ECET220 Laboratory 7
Differential
(Difference) Amplifier and Audio Amplifier
I.
OBJECTIVES
1.
To
demonstrate the operation of a Differential
(Difference) Amplifier
2.
To
demonstrate the operation of an audio amplifier
II. PARTS /
Equipment List:
Equipment:
DMM (digital
multimeter)
Variable dc
power supply
Oscilloscope
Parts:
|
Qty.
|
Component
|
Tolerance Band
|
Wattage Rating, W
|
|
|
1
|
LM324
Quad Op Amp
|
|
|
|
|
1
|
LM386
Audio Power Amplifier
|
|
|
|
|
2
|
10
kΩ Resistor
|
gold
|
¼
|
|
|
2
|
20
kΩ Resistor
|
gold
|
¼
|
|
|
2
|
10
kΩ Trim Potentiometer
|
|
|
|
|
3
|
0.1
μF Capacitor
|
|
|
|
|
1
|
47
pF Capacitor
|
|
|
|
|
1
|
8
Ω Speaker
|
|
|
|
|
1
|
Proto
Board
|
|
|
|
|
|
Hookup
wires of different colors
|
|
|
III. PRE-LAB
a) Differential (difference) amplifier
Figure 1 shows the
circuit of a difference amplifier. Output voltage VOUT can be calculated from the following equation:
Simplify the above
equation using the resistor values given in Figure 1.
VOUT =
Figure 1 – Differential
(Difference) Amplifier
In Figure 1, the input potentiometers are
connected to
V. The LM324 contains four operational
amplifiers within the 14 pin IC. Make sure the positive terminal of the ±15 Vdc
supply is connected to pin 4 of the LM324 IC, the negative terminal is
connected to pin 11, and the common terminal is connected to the common line,
as shown in Figure 1. Be sure you take all meter readings with respect to the
common line of the circuit (and not the −15 V connection of the power source).
The U1D op amp is not used.
Note
that the +/- polarity of U1C is switched in Figure 1 for drawing purposes.
The U1C op amp is the
differential amplifier. U1A and U1B op amps are used as input unity gain buffer
amplifiers and provide 2 different signal inputs (VA, and VB)
to the differential amplifier.
b) Audio amplifier – Design an audio amplifier having
a gain of 20.
Introduction
The last stage in many types of radio gear is an audio amplifier of some type. The audio amplifier provides both voltage and current gain for signals from the detector stage so that a loudspeaker may be driven.
Before the advent of monolithic power ICs,
designing even a simple audio power amplifier circuit involved a fair amount of
labor. Today, ICs are available to deliver powers ranging from the sub-Watt
range (LM386) to more than 100 Watts.
In this experiment, the following will be accomplished:
You will build an audio amplifier using the
LM386 provided in the lab kit and design input/output coupling networks to meet
design specifications.
You will measure the performance of the
LM386, and learn how to measure audio power amplifier output characteristics.
DESIGN INFORMATION:
Figure 2 is the basic circuit for the audio
monitor. Refer to the LM386 data sheet for detailed information on the IC.
Figure
2 – LM386 Audio Power Amplifier
DESIGN SPECIFICATIONS:
The following are the requirements for an operational signoff. The method for
designing in compliance with these specifications will be discussed shortly.
The amplifier frequency response must be
“flat” within ± 3 dB from 100 Hz to 10 kHz. You will plot the amplifier’s
frequency response to demonstrate this.
The amplifier must provide at least 500 mW
into an 8 W load at any frequency between 100 Hz to 10
kHz with no visible distortion on the
amplifier output waveform.
DESIGN STRATEGY FOR ENSURING FREQUENCY
RESPONSE:
There are two primary factors that control
the frequency response of an amplifier circuit: the choice of active components (such as transistors
and ICs), and the design of input and
output coupling networks.
The active components in a circuit tend to
limit the high frequency response. The LM386 IC is capable of amplifying
signals with frequencies well above 20 kHz. Refer to the data sheet to find the
operating frequency range for the LM386. Unless you were to deliberately limit
the high frequency response of the circuit with a low-pass filter, the LM386
would meet the high-end response specification as-is. No designer effort is
required to meet this specification.
