- #Thomas calculus 11th edition solution 1.5 and 1.6 download generator#
- #Thomas calculus 11th edition solution 1.5 and 1.6 download manual#
- #Thomas calculus 11th edition solution 1.5 and 1.6 download full#
The IS level of the germanium diode is approximately 500 times as large as that of the silicon diode. VR(measured) = 5.07 mV IS(calculated) = 4.58 A d. Rm = 9.9 Mohms VR(measured) = 9.1 mV IS(calculated) = 8.21 nA c. Also, the Si has a higher firing potential than the germanium diode. Their shapes are similar, but for a given ID, the potential VD is greater for the silicon diode compared to the germanium diode. VR(V) VD(mV) ID(mA) VR(V) VD (mV) ID(mA) e. Forward-bias Diode characteristics b.1 453. Part 6: Computer Exercise PSpice Simulation 1-1 See Probe Plot page 191.ĮXPERIMENT 2: DIODE CHARACTERISTICS Part 1: Diode Test diode testing scale Table 2.1 Si (mV) 535 OLīoth diodes are in good working order.
707 * 8 mV = 5.66 mV by inspection: Vdc = 0V The signal shifted downward by an amount equal to the voltage of the battery. The shape of the sinusoidal waveform was not affected by changing the positions of the AC-GND-DC coupling switch. The vertical shift of the waveform was equal to the battery voltage. Switch AC-GND-DC switch, make copy of waveform above. 707 = 1.41 Volts V(rms)(measured) = 1.35 Volts [(1.41 1.35)/1.41) * 100 = 4.74% no trace on screen signal is restored, adjust zero level no shift observed the shift is proportional to dc value of waveform g. Part 4: Effect of DC Levels V(rms)(calculated) = 4V * 1/2 *. = 1 ms/cm T(calculated):5 cm* = 5 ms Fig 1.2
#Thomas calculus 11th edition solution 1.5 and 1.6 download generator#
no: there is no voltmeter built into function generator Part 3: Exercises a.
#Thomas calculus 11th edition solution 1.5 and 1.6 download full#
the signal occupied full screen the peak amplitude did not change with a change in the setting of the vertical sensitivity m. obtain its reciprocal that's the frequency. multiply timebase setting by number of cm's occupied by wave. count the number of cm's occupied by the wave 3. adjust timebase to obtain one cycle of the wave 2. 5 ms/cm takes 2 boxes to display total wave 1 ms/cm takes 1 box to display total wave i. (calculated): 1 ms* = l cm (measured): l cm = same h.2 ms/cm takes 5 boxes to display total wave. (calculated): l ms* =2 cm (measured): 2 cm = same g. (calculated): 1 ms* = 5cm (measured): 5 cm = same f. measuring device which reduces loading of scope on a circuit and effectively increases input impedance of scope by a factor of 10. the input impedance of many scopes consists of the parallel combination of a 1 Meg resistance and a 30pf capacitor n. determines mode of triggering of the sweep voltage m. provide for the adjustment of scope from external reference source k. allows for ac or dc coupling of signal to scope and at GND position establishes ground reference on screen h. selects unit of time/screen division on x-axis g. selects volts/screen division on y-axis e. allows the moving of trace in either screen direction d. adjusts the brightness of the beam on the screen c.
#Thomas calculus 11th edition solution 1.5 and 1.6 download manual#
m, b 6 y x 6 22.Solutions for Laboratory Manual to accompany Electronic Devices and Circuit Theory Eleventh EditionĮXPERIMENT 1: OSCILLOSCOPE AND FUNCTION GENERATOR OPERATIONS Part 1: The Oscilloscope a. Perpendicular slope does not exist perpendicular slope 0ġ3. Perpendicular slope perpendicular slope "3 34 Disk (i.e., circle together with its interior points) with center ( ) and radius 3.! ! Circle with center ( ) and radius 2.! ! ! !ħ. Thus by the Principle of Mathematical Induction, S a ak nk k+ n n" " " l l l l k k is true for all n positive integers.ġ. Thus,k k k k k k k k k k k k k k " " " " " " " k k k k k k k+ S a a is also true. k k k k k k " " " 5k k k Since a a and a a, we have a a a a a a a a. Now, assume that S a a is true form some positive integer. Prove S a a for any real number a and any positive integer n.n nn k k k kĪ a a, so S is true. a) 1 = 1 | 1 | = 1 b b " " "l l l ll l l lb b b b b b b b b b b b bĥ4. For x a x > a x a or x 0 for any positive number, a. Section 1.2 Lines, Circles and Parabolas 5ĥ2. Thus, by i), ii), and iii) | a | | a | for any real number. iii) By definition | 0 | 0 and since 0 0, | 0 | 0. Suppose that | x 0 | 0 ii) a 0, | a | a by definition. Suppose that | x 1 | 0 be any positive number and f(x) = 2x + 3. NT = necessarily true, NNT = Not necessarily true.
0 kommentar(er)
