V
\end{equation*}

In decibels, this is $20\log(0.041) = -27.8 \ \mathrm{dB}$

\subsubsection*{4.\quad Compare the measured common mode gain with that calculated in the preliminary work. Explain why it is different.}

\begin{figure}[H]
    \centering
\begin{circuitikz}[scale = 1, transform shape]
    \draw
    (0,0) node[ground]{}
        to[R=$\mathrm{R_6}$, -*] ++ (2,0) node[below]{$\mathrm{V_{B}}$}
        to[R=$\mathrm{R_7}$, -*] ++ (2,0)
        to[R=$\mathrm{R_8}$, -*] ++ (2,0) node[below]{$\mathrm{V_{A}}$}
        to[R=$\mathrm{R_9}$, -*] ++ (2,0)
        to [short, f<=$i$, -o] ++(1,0) node[right]{output}
    ;
\end{circuitikz}
    \caption{The same circuit diagram as in \ref{fig:c2.5}}
    \label{fig:c2.6}
\end{figure}

Here, $\mathrm{V_A} = \mathrm{V_B}$, under the assumptions of ideal op-amps and virtual short circuit by negative feedback. Using nodal voltage equations

\begin{equation*}
    \frac{\mathrm{V_A-V_B}}{\mathrm{R_7+R_8}} = \frac{\mathrm{V_o}}{\mathrm{R_6+R_7+R_8+R_9}}
\end{equation*}

This means that $\mathrm{V_o}$ should be theoretically 0 V. Therefore the ideal $\mathrm{A_{cm}} = 0 \ V/V$. However, this model does not explain how $\mathrm{R_6}$ has a voltage across it.  Nevertheless, the ideal common mode gain is 0 V/V.\newline

The ideal common mode gain is similar to the common mode gain measured in the circuit. Deviations from this can come from external noise such as from the mains power, and resistance values not exact (they have a tolerance level).

\subsubsection*{5.\quad Measure the common‐mode bandwidth.}

\begin{figure}[H]
    \centering
    \includegraphics[width=\textwidth]{4q5.png}
    \caption{Bode plot of the common mode gain using the oscilloscope}
    \label{fig:m1}
\end{figure}

Figure \ref{fig:m1} was obtained by measuring the voltages using the oscilloscope. Unfortunately this figure is incorrect because the results here do not agree with the common mode gain obtained using the prototype board and the gain should not be greater than $1 \ V/V$. In hindsight, we should not have measured from higher frequencies. We simulated this circuit on LTspice and obtained a break frequency at 1.15 Hz.



\subsubsection*{6.\quad Determine the Common‐Mode Rejection Ratio}

\begin{equation*}
    \mathrm{CMRR} = \frac{\mathrm{A_d}}{\mathrm{A_c}} = \frac{10.98}{0.041} = 267.8
\end{equation*}

\subsubsection*{7.\quad Is it better to have an instrumentation amplifier with a high or low CMRR?}

It is better for an instrumentation amplifier to have high CMRR. This ensures that the differences between input voltages are amplified to a much greater extent than equal voltages and noise is amplified. The CMRR is a measure of the rejection of the common mode signal, or the signal that is similar to both inputs. This is because the desired signal is the differential mode. A low CMRR would mean that the common mode signal is amplified to some extent, and is undesirable.

\subsection*{7.3\quad Adjustable Gain}

\begin{figure}[H]
    \centering
    \includegraphics[width=\textwidth]{q73.png}
    \caption{Modified instrumentation amplifier}
    \label{fig:c4}
\end{figure}

\newpage

\subsubsection*{2.\quad Determine the following:}

\paragraph*{a. and b.\quad Differential‐mode gain and bandwidth}\text{}

\begin{figure}[H]
    \centering
    \includegraphics[width=\textwidth]{5q2b.png}
    \caption{Bode plot of differential gain for the modified instrumentation amplifier}
    \label{fig:b1}
\end{figure}

This was the data we obtained from our circuit board and oscilloscope. At low frequency (100 Hz), the gain in decibels is 33.4 dB. Therefore, the differential gain is $\mathrm{A_d} = 46.81 \ V/V$. From figure \ref{fig:b1}, the 3 dB break frequency occurs at roughly 20 kHz. We could have made more measurements to confirm this but we were under time pressure.

