Download 6 Log and AntiLog Amplifiers

5.1
Dr. Yuri Panarin, DT021/4, Electronics
Introduction
Log and Antilog Amplifiers are non-linear circuits in which the
output voltage is proportional to the logarithm (or exponent) of the
input.
It is well known that some processes such as multiplication and
division, can be performed by addition and subtraction of logs.
They have numerous applications in electronics, such as:
6. Log and AntiLog
Amplifiers
•
•
•
•
1
FT221/4 Electronics – 6 Log and AntiLog Amplifiers
Two basic circuits
Multiplication and division, powers and roots
Compression and Decompression
True RMS detection
Process control
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Notes
There are two basic circuits for logarithmic amplifiers
• (a) transdiode and
Ri
(b) diode connected transistor
Ri
Q
Q
vi
vi
vo
vo
Most logarithmic amplifiers are based on the inherent logarithmic
relationship between the collector current, Ic, and the base-emitter
voltage, vbe, in silicon bipolar transistors.
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5.2
Dr. Yuri Panarin, DT021/4, Electronics
Transdiode Log Amplifier
Notes
The input voltage is converted by R1 into a current, which then flows through the
transistor's collector modulating the base-emitter voltage according to the input
voltage.
The opamp forces the collector voltage to that at the noninverting input, 0 V
From Ebers-Moll model the collector current is
I c = I s (e qVbe / kT − 1) = I s (eVbe / VT − 1) ≈ I s ⋅ eVbe / VT
where Is is saturation current, q is the charge of the electron 1.6x10-9 Coulombs,
k is the Boltsman’s constant 1.38x10-23 Joules, T is absolute temperature, VT is
thermal voltage.
I c = I s (e 38.6Vbe − 1) ≈ I s ⋅ e 38.6Vbe
For room temperature 300oK
The output voltage is therefore
i
Vout = −Vbe = −VT ln C
 IS
 v

V
 = − T lg i
2.3  Ri I S

 
Vin 

 =  − 0.0259 ⋅ ln
Ri ⋅ Is 
 
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Dynamic range of Log Amp.
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Notes
Test: Discuss the factors limiting the dynamic range of transdiode
log amplifier. Suggest the methods to increase dynamic range
(a) The output is a perfect log function when IC>>IS. For the small
input Voltage (i.e. current) this limits the lower end of the dynamic
range.
Vbe / V
Vbe / V
qVbe / kT
I c = I s (e
− 1) = I s (e
 I + IS
Vout = −Vbe = −VT ln C
 IS
T
− 1) ≈ I s ⋅ e

I
 ≈ −VT ln C

 IS
T



To extend the lower end of the dynamic range use transistor with
small IS., e.g. for LM394 IS =0.25pA
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5.3
Dr. Yuri Panarin, DT021/4, Electronics
Dynamic range of Log Amp.
Notes
(b) At upper end of the dynamic range the limitations is due to the
bulk resistance of base and emitter regions – rBE. Therefore Vbe
must be corrected to
I +I 
Vbe = VT ln C S  + rBE I C
 IS 
output error is: actual output – ideal output
 I (1 + p / 100) 
I
 − K ln i
OutputError = K ln i
IS


 IS

 = K ln(1 + p / 100)

Typically rBE is in range from 0.25 Ω to 10 Ω
To extend the upper end of the dynamic range use transistor with
small rBE.
e.g. for LM394 rBE =0.5 Ω
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Dynamic range of Log Amp.
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Notes
(c) The second factor is non-idealities of opamp, i.e. input bias
current IOS
 I C + I OS 
Vout = −VT ln

IS


and offset voltage VOS.
 RI + RI OS − VOS 
 V + RI OS − VOS
 = −VT ln i
Vout = −VT ln C
RI S
RI S






