Съдържание

IPhO 2012, Експеримент — Задача 2. Problem E2. Nonlinear Black Box

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IPhO 2012

The 43rd International Physics Olympiad — Experimental Competition · 19 юли 2012 г. · 10 т.

Внимание: Бележка към темата

The complete alternate archive files Experiment/exp.pdf and Experiment/exp_sol.pdf contain the same two experiments and worked solutions as the primary originals. All 18 native pages of both editions were read. The printed question allocates 0.7 points to E1 B.iii, while its solution labels it 0.5 points; both are preserved. Native E1 Part C sign conventions, E2 charging formula Ic, Vmax unit mVs and angular-frequency unit MHz are preserved as printed. The cover mentions blank Answer, Solution and Draft Sheets, which are absent from these standalone source PDFs; no forms or extra marking scheme are invented.

Корица

The examination lasts for 5 hours. There are 2 problems worth in total 20 points. There are two tables in your disposal (in two neighbouring cubicles), the apparatus of Problem E1 is on one table and the apparatus of Problem E2 is on the other table; you can move freely between these tables. However, you are not allowed to move any piece of experimental setup from one table to the other.
Initially the experimental equipment on one table is covered and on the other table is boxed. You must neither remove the cover nor open the box nor open the envelope with the problems before the sound signal of the beginning of competition (three short signals).
You are not allowed to leave your working place without permission. If you need any assistance (malfunctioning equipment, broken calculator, need to visit a restroom, etc), please raise the corresponding flag (“help” or “toilet” with a long handle at your seat) above your seat box walls and keep it raised until an organizer arrives.
Use only the front side of the sheets of paper.
For each problem, there are dedicated Solution Sheets (see header for the number and pictogramme). Write your solutions onto the appropriate Solution Sheets. For each Problem, the Solution Sheets are numbered; use the sheets according to the enumeration. Copy the final answers into the appropriate boxes of the Answer Sheets. There are also Draft papers; use these for writing things which you don’t want to be graded. If you have written something what you don’t want to be graded onto the Solution Sheets (such as initial and incorrect solutions), cross these out.
If you need more paper for a certain problem, please raise the flag “help” and tell an organizer the problem number; you are given two Solution sheets (you can do this more than once).
You should use as little text as possible: try to explain your solution mainly with equations, numbers, tables, symbols and diagrams.
Avoid unnecessary movements during the experimental examination and do not shake the walls of your cubicle - the laser experiment requires stability.
Do not look into the laser beam or its reflections! It may permanently damage your eyes!
The first single sound signal tells you that there are 30 min of solving time left; the second double sound signal means that 5 min is left; the third triple sound signal marks the end of solving time. After the third sound signal you must stop writing immediately. Put all the papers into the envelope at your desk. You are not allowed to take any sheet of paper out of the room. If you have finished solving before the final sound signal, please raise your flag.

Условие

In simple problems, electrical circuits are assumed to consist of linear elements, for which electrical characteristics are directly proportional to each other. Examples include resistance ( $V = RI$ ), capacitance ( $Q = CV$ ) and inductance ( $V = L\dot{I} = L\frac{dI}{dt}$ ), where $R$ , $C$ and $L$ are constants. In this problem, however, we examine a circuit containing nonlinear elements, enclosed in a black box, for which the assumption of proportionality no longer holds.

The setup comprises a multimeter (labelled “IPhO-measure”), a black box that acts as a current source, a black box containing nonlinear elements, and four test leads with stackable connectors for wiring. Be careful not to break the seal on the black box.

The multimeter can measure current and voltage simultaneously. You can store with it up to 2000 data points, each consisting of: voltage $V$ , current $I$ , power $P = IV$ , resistance $R = V/I$ , voltage time derivative $\dot{V}$ ( $= \frac{dV}{dt}$ ), current time derivative $\dot{I}$ ( $= \frac{dI}{dt}$ ) and time $t$ . See multimeter manual for details. If you go beyond 2000 stored data points, the oldest data will be overwritten.

