II.8.1 MEASUREMENT OF THERMAL RELAXATION TIME CONSTANT OF A SERIAL LIGHT BULB
1. INTRODUCTION
Relaxation phenomena occur very commonly in physics. Take a spherical ball of radius a moving with an initial velocity v (0) in a medium with viscosity . The equation of motion for the ball is m (dv/dt) + 6
v=0
a
(II.8.1.1)
Its velocity will decrease exponentially with time as v (t) (t) = v(0) v(0) exp ( t/ ) and it will come to rest eventually.
(II. (II.8. 8.1. 1.2) 2)
called the relaxation time, is given by = m/6
a
(II.8.1.3).
Here is another example. A sample is connected to a bath at temperature T0. It is heated to a temperature T. Then heat is cut off. The sample will cool and its temperature T (t) as a function of time t satisfies the equation ms dT/dt + E (T-T0) = 0
(II.8.1.4)
law of cooling. The second term in the above equation represents the heat lost in in one second by the sample to the bath. The solution to this equation is T (t) = (T-T0) exp exp (-t/ (-t/ ) +T +T0
(II.8.1.5)
The temperature of the sample will approach the temperature of the bath exponentially with a characteristic relaxation time given by = ms/E
(II.8.1.6)
A third example is the following. following. An electric field field E is applied to a polar dielectric medium like acetone. This creates an electric polarization P in the medium. When the electric field is switched off at time t = 0, the polarization decays to zero exponentially with time. There is a characteristic relaxation time associated with this process. The mechanism of relaxation here is more complicated than in the other two cases considered above.
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This is also true of magnetization created in a paramagnetic material on the application of a magnetic field. When the magnetic field is switched off the magnetization in the specimen decreases exponentially to zero with a characteristic relaxation time. Relaxation is a phenomenon characteristic of dissipative mechanical, thermal, electric and magnetic systems. systems. If X is the material property, which relaxes with a characteristic relaxation time , it satisfies the equation dX/dt dX/dt + X/ = 0
(II.8. (II.8.1.7) 1.7)
Let us now consider forcing forcing the system system at a frequency f. For the ball moving in a viscous viscous medium medium this this is is achie achieved ved by apply applying ing a mecha mechanica nicall force force F exp ( t) where where =2 f (II. (II.8. 8.1. 1.8) 8) In the case of the thermal system, forcing is done by applying a periodic heat Q = Q0 exp exp ( t) to the system. system. In the dielectric dielectric example forcing forcing is done by applying applying a periodic electric field E exp ( t). In the magnetic example forcing is done by applying applying a magnetic magnetic field B exp ( t). A relaxation relaxation system system forced forced by a simple harmonic force will satisfy the equation dX/d dX/dtt + X/ = F exp( exp(
t)
(II (II.8.1 .8.1.9 .9))
The solution to this equation is X (t) = [F /(1+
)] exp (
t)
(II.8.1.10)
The response of the system system is also also simple harmonic with the same frequency frequency f. But the response is not in phase with with the force. force. If we write F /(1+ Then and
) = Aexp (-i )
A = F /(1+ tan(
½
)
(II.8.1.11) (II.8.1.12) (II.8.1.13)
, F, the amplitude A of the response decreases with frequency as shown by equation (II.8.1.12). The amplitude, as the frequency tends to zero, zero, is A (0) = F . If >> 1, the amplit amplitude ude A ( ) at freque frequency ncy will will decrea decrease se as A ( ) = A (0)/
(II.8.1.14)
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2. THEORY OF THE EXPERIMENT
Let us consider a serial light bulb. We pass a current I = I0+I1 sin
t
(II.8.1.15)
through the bulb, where I1 is small compared to I0. If we choose I0 and I1 suitably, and the frequency f of the oscillating current is small (about 10 Hz), we can see the light blinking. Let m be the the mass of the filament and s its specific heat. The bulb will lose heat through the thermal conductance of the support connecting the filament to the cap of the bulb and through radiation. We model the heat loss by We measure the temperature difference of the filament over that of the cap, which is at the ambient temperature T0. The heat power input to the filament is 2
2
2
2
Q = I R = (I0 + (1/2)I1 )R + 2I0I1R sin( in( t) (1/2 (1/2)I )I1 R cos(2 cos(2 t ) Q = Q0 + Q1 sin t The temperature
(II. (II.8.1 8.1.1 .16) 6)
Q2 cos cos 2 t
(II (II.8.1 .8.1.1 .17) 7)
of the filament satisfies the equation
msd /dt + E
Q0 + Q1 si sin ( t ) Q2 cos cos (2 t)