this limits the lower end of the dynamic range
To extend the lower end of the dynamic range use ultra-low offset
opamps or special offset nulling techniques.
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5.4
Dr. Yuri Panarin, DT021/4, Electronics
Dynamic range of Log Amp.
Notes
For LM394 rBE =0.5 Ω, and IS =0.25pA (at room temperature).
Estimate the log conformity error at IC=1mA, 100 µA and 10 µA.
For IS=1 mA the output error is 0.5Ω x 1mA= 0.5 mV.
Therefore
0.5mV = 26mV ln(1 + p / 100)
this gives
p = (exp( 0.5mV / 26 mV ) − 1) ⋅ 100% ≈ 1.94%
Estimate the max dynamic range with-in log conformity 1%
outputerror = 26mV ln(1 + 1% / 100%) ≈ 0.26mV
The upper limit is IC=0.26mV x /0.5 Ω =0.52 mA
The lower limit is 0.25 pA / 1% =25 pA.
The dynamic range is 0.52mA/25 pA=0.02 x 109 =2 x 107
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Notes
Thermal and Frequency stability
This equation yields the desired logarithmic relationship over a
wide range of currents, but is temperature-sensitive because of VT
and IS resulting in scale-factor and offset temperature-dependent
errors.
The system bandwidth is narrower for small signals because
emitter resistance increases for small currents.
The source impedance of voltage signals applied to the circuit must
be small compared to Ri. Omitting Ri yields a current-input log
amp.
Using a p-n-p transistor changes the polarity of input signals
acceptable but limits the logarithmic range because of the degraded
performance of p-n-p transistors compared to n-p-n transistors
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5.5
Dr. Yuri Panarin, DT021/4, Electronics
IC Log Amps.
Notes
These basic circuits needs additional components to improve the
overall performance, i.e:
•
•
•
•
•
to provide base-emitter junction protection,
to reduce temperature effects,
bulk resistance error and op amp offset errors,
to accept bipolar input voltages or currents,
and to ensure frequency stability.
Such circuit techniques are used in integrated log amps: AD640,
AD641, ICL8048, LOG100, 4127.
IC log amps may cost about ten times the components needed to
build a discrete-component log amp.
Nevertheless, achieving a 1% logarithmic conformity over almost
six decades for input currents requires careful design.
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Temperature Compensation
 v
vo = −VT ln i
 R1I S
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Notes



The equation for output voltage shows that the scale factor of
the basic transdiode log amp depends on temperature because of
VT and
that there is also a temperature-dependent offset because of IS.
Temperature compensation must correct both error sources.
Figure (next slide) shows the use of a second, matched,
transistor for offset compensation and a temperature-dependent
gain for gain compensation.
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5.6
Dr. Yuri Panarin, DT021/4, Electronics
Temperature Compensation
R1
Q1
vi
R4
D1
Notes
R3
+to
Q2
V1
R2
vo
Ir
 v 
vo = −VT ln i 
 R1 I S 1 
Temperature compensation in a transdiode log amp:
a second transistor (Q2) compensates the offset current (IS) and
a temperature-sensitive resistor (R4) compensates the scale factor
VT
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22
Notes
Temperature Compensation
For transistors Q1 & Q2 we have
FT221/4 Electronics – 6 Log and AntiLog Amplifiers
 v 
 I 
vBE1 = VT ln i  vBE 2 = VT ln r 
R
I
 1 S1 
 IS2 
where Ir is a reference, temperature-independent, current.
The output voltage will be
 R 
 R 
 R   I I R 
vo = v1 1 + 3  = (vBE 2 − vBE1 )1 + 3  = VT 1 + 3  ln r ⋅ S 1 1 
R
R
4 