Photograph of the IPhO-measure multimeter (terminals IN, OUT, GND, LCD display, buttons Start, Stop, Previous, Next, Sample and Power) and its equivalent circuit in a dashed frame labelled “Multimeter”: an ammeter A between IN and OUT and a voltmeter V between OUT and GND.

The constant current source supplies stable current as long as the voltage across its terminals stays between $-0.6125\ \mathrm{V}$ and $0.6125\ \mathrm{V}$ . When switched off, the constant current source behaves as a large (essentially infinite) resistance.

The black box contains an electric double layer capacitor (which is a slightly nonlinear high capacitance capacitor), an unknown nonlinear element, and an inductor $L = 10\ \mu\mathrm{H}$ of negligible resistance, switchable as indicated on the circuit diagram. The nonlinear element can be considered as a resistance with a nonlinear dependence between the voltage and the current [ $I(V)$ is a continuous function of $V$ with $I(0) = 0$ ]. Likewise, for the capacitor, the differential capacitance $C(V) = dQ/dV$ is not exactly constant.

We say that the voltage on the black box is positive when the potential on its red terminal is higher than the potential on the black terminal. Positive voltage will be acquired when the terminals of matching colours on the black box and the current source are connected (you are allowed to use negative voltages).

Circuit diagram of the black box inside a dashed frame labelled “Black box”: between the + and - terminals a capacitor C(V) in parallel with a “Nonlinear device”, with an inductor (coil) in series that can be bypassed by a switch; next to it a photograph of the black box with a red and a black terminal, an on/off switch and a label.

It is safe to discharge the capacitor in the black box by shorting its inputs, either by itself or through the $IN$ and $OUT$ terminals on multimeter: the internal resistance of this capacitor is enough to keep the current from damaging anything.

You are not asked to estimate any uncertainties throughout this problem.

Part A, i. Circuit without inductance (7 points)

In this part, keep the switch on the black box closed (push “I” down), so that the inductance is shorted. Please note that some measurements may take a considerable time, therefore it is recommended that you read through all the tasks of part A to avoid unnecessary work.

(1.0 pts) Confirm that the output current of the current source is approximately $6\ \mathrm{mA}$ , and determine the range within which it varies for voltages between 0 and $+480\ \mathrm{mV}$ . Document the circuit diagram used. [1 т.]

ii. (1.2 pts) Show that the differential capacitance $C(V)$ used in the black box is approximately $2\ \mathrm{F}$ by measuring its value for a single voltage of your choice $C(V_0) = C_0$ . Document the circuit diagram.

Circuit diagram of the current source in a dashed frame labelled “Current source”: terminals - and + with a current-source symbol (arrow downwards) and a switch in series, annotated “I=6mA U=-612.5mV...612.5mV”; next to it a photograph of the current source box with black (-) and red (+) terminals, a switch and a label “Current source 612.5mV 6mA”.

iii. (2.2 pts) Neglecting the nonlinearity of the capacitance [ $C(V) \approx C_0$ ], determine the current–voltage characteristic of the nonlinear element used in the black box. Plot the $I(V)$ curve for obtainable positive voltages on the black box onto the answer sheet. Document the circuit diagram.

iv. (2.6 pts) Using measurements taken from the whole range of obtainable voltages, calculate and plot the $C(V)$ curve for obtainable positive voltages from the black box on the answer sheet. Write down the minimal and maximal values of differential capacitance $C_{\min}, C_{\max}$ . Document the circuit diagram.

Part B Circuit with inductance (3 points)

Enable the inductance by opening the switch on the black box (push “0” down). Using the same method as in pt. A-iii, measure and plot the current-voltage characteristic of the nonlinear element. Describe any significant differences between the curves of parts A and B and suggest a reason using qualitative arguments. You need to know that the nonlinear element also has a capacitance ( $\approx 1\ \mathrm{nF}$ ) which is connected in parallel to the nonlinear resistance. [3 т.]

IPhO-measure: short manual

Photograph of the IPhO-measure display and front panel with yellow numbered callouts 1–9 on the LCD reading “n=006 s=09 t= 386.2s / U= 313.0mV +2.03/s / I= 4.9mA +0.79/s / P= 1519uW R=64.472”; terminals IN, OUT, GND; buttons Start, Stop, Previous, Next, Sample and Power.