Therefore we tried to confirm this on LTspice:

\begin{figure}[H]
    \centering
    \includegraphics[width=\textwidth]{73q2b.PNG}
    \caption{Bode plot of differential gain for the modified instrumentation amplifier on LTspice}
    \label{fig:73q2b}
\end{figure}

\begin{figure}[H]
    \centering
    \includegraphics[width=0.5\textwidth]{73q2b2.PNG}
    \caption{Data points for the bode plot in figure \ref{fig:73q2b}}
    \label{fig:73q2b2}
\end{figure}

In our LTspice simulations, the gain is given by the ratio of the output voltage to the input voltage. The gain we are looking for is not the input voltage but $\mathrm{V_A-V_B}$. This is a simple correction because their difference is essentially $2\times$ input voltage, since the op-amps are unity gain and opposite phase. In log-scale, this is a subtraction of $20\log(2) = 6.02 \ \mathrm{dB}$. Since this is a linear change in log-scale, the 3 dB break frequency does not change. Lastly, this correction only applies to differential mode gain, and not common mode gain. This calculation will be done automatically without further mention.\newline

From LTspice, the differential mode gain is $(35.8 - 6.0) \ \mathrm{dB} = 29.8 \ \mathrm{dB}$ which is 31.0 V/V. Also, the break frequency is 20.8 kHz. These values are close to our experimental values.

\paragraph{c. and d.\quad Common mode gain and bandwidth}\text{}

\begin{figure}[H]
    \centering
    \includegraphics[width=\textwidth]{5q2d.png}
    \caption{Bode plot of common gain for the modified instrumentation amplifier}
    \label{fig:5q2d}
\end{figure}

Again, here we made the mistake of looking at extremely high frequencies for the common mode bandwidth. Unfortunately we spent too much time on this part and were not able to complete the remaining sections.

\begin{figure}[H]
    \centering
    \includegraphics[width=\textwidth]{73q2d.PNG}
    \caption{Bode plot of common gain for the modified instrumentation amplifier on LTspice}
    \label{fig:73q2d}
\end{figure}

\begin{figure}[H]
    \centering
    \includegraphics[width=0.5\textwidth]{73q2d2.PNG}
    \caption{Data points for the bode plot in figure \ref{fig:73q2d}}
    \label{fig:73q2d2}
\end{figure}

We simulated the circuit on LTspice. The common mode gain here is -86.8 dB at low frequencies. Therefore $\mathrm{A_c} = 4.59\times 10^{-5} \ V/V$. The break frequency occurs at 1.15 Hz. This is because the common mode gain remains roughly constant from 0 Hz to this break frequency. At frequencies higher than this, the common mode gain increases, plateaus and has a weird pattern thereafter.

\paragraph{e.\quad CMRR}

This is given by the ratio of the differential mode gain to the common mode gain.

\begin{equation*}
    \mathrm{CMRR} = \frac{\mathrm{A_d}}{\mathrm{A_c}} = \frac{46.81}{4.59 \times 10 ^{-5}} = 1.02 \times 10 ^6
\end{equation*}

In decibels, the CMRR is 120.2 dB.

\subsubsection*{3.\quad What effect does $\mathrm{R}_{10}$ have on the circuit?}

\begin{figure}[H]
    \centering
\begin{circuitikz}[scale = 1, transform shape]
    \draw
    (0,0) node[ground]{}
        to[R=$\mathrm{R_6}$, -*] ++ (2,0) node[below]{$\mathrm{V_{B}}$}
        to[R=$\mathrm{R_7}$, -*] ++ (2,0)
        to[R=$\mathrm{R_8}$, -*] ++ (2,0) node[below]{$\mathrm{V_{A}}$}
        to[R=$\mathrm{R_9}$, -*] ++ (2,0)
        to [short, f<=$i_1$, -o] ++(1,0) node[right]{output}
        (6,0) to [short] ++ (0,1)
            to [R, l_=$\mathrm{R_{10}}$] ++ (-4,0) -- ++ (0,-1)
    ;
\end{circuitikz}
    \caption{Simplified circuit diagram of modified instrumentation amplifier}
    \label{fig:c2.7}
\end{figure}

As similar to the previous theoretical questions, this circuit has been created under the assumptions of ideal op-amps, and negative feedback leading to virtual short circuit.

