\input{labpreamble}
\begin{document}

\subsubsection*{PHSX 222L\hfill $RC$ Period\hfill}

% Equipment:
% Capacitors
% Resistors
% Neon Bulbs
% Jumper Wires
% Breadboards
% Function Generators
% Oscilloscopes
% Multimeters

\underline{Object:} (a) To study the voltage-time relationship in a
charging and discharging capacitor and to measure the circuit's time
constant, $\tau$.  (b) To use an $RC$ circuit with a neon bulb to
measure the capacitance.  (c)  to study two special $RC$ circuits, the 
differentiator and the integrator.

\underline{Theory:} 

\vspace{.1in}

\underline{Determining $C$ with a Neon bulb}:
When a voltage is applied to a neon bulb, no current will flow until the voltage 
has been increased to the point where the neon gas has become ionized.  The 
ionizing voltage will be called $V_{\rm on}$.  Once the gas has been ionized it will
remain ionized and continue to conduct even though the applied voltage is somewhat
reduced.  The voltage at which the gas returns to its unionized state and at
which the bulb ceases to conduct will be called $V_{\rm off}\ (<V_{\rm 
on})$.

If a neon bulb is connected in parallel with the capacitor in an RC
circuit powered by a voltage source as shown in figure \ref{bulb}, the
voltage will gradually build up until the ionizing potential of the
neon bulb is reached.  At this point the bulb suddenly begins to
conduct---thereby discharging the capacitor.  This is seen as a brief
flash in the bulb.  After the discharge, the capacitor will begin to
charge up again and another flash will occur.  This process repeats
over and over again causing the bulb to flash or blink at regular
intervals.  See problem P28.62 in Serway and consider our neon bulb as
the voltage-controlled switch.  The time between flashes $T$ can be
calculated mathematically as follows:

\goodbreak
\begin{equation}
V_C = V_0(1-e^{-t/RC})
\end{equation}
\begin{equation}
t=RC\ln\left({V_0\over V_0 - V_C}\right)
\end{equation}
\begin{equation}
T = t_{\rm on} - t_{\rm off} = RC \ln\left({V_0 - V_{\rm off}\over V_0 - V_{\rm on}}\right).
\end{equation}

Where $t_{\rm on}$ and $t_{\rm off}$ are the time from the completely 
uncharged state.  This works as long as the resistance of the 
conducting bulb is negligible, which it is.  However, a  more 
complete derivation is patterend after Serway P28.62.

The period $T$ is from one peak to the next and consists of a 
discharging interval, $t_{1}$, and a charging interval, $t_{2}$.  
Then $T = t_{1} + t_{2}$.

During $t_{1}$, $V_{C}(t) = V_{\rm on} e^{-t/R_{B}C}$ and continues 
until $V_{C}=V_{\rm off}$.  So $V_{\rm off} = V_{\rm on} e^{-t/R_{B}C}$ 
which gives $t_{1} = R_{B}C \ln (\frac{V_{\rm on}}{V_{\rm off}})$.

During $t_{2}$, the charging interval, $V_{C}(t)=V_{0} - (V_{0} - V_{\rm 
off})e^{-t/(R_{a}+R_{B})C}$.  When solved for $t_{2}$ it yields 
\begin{equation}
	t_{2}=(R_{a}+R_{B})C \ln(\frac{V_{0}-V_{\rm off}}{V_{0} - V_{\rm 
	on}})
\end{equation}



\setlength{\unitlength}{2mm}
\begin{figure}[h]
\begin{center}
\begin{picture}(65,15)(0,3)
\put(0,0){\framebox(15,15){\shortstack{voltage\\ source}}}
\put(13,13){\makebox(0,0){+}}
\put(13,2){\makebox(0,0){--}}
\put(13,8){\makebox(0,0){$V_{0}$}}
\put(15,13){\line(1,0){10}}
\put(25,10){\begin{picture}(14,5) %resistor
\put(0,3){\line(1,0){1}}
\put(1,3){\line(1,2){1}}
\multiput(2,5)(4,0){3}{\line(1,-2){2}}
\multiput(4,1)(4,0){2}{\line(1,2){2}}
\put(12,1){\line(1,2){1}}
\put(13,3){\line(1,0){1}}
\end{picture}}
\put(39,13){\line(1,0){20}}
\put(15,2){\line(1,0){44}}
\put(48,2){\line(0,1){5}}
\put(48,8){\line(0,1){5}}
\put(46,7){\line(1,0){4}} % capacitor
\put(46,8){\line(1,0){4}}
\put(59,2){\line(0,1){4}}
\put(59,9){\line(0,1){4}}
\put(59,7.5){\circle{3}}  % bulb
\put(59,7.5){\circle*{.5}}
\put(31,10){\makebox(0,0){$R$}}
\put(44,7.5){\makebox(0,0){$C$}}
\put(63,7.5){\makebox(0,0){bulb}}
\end{picture}
\end{center}
\caption{Circuit for determining $C$ with a Neon Bulb}
\label{bulb}
\end{figure}
\setlength{\unitlength}{.3mm}


