PHYSICS 213
PHYSICS 211L
Exp. 1: Basic Oscilloscope Operations
Grade:Date: March 8, 2009Name: Ali AlawiehPartners name: KHODOR
ABOU DAYASection number & Name of Instructor: Section 5 Lamiss
Zaidouny1. DataTable 1: Amplitude MeasurementsAmplitude control
Function GeneratorCH1
VOLTS/DIV Number of divisionsPeak to Peak Voltage
Fully CW5 V/Div5.9 Div29.5 V
- Way CW5 V/Div3 Div15 V
- way CW5 V/Div2.5 Div7.5 V
Fully CCW0.2 V/Div7 Div1.4 V
Fully CW Square wave: Fully CW Triangular wave: Table 2:
Frequency MeasurementsSet the frequency using the frequency counter
(counters are accurate to within about 10 parts per million)
Frequency of
Function GeneratorFrequency of
CounterTime/DIV
Number of divisionsPeriod
( # of divs * time/div)Frequency
(from scope)(1/T)% of difference
100 Hz100 Hz4.9 div102 Hz2.0
2.00 kHz2.00 kHz2.5 div2.00 kHz0.0
40.0 kHz40080 Hz5 div40.0 kHz0.2
150 kHz150863 Hz3.4 div147 kHz2.5
1.30 MHz1.26 MHz4 div1.25 MHz0.8
Comment on the accuracy of the scope measurement.
According to the difference in frequencies that range between 0
% and 2 % ( Average = 1.1%) between the actual and measured
frequency; this mean that the oscilloscope is accurate and reliable
in its measurement.Describe what happens when you vary the trigger
and when you press the +/-trigger control. What is the use of the
trigger on the oscilloscope?
When we vary the trigger value, the waveform displayed on the
scope will move on the screen until its stable at one point;
therefore, the aim of the trigger is to stabilize the displayed
waveform to be able to study its characteristics.
When we press the +/- trigger, the waveform is inverted on the
screen. Table 3.
Wave form Unit: Sketch and Explanation
(a) K1 and K2 open
Sketch of Wave FormExplanation
+
Waveform of 1st branch + waveform of 2nd branchThe resulting
waveform:
In this case, K1 is open where K2 is open so no current pass in
the lower branch and a current pass in the upper one through the
diode. The diode allow only positive current to pass and since the
initial current here is sinusoidal having a negative and a positive
part, the result is that only the positive part passes through the
diode, arrive at the load and appear on the screen while the
negative part dont pass and appears as a zero line.
(b) K1 open and K2 closed
Sketch of Wave FormExplanation
In the first branch
In the second branch
These 2 added will give us the following waveform:
In This case where K1 is open and K2 is closed, current will
pass through the two branches of the circuit. In the first branch
only the positive half of the sinusoidal current will be allowed to
pass through the diode and in the second branch the positive part
is only allowed to pass through the diode, but the two waveforms as
shown in the figure have a pie difference in the phase angle so the
positive part of the current in the first branch is negative in the
second and vice versa. Thats why the curve will look so due to the
superposition of the two at the level of the resistor whose voltage
is being displayed.
(c) K1 closed and K2 open
Sketch of Wave FormExplanation
Waveform of first branch:
Waveform of second branch:
Resulting waveform across resistor:
In this case we are having a short circuit on the diode in the
first branch so the current will pass through the wire not the
resistor since it has less resistivity so the whole sinusoidal
current will pass to the resistor undistorted or changed as if the
resistor is directly connected to the generator. However, the open
switch K2 will allow no current to pass through the second branch
so the only waveform displayed is that of first branch.
(d) K1 and K2 closed
Sketch of Wave FormExplanation
Waveform of the 1st branch:
Waveform of 2nd branch:
The resulting waveform across the resistor is:
For closed K1, as before the current in the upper branch will
pass through the branch of lower resistance that is through the
wire not the diode; therefore, the current waveform will arrive
undistorted at the level of the resistor.Simultaneously, closing K2
will allow current to pass through the lower branch but here
through the diode that will allow only the positive part to pass
resulting in a waveform that is positive in half period and null in
the other half.However, the phase difference between the current of
the two branches is pie () so the superposition of the two currents
at the level of resistor will result in the shown waveform since by
the time the 1st branchs current is positive that of 2nd branch is
null resulting in a positive waveform. At the other half of period,
the 1st branchs current is negative and that of 2nd is positive
resulting in approximately zero since the two waveforms come from
identical origin.What is noted here is that the null line is just
below the graticule line even if the two currents carry equal
voltage, the reason for that is that the diode requires an
activation voltage that it takes for itself and reduce the voltage
thats why the superposition is less than what is expected.
Note:We have referred to the previous waveform as a
representation of both current and voltage since it measures the
voltage across the resistor and that is directly proportional to
the current (by ohms law (U=RI) so the waveform represents both
across the resistor.Table 3. X-Y Operation
Phase Measurements: 5 kHz
Phase shift unit2A2DPhase angle ((Measured)
A6441.81
B61.312.51
To get the phase angle we apply: sin=D/A that is
=sin-1(D/A)Table 4. X-Y Operation
Phase Measurements: 10 kHz
Phase shift unit2A2DPhase angle ((Measured)
A62.625.68
B60.87.66
Work out the derivation steps between eq(1) and eq(2). What
happens when (=0 and (=90?
Before studying the two given cases (=0 or 90), its better to
rearrange the equation that we have reached in order for not to
encounter the problem of having zero in the denominator:
We can hereby multiply by sin2 on both sides, this will
give:
Then, we consider the first case:
1)
Upper branch (1st)
Lower branch (2nd)
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