This circuit can be simplified further as there are parallel branches within this circuit.

\begin{figure}[H]
    \centering
\begin{circuitikz}[scale = 1, transform shape]
    \draw
    (0,0) node[ground]{}
        to[R=$\mathrm{R_6}$, -*] ++ (2,0) node[below]{$\mathrm{V_{B}}$}
        to[R=$\mathrm{R_X}$, -*] ++ (4,0) node[below]{$\mathrm{V_{A}}$}
        to[R=$\mathrm{R_9}$, -*] ++ (2,0)
        to [short, f<=$i$, -o] ++(1,0) node[right]{output}
    ;
\end{circuitikz}
    \caption{Simplified circuit diagram of figure \ref{fig:c2.7}}
    \label{fig:c2.8}
\end{figure}

\begin{equation*}
    \frac{1}{\mathrm{R_X}} = \frac{1}{\mathrm{R_7+R_8}} + \frac{1}{\mathrm{R_{10}}}
\end{equation*}

Doing nodal voltage analysis:

\begin{equation*}
\begin{aligned}[c]
\frac{\mathrm{V_B}}{\mathrm{R_6}} &= i\\
\Rightarrow\mathrm{V_B}&=\mathrm{R_6}i
\end{aligned}
\qquad\qquad
\begin{aligned}[c]
\frac{\mathrm{V_A}}{\mathrm{R_6+R_X}} &= i\\
\Rightarrow\mathrm{V_A}&=\left( \mathrm{R_6+R_X}
\right)i
\end{aligned}
\qquad\qquad
\begin{aligned}[c]
\frac{\mathrm{V_o}}{\mathrm{R_6+R_X+R_9}} &= i\\
\Rightarrow\mathrm{V_o}&=\left( \mathrm{R_6+R_X+R_9} \right)i
\end{aligned}
\end{equation*}

Therefore the differential mode gain is:

\begin{equation*}
    \frac{\mathrm{V_o}}{\mathrm{V_A-V_B}} = \frac{\left( \mathrm{R_6+R_X+R_9} \right)i}{\left( \mathrm{R_6+R_X}
\right)i - \mathrm{R_6}i} = 1 + \left(\mathrm{R_6+R_9} \right) \times \frac{1}{\mathrm{R_X}} = 1 + \left(\mathrm{R_6+R_9} \right) \left( \frac{1}{\mathrm{R_7+R_8}} + \frac{1}{\mathrm{R_{10}}} \right)
\end{equation*}

Substituting all the resistor values except $\mathrm{R_{10}}$:

\begin{equation*}
    \mathrm{A_d} = 11 + \frac{200}{\mathrm{R_{10}}}
\end{equation*}

This equation is for  $\mathrm{R_{10}}$ in the units of k$\Omega$. When $\mathrm{R_{10}} = 10 \ \mathrm{k}\Omega$, the theoretical gain is 31 V/V. When $\mathrm{R_{10}} = 5 \ \mathrm{k}\Omega$, the theoretical gain is 51 V/V.  It follows that for small values of $\mathrm{R_{10}}$, the gain would be extremely high; for large values of $\mathrm{R_{10}}$, the differential mode gain would be smaller and tend to 11 V/V (as is the case for if $\mathrm{R_{10}}$ is not present).

\subsubsection*{5.\quad Repeat measurements for the second modified circuit.}

The circuit is the same as in figure \ref{fig:c4}, except $\mathrm{R}_{10} = 5 \ \mathrm{k}\Omega$ instead of $10 \ \mathrm{k}\Omega$.

