[Solved] Formatting and centering spacing issues between figures and tables

Daniel Webb Asks: Formatting and centering spacing issues between figures and tables

Code:
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begin{document}

title{ECEN 347 - LAB 1}
author{Daniel Webb }
date{3/01/21}
maketitle

section{Abstract}
%Brief summary of report & Highlight purpose, findings and conclusion
The lab report highlights the use of half-wave and full-wave rectifiers by exploring the effect of diode and capacitor properties. Rectification smoothing of the voltage ripple will be observed and manipulated by diodes and capacitors.   
section{Introduction Theory}
%Contains essential Background info & Appropriate theory and eqs
Converting an AC to DC voltage is useful because DC voltage devices operate on that supply. In order for an AC power supply to deliver a DC voltage, a rectifier must created. The main piece of a rectifier circuit is selecting the diode to manipulate the output. A diode's main purpose is to block current flow in one direction. This trait allows for AC source signals to be modified. Theoretically diodes will have zero voltage drop when in a forward bias(positive voltage to anode) which means diodes should be treated as zero resistivity or zero current.  \ \
Two types of rectifiers demonstrated in this lab are the half-wave and full-wave. 
The voltage ripple is defined as the difference between max and min of the output voltage which is pictured in later figures. The goal of building these circuits is to decrease the ripple for a more stable DC like effect. "Smoothing" the voltage ripple is obtained by placing a capacitor in parallel with load resistance. By adding this component, the voltage charges up the capacitor at max values. Then discharges when voltage decreases in the opposite direction of capacitor orientation. Given equation (1) an approximation for the ripple can be calculated. \
begin{figure}[h!]
centering
includegraphics[scale=.5]{voltage ripple.JPG}
caption{Voltage ripple}
label{fig}
end{figure}
begin{equation}
label{equation:(1)}
Delta V_0 approx V_m(2 pi/ omega RC)= V_m/fRC
end{equation}

subsection{Half wave rectifier}
Half wave rectifiers will limit the source halfway, effectively cutting off any residual negative voltage because of the way the diode limits current in the opposite direction.
begin{figure}[h!]
centering
includegraphics[scale=.85]{Halfwave Rectifier.PNG}
caption{Half-Wave Rectifier}
label{fig}
end{figure}


begin{figure}[h!]
centering
includegraphics[scale=.65]{halfwaave rect.PNG}
caption{Half-Wave Rectifier result}
label{fig}
end{figure}
The resulting output voltage remains positive because the diode blocks the rest of current flowing in the negative direction.

subsection{Full-wave rectifier}
The Full-Wave rectifier is used to rectify the full wave or convert the entire signal or partial signal to DC. This circuit positions the diodes in such a way that total current is redirected to cancel each other resulting in a netzero current. The positive cycle is rectified first through the path of D1 and D2. The negative cycle is then rectified after traveling through D3 and D4. 
begin{figure}[h!]
centering
includegraphics[scale=.8]{full wave.PNG}
caption{Full-Wave Rectifier}
label{fig}
end{figure}


section{Experimental/Procedure}
%give brief description of steps of experiment, detail ambiguous procedures. clear and complete
subsection{Step1: Half Wave Rectifier}
begin{enumerate}
    item Set up a forward biased half-wave rectifier modeled in Figure 1. Using the following components and their values. 
        begin{itemize}
            item $V_s =$ 20 $V_{pp} (V_m = 10V)$ 
            item $f =$ $100 Hz$
            item $R=$10 $ohms$
            item $Diode= 1N4001$
        end{itemize}
    item Use the Oscilloscope to measure source Voltage $V_m$ and $V_o$ then determine forward voltage of the diode through the relationship of $V_m$ and $V_0$
    item Implement a 20 % ripple of $V_m$ by placing a 47 uF capacitor in parallel with R from Fig. 1 and compare experimental vs theoretical ripple value.
    item Replace C to be 22 uF then 100 uF. Calculate the percentage of $V_m$ (Voltage Ripple) for each capacitor experiment.
end{enumerate}