The low frequency response is limited by
the coupling capacitors. As frequency
is reduced, capacitive reactance (XC) increases. Therefore, loads
that are in series with capacitors receive less voltage and power at lower
frequencies. In order to minimize this problem, capacitive coupling networks
need to be designed for tight coupling
at the lowest frequency of interest.
For our audio amplification circuit, the lowest frequency of interest would be
100 Hz.
When a capacitive coupling circuit is tightly
coupled, it means that the capacitive reactance, XC, is much smaller
than the Thevenin resistance of the circuit. The result is that very little of
the total voltage drops across the capacitor. The voltage divider rule will be
used to calculate the effect that the coupling capacitor has on the signal
voltage.
With the voltage divider rule equation, you
can observe the influence of the capacitive reactance on the circuit. To
minimize the voltage drop across the capacitor, the capacitive reactance must
be much smaller than the Thevenin resistance. The design rule used to
accomplish this states that the capacitive reactance must be less than or equal
to one tenth of the Thevenin resistance.
Using the lowest frequency of interest (100
Hz) and the circuit Thevenin resistance along with the design rule equation,
the value of the capacitance can be calculated. Select a standard value
capacitor equal to or larger than the calculated capacitance.
For example, C6 is the output coupling
capacitor. The input to C6 is the IC chip output terminal. Due to the negative
feedback (internal to the IC), the output terminal has a resistance of about 0.1W.
The 8 W speaker is connected as the C6 output load.
The equivalent Thevenin resistance is therefore 8.1W.
For tight coupling, the reactance of C6 must be less than 0.81 W
at 100 Hz. knowing this, we can solve for C6. Now, select a capacitor value
equal to or greater than the calculated value for C6. (Note: The capacitor
voltage rating must be greater than the output signal peak amplitude.)
C2 is the input coupling capacitor. The input
to C2 is the circuit input. We will assume that the circuit input voltage
source has a source impedance of 0 W. The 10 kW
volume control potentiometer in parallel with the amplifier input resistance is
connected as the C2 output load. By consulting the LM386 data sheet for RIN,
the equivalent Thevenin resistance can be calculated. Using the design rule and
the capacitive reactance equation at 100 Hz, you can select a capacitor value
for C2.
DESIGN STRATEGY FOR POWER OUTPUT:
The power output available from a
direct-coupled, push-pull output amplifier like this one depends on primarily
two factors. The factors are the power supply voltage and the load resistance.
The load resistance is the 8 W
speaker.
In the LM386 and other amplifiers, the power
output is limited by the supply voltage and the available output current. To control the maximum output power, you
choose a value for VCC. This value is calculated based on the
power required and the load resistance to be driven. In this amplifier circuit,
no negative supply is used, so VRAIL-RAIL = VCC. The desired
amplifier output power is 500 mW. Using the power equation, the signal voltage
amplitude required to deliver the 500 mW can be calculated in Vrms.
The supply voltage VRAIL-RAIL must
be greater than the output signal peak to peak voltage. A design rule equation
that can be used to calculate the supply voltage is:
In your report, calculate the minimum VCC required to
obtain the specified output power. Your report should also include your
measurement of how much undistorted power was actually available with the
calculated value of VCC applied.
OTHER CIRCUIT COMPONENTS
There are other passive components that are
used in the circuit to either improve stability (prevent undesired
oscillation), or prevent the amplification of unwanted signals. They are:
C3: An RF bypass capacitor. C3 shorts RF
signals to ground before they can be passed into the LM386. Without C3, the
LM386 might amplify or detect the RF signals producing audible radio stations
at the speaker.
C4: A power supply bypass capacitor for the
LM386 internal bias network.
C1: A bypass or filter capacitor for the MAIN
power supply. C1 should be placed very close
to the IC.
R2, C5: An output “snubber” network. This
network helps to prevent the LM386 from oscillating as a result of driving an
inductive load such as the speaker. It is designed to be effective only on
signal frequencies above the audio frequency range. For design purposes, the
resistance value for R2 is usually set to RL (8 W)
and C5 is set to 0.1 µF. The R2-C5 network appears as an open circuit for
signals below 10 kHz. For signal frequencies above 10 kHz, where speaker
inductance is increasing, the snubber network appears resistive, causing the
amplifier to see a largely resistive load.
