5.1 Dr. Yuri Panarin, DT021/4, Electronics Introduction Applications of operational amplifiers for signal shaping/ transformation require the use of nonlinear feedback networks. The amplifier with such feedback can • • • •
5. Signal Shaping
approximate transfer curves, linearize transducers, limit the amplitude of signals, perform mathematical operations, etc...
Basic to most of these nonlinear feedback networks is the use of non-linear voltage-to-current characteristics of semiconductor junctions: diodes, zener diodes, and transistors. In some applications, the large-signal switching properties of such elements are used, whereas in others the nonlinearity of the junction itself is utilized. FT221/4 Electronics – 5 Signal Shaping
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Approximation The most general way of generating such functions is through the use of piecewise linear approximation, as shown below. The accuracy of such an approximation is determined by the number of line segments used.
In this chapter we present a discussion of • feedback limiters, • diode function generators, • sine-wave shaping Recommended Text:
1. Honeycutt R.A., OpAmps and Linear Integrated Circuits, Delmar Publisher Inc. (1988) pp. 389-396 2. Tobey G.E., Graeme J.G., Huelsman L.P., Operational Amplifiers: Design and Application, McGraw-Hill Book Company (1971) pp. 251-258 3. Sedra, A.S. & Smith, K.C., Microelectronic Circuits (4th Edition), Oxford University Press (1998), pp. 880-883
The complete piecewise curve is obtained by the summation of individual line segments whose "breakpoint” voltages and slopes are determined separately for each segment.
Piecewise approximation of a nonlinear function.
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5.2 Dr. Yuri Panarin, DT021/4, Electronics Notes
Segments Such segments may be generated with simple limiter circuits and summed by the operational amplifier. The amplifier summing junction is a summation point for the currents from the breakpoint networks and the resistor RF provides the scaling function
A more practical means of obtaining the desired line segments is through the use of series and shunt limiters. FT221/4 Electronics – 5 Signal Shaping
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Notes
Shunt Limiter Method Note that this approach is used to obtain transfer function with decreasing slope
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5.3 Dr. Yuri Panarin, DT021/4, Electronics Series Limiter Method
Notes
Note that this approach is used to obtain transfer function with increasing slope
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FT221/4 Electronics – 5 Signal Shaping
Notes
Inverse function
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Note that each of the diode breakpoint circuits can be represented as a nonlinear transconductance.
By using such networks as feedback elements we obtain the inverse function.
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5.4 Dr. Yuri Panarin, DT021/4, Electronics Notes
Series Limiter in Feedback This is example of Shaper with a Series Limiter in Feedback to produce a transfer function with decreasing slope Resistors Rt1 and Rs form a voltage divider of the output Vo Vd = Vo
Rs or Vo = Vd (Rs + Rt1 ) / Rs Rs + Rt1
If Vo < 0.6V (Rs + Rt1 ) / Rs then diodes are not conducting and the amplifier gain is –Rf/Ri Otherwise one of the diode is conducting and the gain is: AV = −
R f || Req1 Ri
=−
R f Req1 ( R f + Req1 ) Ri
Req1 =
Rt21 − Rt1 Rs1 Rs1
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Notes
Temperature Compensation The main lack of these circuits is the temperature dependence of circuit parameters This problem always persists whenever use of p-n junction The forward conduction characteristics of the silicon diode are somewhat temperature-sensitive and can cause changes in the breakpoints of the curve. This effect can be compensated partially by a similar voltage drop in temperature-compensating element
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5.5 Dr. Yuri Panarin, DT021/4, Electronics Temperature Compensation (shunt)
Notes
In the shunt limiter, the base-to-emitter voltage drop offsets much of the temperature sensitivity of the diode forward voltage
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FT221/4 Electronics – 5 Signal Shaping
Temperature Compensation (series)
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Notes
In the series limiter, a second diode acts as a temperaturecompensating element for the breakpoint diode
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5.6 Dr. Yuri Panarin, DT021/4, Electronics Shunt Limiter Diode Function Generation
Notes
A more flexible approach is shown below, wherein both the locations of the breakpoints and the slopes of the line segments are individually adjustable. Note that the slopes can be positive, negative, or zero. The breakpoints can easily be made variable or can be temperaturecompensated
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FT221/4 Electronics – 5 Signal Shaping
Series Limiter Diode Function Generation
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Notes
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5.7 Dr. Yuri Panarin, DT021/4, Electronics Precision limiter
Notes
The precision limiter, with its ability to simulate ideal diodes, can be used to generate line segments whose breakpoints are precisely known and which are temperature-insensitive.