IPhO-measure is a multimeter capable of measuring voltage $V$ and current $I$ simultaneously. It also records their time derivatives $\dot{V}$ and $\dot{I}$ , their product $P = VI$ , ratio $R = V/I$ , and time $t$ of the sample. Stored measurements are organized into separate sets; every stored sample is numbered by the set number $s$ and a counter $n$ inside the set. All saved samples are written to an internal flash memory and can later be retrieved.

Electrical behaviour

The device behaves as an ammeter and a voltmeter connected as follows.

Small photograph of the multimeter next to its equivalent circuit: an ammeter A between IN and OUT and a voltmeter V between OUT and GND, dashed frame labelled “Multimeter”.

	Range	Internal resistance
Voltmeter	$0 \ldots 2\ \mathrm{V}$	$1\ \mathrm{M\Omega}$
Voltmeter	$2 \ldots 10\ \mathrm{V}$	$57\ \mathrm{k\Omega}$
Ammeter	$0 \ldots 1\ \mathrm{A}$	$1\ \Omega$

Basic usage

Push “Power” to switch the IPhO-measure on. The device is not yet measuring; to start measuring, push “start”. Alternatively, you can now start browsing your stored data. See below.
To browse previously saved samples (through all sets), press “Previous” or “Next”. Hold them down longer to jump directly between sets.
While not measuring, push “Start” to start measuring a new set.
While measuring, push “Sample” each time you want to store a new set of data (i.e. of the readings shown on the display).
While measuring, you can also browse other samples of the current set, using “Previous” and “Next”.
Press “Stop” to end a set and stop measuring. The device is still on. You are ready to start a new measuring session or start browsing stored data.
Pushing “Power” turns the device off. The device will show text ”my mind is going ...”; do not worry, all the data measurements will be stored and you will be able to browse them after you switch the device on again. Saved samples will not be erased.

Display

A displayed sample consists of nine variables:

index $n$ of the sample in the set;
index $s$ of the set;
time $t$ since starting the set;
voltmeter output $V$ ;
rate of change of $V$ (the time derivative $\dot{V}$ ); if derivative cannot be reliably taken due to fluctuations, “+nan/s” is shown;
ammeter output $I$ ;
rate of change of $I$ (the time derivative $\dot{I}$ ); if derivative cannot be reliably taken due to fluctuations, “+nan/s” is shown;
product $P = VI$ ;
ratio $R = V/I$ .

If any of the variables is out of its allowed range, its display shows “+inf” or “-inf”.

Отговори

Решение

Part A. Circuit without inductance (7 points)

It is possible to make all the measurements needed for this problem with a single circuit as shown in the figure. While the current source is switched on, we are charging the capacitor in the black box, until the current $I(V_{\max})$ through the nonlinear element equals to the output current $I_0$ of the current source. $V_{\max} = 540 \pm 40\ \mathrm{mVs}$ varies from one experimental setup to another. When the current source is switched off or disconnected, the capacitor will discharge through the nonlinear element.

Wiring diagram: the Multimeter (terminals IN, OUT, GND), the Current source (terminals − and +, switch table O/I with I ticked) and the Black box (switch table with I ticked, red and black terminal). Red wires join current source + to multimeter IN and multimeter OUT to the black-box red terminal; black wires join current source −, black-box black terminal and multimeter GND.

i. (1 pt) During charging of the capacitor from $V = 0$ to $V = V_{\max}$ we note that the output of the current source is constant ( $I_0 = 6.0\ \mathrm{mA}$ ) close to the precision of the multimeter.

ii. (1.2 pts) Using the definition of differential capacitance, we can calculate the current through the capacitor in the black box from the time derivative of the voltage on the black box.