 R4   I S 2 vi 
4 
Matched transistors (IS1 = IS2) will cancel offset.
In order to compensate the gain dependence on temperature, R4 must be much
smaller than R3 and such that d(VT/R4)/dT = 0.
This requires dR4/R4 = dVT/VT (= l/T).
At T = 298 K, the temperature coefficient of R4 must be 3390 x 10-6K.
D1 protects the base-emitter junction from excessive reverse voltages.
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5.7
Dr. Yuri Panarin, DT021/4, Electronics
Notes
Stability Considerations
Transdiode circuits have a notorious tendency to oscillate due to the presence of
an active element in the feedback that can provide gain rather than loss.
Consider the voltage-input transdiode. Ignoring op amp input errors, we have
Vn = Vi − R ⋅ I c and
Vo = −VBE
The feedback factor β for a given value of Vi,
R
Q
is determined as
β = dVn / dVo = R ⋅ dI c / dVBE
vi
Differentiating IC and using the fact that
Ic = Vi/R, we obtain
β = R ⋅ I c / VT = Vi / VT
vo
indicating that β can be greater than unity.
For instance, with Vi = 10 V we have β = 10/0.026 = 400 = 52 dB, indicating
that in the Bode diagram the |1/β | curve lies 52 dB below the 0 dB axis.
Thus, the |1/β| curve intersects the |a| curve at fc >> ft, where the phase shift
due to higher-order poles is likely to render the circuit unstable; an additional
source of instability is the input stray capacitance Cn
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Notes
Range Considerations
The transdiode circuit is compensated by means of an emitter
resistor RE to decrease the value of β and a feedback capacitor CF
to combat Cn, as shown.
To investigate its stability, refer to the incremental model, where
the BJT has been replaced by its common-base small-signal model.
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5.8
Dr. Yuri Panarin, DT021/4, Electronics
Range Considerations
Notes
Transistor parameters re and ro depend on the operating current Ic,
re = αVT / I C ≈ VT / I C
ro = V A / I C
where VA is called the Early voltage (typically ~ 100 V). Cµ is the basecollector junction capacitance. Both Cµ and Cn are typically ~10 pF range.
R1 = R || ro || (rd + R ) and R 2 = re + RE
KCL at the summing junction yields
Eliminating ie and rearranging yields
vn (1 / R1 + jω (Cn + C µ ) ) + α ⋅ ie + jωC F (vn + vo )
vn
1 + jωR1C1
1 + jωR 2C F
= vo
R1
R2
where ie=-vo/R2 and C1=Cn+Cµ+CF
1 vo R 2 1 + jωR 2C F
≡ =
β vn R1 1 + jωR1C1
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Notes
Range Considerations
1 vo R 2 1 + j ( f / f z )
1
1
where f z =
and f p =
= =
⋅
β vn R1 1 + j ( f / f p )
2πR1C1
2πR 2C F
The | 1/b| curve has a low-frequency asymptote at R2/R1, a high-frequency
asymptote at C1/CF, and two breakpoints at f=fz and f= fp.
While C1/CF and fz are constant, R2/R1 and fp depend on the operating current
IC. As such, they can vary over a wide range of values.
The hardest condition is when Ic =
Ic(max), since this minimizes the value
of R2/R1 while maximizing that of fp,.
As a rule of thumb, RE is chosen to make
R2(min)/R1 ~ 0.5 for a reasonably low
value of |β |max,
CF is chosen to make fp(max) ~ 0.5 fc for
reasonable phase margin.
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5.9
Dr. Yuri Panarin, DT021/4, Electronics
Speed of Response
Notes
As the input level is decreased, we witness an increasing dominance of fp ,
which slows down the dynamics of the circuit.
Since at sufficiently low current levels re>>RE, we have fp=1/(2πreCF)
The corresponding time constant is τ = reCF=(VT/IC)CF =(VT/ Vi)RCF
indicating that τ is inversely proportional to the input level, as expected.
For instance, with Ic = 1 nA and Cp = 100 pF, we have τ = (26 x 10-3/10-9) x
100 x 10-12 = 2.6 ms.
It takes 4.6 τ for an exponential
transition to come within 1 percent of its
final value, therefore our circuit will
take about 12 ms to stabilize to within 1
percent.
This limitation must be kept in mind
when operating near the low end of the
dynamic range.
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Diode-connected Log Amp
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Notes
In the second circuit a BJT connected as a diode to achieve the logarithmic
characteristic.
The analysis is the same as above for the transdiode connection, but the
logarithmic range is limited to four or five decades because the base current adds
to the collector current.
On the pro side,
• the circuit polarity can be easily changed by reversing the transistor,
• the stability improves, and
• the response is faster.
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5.10
Dr. Yuri Panarin, DT021/4, Electronics
Notes
Input Current Inversion
The basic log amp in only accepts positive input voltages or currents.
Negative voltages or currents can be first rectified and then applied to the log
amp, but this adds the errors from the rectifier.
Alternatively, the log amp can be preceded by a precision current inverter.
The current inverter in Figure below uses two matched n-p-n transistors and a
precision op amp to achieve accurate current inversion.
The collector-base voltage in both Q1 and Q2 is 0 V, so that the Ebers-Moll
model for BJT transistors leads to
ie1 = I ES 1 (e vBE 1 / VT − 1) 

ie 2 = I ES 2 (e vBE 2 / VT − 1) 
where IES1 and IES2 are the respective
emitter saturation currents of Q1 and Q2.
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Notes
Input Current Inversion
From circuit inspection, assuming an op amp with infinite open-loop gain but
finite input currents and offset voltage,
ie1 = ii + I b1

io = ii


= v BE1 + Vio 

ie 2 = io + I b 2
Solving for the output current in terms
of the input current yields
v BE 2

I 
I ES 2 Vio / VT
+ I ES 2 1 + b1 eVio / VT − I ES 2 − I b 2
e
I
I ES 1
ES 1 

which shows that, in order to have
small gain and offset errors, the
offset voltage must be small
compared to VT,
the op amp offset current must be
small compared to the input current,
and Q1 and Q2 must be matched.
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5.11
Dr. Yuri Panarin, DT021/4, Electronics
Exponential Amplifiers
Notes
An exponential or antilogarithmic amplifier (antilog amp), performs the
function inverse to that of log amps:
its output voltage is proportional to a base (10, e) elevated to the ratio
between two voltages.
Antilog amps are used together with log amps to perform analog
computation.
Similar to Log Apms there are two basic circuits for logarithmic
amplifiers
• (a) transdiode and
• (b) diode connected transistor
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Notes
Antilog Amplifier
Interchanging the position of resistor and transistor in a log amp yields a basic
antilog amp.
The base-collector voltage is kept at 0 V, so that collector current is given by
ic ≈ I s ⋅ exp(v BE / VT )
vo = iC R1 = I S R1 exp(−vi / VT )
and for negative input voltages we have:
There is again a double temperature dependence because of IS and VT.
Temperature compensation can be achieved by the same technique shown for
log amps.
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5.12
Dr. