2013 Checkpoints Chapter 10
AMPLIFIERS
Question 378
If the gain needs to be 500, the gradient must be 500. The upper limit on the linear region (angled straight line
section) is 4mV peak. This means that the range of inputs is 2 × 4mV = 8mV.
The linear region needs to be centred on 0 ± 4mV
The output voltage needs to be ×500, ∴4mV × 500 = 2V, and -4mV × 500 = -2V.
Hence the scale on the graph needs to be as follows:
Output Voltage (V)
2
0
-2
-4 -2
Input Voltage (mV)
0 2 4
Question 379
The output signal will have exactly the same shape, but the scale on the vertical axis will be ×500 the original.
From the original graph the input voltage varies from +2mV to -2mV (assumed).
-3
-3
∴ the output voltage will range from (+ 2 × 10 × 500) = 1V to (- 2 × 10 × 500) = -1V
Output Voltage (V)
1
0
2
4
6
8
time (ms)
-1
Question 380
The gain of the amplifier is given by
VOUT
4divisions
=
= ×2. The voltage is also inverted since the output
2divisions
VIN
graph is upside down compared with the input graph.
Question 381
If VOUT (Peak to Peak) = 2V, then VIN (Peak to Peak) = 1V. ∴ Each division of the graph = 0.5V.
This means that VIN is centred on 1.5V with ±0.5Vpeak.
∴ DC component = 1.5V, AC = ±0.5Vpeak
Question 382
The voltage gain is the gradient of the graph, in the linear region.
−12
The gradient is
= -6
∴ gain is ×6 (inverting) (ANS)
2
Question 383
Input Voltage
1.0
0
-1.0
Time
Question 384
The linear region of the graph includes input voltages between -1.0 and +1.0V.
Since the gain is ×6, this means that the output voltage will be not be clipped and will be between ±4.8 V.
Output Voltage
4.8
0
-4.8
Question 385
Output Voltage (V)
8.0
0
-60 -40 -20
0
Input Voltage (mV)
20 40 60
-8.0
Question 386
The voltage gain is the gradient of the graph, in the linear region.
-16
The gradient is
= -200
∴ gain is ×200 (inverting) (ANS)
80 x 10−3
Question 387
The maximum AC voltage that can be amplified without distortion needs to be within the range of ±40mV, as
long as the signal is centred on 0V.
This would give a range of 80mVpeak-peak. (ANS)
Question 388
The maximum input is 10mV, this is well within the linear range of the amplifier, so there will not be any distortion
(clipping) of the input signal. Therefore the output signal will be a sinusoidal wave.
The amplification is ×200 and inverting. This means that 10mV ×-200 = -2.0V, and -10mV ×-200 = 2.0V.
Output Voltage (V)
2.0
1.0
0
-1.0
-2.0
0
10
20
30
40
time (ms)
Question 389
To amplify 2.4mV to 3.6V the signal needs to be increased by a factor of ×1500.
∴B
(ANS)
Question 390
To eliminate the distortion, due to clipping, the output of amplifier A1 must be no larger than 48mV.
If the input signal is 2.4mV then
A1 = ×20 (ANS)
Question 391
The RMS voltage generated is actually the input to the amplifier. Therefore the RMS output has been multiplied
by ×150, from the gain of the amplifier.
This means that the input signal needed to be 2.1V ÷ 150 = 14mV.
From the table of values given, this will be equivalent to a speed of 1.4m/s. ANS
Question 392
To convert RMS to peak-peak, multiply RMS × 2 2
∴ 2.1 × 2 2 = 5.9V (ANS)
Question 393
The range of voltages, that can be amplified without clipping (distortion), is from -12mv to 12mV.
Therefore, 24mV. This means that the peak to peak range of the AC must be 24mV.
To convert peak to peak to RMS divide by 2 2
∴ 24 ÷ 2 2 = 8.5mV (ANS)
Question 394
-3
If the gain is ×120, then a 10mV input will lead to a 10 × 10 × 120 output. ∴Output = 1.2V
Output Voltage (V)
1.2
0
-10
10
Input Voltage (mV)
-1.2
Note that the vertical scale used in the answer is a little misleading.
Question 395
Amplification = ×200, ∴23mV (within linear range) will be amplified by a factor of ×200, producing a 4.6VRMS
signal. 4.6VRMS = 4.6 × 2 2 Vpeak-peak = 13 Vpeak-peak. (ANS)
Question 396
If VOUT = ±6V, then VIN = ±6 ÷ 200 = ±30mV
VOUT
6V
-30 mV
30 mV
VIN
-6V
Question 397
The signal generator has a sinusoidal output with a peak to peak voltage of 50.0 mV.