IV. PROCEDURE
A. Construction of the difference Amplifier
1.
Construct
the circuit exactly as shown in Figure 1on a breadboard. Enter the simplified
equation for VOUT that you found in Pre-Lab on the worksheet.
2.
Measure
voltages at pins 4 and 11with respect to ground and record the values in Table
1on the worksheet.
3.
Adjust
the potentiometers R11 and R13 until the voltages VA
= VB = +3 VDC. Record the DMM measurements for these
voltages in Table 2 on the worksheet.
4.
Calculate
the output voltage VOUT using the simplified formula found in part
IV.A.1. Record both the calculated value and DMM measurement of VOUT in
Table 3 on the worksheet. Also, determine the difference between the measured
value and calculated value.
5.
Adjust
the potentiometers R11 and R13 until the voltages VA
= VB = -3 VDC. Record the DMM measurements for these
voltages in Table 4 on the worksheet.
6.
Calculate
the output voltage VOUT using the simplified formula found in part IV.A.1.
Record both the calculated value and DMM measurement of VOUT in
Table 5 on the worksheet. Also, determine the difference between the measured
value and calculated value.
7.
Adjust
the potentiometers R11 and R13 to obtain VA =
-1 VDC and VB = +3 VDC. Record the DMM measurements for
these voltages in Table 6 on the worksheet.
8.
Calculate
the output voltage VOUT using the simplified formula found in part
IV.A.1. Record both the calculated value and DMM measurement of VOUT in
Table 7 on the worksheet. Also, determine the difference between the measured
value and calculated value.
9. Answer the questions in section IV.A.9 on the
worksheet.
B. Construction
of the Audio Amplifier
1. Design the audio monitor circuit. Show all
calculations clearly in your “Design Method” on the worksheet.
Note: The pin 1 and pin 8 terminals are not
connected at this time. These pins are used to provide additional feedback to
the IC as described in the data sheet.
2. Connect a signal function generator as the V1
input. The circuit input signal should be a sine wave initially set at 1 kHz.
Turn on the circuit power and then the generator power. While monitoring the
amplifier output (test point D), adjust the generator amplitude and the volume
control potentiometer until the speaker output waveform clips. Now, reduce the
amplifier input signal amplitude until no clipping occurs on the amplifier
output.
Briefly enjoy the sound produced from your
speaker. Vary the generator frequency and appreciate the “music.” Indicate on
the worksheet if you could hear the sound as you changed the frequency.
3. Now, disconnect the 8 W
speaker replacing it with an 8 W, 1 W (or higher)
non-inductive resistor. This prevents the inductive speaker from affecting your
readings.
Measure the frequency response of the
amplifier from 10 Hz to 100 kHz inclusive. A sequence of 10, 20, 50, and 100 …
is recommended. Prior to recording your measurements, check to make sure that
the amplifier output is not distorted. Record your oscilloscope measurements
taken at test points A (signal input), B (volume control input), C (amplifier
input), D (amplifier output), and E (load voltage). Record all data in Table 8
on the worksheet.
4. Using the measured data, calculate the gain
for each stage both in units of Volts per Volt and dB (decibels). The input
coupling capacitor stage gain is calculated by dividing the voltage reading at
test point B by the voltage reading at test point A. The volume control gain is
calculated by dividing the test point C voltage reading by the test point B
voltage reading. The LM386 amplifier gain is found by dividing the test point D
voltage by the test point C voltage. The output coupling capacitor stage gain
is determined by dividing the voltage at test point E by the voltage at test
point D. The total system gain of the amplifier circuit is calculated by
dividing the output signal amplitude at test point E by the input signal
amplitude at test point A.
5. Graph the overall gain (in dB) as a function
of the signal frequency using semi log graph provided on the worksheet (you may
copy the template provided in the worksheet, paste it onto MS Paint, and add
your graph).
6. Using the capacitive coupling design rule,
calculate the coupling capacitor output voltage for an input voltage of 1 Vrms
at the lowest frequency of interest (100 Hz).
Calculate the coupling capacitor output
voltage for an input voltage of 1 Vrms at 10 kHz.
Compare the measured
amplifier voltage gain to the LM386 data sheet voltage gain. Subtract the
theoretical value from the measured value, expressing the result in dB.
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