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FT221/4 Electronics – 5 Signal Shaping
Assessment (2004) 1. 1.
2.
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Notes
Describe two different approaches for signal shaping/transformation. State briefly the appropriate examples. The first approach is so-called peace-wise linear approximation where nonlinear function is approximated by several linear segments. This approach is based on non-linear networks utilising non-linear voltage-to-current characteristics of semiconductor junctions of diodes and zener diodes. Such non-linear networks can be placed in the feedback of OpAmp circuits or used straightforward in passive R-D breakpoint circuits. The second approach utilises nonlinearity of the large-signal switching properties of the p-n junction itself. Examples: (i) R-D shaper based on diode V-I characteristic and (ii) transistor amplifier with a nonlinear large-signal transfer characteristic.
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5.8 Dr. Yuri Panarin, DT021/4, Electronics Exercise A (1999, 2004)
Notes
The circuit shown in Figure is to be used to generate the following piece-wise linear function: dVO/dVS = -1; -4 V ≤ VS ≤ +4V dVO/dVS =-2; |VS| ≥ 4V Assuming ideal components and that R1 = 10 kΩ, determine values for R2, R3, R4 and the d.c. reference sources VRX and VRY. VRX R2
R2
Solution:
D1
The gain of the circuit for small input voltages is G=-R4/R1=-1, i.e. R4=R1 The gain of the circuit for high negative input voltages is G=-R4/(R1||R3)=-2, i.e. R4=2(R1||R3) i.e. R3=R1. Similar for positive input: R2=R1. Therefore R1=R2=R3=R4=10 kΩ. Breakpoint at ±4V. i.e. VRX = VRY = 4 V
R4
R1
R3 VS
D2
R3
VO
-VRY
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Exercise B (1999, 2004) If the output of this functional generator (VO) now is used to provide an input (VS1) to the circuit shown in Fig. b, which contains a four-quadrant analogue multiplier (X) with a scaling factor of 0.1, sketch to scale the voltage transfer characteristic (VO1 vs VS) of the overall system over the input range : 0 V ≤ VS ≤ +5 V.
Solution:
VX V 0.4 ⋅ VO1 =− O = R R R
Vx=(0.1) (4) VO1 =0.4 VO1
VO1 =
VO = 2.5 ⋅ VO 0. 4
5.1 Sine-Wave Shaping
VO1
X
+4V
R
15 V
10 V
R
VS1
VS1
VO1 2V
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4V
5V
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5.9 Dr. Yuri Panarin, DT021/4, Electronics Sine-Wave Shaping
Notes
We show that the Diodes can be combined with resistors to synthesize two-port networks having arbitrary nonlinear transfer characteristics. Such two-port networks can be employed in waveform shaping— that is, changing the waveform of an input signal in a prescribed manner to produce a waveform of a desired shape at the output. In this section we illustrate this application by a concrete example: the sine-wave shaper. This is a circuit whose purpose is to change the waveform of an input triangular-wave signal to a sine wave.
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Notes
Sine-Wave Shaping The sine-wave shaper is a practical building block used extensively in function generators. This method of generating sine waves should be contrasted to that using linear oscillators (Wein Bridge , etc…) Although linear oscillators produce sine waves of high purity, they are not convenient at very low frequencies. Also, linear oscillators are in general more difficult to tune over wide frequency ranges. In the following we discuss two distinctly different techniques for designing sine-wave shapers.
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5.10 Dr. Yuri Panarin, DT021/4, Electronics Sine-Wave Shaping Techniques
Notes
The sine wave shape can be obtained by using standard Function Approximation techniques as described above. In the following we discuss three distinctly different techniques for designing sine-wave shapers. • 1.Breakpoint method, • 2.Non-linear Amplification • 3.Non-linear Diode V-I characteristic
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1. Breakpoint method
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Notes
In the breakpoint method the desired nonlinear transfer characteristic (sineshape) is implemented as a piecewise linear curve. Diodes are utilized as switches that turn on at the various breakpoints of the transfer characteristic, thus switching into the circuit additional resistors that cause the transfer characteristic to change slope. Exercise: The circuit in Fig. is required to provide a three-segment approximation to the nonlinear i-v characteristic, i = 0.1v2, where v is the voltage in volts and i is the current in milliamps.