FloatBarrier
subsection{Full-Wave Rectifier}
begin{figure}[h!]
centering 
includegraphics[scale=.7]{full-wave.JPG}
caption{Full-wave Rectifier}
label{fig}
end{figure}
FloatBarrier
begin{enumerate}
    item Set up a full-wave diode rectifier pictured in Fig. 5. Setting $V_s=$ 20 $V_{pp}$
    item Measure $V_{o-Ch1}, V_{o-Ch2},$  $V_m$
    item Place a 100 uF Capacitor in parallel with R from Fig. 5. Record the voltage ripple.
    item Replace C with a larger and smaller capacitance and record the effects of the voltage ripple.
end{enumerate}
%------------------------------RESULTS-------------------------------------------------------------------------------
newpage
section{Results and Discussion}
subsection{Half-wave Rectifier}     %%PART 1 ####

begin{figure}[h!]
centering
includegraphics[scale=.09]{Part1 half-wave.jpg}
caption{$V_m$ and $V_0$ measurements}
label{fig}
end{figure}
Clear half wave rectification is achieved.
FloatBarrier

begin{table}[h]
begin{center}
begin{tabular}{|c|c|c|} 
hline
 Voltage & V \ [0.5ex] 
 hlinehline
 $V_m$ & 10 \
 $V_0$ & 9.6 \
 hline
end{tabular} 
end{center}
caption{$V_m$ and $V_0$ measurements}
label{tab:table-name}
end{table}


begin{center}
underline{Theoretical Diode voltage:} \
$V_m = V_d + V_0$ \
$10 = V_d + 9.6$ \
$V_d =.4 V$ \  
end{center}


begin{table}[h]
begin{center}
begin{tabular}{|c|c|c|} 
hline
 Voltage term & Theoretical & Experimental \ [0.5ex] 
 hline
 $V_m$ & 10 & 11.6\
 hline
 $V_d$ & .4 & 1.2 \
 hline
end{tabular}
end{center}
caption{$V_m$ and $V_d$ measurements}
label{tab:table-name}
end{table}

begin{figure}[h]
centering
includegraphics[scale=.13]{Diode voltage.jpg}
caption{Experimental Diode Voltage}
label{fig}
end{figure}
FloatBarrier

subsection{Capacitor values 22 uF,47 uF,100 uF}
begin{figure}[h]
centering
includegraphics[scale=.25]{20perc ripple.jpg}
caption{20 % ripple voltage 47uF Capacitor}
label{fig}
end{figure}
FloatBarrier
begin{table}[h]
begin{center}
begin{tabular}{|c|c|c|} 
hline
   Theoretical & 2V \ [0.5ex] 
 hline
 Experimental & 1.79 V \
 hline
 Error & .21 V \
 hline
end{tabular}
caption{20 Percent Ripple Error}
label{tab:table-name}
end{center}
end{table}
  

FloatBarrier
The small percentage difference resulting in the experiment capacitor is from the fact that capacitors are not perfect at managing a ripple voltage. Ripple voltages themselves are analog from converting AC to DC. There is bound to be error in calculated vs experimental procedures. Another small factor observed was when the diode voltage was calculated to be .4V it was measured at 1.2 V which is unexpected and could be due to faulty oscilloscope connection techniques.  

begin{figure}[h!]
centering
includegraphics[scale=.2]{Part1 22uF.jpg}
caption{22 uF Capacitor}
label{fig}
end{figure}


begin{figure}[h]
centering
includegraphics[scale=.215]{Part2 100 uF.jpg}
caption{100 uF Capacitor}
label{fig}
end{figure}