I_c = \frac{dQ}{dt} = \frac{dQ}{dV}\frac{dV}{dt} = C(V)\dot{V}

Graph of I (mA) from 0 to 6 versus V (V) from 0 to 0.5: Part A curve (blue, crosses) rises to about 4.7 mA at 0.06 V, falls smoothly to about 0.7 mA near 0.33 V and rises steeply above 0.45 V to about 5.5 mA at 0.52 V; Part B curve (red) follows it except between about 0.07 and 0.33 V, where it drops to about 1.8–2.3 mA and then jumps down to about 0.6 mA at 0.33 V.

There are several ways to determine the capacitance used in the black box based on chosen voltage.

When the voltage on the black box is close to zero, the current through the nonlinear element is also close to zero, because $I(V = 0) = 0$ . After switching the current source on, most of the input current $I_0$ will at first go through the capacitor.
$C_0 = I_0/\dot{V}_{\uparrow}(V = 0)$
This can be measured more precisely after first reversing the polarity of the current source and charging the capacitor backwards, because the multimeter does not display derivatives when they change sharply (as in few moments after switching the current source on).

Example measurements taken this way follow.

$V_{\uparrow}(0)$ (mV/s) 3.51 3.32 3.55
$C_0$ (F) 1.71 1.81 1.69
$C_0 = 1.74\ \mathrm{F}$
When the voltage on the black box is $V_{\max}$ , the current through the nonlinear element is $I_0$ . Switching the current source off, we will have the capacitor discharging with the same current.
$C_0 = -I_0/\dot{V}_{\downarrow}(V = V_{\max})$
We can also measure the capacitance for any intermediate voltage as in A-iv.

$V_{\uparrow}(0)$ (mV/s)	3.51	3.32	3.55
$C_0$ (F)	1.71	1.81	1.69

iii. (2.2 pts) If we neglect the nonlinearity of the capacitor, there are (at least) two ways to obtain the current–voltage characteristic of the nonlinear element in the black box.

Applying Kirchhoff’s I law to the charging capacitor,
$I(V) = I_c - C_0\dot{V}_{\uparrow}(V).$
An $I(V)$ characteristic obtained by charging the capacitor is shown on the following figure.
Applying Kirchhoff I law to the discharging capacitor,
$I(V) = -C_0\dot{V}_{\downarrow}(V).$

iv. (2.6 pts) In order to obtain the differential capacitance, we solve a system of linear equations by eliminating $I(V)$ :

\begin{cases} I_0 = \dot{V}_{\uparrow}C(V) + I(V) \\ I(V) = -\dot{V}_{\downarrow}C(V); \end{cases} \implies C(V) = \frac{I_0}{\dot{V}_{\uparrow} - \dot{V}_{\downarrow}}.

Therefore we need to take measurements during both charging and discharging the capacitor in the black box at the same voltages. A graph of measurement results follows.

Part B. Circuit with inductance (3 points)

Graph of C (F) from 1.7 to 2.1 versus V (V) from 0 to 0.5: a noisy line with crosses at about 1.85 F near 0–0.12 V (dip to 1.82 at 0.12 V), rising roughly linearly to about 2.02 F at 0.46 V, then fluctuating and dropping sharply below 1.7 F at about 0.515 V.

Measuring and plotting the current–voltage characteristic of the nonlinear element in the same way as in part A-iii, we obtain a graph that differs only in the negative differential resistance ( $I'(V) < 0$ ) region, in our case $70\ \mathrm{mV} < V < 330\ \mathrm{mV}$ . This is the region where, when we look at small-signal oscillations, the nonlinear element behaves as a negative-valued Ohmic resistance. After enabling the inductance we have a LC circuit whose oscillations are amplified (instead of being dampened) by the negative differential resistance. Because the resonant frequency $\omega = \sqrt{\frac{1}{LC_p}} \sim 30\ \mathrm{MHz}$ (with $C_p$ being the capacitance of the nonlinear element) is high, we actually measure the average current through the nonlinear element, while the real current oscillates all over the region of negative differential resistance.

Оригинал в Архива: 2012-08-09 IPhO2012_Experimental_problems.pdf · официални решения: 2012-08-09 IPhO2012 Experiment_Solutions_ENG.pdf · допълнителни данни: exp.pdf · допълнителни данни: exp_sol.pdf