\paragraph*{a. and b.\quad Differential‐mode gain and bandwidth}\text{}

\begin{figure}[H]
    \centering
    \includegraphics[width=\textwidth]{73q5b.PNG}
    \caption{Bode plot of differential mode gain for the second modified instrumentation amplifier on LTspice}
    \label{fig:73q5b}
\end{figure}

\begin{figure}[H]
    \centering
    \includegraphics[width=0.5\textwidth]{73q5b2.PNG}
    \caption{Data points for the bode plot in figure \ref{fig:73q5b}}
    \label{fig:73q5b2}
\end{figure}

At low frequencies, the differential mode gain is 34.2 dB. Therefore $\mathrm{A_d} = 51.0 \ V/V$. The occurence of the break frequency is when the gain is 3 dB lower than this. This occurs at 14.4 kHz. Therefore the bandwidth is 14.4 kHz, as the lower bound is 0 Hz.

\paragraph{c. and d.\quad Common mode gain and bandwidth}\text{}

\begin{figure}[H]
    \centering
    \includegraphics[width=\textwidth]{73q5d.PNG}
    \caption{Bode plot of common mode gain for the second modified instrumentation amplifier on LTspice}
    \label{fig:73q5d}
\end{figure}

\begin{figure}[H]
    \centering
    \includegraphics[width=0.5\textwidth]{73q5d2.PNG}
    \caption{Data points for the bode plot in figure \ref{fig:73q5d}}
    \label{fig:73q5d2}
\end{figure}

From the bode plot, at low frequencies, the common mode gain is -82.4 dB. Therefore, $\mathrm{A_c} = 7.55 \times 10^{-5} \ V/V$. The break frequency occurs 3 dB above this. This occurs at 1.15 Hz. Therefore the bandwidth is from 0 Hz to 1.15 Hz, so the bandwidth is 1.15 Hz.

\paragraph{e.\quad CMRR}

This is given by the ratio of the differential mode gain to the common mode gain.

\begin{equation*}
    \mathrm{CMRR} = \frac{\mathrm{A_d}}{\mathrm{A_c}} = \frac{51.0}{7.55 \times 10 ^{-5}} = 6.76 \times 10 ^5
\end{equation*}

In decibels, the CMRR is 116.6 dB.

\subsection*{8\quad Investigation}

\begin{figure}[H]
    \centering
    \includegraphics[width=0.8\textwidth]{8q.jpg}
    \caption{Interference source with out signal}
    \label{fig:8q}
\end{figure}

The oscilloscope in figure \ref{fig:8q} has been kindly given to us by Yunzhong, Tiffany and Josh. The waveform is similar in shape to that obtained in workshop one. It demonstrates noise interference at the mains frequency of 50 Hz superimposed onto the output signal of 1 kHz. It demonstrates the importance of considering the impacts of noise interference on signals.

\subsection*{9\quad Report}

\subsubsection*{1.\quad What is the purpose of an instrumentation amp? What applications does it have?}

Instrumentation amplifiers are used when a precise signal needs to be read and amplified. The characteristics of an instrumentation amp greatly suit this device for this purpose. Its high input impedance allows for less noise interference; its differential mode gain can be changed by replacing one resistor in its circuit; large common mode signals are not amplified. As a result, its CMRR is very high.\newline

Instrumentation amplifiers are used for a number of applications. They can be used as a testing and measuring tool of signals. Of importance would be its biomedical application. Physiological signals tend to be very small in nature and heavily susceptible to noise; an EKG is an example. Therefore there is a need for a device that can amplify these small differential signals, and not be distorted by noise or common mode signals: an instrumentation amplifier.