Therefore the RMS output will be 50.0 ÷ 2 2 = 17.7 mV (ANS)
Question 398
The gain of an amplifier is the ratio of
Vout
.
Vin
In this case that is equal to
5
50 ×10-3
= 100 (ANS)
The value for Vout is read from the CRO screen. The graph is 5 squares high (peak to peak) and the scale is one
square = 1.0 volt.
Question 399
1
.. In this case T = 4 squares on the CRO. On the time axis each
T
square is equal to 0.5 ms. So the period is 2.0 ms.
1
1
∴f=
=
= 500 Hz (ANS)
T
2 × 10 −3
The frequency of the output is given by f =
Question 400
∆Vout
gain =
50 =
∆Vin
12
∆Vin
∴∆Vin =
12
50
∴∆Vin = 0.24 Vpeak
0.24
∴ ∆Vin RMS =
2
= 0.17 Vrms (170 mV) (ANS)
Note: this is answer 401 in the book.
Question 401
There will be no difference in the sound.
Question 402
Find the ratios of the peak values of both output and input voltages.
∆Vout
gain =
∆Vin
gain =
1.0
−0.2
∴ gain = × -5 (ANS)
Question 403
The frequency of the output is given by f =
∴f=
1
. From the graph the period (1 cycle) is 20 ms.
T
1
1
=
= 50 Hz (ANS)
T
20 × 10−3
Question 404
The peak output is 1.0 V. To convert Vpeak to VRMS divide by
∴ 1.0 ÷
2.
2 = 0.71V (ANS)
Question 405
-3
Power is given by P = VI. To find the PRMS then we need to use VRMS × IRMS. (Use 10mA as 10 × 10 A)
-3
∴ PRMS = 0.71 × (10 × 10 ÷ 2 ) = 5mW (ANS)
Question 406
To convert peak to peak to RMS divide by 2 2 .
∴ 2.5 ÷ 2 2 = 0.88 V (ANS)
Question 407
The frequency of the output is given by f =
∴T=
1
.
T
1
1
-4
=
= 8.3 × 10 s = 0.83 ms
f
1200
(ANS)
Question 408
The maximum gain possible is given by 14 ÷ 2.5 = 5.6
∴ B, C, D are all possible (ANS)
Question 409
∆Vout
gain =
∆Vin
2-0
gain = 0.04 - 0
gain = × - 50
∴ magnitude of gain is × 50 (ANS)
Question 410
The amplification is linear and inverting.
∴ A (ANS)
The statement in the answers that “distortion’ is not on the syllabus, is misleading, as ‘clipping’ and the resultant
distortion of the signal IS on the course.
Question 411
The magnitude of the amplification is given by the gradient of the graph.
=
6
120 ×10
-3
= × 50 (ANS)
Question 412
The amplifier has a region of linear amplification when the input voltage is between ±60mV.
The negative slope demonstrates that the output signal is inverted from the input signal.
The horizontal sections vin >+60 mV,
vin < -60mV will clip the output signal, since in these regions the amplification is zero and
vin > +60mV will clip the output signal, since in these regions the amplification is zero.
The output signal is inverted, amplified by 50, and clipped when the magnitude of the input signal is greater than
60mV.
The maximum value is 60 mV × 50 = 3 V.
3
1
t (ms)
1.0
-3
2.0
Question 413
The voltage gain of the amplifier is the gradient of the linear region of the VOUT – VIN graph. Be sure to check for
units.
∆Vout
∆Vin
gain =
−
gain =
9−9
0.03 − − 0.03
gain = - 300 (ANS)
Question 414
Any input voltage that is greater than 30 mV will result in an output voltage of 9.0 V. Any input voltage that is
less than 30 mV will result in an output voltage of 9.0 V. From this we will get an output voltage graph of:
VOUT (V)
9
0
-
0.02
0.01
time (s)
9
Question 415
The linear region is before the clipping occurs.
Use Gain =
Vout
Vin
Use the points (on the graph) VOUT = -10,
When VIN = 200 mV
∴ Gain =
10
0.2
∴ Gain = 50
(ANS)
Question 416
If you are in doubt about how to draw this, just plot a few points from the previous two graphs.
i.e. At t = 0, the voltages are (0, 0),
at t = 5, the voltages are (200, -10)
clipping occurs once VIN > 200 mV
V
10
-200
200
-10
mV
Question 417
The signal will have the same shape, with a maximum value of 2V.
2
0
-2
Question 418
Clipping will occur when the input voltage is outside the linear region of the amplifier. This is when the input
voltage is greater than 10 mV or less than -10 mV.
If the input signal is clipped the output will be a flat line at ± 4 V.