AEC Lab Manual Jan-2016

List of Experiments.1. BJT Current Series Feedback.2. Op-Amp Voltage Shunt Feedback.3. Wien Bridge Oscillator with Op-amp.4. Hartley Colpitt’s Oscillators with Op-amp.5. Square & Triangular Wave Generator with Op-

Amps.6. First Order Band Pass Active Filter.7. Class-AB Push-Pull Power Amplifier.8. Buck Switch mode Power Supply.9. Boost Switch mode Power Supply.

10. Analog to Digital Converter using ADC 0808.11. Digital to Analog Converter DAC 0808.

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CURRENT-SERIES FEEDBACK AMPLIFIER

AIM: To measure the voltage gain of current - series feedback amplifier.

CIRCUIT DIAGRAM:

R = 100, RE = 1.5K, RS= RL = 4.7K R1 = 1M (Set Ic = 1ma), CC = 10µF,

CE = 100µF, VCC = 10V, Transistor - BC 547

1. To remove feedback, Short R.

2. By varying R, one can change the feedback.

Procedure

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1. Determine the Voltage Gain, Input, Output Impedances and BW,

with and without Feedback. Observe the changes in values.

2. Find the gain at 5 KHz. Multiply it with 0.707. Increase the

frequency till you get the same value. It is f2. Repeat the same till

you get low frequency f1.

Current Shunt Feedback

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R = 100, RE = 1.5K, RS = RL = 4.7K, R1 = 1M (Set Ic = 1ma), CC = 10µF, CE

= 100µF, VCC = 10V, Transistors - BC 547

Follow the same procedure as shown above.

Voltage-shunt feedback amplifier(Inverting Amplifier with Feedback).

You cannot find the Gain without Feedback in this case. But you can varythe amount of feedback by changing RF. Ri& Ro cannot be determined.

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Voltage Shunt

RF = 47K,RE = 1.5K, RS = RL = 4.7K, R1 = 1M (Set Ic = 1ma)

CC = C = 10µF, CE = 100µF, VCC = 10V

Follow the same procedure as shown above.

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Wien- Bridge Oscillator

R1 = R2 = 1.5k, C1 = C2 = 0.1µF, f0 = 1KHz. Ri= 1K, RF = 4.7K Pot.

1. Adjust the Pot till you get a clean Sine Wave.2. For various values of R or C determine the frequency of oscillation.3. The Wien Bridge will produce “Zero” phase shift at only at

resonance frequency determined from above formula. At all otherfrequencies, a different phase angle which will not satisfy theBarkhausen criteria hence all other frequencies oscillations die down.

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The Colpitts oscillator

The Hartley oscillator

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Triangular/rectangular wave generator.

R3 = 1K, R1 = 12K, R2 = 180 , C = 0.1µF, f = 1KHz.

=

A is a comparator. Assume that the comparator output is +Vsat. The voltage atpoint P = Vsat×(R2/R1). During this time, the integrator output is steadily falling.When the ramp voltage at P equals with Vsat X R2/R1, comparator switches toopposite saturation level.

(R2/R3) = P/P Vramp = 2VSat× . The P/P swing of both the waveforms is as shown infig. Square wave between +Vsat to –Vsat, & Triangular wave is between

Vsat to – Vsat.

Sawtooth wave generator:

P

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The difference between the triangular and sawtooth waveform is that the rise time of thetriangular wave is always equal to its fall time while in sawtooth wave generator, rise time maybe much higher than its fall time, or vice versa. The triangular wave generator can beconverted into a sawtooth wave generator by injecting a variable dc voltage into the non-invertingterminal of the integrator. This can be done by using a potentiometer as shown in figure. When thewiper of the potentiometer is at the Centre, the output will be a triangular wave since the duty cycleis 50 %. If the wiper moves towards -V, the rise time of the sawtooth becomes longer than the falltime. If the wiper moves towards +V, the fall time becomes more than the rise time.

Procedure:

1. Vary C & find out the frequency.2. Vo1 is Square wave & Vo2 is Sawtooth/Triangular Wave.

P

R3=1K

R1

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5 Volt Output Switching Regulator using LM2575

Aim: Design and verify the operation of 5 volt output switching regulatorusing LM2575.

Apparatus: LM2575, 47 uF, 330 uF, 120 uH, varactor diode ERB81-004, 0-30 v power supply.

The LM2575 series of regulators are monolithic integrated circuits that

provide all the active functions for a stepdown (buck) switching regulator,capable of driving a 1A load with excellent line and load regulation. Thesedevices are available in fixed output voltages of 3.3V, 5V, 12V, 15V, and anadjustable output version. Requiring a minimum number of externalcomponents, these regulators are simple to use and include internalfrequency compensation and a fixed-frequency oscillator.

The LM2575 series offers a high-efficiency replacement for popular three-terminal linear regulators. It substantially reduces the size of the heat sink,and in many cases no heat sink is required. A standard series of inductorsoptimized for use with the LM2575 are available from several differentmanufacturers. This feature greatly simplifies the design of switch-modepower supplies.

Other features include a guaranteed ±4% tolerance on output voltagewithin specified input voltages and output load conditions, and ±10% onthe oscillator frequency. External shutdown is included, featuring 50 A(typical) standby current. The output switch includes cycle-by-cyclecurrent limiting, as well as thermal shutdown for full protection underfault conditions.

