Lab 2 - ECE 420L 

Authored by Silvestre Solano,

Email: Solanos3@unlv.nevada.edu

2-6-2015

 

 

 

This lab report will document the effort made to perform lab 2.

 

 

Show scope waveforms of a 10:1 probe undercompensated, overcompensated, and compensated correctly.

Undercompensated

 
Overcompensated

 
Compensated Correctly

 
 
 

 
Comment on where the type of scope probe (i.e., 1:1, 10:1, 100:1, etc.) is set on your scope (some scopes detect the type of probe used automatically).
 
The scope probes used for this lab have switches that allow the scope prove to switch between 1:1 and 10:1. For the entire lab experiment, the 1:10 setting was used. One of the probes used is shown below. The switch is highlighted.
 

 
 

 
Draft the schematic of a 10:1 scope probe showing: the 9 MEG resistor, 1 MEG scope input resistance, capacitance of the cable, scope input capacitance, and capacitance in the probe tip.
 

 
 
 

 
Using circuit analysis, and reasonable/correct values for the capacitances, show using circuit analysis and alegbra (no approximations), that the voltage on the input of the scope is 0.1 the voltage on the probe tip.
 
To begin the algebra, the 88 pF and 20pF capacitors are added together to form a 108pF capacitor since they are in parallel. The probe tip capacitance and the 9MEG resistor form impedence Z1 and the 108pF capacitor along with the 1MEG resistor will form Z2 as shown below.
 


 
Then, the circuit becomes a basic voltage divider problem as shown below.
 



 
The bottom fraction gets inverted.
 

 
A nine can be factored out of the bottom fraction.
 

 
The bottom fraction enclosed in parentheses cancels out and the final result is shown below.
 

 
 

 
Devise an experiment, using a scope, pulse generator, and a resistor, to measure the capacitance of a length of cable. Compare your measurement results to the value you obtain with a capacitance meter. Make sure you show your hand calculations.
 
The basic premise of the experiment is to construct a basic RC circuit below using the cable as a capacitor. The schematic for the circuit is shown below. The circuit will have a 1 Volt pulse at 1KHz frequency. The time delay is measured and will be used with the equation Td=0.7*RC to determine the capacitance.
 

 
The cable used for this experiment is shown below.
 

 
The waveform captured from this experiment is shown below.
 

 
According to the picture above, the time delay Td is approximately 77 uS. Using this information, the calculated capacitance of the cable is approximately 77u / (1MEG*0.7) = 0.11 nF.
 
Using the digital multimeter found in the lab, the capacitance of the cable is measured to be about 0.120 nF. The multimeter's output is shown below.
 

 
The experiment was apparently very accurate. The calculated value of 0.11 nF closely matches the experimental value of 0.12 nF.
 
 
 
 
Build a voltage divider using two 100k resistors. Apply a 0 to 1 V pulse at 1 MHz to the divider's input. Measure, and show in your report, the output of the divider when probing with a cable (having a length greater than or equal to 3 ft) and then a compensated scope probe. Discuss and explain the differences.
 
Unfortunately, my so-called partner did not give me the locker combination for the breadboard, so I had to use a very small novelty breadboard to perform this experiment. The setup for the experiment is shown below.
 

 
Using a compensated 1:10 scope probe, the following waveform was captured.

Compensated

 
Using the same cable whose capacitance was determined earlier, the following waveform was captured for the uncompensated probe.
 
Uncompensated

 
The most important difference between the uncompensated and compensated waveforms is the time delay, highlighted in red rectangles. The time delay for the compensated circuit is about 82.6 nS and the time delay for the uncompensated system is about 285 nS. Clearly, there is a greater time delay in the uncompensated system because the cable adds an extra unwanted capacitance, which slows down the signal being probed. The compensated seems to do its job by reducing the time delay of the circuit to 82.6 nS, which implies that the capacitance of the cable was compensated for. However, the compensated  voltage should have been about 50 mV, but the waveform shows 176 mV, which is about 3.5 times bigger than expected. The uncompensated waveform voltage seems a little off as well, but not as much as the compensated waveform. The compensated waveform shows a voltage of about 340 mV, when it sould have been closer to 500 mV.
 

 
 
Finally, briefly discuss how you would implement a test point on a printed circuit board so that a known length of cable could be connected directly to the board and not load the circuitry on the board.
 
I think that the solution is quite obvious given the structure of the compensated scope probe. The solution is to add a capacitor in parallel with a resistor, whose values will be determined using the resistance and capacitance of the cable, on the test point where the cable will make contact with the pcb.
 
 
 
 
As always, I will back up my stuff to my flash drive/ laptop as shown below.
 

 

 

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