Find the values of R1, R2 and R3 such that the approximation is perfect at v = 2 V, 4 V, and 8 V. Calculate the error in current value at v = 3 V, 5 V, 7V, and 10 V. Assume ideal diodes
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5.11 Dr. Yuri Panarin, DT021/4, Electronics Notes
Breakpoint method (solution) i = 0.1v2, At v=2V, i=0.4 mA , thus R1=2/0.4=5 kΩ. For 3V V1, R5 R 4 + R5 This implies that as the input continues to rise above V1 the output follows but with a reduced slope. This gives rise to the second segment in the output waveform. Next consider what happens as the voltage at point B reaches the second breakpoint determined by V2, At this point D1 conducts, thus limiting the output Vout to V2 This gives rise to the third segment, which is flat, in the output waveform. The result is to “bend'' the waveform and shape it into an approximation of a sine wave. VOUT = V1 + (VIN − V1 )
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Notes
2. Nonlinear-Amplification Method The other method we discuss for the conversion of a triangular wave into a sine wave is based on feeding the triangular wave to the input of amplifier having a nonlinear transfer characteristic that approximates the sine function.
One such amplifier circuit consists of a differential pair with a resistance connected between the two emitters, as shown.
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5.14 Dr. Yuri Panarin, DT021/4, Electronics Notes
Design Exercise With appropriate choice of the values of the bias current I and the resistance R, the differential amplifier can be made to have a transfer characteristic that closely approximates . A detailed analysis of this circuit shows that its optimum performance when I.R=2.5VT ~ 62.5 mV. (VT ~ 25 mV at room temperature) For this design the amplitude of the input triangular voltage must be 6.6 VT (~165 mV), and the corresponding sine wave across R has a peak value 2.42 VT (~ 61 mV). Exercise: For I=0.25 mA and R=10 kΩ find the peak value of the sine wave output vo.
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Notes
Solution From I.R=2.5VT ~62.5 mV we find R=62.5 mV/0.25mA=250Ω The emitter current of the first transistor Q1 is I E1 = I + ( 2.42VT ) / R = I [1 + ( 2.42VT ) / IR] = I [1 + ( 2.42VT ) / 2.5VT ] = I (1 +
2.42 ) 2.5
The collector currents are: 2.42 2.42 I C1 = I 1 + and I C 2 = I 1 − 2.5 2.5
Finally the output (differential) voltage is: vO = (VCC − I C 2 ⋅ RC ) − (VCC − I C1 ⋅ RC ) = ( I C1 − I C 2 ) ⋅ RC = = I ⋅ RC ⋅
2 ⋅ 2.42 2 ⋅ 2.42 = 0.25 ⋅10 ⋅ = 4.84V 2.5 2.5
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5.15 Dr. Yuri Panarin, DT021/4, Electronics Notes
3.Non-linear V-I characteristic Non-linear Diode V-I characteristic can be used directly to form a non-linear shape closed to sine wave. The circuit of simple diode sine wave shaper utilizes a series resistance R and two shunting diodes connected to ground, one with anode grounded, and the other with cathode grounded. The input is a triangle wave whose amplitude is such that its zero-crossing slope equals that of the sine wave. The following exercise demonstrates the design and performance of this circuit
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Exercise
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Plot
Design a sine-wave shaper using a series resistance R and two shunting diodes. The input is a triangle wave whose amplitude is such that its zerocrossing slope equals that of the sine wave. The diodes are characterized as conducting 1 mA at 0.7 volts, with n = 2. Find the triangle-wave peak voltage, and a suitable value for R. Then find the angles θ (θ = 90° at the wave peak) where the output of the circuit is 0.7, 0.65, 0.6, 0.55, 0.5, 0.4, 0.3, 0.2, 0.1 and 0 V. Use these angles to find the values of the prototype sine wave (i.e., vo = 0.7 sin θ) and the corresponding errors. Present your results in tabular form.
