Practices with the Oscilloscope

As first practice in GOING, we are going to see the operation basic and applied to the radio communications. We must understand that the analytical one of the signal is very important to see its basic parameters. The oscilloscope is a visualization and measuring instrument of the electrical signals. We have a brief manual in youtube that we will begin to see it soon to happen to make some tasks with the same

We have a good tutorial that we are going to see before following with the Good https://www.electronicafacil.net/tutoriales/Uso-del-osciloscopio.php tasks, we happened to the tasks

•   Vamos to calibrate our oscilloscope.

•  First that we must do it is a calibration of the sounding that we used in the measures. The sounding usually has a commutator of X1 (enters the signal so what) and X10 (it reduces to the signal 10 times). We put in X1 and with the sounding we looked for the exit that the own oscilloscope must to calibrate it.

• Following the model it can give to a signal of 0,5 volts tip to tip, 2 Vpp, etc.

• The frequency of the signal of that generating intern usually is on 1 KH

Fitting to the base of times and the vertical scale we must obtain a perfect square signal. If it is not it, we fit the sounding with a screwdriver until obtaining totally straight the horizontal lines.

In the first case, we have a perfect signal (square). We must try to obtain a signal but the similar thing to her. Perhaps we are with 2º signal (little compensated) or 3ª signal, that it would excessively be compensated.

The first practice consists of making screenshot of the viewfinder of the oscilloscope where it is how it is the squared signal, once fit in the sounding

•     Practices with the Proteus simulator. He is a group of several practices only simulated.

• In the folder of DRIVE and with the name you practice/oscilloscope, we have several practices to practice with our simulator.  IMPORTANT. Each activity that becomes must be reflected in the memory with a screenshot and description of the activity.

• 2.1 First of all, to do regarding the condenser. In this case, only we are going it to simulate with proteus.

• To mount the circuit with the given values that we have actually.  To visualize the signal and to make screenshot of the load and unload.

• To measure the time τ and to compare it with τ theoretical 2.3.

•  To calculate R of load so that τ is of the double. To replace in proteus and to visualize the signal of load and unloading 2,4 In document pdf to give, we must locate the signals of entrance and the two of load and unloading

• Actually of the diodes we have a document of theory on the operation of the diode and typical circuits to work as rectifier of average wave and complete wave.

• 3.1 Montar in proteus the circuit where we took values from Tension and current of the diode. To pass the values to the table that appears in drive (excel).

•  When adding the values, will be the characteristic line of the diode.

• 3.2 Montar a circuit of average wave. To take the tensions to the entrance and when coming out to verify the difference

• 3.3 Montar both circuits of complete wave (one with transformer and other that has 4 diodes). To take the tensions to the entrance and exit

3.4 In some of the smoothing circuits, to put in parallel a diode so that the exit tension is but smooth (we reduce the tension of curly)

1.  3º Circuits RLC. In this case we are going to work with proteus and we will mount circuit RLC, with the components that we have in the factory.
2.  Theoretical activity.  To design a circuit RLC series so that the resistance is of 100 Ohms, the inductive impedance of 80 and the capacitive one of 50. The generator that we are going to use has a frequency of 10 KHz and one peak voltage of 5 volts.
3.  To calculate the values of L and C.
4.  To mount and to take the measures from tensions Vr, VL and Vc. In order to measure the tension in the coil and the condenser, we must put a mass between L and C, as one is in the figure.
5. To calculate the phase angle that is between the current and the tension in each component
6. It is necessary to consider, that stops a little while certain, the current it leaves by the high part of the generator and happens through R, L and C. If the mass is after passing L, the tension she is correct, but for C, the reference (mass) it is before passing the condenser. Therefore, we must invest in the oscilloscope, the signal of the C1 so that the measures are taking the same criterion
7. We change the values of L by 47 µHr and C by 330 KpF. To calculate the resonance frequency

To fit our circuit with the values of fo, R, L and C to verify that the tension in the coil is equal to the tension of the condenser (in resonance). To verify that Fo leaves 40412 Hertz, approx, when we applied the formula:

• Pasamos to mount the circuit in factory. We will use:
• A coil of 47 µHr and a 330 condenser of KpF.
• This of the KpF sounds rare, but it is thus. they are Kilo and soon tip, reason why we remained in nF. Therefore, the condenser is equivalent 330nF = 0,33 µF
• We put a resistance of 2.2 in series kΩ with L and C.
• To take the values from the tension in the coil and the condenser. To see scale of the oscilloscope.

