Bài báo cáo thực hành 3 môn lý thuyết mạch (Circuit Theory)

EE233 lab report 3

The University of Danang

Danang University of Science and Technology

LAB REPORT

Instructor : Nguyen Tri Bang

Group members: Tran Viet Tu

Nguyen Cong Thien

Dinh Ngoc Tien

Le Dinh Hoai Nam

Danang 2017 Danang 2016

I Objectives

analysis and design

II Equipment used

 Oscilloscope:

Function: display the waveform of the Vin and Vout in the electrical circuit

 Multi-meter :

Function: measuring the value of resistors

 Arbitrary waveform generator :

 Function: electric source, we use this device in order to generate the

waveform



Section 6: Pre-lab

Section 6.1 Designs of simple amplifiers

For the circuit in figure 1 with power supplies VCC = 12.0 V, VEE = -12.0 V, and assuming the op-amp is ideal, answer the following questions:

1 Design an inverting amplifier using one op-amp and two or more resistors Design it such that it has a gain of -10 (this gain is negative) Pick resistor values that you have in the lab kit Include a schematic of this circuit with the component

values labeled with your completed lab assignment

pre-Set V=V + =V - =0 (V + and V - : voltage at non-inverting and inverting node respectively.)

Gain is -10 so we choice R2=10KΩ and R1=1KΩ.

2 Simulate this inverting amplifier circuit with SPICE to make sure the circuit works as designed.

3 Design a non-inverting amplifier such that it has a gain of +11 (this gain is positive) Pick resistor values that you have in the lab kit Include a schematic

of this circuit with the component values labeled with your completed pre-lab assignment.

The op-amp is ideal: V+=V-=V3=Vi

Vi =1V and Vo=11V, the gain is 11 therefore the circuit works as designed.

Section 6.2 Analysis of integrators and differentiators

For the circuit in Figure 2 with power supplies VCC = 12 V, VEE = -12 V, and assuming the op-amp is ideal, answer the following question.

1 Derive the time-domain equation for Vo (t) in terms of Vi (t) Show that the circuit performs the function of an integrator.

Vc = Q/C, Vc = Vx – Vout = 0 - Vout

∆ - (d Vout )/(dt) = (dQ)/(Cdt)

We have dQ/dt is electric current and the node voltage of the integrating op-amp atits inverting input terminal is 0, X = 0, the input current Iin flowing through the input resistor, Rin is given as:

Iin = (Vin-0)/Rin= Vin/ Rin

The current flowing through the feedback capacitor C:

If = (CdVout)/ (dt) = (CdQ)/(Cdt) = (dQ)/(dt) = (dVoutC)/(dt)

Assuming we have ideal op-amp, no current flows into its terminal Hence, the nodal equation at the inverting input terminal:

Iin = If = Vin/ Rin = (dVoutC)/(dt)

∆ Vin/ Vout x (dt)/ RinC = 0

The ideal voltage output for the OP-amp Integrator as:

2 For the circuit in Figure 3 with power supplies VCC = 12 V, VEE = -12 V, and assuming the op-amp is ideal, answer the following question Derive the time-domain equation for Vo(t) in terms of Vi(t).Show that the circuit

performs the function of a differentiator

Iin = IF and IF = - Vout / RF

The charge on the capacitor equals Capacitance x Voltage across the capacitor

inverting input terminal of the operational amplifier

For the circuit in Figure 4 with power supplies VCC = 12 V, VEE = -12 V, and assuming the op-amp is ideal, answer the following questions:

3 What is the low-frequency gain of this circuit?

Low frequency gain: Av = -R2/R1= -20

The circuit performs nearly like the circuit in figure 2 which is an integrator circuit.

5 Use SPICE transient analysis to simulate this circuit in the time domain using a sine wave input with amplitude 300 mV and frequency 300 Hz From the SPICE output plot of the input and output waveforms, confirm that this circuit is an integrator

At frequency of 300Hz, voltage gain is reduced and can be seen significantly

Figure: Output (blue) and Input (green) signals

If Rb increases, the voltage gain also increase

Hence, Rb has a purpose to increase the voltage gain

Rb is used to prevent the capacitor from storing charge because of small offset current and voltage at input

When Rb increases, the current through capacitor is gradually decreased, then capacitor spends more amount of time to reach the steady state

Moreover, a trade-off between voltage gain and the time taking for circuit operates

in steady state

However that amount of time is really not clear

We can observe the longer amount of time for circuit to operate in steady state in the case of Rb = 100kΩ.

Rb = 20kΩ

Rb = 100kΩ

Section 6.3 Analysis and simulation of an active low-pass filter

1 Derive the equation for the amplitude of Vo(t) in terms of the input

amplitude A, input frequency ω, and circuit components R and C Do not use any numerical values Note: there is no need to derive the output phase

Use SPICE AC analysis to simulate this circuit and generate the gain plot Compare the SPICE

gain plot with your plot in item 3 above Explain any differences

3) From the formula of gain :

We can see that, value of gain depends on the value of ω = 2πf

If input signal has low frequency, the value of gain will become larger In other words, the smaller the frequency is, the bigger the value of gain is

If input signal has high frequency, the value of gain will be smaller This will attenuate the signal the input