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5.16 Dr. Yuri Panarin, DT021/4, Electronics Notes
Solution For a sine-wave of peak output V0, i.e. v=V0 •sin ωt, the zerocrossing slope is dv = V0 ⋅ ω cos(ωt ) t =0 = V0 ⋅ ω [V/s] dt
t =0
Thus the tri-angle wave reaches V0ω × T / 4, or V0 (2πf ) × (1 / 4 f ) = 2πV0 / 4 = 1.57 ⋅ V0 = 1.57 ⋅ 0.7V ≈ 1.1V ,
at the peak. Though the choice is arbitrary, let us assume a sine wave peak of V0=0.7V, with Id peak = 1mA, such that the triangle input peak =0.7 x 1.57 = 1.10V, with the drop across R being: (0.57) (0.7) = 0.4V. Thus R = 0.4 V/l mA = 400Ω. Now, in general, vi = 1.10 (θ/90), or θ = 81.8 vi, over the range 0° to 90°. 61
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FT221/4 Electronics – 5 Signal Shaping
Solution The forward current through the diode is i = id ≈ I s ⋅ exp( vd / nVT )
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Notes vi
400Ω
vo=vd
Thus the ratio of the currents is
i / I o ≈ exp[(vd − V0 ) / nVT ]
For given values: VT = 25 mV, n=2,V0=0.7V, I0=1mA, we have i = 1mA ⋅ exp((vd − 700) / 50 ) The input voltage can be presented as function of output as: vi = vR + vd = i ⋅ R + vo = I 0 exp[(vd − V0 ) / VT ]⋅ R + vd
And NOT vice versa
vo ≠ F (vi )
Therefore we use a following procedure FT221/4 Electronics – 5 Signal Shaping
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5.17 Dr. Yuri Panarin, DT021/4, Electronics Procedure
Notes
1. Fix the output vo=vd and find the current in the circuit i: i = 1 mA ⋅ exp((vd − 700 mV) / 50 mV )
2. Find the input voltage (vi) for fixed vo vi = vR + vd = i ⋅ R + vo = i ⋅ 400 Ω + vo 3. Find the corresponding angle θ. In general, vi = 1.10V (θ/90o) over the range from 0° to 90°, therefore θ = 81.8o ⋅ vi 4. Find the value of ideal sinusoidal signal: vsin = 0.7 V ⋅ sin(θ ) 5. And finally the output error is Error = (vo - vsin )/vo ×100%
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Solution
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Notes
Following this procedure we get: For vo = 0.65V , i = 1 exp((650 − 700) / 50 ) = 0.368mA, vi = 0.65 + 0.368 ⋅ 0.4 = 0.79V ,
θ = 81.8o ⋅ 0.79 = 65.2 o , 0.7sin (65.2o ) = 0.635V For vo = 0.55V , i = 1 exp((550 − 700) / 50 ) = 0.05mA, vi = 0.55 + 0.05 ⋅ 0.4 = 0.58V ,
θ = 81.8o ⋅ 0.58 = 46.6o , 0.7 ⋅ sin(46.6 o ) = 0.509V For vo = 0.5V , i = 1 exp((500 − 700) / 50 ) = 0.018 mA, vi = 0.5 + 0.018 ⋅ 0.4 = 0.507V , θ = 81.8o ⋅ 0.507 = 41.5o , 0.7 ⋅ sin(41.5o ) = 0.464V For vo = 0.45V , i = 1 exp(( 450 − 700) / 50) = 0.007 mA, vi = 0.45 + 0.007 ⋅ 0.4 = 0.453V ,
θ = 81.8o ⋅ 0.453 = 32.8o , 0.7 ⋅ sin(32.8o ) = 0.379V For vo = 0.35V , i = 1 exp((350 − 700) / 50 ) = 0.0009mA, vi = 0.45 + 0.0009 ⋅ 0.4 = 0.35V , θ = 81.8o ⋅ 0.35 = 28.6o , 0.7 ⋅ sin(28.6o ) = 0.335V
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5.18 Dr. Yuri Panarin, DT021/4, Electronics Summary
Notes
Present the result in tabular form
θo vo,V 0.7sinθ, V ε, mV ε, %
90 65.2 53.5 46.6 41.5 32.8 24.5 16.3
8.2 0
0.7 0.65 0.60 0.55 0.50 0.40 0.30
0.1 0
0.2
0.7 0.635 0.563 0.509 0.464 0.379 0.291 0.197 0.099 0 0
15
37
41
36
21
9
3
1 0
0
2.4
6.6
8.1
7.8
6.4
3.1
1.5
1 0
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FT221/4 Electronics – 5 Signal Shaping
Exercise 1
The circuit in Fig. is used as a sinewave shaper The input is a triangle wave whose amplitude is 5V, VCC = ± 5 V, R1 = R6 = 2 kΩ, R2 = R4 = R5 = 4 kΩ, R3 = 12 kΩ . Assume an ideal diode with voltage drop = 0.5V.