FloatBarrier

begin{table}
begin{center}
begin{tabular}{|c|c|c|} 
hline
 Capacitor & Ripple voltage & Percent \ [0.5ex] 
 hlinehline
 22 uF & 3.0 & 30 \
 47 uF & 1.79 & 17.9 \
 100 uF & 1.39 & 13.9 \
 hline
 end{tabular}
caption{Capacitor Ripple voltage percentages}
label{tab:table-name}
end{center}
end{table}

The larger the Capacitor results in a smaller voltage ripple due to the fact that capacitors absorbs the majority of the ripple.In theory with a large enough capacitance the voltage ripple could be diminished to zero.      
FloatBarrier

subsection{Full-wave Rectifier} %PART2#####-0-------------
  
    begin{figure}[h]
     centering 
        includegraphics[scale=.3]{Full wave waveform.jpg}
        caption{Full wave: $V_m$, $V_o-Ch1$, and $V_o-Ch2$ }
        label{fig}
    end{figure}
    
       
    begin{figure}[h]
    centering
        includegraphics[scale=.1]{10 perc.jpg}
        caption{10 % Ripple Value}
        label{fig}
    end{figure}
    
    begin{table}
        begin{center}
        begin{tabular}{|c|c|} 
        hline
        Measurement & Voltage \ [0.5ex] 
        hlinehline
        $V_m$ & 10V  \
        $V_{o-ch1}$  & 10.1 V  \
        $V_{o-ch2}$ & 10.1 V  \
        hline
        end{tabular}
        caption{Full-Wave }
        label{tab:table-name }
       end{center}
    end{table}

   FloatBarrier
    

 
    
    
    begin{table}
        begin{center}
        begin{tabular}{|c|c|c|} 
        hline
         Measurement &  Theoretical Voltage(V) & Experimental Voltage(V) \ [0.5ex] 
         hlinehline
         $V_m$ & 10 & 10  \
         $V_{o-Ch1}$  & 10 & 14.1  \
         $V_{o-Ch2}$ & 10 & 13.2  \
         Ripple & 1.0 & 1.0 \
         hline
        end{tabular}
        caption{Table: 10 % Ripple voltage 100 uF}
        label{tab:table-name}
        end{center}
    end{table}
    
    
    begin{center}
    begin{figure}[]
        centering
        includegraphics[scale=.15]{20210222_163104.jpg}
        caption{147 uF Ripple}
        label{fig}
    end{figure}
    end{center}
    
    begin{center}
    begin{figure}[]
        centering
        includegraphics[scale=.15]{22uf part2 crop.jpg}
        caption{22uF Ripple}
        label{fig}
    end{figure}
    end{center}

    begin{table}
        begin{center}
        begin{tabular}{|c|c|c|} 
        hline
        Capacitor &  Theoretical Ripple(V) & Experimental Ripple(V) \ [0.5ex] 
        hlinehline
         22uF & 4.54 & 3.59  \
         100uF & 1.0 & 1.0  \
         147uF & .68 & .304   \
         hline
        end{tabular}
        caption{Capacitor Ripple voltage}
        label{tab:table-name}
        end{center}
    end{table}

section{Conclusion}
%Summarize results, emphasize key findings, make conclusion, outloook feedback
Changing capacitance will directly affect ripple voltage as shown in Equation ref{equation:(1)}. By increasing the capacitance the voltage ripple will decrease. Voltage ripple will increase when capacitance decreases. To summarize, ripple voltage is inversely related to capacitance. Voltage ripple decreases because as Capacitors absorb charge they consequently increase their voltage making the charge higher. When Testing the theory of zero voltage drop across diodes in forward bias its not fully accurate due to device resistance. It was observed that theoretical voltages are calculated to be higher than their experimental result due the dissipation of capacitors and diodes. By adding a diode to a AC source the signal will be reduced to DC signal on the negative cycle as pictured in the Half-wave rectified signals. However when a full wave Rectifier is in place the positive and negative cycle will be reduced to DC. The goal is limit the amount of AC like behavior contained in the signals
end{document}

Im having trouble formatting the full wave rectifier in the results sections between the figures still

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