\subsubsection*{2.\quad Present the results in a table.}

\begin{center}
 \begin{tabu}{c|[2pt] c| c|| c| c|| c}
  & $\mathrm{A_d}$ (V/V) & $\mathrm{BW_d}$ (kHz) & $\mathrm{A_c}$ (V/V) & $\mathrm{BW_c}$ (Hz)& CMRR (dB) \\\tabucline[2pt]{-}
 no $\mathrm{R_{10}}$ & 10.98 & 30 & 0.041 & 1.15 & 48.6 \\
 \hline
 $\mathrm{R_{10}} = 10 \ \mathrm{k}\Omega$ & 46.81 & 20 & $4.59\times 10^{-5}$ & 1.15 & 120.2\\
 \hline
$\mathrm{R_{10}} = 5 \ \mathrm{k}\Omega$ & 51.0 & 14.4 & $7.55\times 10^{-5}$ & 1.15 & 116.6\\
\end{tabu}
\end{center}

The results above were measured on the circuit except for the common mode bandwidth. Also the differential mode was not done for $\mathrm{R_{10}} = 5 \ \mathrm{k}\Omega$.\newline

From our simulations on LTspice:

\begin{center}
 \begin{tabu}{c|[2pt] c| c|| c| c|| c}
  & $\mathrm{A_d}$ (V/V) & $\mathrm{BW_d}$ (kHz) & $\mathrm{A_c}$ (V/V) & $\mathrm{BW_c}$ (Hz)& CMRR (dB) \\\tabucline[2pt]{-}
  no $\mathrm{R_{10}}$ & 11.0 & 33.8 & $1.63\times 10^{-5}$ & 1.15 & 116.6\\
 \hline
 $\mathrm{R_{10}} = 10 \ \mathrm{k}\Omega$ & 31.0 & 20.8 & $4.59\times 10^{-5}$ & 1.15 & 116.6\\
 \hline
$\mathrm{R_{10}} = 5 \ \mathrm{k}\Omega$ & 51.0 & 14.4 & $7.55\times 10^{-5}$ & 1.15 & 116.6\\
\end{tabu}
\end{center}

\subsubsection*{3.\quad What is the effect of increasing gain on CMRR and bandwidth? Do you think this should be an important design consideration?}

From our results, increasing $\mathrm{A_d}$ yields a smaller bandwidth for differential mode gain. We only measured one common mode gain, and the signals were heavily influenced by noise. From the LTspice simulation, the common mode gain increases as the differential mode gain increases. Also from the simulation, the common mode bandwidth stays constant at 1.15 Hz. Lastly, our results for CMRR show a weird pattern. First it increases then decreases slightly. From the LTspice simulations, the CMRR remains constant regardless of the differential mode gain.\newline

The effect of differential mode gain on bandwidth is definitely an important design consideration; from our results, CMRR remains constant so there is no consideration regarding this. Having a high differential mode gain is in itself a benefit, but this would yield a low differential mode bandwidth. This means that for frequencies above this bandwidth, the differential gain is no longer constant but a smaller and different value. As a result, the output signal would be distorted, and not representative of the input signals (as the output is affected by frequencies outside the bandwidth). The user has to balance a high enough differential mode gain and a large enough bandwidth for an effective circuit.

\section*{Appendix}

Original circuit.

\subsection*{7.1\quad Differential Mode Bandwidth}

\begin{figure}[H]
    \centering
    \includegraphics[width=\textwidth]{71q7.PNG}
    \caption{Differential Mode Bode plot}
    \label{fig:71q7}
\end{figure}

\begin{figure}[H]
    \centering
    \includegraphics[width=0.5\textwidth]{71q7b.PNG}
    \caption{Data points of differential mode gain}
    \label{fig:71q7}
\end{figure}

\subsection*{7.2\quad Common Mode Bandwidth}

\begin{figure}[H]
    \centering
    \includegraphics[width=\textwidth]{72q5.PNG}
    \caption{Common Mode Bode plot}
    \label{fig:71q7}
\end{figure}

\begin{figure}[H]
    \centering
    \includegraphics[width=0.5\textwidth]{72q5b.PNG}
    \caption{Data points of common mode gain}
    \label{fig:71q7}
\end{figure}

Phase shift at the break frequency is 45.2\degree.

\printbibliography

\end{document}