%Notes: V_2 is about 70 V
%V_1 is about 50-60 V.
%In proc 3 use resistor of a couple of megaohms and a capacitor of a few 
%microfarads.
% 
% PHSX 223 Expt 2
% 
% When first measuring the turn-on and turn-off voltages: The neon bulbs 
% are rated for about .5 W so when the turn-on voltage is 70 V then a 
% currint-limiting resistor needs to be inserted.  The current can be no 
% more than .5 W/70 V = 7 mA. Therefore the resistor must have a value 
% of at least 70 V/7 mA = 10 kohm.
% 
% When using the bulb and the capacitor in parallel, the R must be very 
% large so the voltage drop across it will bring the voltage at the lamp 
% down below the turn-off voltage.  2 Mohm is about right with a small 
% capacitor (2 microF) when using the oscilloscope.  An even larger 
% resistor with a very large capacitor (500 microF) will allow visual 
% timing.
% 
%  Another combo that worked was R=110000 ohm and C=1E-5 F
%
% The white Instek power supplies don't put out enough voltage, but the 
% cheap discharge tube power supplies (also used for e/m) should.  The 
% new Daedalon high-voltage put out plenty of volts for this--in fact 
% students should be careful to not turn them up very high (they need 70 
% V and the supplies can do 5k V).

\newpage

\underline{Procedure:} Use a breadboard to wire up the circuits (see my
website for pictures and explanation).
% \vspace{-.1in}

% Old instructions for using the R of the voltmeter instead of a 
% separate R.

% \item Connect a voltage source $V_{0}$ in series with a capacitor and
% a voltmeter and measure the voltage across the internal resistance of
% the voltmeter as a function of time.  The difference between this
% voltage and that of the power source will be the voltage on the
% charging capacitor, which is to be plotted as a function of time and
% compared to the theoretical curve (see equation \ref{charge}).
% 
% \item Remove the voltage source from the circuit and using the
% internal resistance in a voltmeter as the resistance in an $RC$
% circuit, measure capacitor voltage as a function of time for a
% capacitor discharging through the meter resistance.  Make a plot of
% this voltage as a function of time and compare your result with the
% theoretical curve (see equation \ref{discharge}).


\underline{Determining $C$ with a Neon bulb}:
\vspace{-.1in}
\begin{enumerate}
	
\item First determine $V_{\rm on}$ and $V_{\rm off}$ directly without
the capacitor in the circuit.  Make sure you have at least a 10
k$\Omega$ current-limiting resistor in series with the bulb so as to
not blow out the bulb.  Use a voltmeter across the bulb to measure
$V_{\rm on}$ and $V_{\rm off}$ as you slowly ramp up the voltage on
your power supply, then back down again.

\item Connect up an $RC$ circuit with the neon bulb as per figure
\ref{bulb} ($V_0$ must be greater than $V_{\rm on}$).  Use a resistor
with a large $R$, in the M$\Omega$ range, and a capacitor with $C$ in
the $\mu$F range or slightly lower.  Measure the period $T$ by
counting the number of flashes in a given time ($T=1/f$).  From the
period $T$, the resistance $R$, the applied voltage $V_0$, and the bulb
voltages $V_{\rm on}$ and $V_{\rm off}$, calculate the value of the
capacitance used in the circuit, and compare with the stated value. 
(For time periods that are too short to count visually, use an
oscilloscope to display the waveform of the capacitor voltage.  The
period $T$ can be determined from the oscilloscope display.)

\item Repeat the previous procedure with two capacitors in series or
in parallel to see if they obey equations 26.8 and 26.10 in Serway.

\end{enumerate}



\end{document}