Circuit is named “low-pass” because with low frequency, it will permit the signal to pass through, particularly, the gain of the circuit will be become larger.4) We have: Vi(t) = Acos( ωt)

In the integrator circuit,

7.1 Instruments needed for this experiment

The instruments needed for this experiment are: a power supply, an arbitrarywaveform generator, a multimeter, and an oscilloscope

7.2 Inverting amplifiers

1 Build the circuit in Figure 1 using power supplies ±12 V and the resistor values from your design in the Pre-lab, section 6.1 item 1

The real value of resistor R1= 1.012 Ω and R2 = 10.114 Ω

1 Use a sine wave input with small amplitude so that the output is not affected by

varying it using 1 2 5 sequence up to 1 MHz(i.e.set input frequency to 10 Hz, 20

Hz, 50 Hz, 100 Hz, 200 Hz, … up to 1 MHz), measure the experimental values of the gain of this circuit at each frequency Record them in a table for later data analysis

2 Display Vo on Channel 2 and adjust the time base to display 2 to 3 complete cycles of the signals.

3 Get a hardcopy output from the scope display with both waveforms to confirm that the circuit is an integrator Turn this hardcopy in as part of your lab report

Real values: R = 1.034 kΩ

R b = 20.247 kΩ

C = 0.2 µF

4 Change the input signal to a square wave with the same 300 mV amplitude (300

mV to +300 mV) and frequency 300 Hz Get a hardcopy output from the scope

display with both waveforms to confirm that the circuit is still an integrator Turn this hardcopy in as part of your lab report

5 Repeat item 4 above using a triangular input signal with same amplitude and frequency Turn this hardcopy in as part of your lab report

yellow) waveforms with triangular input signal

6 Now change the input back to a sine wave as in item 1 Remove the resistor Rb What happens to the output signal? Explain the phenomenon you observe on the oscilloscope Reinsert the resistor Rb and verify that the circuit functions as

designed.

Figure: Output (blue)

and input (yellow) signals

when R b is removed

- Explain: when Rb is removed out of the circuit, there is no DC current go through negative feedback that makes the DC offset become unstable The output signal will oscillate corresponding to the instability of the DC offset

7.4 Low-pass filters

1 Build the circuit in Figure 5 with power supplies ±12 V Use a sine wave of amplitude 100 mV as an input signal (see item 2 below for frequency) and display both the input and output signals on the oscilloscope (2 to 3 complete cycles)

2 Vary the input signal frequency in 1-2-5 sequence from 1 Hz to 100 KHz At each frequency, measure the gain of the circuit, using the data from the

oscilloscope display Keep this data in a table for later plotting

Real values: R 1 = 1.173 kΩ

R 2 = 10.514 kΩ

Frequency (Hz) V in (Vpp - V) V out (Vpp - V ) Gain (ratio)

8 Data analysis:

8.1 Inverting Amplifiers.

1 Inverting amplifiers

1.1 Compare the experimental gain measured in section 7.2 item 1 with the

calculated gain in the pre-lab and with the gain as simulated by SPICE Explain any difference between these values

The calculation gain is 10 and it is nearly 10 with small error

Measured gain Calculated gain %Error

The difference is due to

- The real values of resistors are not equal the values in calculations The real

value of resistor R1= 1.173 Ω and R2 = 10.514 Ω

- The op-amp is supposed to be ideal but actually it is not

1.2 From the table of data in section 7.2 item 1, plot the gain of this circuit as dB versus frequency, using the technique described in the Discussion section

Frequency Gain(ratio) Gain(dB)

Figure: Input (green) and Output (blue) signal in section 6.2.5

Difference: There is only small different between two pictures at the amplitudes Explanation: In the oscilloscope, the signals on the screen only display stable form

of the output signal after time while the LTspice shows us the output waveform from the beginning when the output voltage needs a small time to reach the stable state

2 With the experimental observation in section 7.3 items 6, explain the function ofthe resistor Rb

With high frequency of Vi, Rb restricts the value of Vo Without Rb, the Value of Vo will increase greatly until it reaches the value of Vcc

8.3 Low-pass filters

1.From the data in section 7.4 item 2, plot the gain (in dB) of the circuit as function

of frequency(using the technique described in the Discussion section) and compare

it with the plots in section 6.3 item 1 and in section 6.3 item 2 (SPICE plot)

Explain any differences between these 3 plots

Frequency (Hz) Gain

(ratio)

Gain (dB)

50K 0.824 -1.681

The table of Gain vs Frequency in prelab 6.3.1

There are some small differences between these graphs since:

- The real values of resistors and capacitor are not equal the theoretical ones

- The op-amp in fact is not ideal

- Some errors in the measuring instruments.

Conclusion:

This report has discussed about characteristics of circuits built with op-amps Italso pointed out how important of using ltspice simulation is to verify circuit design, to test whether the designed circuits work properly, then we can adjust our circuits to work more accurately

Through this lab we will do :

1 Design an inverting amplifier for a given gain

2 Using LTspice to simulate and analyze electrical circuits

3 Understanding about the operation of integrator, differentiator, low-pass filter and the functions of some of their components such as the feedback resistor

4 Explaining some differences between calculated and real data.The error in experiments is the result of many subjective reasons and objective reasons Thedifferences between the theoretical and practical values is the major reason which leads to error and mistakes in measurement process and some

mechanical causes are also lead to error