1. Find the breakpoints of this circuit and sketch the transfer function . 2. Compare the output signal with the prototype sine wave of the same amplitude and find the corresponding errors at breakpoints.
Notes +VCC D2
D1
R2 VS
70
R5 R3
R6
R4
R1
VO
D1 D2
R4 R3
R6 R5
-VCC FT221/4 Electronics – 5 Signal Shaping
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5.19 Dr. Yuri Panarin, DT021/4, Electronics Solution 1 1. Find the breakpoints of this circuit and sketch the transfer function .
The first breakpoint occurs when R3 R4 VOB1 = ECC + VD R3 + R4 R3 + R4 Or
VOB1 = ECC
VOB1 = 5V ⋅
+VCC D2
D1
R + R4 R4 + VD 3 R3 R3
R2
4k 16k + 0.5V = 5 / 3V + 2 / 3V = 7 / 3V ≈ 2.33V 12k 12k
GL = −
Notes
VS
The input (triangular source) voltage is
R6
R4
VO
D1
The gain is
R3
R1
R2 = −2 R1
R5
D2
R4 R3
R6 R5
VSB1 = VOB1 / GL = −7 / 6V ≈ −1.167V
-VCC 73
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Solution 1 1. Find the breakpoints of this circuit and sketch the transfer function .
The second breakpoint occurs when R R + R6 VOB 2 = ECC 6 + VD 5 R5 R5 2k 6k Or VOB 2 = 5V ⋅ + 0.5V ⋅ = 2.5V + 0.75V = 3.25V 4k 4k for VOB1 < VO < VOB2 the gain is
VS
+VCC D2
D1
R2
R5 R3
D2
VOB 2 − VOB1 3.25 − 2.33 = −2.083V = −1.167 + GM −1 FT221/4 Electronics – 5 Signal Shaping
R6
R4
VO
D1
The input (triangular source) voltage is
VSB 2 = VSB1 +
Notes
R1
R || R 4k || 4k GM = − 2 4 = − = −1 R1 2k
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R4 R3
R6 R5
-VCC 75
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5.20 Dr. Yuri Panarin, DT021/4, Electronics Solution 1 1. Find the breakpoints of this circuit and sketch the transfer function .
Notes +VCC D2
for VO > VOB2 the gain is D1
R || R || R 2k || 2k = −0.5 GH = − 2 4 6 = − R1 2k
R2
VS
And the output at maximum is
R5 R3
R4
R1
VO max = VOB 2 + GH (VS max − VSB 2 ) =
VO
D1
= 3.25V + 0.5(5V − 2.083V ) = 4.71V
R6
D2
R4 R3
R6 R5
-VCC 77
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Solution 1 1. Find the breakpoints of this circuit and sketch the transfer function . B C
5
Solution 2 +VCC D2
D1
4
R2 VS
2. Compare the output signal with the prototype sine wave of the same amplitude and find the corresponding errors at breakpoints
R5 R3
R6
R1
VO
2
D1
R4 R3
1
2
3
4
5
Vin
)
At first breakpoint VSB1 = −1.167V
D2
R2 VS
R3
At second breakpoint
R6
R4
VO
Err = ( 2.33 − 1.686) / 1.686 ⋅100% = 38.3% D1 D2
Err = (3.25 − 2.86) / 2.86 ⋅ 100% = 13.4%
R5
R5
R1
-VCC
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+VCC
D1
Vsin 2 = 4.71V ⋅ sin (90 ⋅ 2.083V / 5V ) = 2.86V
R6
0 0
(
o
Vsin1 = 4.71V ⋅ sin (90 ⋅1.167V / 5V ) = 1.686V
1
D2
The ideal prototype signal is
Vsin = 4.71V ⋅ sin 90 ⋅Vs / 5V R4
Vout
3
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R4 R3
R6 R5
-VCC 79
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