(II. (II.8. 8.1.1 1.18) 8)
E is the cons Writin Writing g E/ms E/ms = 1/ , Q j/ms =
j the
d /dt + / =
0 +
above equation can be written as sin 1 si
Integrating this equation from from t = 0, when (t) =
0(1 2/
t
2
sin sin 2 t
= 0, we get
exp ( t/ ) + ( 1 /[1+ [1+4
(II (II.8.1 .8.1.1 .19) 9)
2 2 1/2
] si sin(2 t +
2 1/2
]
sin( t+ 1)
2)
(II.8.1.20)
where tan ( 1) =
and tan ( 2) = 2
-3
If I1/I0 = 0.1, then 1/ 0 0.2 and / 0 5 x10 . So the second harmonic signal is less than 1% of the first harmonic. We shall neglect the second harmonic term. The intensity of the radiation emitted from the filament will vary as nearly four. So the intensity of the emitted light will vary as
where
is
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IL = IL0 + IL1 sin sin ( t) where IL1 will be proportional to
0
(
)
(II. (II.8. 8.1. 1.22) 22) 2 2 1/2
1[1/(1+
) ] sin( t +
1)
We pick this intensity fluctuation with a photodiode circuit circuit in the linear linear mode. The photodiode BPW 34 will give an output voltage proportional to the intensity of the light. The rms AC output voltage Vout of the photo-detector is measured with a multimeter. The frequency, f, of the AC signal is changed, keeping the AC voltage across the bulb constant. The resulting set of measurements are fitted to a formula 2
VAC (f) = V (0) /[1+ /[1+((2 f ) ] The value of V(0) V(0) and
0.5
(II.8.1.23)
are found as described below.
3 THERMAL RELAXATION SET UP:
The photodiode is BPW 34. The photodiode circuit circuit is shown in Figure II.8.1.1. II.8.1.1. R
PD
Output +
Figure II.8.1.1: Photodiode circuit
The blinking light source is a 3 V serial light bulb. The connection to the light bulb is as shown in Figure II.8.1.2. II.8.1.2. A resistance resistance of 6 Ohms (1 W) and a resistance resistance of 2 Ohms (1W) are connected in series with with the bulb and a 5 V regulated DC supply as shown in Figure II.8.1.2. II.8.1.2. When the switch SW1 SW1 is thrown to to one side, the DC voltage across the 6 Ohm resistance can be measured by a DMM2 in the DC 2 V range connected between the terminals T and G. This voltage divided by 6 gives the DC current through the bulb. With the switch SW2 is in the LC position, the bulb glows a dull red. One can connect an external plug pattern resistance box (0.1 to 10 Ohm) across the two Ohms resistor resistor and can vary the current through the bulb by adjusting the resistance in the box. When the switch SW2 is in the HC position it shorts the resistance resistance of 2 ohms and increases the current through the the bulb. The bulb
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glows brightest when SW2 is in HC position. The distance of the photodiode from the bulb is adjusted so that the output DC voltage measured by DMM1 at the terminals marked PD O/P is about 4.5 to 5 V in the high current position of SW2. A signal generator is connected through two 50 Ohm resistances and two 1000 f capacitors to the bulb.
SG/Audio Amp
O
Gd
Rheostat
0 5
SW2 0 5
0 f 0 0 1
SW1
0 0 f 0 1
PD
PD Circuit AMP
6
DMM2
Stabilized +5V DC Supply
DMM1
Figure II.8.1.2 Connections for Relaxation Experiment The signal generator is switched on and its frequency is adjusted to about 20 Hz. This frequency can be measured on DMM2 when it is put in Hz position and the switch SW1 is is thrown in the AC position. The rms AC voltage across 50 Ohms can be measured by changing the range switch on the DMM to AC 2 V position. The output of the signal generator is adjusted so that the RMS AC voltage across the 50 Ohm resistor as measured by DMM2 is 1 Volt. The AC current current is about 10 to 20% of the DC current. AC frequency is changed keeping the AC voltage across the 50 Ohm resistor constant by adjusting the signal amplitude knob on the signal generator. The AC current causes the the temperature of the filament in the bulb to oscillate about a mean temperature determined by the the DC current. This produces a fluctuating AC output of the photo-diode at the frequency of the AC current. The RMS value of the fluctuating voltage is measured by connecting DMM1 set to measure AC 200 millivolts to the terminals marked PD O/P. The frequency of the AC voltage is varied in steps and the AC output voltage is measured.
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For serial light bulbs the relaxation time time is of the order of 0.1 s. So even at 20 2 Hz frequency, is more than a hundred times larger than 1. So equation II.8.1.23 becomes to a good approximation VAC(f) = V(0)/ (2 f )
(II.8. (I .8.1.2 1.24)
The Front panel diagram of the relaxation set up is shown below.