## Important. The him and signal generator oscilloscope has the common mass when connecting them to the network. In order to isolate the instruments, we must put a little insulating tape in the plugs. Associate three images of how they are due to put, connection of the circuit and result

You practice of Filters. Simulation in proteus

The radioelectric space is made up of an ample variety of signals of RF, each with its corresponding frequency. The filters are circuits that are going to prevent the passage of certain frequencies and is going to let pass the frequencies that interest to us.

This objetito can be made with components liabilities (R, L, C) or with active components (transistors, op amps, etc). These last ones are more necessary.

1. The filters are classified in:
2. Filters happen low, when it lets pass the low frequencies
3. Filters high step, that let pass the high frequencies in prevent that they pass the losses
4. Band-pass filters, that allow to pass the frequencies between some certain values.
Filter elimination of band, right in opposition to the previous one

To the left of the image, we have a simple filter made up of a resistance and a condenser.

Knowing the answer of the condenser to the frequency, what you think that it is going to let pass, the low frequencies or high?

Development of the practices. Until we do not make the three tasks, not to raise it Drive. As always, in a pdf, with the name of Filtros.pdf

1.
2.   We mount the circuit of the figure with the program proteus. The S-values 1 KΩ and 10 nF
3.  We are going to calculate the cutoff frequency, that is defined as that one where the ohmic value of the resistance agrees with the ohmic value of the capacitive reactance.
4. We calculate the gain in tension for the cutoff frequency, that comes given by the quotient between the tension from exit and the tension of entrance to that frequency. We have: .
5. If we replaced R by Xc, we have:D
6.  to emostrar that to the cutoff frequency, the gain in dB (mV) has fallen approximately in 3 dB.
7. We are going to make our analysis in the frequency domain to show the gain with respect to f. For it we must construct our circuit and add an Analyzer of frequency (to see eyelash Graph Mode, and to take the instrument Frequency). The entrance that we give is a sine generator of tension 10 volts. The sounding of tension we called it Vo. Once finished, we must select in the frequency measurer the signal of the generator as entered and to take the sounding within the screen of the measurer. When giving to the spacebar, we must have our finished practice, showing the frequency response of this filter. To see image

Finally, to verify that the value of the cutoff frequency agrees with the given one in the graph. Double click in the green bar becomes and it becomes but great

High-pass filter

1. We will do the same thing, but this time, using a coil. He would be the same for a condenser, only that the position they have varies. To see image superior
2. To mount in proteus a high-pass filter where R = 100 Ω and L = 1 mHr
3. To calculate the cutoff frequency, knowing that it happens when the value of the inductive impedance agrees with the resistance

To add the frequency analyzer and to verify that the theoretical Fc, agrees with the shown one in the graph of the frequency response

Band-pass filter

Theory. Now it is called on to us “to unite” both previous filters to secure the two functions (to let pass the losses and the discharges).

The idea is that if the low one lets pass until certain Fcs frequency and the stop lets pass from a frequency the ICF,   the result is that pass the frequencies from the ICF to Fcs.

We are going to design a circuit that lets pass a rank of frequencies (the ICF of 1000Hz and one fcs of 10kHz. We must consider the impedances of entrance and exit of the filter. We suppose that the sign source (Entered) has an impedance of 50Ω and that connects to a load 47kΩ.

1. In order to come with the calculations, we must consider the theoretics of the 10, that is applied in electronics, by which, it is possible to be considered that an impedance branch is despicable, if is the 10%. For example, if we have a resistance of 100, in parallel with another one of thousands, the one of 1000 can be despised to make estimates.
2. We go with the filter low step, R2-C2. In this case, C2 has a resistance of load of 47kΩ, therefore, the impedance of the condenser must be 10%, giving rise to a value of the condenser of:
3. We make accounts and nF gives 3,38 C2 us =
4. R2 will be the same the one of the reactance of C2 to fcs.  => R2 = 4.7 kΩ.
5. We add the circuit again to show the impedance that “sees” the R1 resistance. It would be the load that it has connected in parallel. Since we have calculated that R2 is 4700 Ω, we can consider that value, as permanent value of load impedance, since it does not vary with the frequency, something that yes makes C2. Therefore, R1 = 470 Ω
6. To the equal way that before, the impedance of C1 must be equal to R1, therefore:
7. In this case, C1 is 338 nF

We come to mount our circuit in proteus.