2.0 Procedure
2.1 Oscilloscope and Signal Generator
1) We used “PROBE COMP” signal to insert a 1 kHz, 5V p-p (peak to peak) square
wave into the first input, Ch1. We set that channel for AC coupling. We got the following graph
on the screen:
As it is seen, the frequency is 1KHz and peak to peak voltage is about 5 voltage. After
adjusting the probe we got different graphs with distortions. The switch on the probes was in the
10x setting.
An example of an overcompensated signal:
An example of an undercompensated signal:
2) We did the same things with the channel 2 and obtained the similar graphs. I am
not including them in my report for the sake of simplicity.
3) We now connected a DC power supply to Ch1 and changed its voltage. We
checked the accuracy of the panel meter on the power supply using the scope readings. The
maximum voltage we got from power supply was 30.6 V
But on the oscilloscope we read the value as 30.8 V
The error = ( ( 30.8 – 30.6 ) / 30.6 ) * 100 = 0.65
There is only a 0.65 % error in the reading of the oscilloscope. This is a neglible error.
We varied the voltage and tried to get the minimum value. Minimum value of the
generator was 3.6 V.
On the oscilloscope we read 3.71V.
The error = ( ( 3.71 – 3.6 ) / 3.6 ) * 100 = 3.05
There is only a 3.05 % error in the reading of the oscilloscope.
Then we changed the coupling mode from DC to AC. But we couldn’t get the same
result with AC coupling. This experiment cannot be done in AC coupling because AC coupling is
for AC sources and DC coupling is for DC sources. The source is DC in this experiment so we
cannot do AC coupling.
4) We connected the FG-7005C signal generator to Ch 1 and set the generator to
sinusoid mode with 999.6 KHz frequency. We obtained the following graph on the
oscilloscope:
We couldn’t exactly set the frequency to 1 KHz in the generator but we exactly
obtained 1Khz in the oscilloscope.
The error = ( (1 – 0.996) / 0.996 ) * 100 = 0.4
There is only a 0.4 % error in the reading of the oscilloscope.
We then changed the frequency to 9.960 KHz in the generator. This time we got the
exact value in the oscilloscope. And at this frequency value we tried to reach the
maximum peak to peak to voltage value which was 230 V. It is illustrated below:
And for the minimum voltage value we got 2.1 V as shown below:
Note that we again set the frequency back to 1 KHz. However the oscilloscope value
was 1.021 KHz. If we calculate the error again:
The error = ( ( 1.021 – 1 ) / 1 ) * 100 = 2.1
There is only a 2.1 % error in the reading of the oscilloscope.
5) We this time set the generator to output a square wave. And again we compared the
period and frequency readings on the generator to those actually observed on the
oscilloscope.
We set the frequency of the generator as 998.8 Hz as shown below:
Its period : T = 1/f = 1 / 998.8 = 1.001 us
The oscilloscope capture is shown below:
The frequency observed in the scope is 997.8 Hz and therefore the period is:
T = 1 / 997.8 = 1.002 us
The error = ( (1.002 – 1.001) / 1.001 ) * 100 = 0.099
There is only a 0.099 % error in the period reading of the oscilloscope.
The minimum pulse width is obtained by maximizing the frequency and therefore
minimizing the period. Because we know that the pulse width of a square wave is
simply the half of the period.
The minimum pulse width is 91.8 ns and it is shown below:
The maximum pulse width is obtained similiary by minimizeing the frequency and
thus maximizing the period.
The maximum pulse width is about 6 us and shown below:
2.3 The Digital Multimeter (DMM)
1) We set up the following circuit :
Vsupply = 10 V.
2) Multimeter value for V(1) = 10.03 V and for V(2) = 6.67 V
Oscilloscope value for V(1) = 9.92 V and for V(2) = 5.99 V
Error for V(1) = ( ( 10.03 – 9.92 ) / 10.03 ) * 100 = 1.097
There is only a 1.097 % error in the reading of the oscilloscope.
Error for V(2) = ( ( 6.67 – 5.99 ) / 6.67 ) * 100 = 10.19
There is a 10.19 % error in the reading of the oscilloscope. This error is not
negligible. I guess we didn’t wait enough for the multimeter to set its real value or
there was an unpredictable experimental error.
3) V(2) = I * R(b) = ( Vsupply / ( R(a) + R(b) ) ) * R(b)
4) I = ( Vsupply / R(a) + R(b) ) = ( 10 / ( 10*10^3 + 20*10^3 ) ) = 3.3 * 10^-4 A
With multimeter we measured the current as 0.34 mA = 3.4 * 10^-4 A
The error = ( (3.4 * 10^-4 - 3.3 * 10^-4) / 3.4 * 10^-4 ) * 100 = 2.94
There is only a 2.94 % error in the reading of the oscilloscope.
5) Since the resistor of an ammeter is nearly zero , when we connect an ammeter
across the terminals of a voltage source the current goes to infinity from the Ohm’s
law => I = V / R
Infinite current will burn the ammeter.
2.4 Resistor Current-Voltage (I-V) characteristics
1) With multimeter we measured the 10 kOhm resistor as 9.97 kOhm
2)
YORUM
2.5 Transient Analysis
1) We constructed the following circuit
2) We set Vs a square wave with a frequency of 1kHz and a 50% duty cycle.
3) Place channel 1 of the oscilloscope at Vs and channel 2 at Vout.
4)
5) Square wave with the frequency 100 Hz:
Square wave with the frequency 10 kHz:
Square wave with the frequency 100 kHz:
2.6 Second order circuit: frequency response