AN67

A Product Line of
Diodes Incorporated
AN67
Designing with shunt regulators – mixing, adding or
summing
Peter Abiodun A. Bode, Snr. Applications Engineer, Diodes Incorporated
Introduction
This application note demonstrates how a three-terminal shunt regulator may be used to
implement a simple summing circuit or mixer. It is an extension of the subject first introduced in
AN66 which shows how a shunt regulator can be used as an AC amplifier.
The proposal
Figure 1 shows the AC amplifier. Because feedback through R1 maintains the reference pin at a
constant DC value, this point represents an AC virtual earth or “ve”. It means that this point can
be used as a summing junction for several independent inputs. This is shown in Figure 2.
C2
R3
Vcc
Vout
10k
1μF
R1
100k
REF1
C1
R4
1μF
10k
Vin
ZR431
R2
100k
ve
GND
Figure 1 - AC amplifier using a reference
C2
R3
Vcc
Vout
10k
1μF
R1
100k
Cg1
V1
1μF
10k
Cgn
Rgn
1μF
10k
Vn
ve
Rg1
REF1
ZR431
R2
100k
GND
Figure 2 - Shunt regulator as a general multi-input summing amplifier
The transfer function of the circuit is given by
⎛ v
v
v ⎞
v out = R1⋅ ⎜⎜ 1 + 2 + ... + n ⎟⎟
Rgn ⎠
⎝ Rg1 Rg2
This is the basic idea of the summing amplifier. The nature of the output depends on the nature
of the inputs. Consider, for example, the 2-input amplifier shown in igure 3
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C2
R3
Vcc
Vout
5k
1μF
R1
100k
ve
C1
R4
1μF
10k
C3
R5
1μF
10k
V1
V2
REF1
ZR431
R2
100k
GND
Figure 3 - Two-input amplifier
f1 = f2
If both v1 and v2 are of similar bandwidth then the output is a straightforward amplified phasor
sum of the two inputs.
For example, suppose v1 and v2 are given by:
v1 = V1 ⋅ sin ωt
v 2 = V2 ⋅ sin(ωt + α )
The output voltage, vO, is of the form
v O = −VO ⋅ sin(ωt + θ )
where
and
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VO = G AC ⋅ V12 + V22 + 2V1V2 . cos α
⎛
⎞
⎟
⎜ V 2 + V 2 + 2V V . cos α ⎟
2
1 2
⎝ 1
⎠
θ = cos −1 ⎜
V1 + V2 . cos α
2
Equation 1
Equation 2
Equation 3
(see Appendix)
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The result is shown in Figure 5, based on a simulation of Figure 4:
1u
v out
R3
5k
1u
v in1
R1
100k
10k
R5
C3
1u
v in2
10
V cc
C2
10k
0 Sine(0 100m 1k -250u 0)
U1
Z R431
R4
C1
V1
Load
10k
R2
100k
V2
0 Sine(0 50m 1k 0 0)
Figure 4 - Simulation circuit demonstrating summing or adding
100
80
60
40
mV
20
0
-20
-40
-60
vo u t
vin 2
vin 1
-80
1
0.8
0.6
v out / V
0.4
0.2
-0
-0.2
-0.4
-0.6
-0.8
-1
168
169
170
171
Tim e/m S ec s
172
1m S ec s /div
Figure 5 - Simulation result of figure 4
Figure 5 shows
And
AC gain,
Therefore,
v in1 = 100mV ⋅ Sinωt
v in 2 = 50mV ⋅ Sin(ωt +
π
2
)
- blue trace (f = 1kHz)
- black trace (f = 1kHz)
GAC = 10
VO = (10 ⋅ 0.1)2 + (10 ⋅ 0.05 )2
- red trace (f = 1kHz)
12 + 0.5 2 = 1.118V
⎞
⎟ = 1.107Rads
⎜ 12 + 0.5 2 ⎟⎟
⎠
⎝
⎛
θ = cos −1 ⎜⎜
Hence
0 .5
v O = −1.118Sin(ωt + 1.107 )
i.e. vO leads vin1 by 1.107 radians
or about 63.43° and is inverted.
If v1 and v2 are of different frequencies, one of two things will happen as follows.
f2< f1 < 2.f2
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If f1 and f2 are different but the ratio of separation is less than 2, the two frequencies will “beat”
together. “Beating” is interference between two slightly different frequencies which manifests as
a periodic variation in amplitude of a higher frequency. This is illustrated in the simulation results
in Figure 7
v 1 = V sin ω1t
If
and
v 2 = V sin ω 2 t
The output voltage vO is given by;
⎛ ω + ω2 ⎞
⎛ ω − ω2 ⎞
= −2V cos⎜ 1
⎟t
⎟t ⋅ sin⎜ 1
⎝ 2 ⎠
⎝ 2 ⎠
Equation 4
The cosine term contains half the frequency difference between f1 and f2 but, due to its
interaction with the sine term, the waveform envelope it produces is that of f1-f2, or beat
frequency. The sine term behaves like a carrier signal (for the beat frequency) whose frequency
is the average of f1 and f2.