PD/OP
HC
T
SW2
LC
LN
RH
AC
SG
SW1
DC
G
Fig. II.8.1.3 Front panel diagram of thermal relaxation set u p 4. APPARATUS REQUIRED:
Thermal relaxation set up, a signal generator, a DMM which will measure DC Volts, AC Volts in the 200 mV mV range and frequency correct to one decimal place in the range 20 to 100 Hz. 5. PROCEDURE: (a) AC PART:
Switch on the thermal relaxation set up., Put the DPDT switch SW2 at the right top to position HC. Put the DPDT switch SW1 at the right bottom to position DC. Connect a DMM between the banana terminals terminals marked T and G on the right. right. Set it in DC 2 volts range to measure the voltage to three decimal places.. The meter will show a reading reading around 1.8 V. This is the DC voltage voltage across 6 Ohms in series with the bulb. If the same meter (or another DMM1) is connected between terminals
229
marked P/D OP on the top left of the panel, the meter will show an amplified Photo diode output DC voltage. voltage. The distance distance between the photodiode and the bulb can be changed by releasing the locking nut LN and turning the screw at the center of the panel. The reading on the t he meter will increase as the photodiode is brought brough t closer to the bulb and it will decrease as the photodiode is moved farther from the bulb. Adjust the screw so that the DC output of the photodiode is between 4 and 5 volts in magnitude. Lock the nut. Connect a signal generator between the two banana plugs at the bottom left marked SG. The bottom bottom banana terminal should be connected to the earth earth of the signal signal generator. Set a frequency of about 20 Hz on the signal generator and the output voltage around 2 V AC. Connect the DMM2 between the terminals T and G and put the DPDT switch SW1 at the bottom right corner to AC. The meter should be selected to read AC volts. The meter will read the voltage across 50 Ohms in series with the bulb. Adjust the output amplitude of the signal generator so that the meter reads 1 V AC. RMS value of the AC current through through the bulb is 1/50 = 0.020 A. Turn the knob on the DMM2 to the frequency position and measure the frequency of the AC voltage. Adjust this frequency frequency to 20 Hz by turning the frequency adjusting adjusting knob on the signal signal generator. Again check that the AC voltage voltage read by the DMM is is 1 V. Then read the AC voltage at the terminals marked P/D OP either using DMM2 (or DMM1in the AC mode). Note this reading. Again connect the DMM2 to the terminals T and G, set it to measure frequency and adjust the AC frequency from the signal generator to be 22 Hz. Check that the AC voltage read by DMM2 is 1 V. Measure the AC output voltage of the photodiode now. Repeat these measurements as the frequency increases from 20 to 100 Hz and as the frequency decreases from ALWAYS KEEP THE AC VOLTAGE ACROSS T-G 100 to 20 Hz. TERMINALS TO 1 V AT EACH FREQUENCY. The average signal signal output in mV is taken at each frequency. frequency. Table II.8.1.1 II.8.1.1 shows a sample set of readings. This is VAC av in column 4 of the table. 1/VAC av is given in column 5.
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Table II.8.1.1 RELAXATION MEASUREMENT
pd dc 5.2 V o/p dc Voltage across 6 Ohms AC voltage across 50 Ohms
1.862 1
freq
V AC mV
V AC mV
V AC mV
f decreasing 119 106 102 92 85 81 67 58 52 47 38 33 28 26 23
average
20 22 24 26 28 30 35 40 45 55 60 70 80 90 100
f increasing 117 108 100 92 83 81 65 58 52 47 38 33 28 26 23
118 107 101 92 84 81 66 58 52 47 38 33 28 26 23
V V 1/V ACav
V AC cal mV
0.00848 0.00935 0.0099 0.0109 0.0119 0.01234 0.01515 0.01724 0.01923 0.02128 0.02632 0.0303 0.03571 0.03846 0.04348
113 103 94 87 81 75 65 56 50 41 38 32 28 25 23
1/VACav is plotted against against f. f. Such a plot is shown in Figure II.8.1.3. II.8.1.3. The slope slope of this -1 graph 2 /Vrms (0) is 4.39x10 s/V. The AC voltages measured on the DMM are RMS voltages. To find we should find Vrms (0). This is done by doing a DC experiment experiment described below.