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5V to 12V Boost Converter using LM2577

Aim: Design and verify the operation of 5v to 12v boost converter usingLM 2577.

Apparatus: LM2577, IN5822, 220 uF/35v, 0.33 uF, 2.2 K, 100 uH, 18 K,2K, 1000uF/50v, 0.1 uF, Power supply.

The LM1577/LM2577 are monolithic integrated circuits that provide all ofthe power and control functions for step-up (boost), flyback, and forwardconverter switching regulators. The device is available in three differentoutput voltage versions: 12V, 15V, and adjustable. Requiring a minimumnumber of external components, these regulators are cost effective, andsimple to use.

Pin Diagram

Operation

The LM2577 turns its output switch on and off at a frequency of 52 kHz,and this creates energy in the inductor (L). When the NPN switch turns on,the inductor current charges up at a rate of VIN/L, storing current in theinductor. When the switch turns off, the lower end of the inductor fliesabove VIN, discharging its current through diode (D) into the outputcapacitor (COUT) at a rate of (VOUT VIN)/L. Thus, energy stored in the

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inductor during the switch on time is transferred to the output during theswitch off time. The output voltage is controlled by the amount of energytransferred which, in turn, is controlled by modulating the peak inductorcurrent. This is done by feeding back a portion of the output voltage to theerror amp, which amplifies the difference between the feedback voltageand a 1.230V reference. The error amp output voltage is compared to avoltage proportional to the switch current (i.e., inductor current during theswitch on time). The comparator terminates the switch on time when thetwo voltages are equal, thereby controlling the peak switch current tomaintain a constant output voltage.

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Digital to Analog Converter

Typical DACs are available in either current output (IDAC) or voltageoutput (VDAC) configurations. While the voltage output VDACs are moreconvenient to implement, they tend to be slower and more expensive thantheir current output counterparts (refer to figure 3-1 for a VDAC example).Therefore, for high-speed applications, circuit designers usually choosecurrent output IDACs and then use a high-speed op-amp to provide the I-Vconversion at the output of the DAC. For some low cost applications, anIDAC with a simple RC filter on its output is often enough to meet certainnon-demanding, high-input impedance applications.

Mathematical analysis of the DAC circuit is as follows:Resolution of the DAC:

1. Number of bits2. Vref

Equation #1: DAC resolution = Vref/( 2n-1)Vref = the DAC reference voltagen = No. of bits (e.g., 8-bit DAC)

Resolution = 5V / 256 = 19.5 mVThis means that the smallest analog voltage step size that can berepresented by the DAC with aVref = 5V is 19.5 mV.

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To determine Vout for any binary value input:

Equation #2: Vout = Vref (N / 2n)

Where: Vout is the DAC’s output voltage(After I–V conversion with an IDAC or the Vout of a VDAC)Vref = the DAC reference voltageN = is the decimal equivalent to the binary input valuen = No. of bits accommodated by the DAC (e.g., 8-bit DAC=256)

If we close the switch for only the LSB (S0 in figure 3-1, binary value =00000001) we should see approximately 20 mV at the output, because:

5V (1/256) = 19.5 mV

With all 8 switches closed (i.e., binary value = 11111111) we should seeapproximately5V (255 / 256) = 4.98V at the output (this is illustrated in figure)

Result:

Review Questions:

1) What is the name for this type of DAC?

2) The output voltage is the ---------- sum of all the input voltages in this circuit.

3) List the other types of DAC’s that are available?

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Analog to Digital Converter

Aim: To design and verify Analog to digital converter operation

Apparatus: ADC 0808, LED’s, Function generator, connecting wires

Theory: The ADC 0808 is an 8-bit A-to-D converter, having data lines D0-D7. It works on the principle of successive approximation. It has a total ofeight analogue input channels, out of which any one can be selected usingaddress lines A, B and C. Here, in this case, input channel IN0 is selectedby grounding A, B and C address lines.

Usually the control signals EOC (end of conversion), SC (startconversion), ALE (address latch enable) and OE (output enable) areinterfaced by means of a microprocessor. However, the circuit shown hereis built to operate in its continuous mode without using anymicroprocessor. Therefore the input control signals ALE and OE, beingactive-high, are tied to Vcc (+5 volts). The input control signal SC, beingactive-low, initiates start of conversion at falling edge of the pulse, whereasthe output signal EOC becomes high after completion of digitization. ThisEOC output is coupled to SC input, where falling edge of EOC output actsas SC input to direct the ADC to start the conversion.As the conversion starts, EOC signal goes high. At next clock pulse EOCoutput again goes low, and hence SC is enabled to start the next conversion.

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Thus, it provides continuous 8-bit digital output corresponding toinstantaneous value of analogue input. The maximum level of analogueinput voltage should be appropriately scaled down below positive reference(+5V) level. The ADC 0808 IC requires clock signal of typically 550 kHz,which can be easily derived from an astable multivibrator, constructedusing 7404 inverter gates.

In order to visualize the digital output, the row of eight LEDs (LED1through LED8) have been used, wherein each LED is connected torespective data lines D0 through D7. Since ADC works in the continuousmode, it displays digital output as soon as analogue input is applied. Thedecimal equivalent digital output value D for a given analogue inputvoltage Vin can be calculated from the relationship.

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AEC Lab Manual Jan-2016

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