The beat frequency can produce interesting acoustic effects when used for mixing audio
frequencies when it is perceived as a third tone. This is because beating can also occur with
complex waveforms due to harmonics of one signal interacting with close harmonics of another
– known as inter-modulation distortion.
1u
v out
R3
5k
1u
v in1
R5
1u
v in2
C1
V1
0 Sine(0 100m 1.1k 0 0)
R1
100k
50k
C3
10
V cc
C2
50k
U1
Z R431
R4
Load
10k
R2
100k
V2
0 Sine(0 200m 1k 0 0)
Figure 6 - 2-input shunt-regulator mixer illustrating beat frequency phenomenon
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v in1 / m V
80
40
20
-20
-40
-80
v in2 / m V
80
40
20
-20
-40
-80
1.5
v out / V
1
0.5
0
-0.5
-1
-1.5
150
155
160
165
170
175
Tim e/m S ec s
5m S ec s /div
Figure 7 - Beat frequency output
In the above example v1 has a frequency of 1.1kHz and v2 1kHz. This generates a beat frequency
of 100Hz. In audio processing, these non-harmonic tones are sometimes referred to “off-key
notes”.
f1 >2 f2
If the two signals have widely different frequencies, then they simply add together in a manner
where the two signals are visibly combined.
This is illustrated in Figure 8 and Figure 9.
1u
v out
R3
5k
1u
v in1
R5
1u
v in2
C1
V1
0 Sine(0 100m 10k 0 0)
R1
100k
10k
C3
10
V cc
C2
10k
U1
Z R431
R4
Load
10k
R2
100k
V2
0 Sine(0 50m 1k 0 0)
Figure 8 - Shunt regulator summing amplifier – f1 > 2f2.
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80
60
40
mV
20
0
-20
-40
-60
vo u t
vin 2
vin 1
-1 0 *:V2 _ P
-80
1
V
0.5
0
-0.5
-1
119
119.5
120
120.5
121
121.5
122
Tim e/m S ec s
500uS ec s /div
Figure 9 - Simulation result of summing amplifier – f1 > 2f2 - Figure 8
The two input signals v1 and v2 ([email protected] and [email protected] respectively) are shown
together on the top trace (blue and black). An inverted copy of v2 is displayed on the output to
show the relationship between the output and the inputs.
Conclusion
This application note shows that a shunt regulator can be used as a summing amplifier or mixer
using the same basic configuration. This demonstrates the flexibility of a shunt regulator.
Recommended further reading
AN66 - Designing with Shunt Regulators – AC Amplifier
AN57 - Designing with Shunt Regulators – Shunt Regulation
AN58 - Designing with Shunt Regulators – Series Regulation
AN59 - Designing with Shunt Regulators – Fixed Regulators and Opto-Isolation
AN60 - Designing with Shunt Regulators – Extending the operating voltage range
AN61 - Designing with Shunt Regulators – Other Applications
AN62 - Designing with Shunt Regulators – ZXRE060 Low Voltage Regulator
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Appendix - Proof of Equation 1
v 1 = V1 ⋅ sin ωt
Given
v 2 = V2 ⋅ sin(ωt + α )
v O = −(v 1 + v 2 ) = −VO ⋅ sin(ωt + θ )
and
Determine VO and θ
Solution
Represent v1, v2 and vO on a phasor diagram as shown below.
V
VO
V2
φ
θ
α
V1
V
Figure 10 - Phasor diagram representation of v1, v2 and vO
VO2 = V12 + V22 − 2V1V2 cos φ
- applying cosine rule
cos φ ≡ cos(π − α ) ≡ − cos α
- identity
Gives
VO2 = V12 + V22 + 2V1V2 cos α
Equals
VO = V12 + V22 + 2V1V2 cos α
cos θ =
After substitution
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- as required.
V1 + V2 cos α
VO
⎤
⎥
⎢⎣ V + V + 2V1V2 cos α ⎥⎦
⎡
θ = cos −1 ⎢
V1 + V2 cos α
2
1
- as required.
2
2
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The circuits in this design/application note are offered as design ideas. It is the responsibility of the user to ensure that the circuit is fit for
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or
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