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0.045 0.040 0.035 0.030 0.025 0.020 0.015 0.010 0.005 20
40
60
80
100
Frequency in Hz
Figure II.8.1.3 Plot of 1/VACav against frequency b) DC PART
Remove the signal generator connections. Put the DPDT DPDT switch SW2 on the right right top to LC. Connect a plug pattern resistance box (0.1 (0.1 to 10 Ohms) to the terminals marked RH. Put the bottom DPDT switch SW1 to position DC. Connect the DMM in the DC Volts mode to the terminals T and G. Pull out 0.1 Ohm from the the plug pattern box. Measure the voltage V on the DMM2. The current through through the bulb is now V/6 A. Connect the same DMM2 (or use another DMM if available) to the terminals P/D OP and measure the output DC Voltage VDC,PD of the photodiode. Repeat the measurement as the resistance in the box is increased in small steps of 0.1 or 0.2 ohms till the P/D output voltage comes down to 2.5 V. A sample set of readings are given in the following table II.8.1.2. First column gives the DC voltage measured at terminals terminals T and G. This voltage divided by 6 is the current current I through the bulb given in the second column. The third column gives the square of I. The fourth column is the DC voltage measured at the terminals PD OP. A plot of the 2 photo-diode voltage against I is given in Figure II.8.1.4.
232
Table II.8.1.2 DC Photodiode output as a function of DC Current through the bulb
DC Volts across 6 Ohms
Current I in amp
I^2
DC O/P of PD in Volts
1.843 1.837 1.817 1.771 1.761 1.742 1.728 1.72 1.704 1.696
0.307 0.306 0.303 0.295 0.294 0.290 0.288 0.287 0.284 0.283
0.09435 0.09374 0.09171 0.08712 0.08614 0.08429 0.08294 0.08218 0.08047 0.07990
3.459 3.321 2.858 1.993 1.824 1.559 1.331 1.232 1.045 0.953
3.5
3.0
2.5
2.0
1.5
1.0
0.078
0.080
0.082
0.084
0.086
0.088
0.090
0.092
0.094
0.096
2
I in amp^2
Figure II.8.1.4 Plot of DC Output voltage of photodiode against the DC current through the bulb. The equation to this line is
2
VDCOP = A +BI
(II.8.1.25)
233
where A, the intercept, is -13.06 -13.06 and B, the slope, is 174.1. This may be written as 2
2
VDC O/P = B(I + A/B) = 174.1(I 0.075) Written this way the equation shows that a minimum c urrent is required through the bulb for the photo diode to respond. In the AC experiment the rms AC current through the bulb is 1/50 = 0.02 A. So peak amplitude of AC current current is 1.414*0.02 = 0.028 amp. This is superposed on the DC current of 1.862/6 = 0.310 Amp Superposition of AC AC with with DC currents leads to to a maximum current of 0.310+0.028 = 0.338 Amp through the bulb and to a minimum current of 0.310-0.028 = 0.282 amp through through the bulb. As the frequency tends to zero, these currents will produce produce an output of the photo diode of 6.83 and 0.78 V respectively when substituted substituted in equation (II.8.1.25). (II.8.1.25). In the the limit of low frequencies (f tending to zero) this excursion of the photo diode output voltage, (6.83-0.78) V, is equal to 2V(0) peak . So V(0) peak = = 3.02 V. In the AC part of the experiment we measured RMS AC voltages on the multimeter. So we must find V(0)rms, which is equal to V(0) peak *0.707. This means that the rms value, Vrms(0) in the slope of AC experiment, is 3.02x0.707 = 2.14 V. can be calculated calculated from the slope slope of the line in Figure Figure II.8.1.3 II.8.1.3 ( 2 /V(0)rms = -1 4.39x10 ) s/V and V(0)rms = 2.14 V. comes out to be 0.15 s. Using the values of Vrms (0) and
and the Debye formula 2 2 2 0.5
V(f) = Vrms (0)/[1+4 f ]
we calculate the values of V at the frequencies of measurement. These values are shown in the last column of Table II.8.1.1 under VAC cal in mV. Figure II.8.1.5 shows the experimental values of VAC as a function of frequency and the values calculated from Debye formula are plotted as a continuous curve. The agreement is good. good.
234
120
Points are from expt. Fit to Debye formula is continuous curve
100
80
60
40
20
20
40
60
80
100
Frequency in Hz
Figure II.8.1.5 II.8.1.5 Comparison of experiment with Debye Formula
NOTE: Different makes of the serial light bulbs have different values of the relaxation time and the photodiode output will vary with the make of the bulb for the same current through the bulb bulb and the distance of photo-diode from the bulb. So the above readings must be taken as a sample only.
Questions: 1. Give examples of relaxation in physics. 2. A capacitance C is connected in series with a resistance resistance R. A DC voltage V is applied at time t = 0. How will the voltage across the capacitor change with time? On what factors does the relaxation time depend? 3. There are two thermal systems. The first system has a large thermal capacity and is connected to a thermal reservoir at temperature T through a weak link. The second system has a low thermal thermal capacity and is connected to the reservoir with a strong thermal link. Which system will relax faster and why?
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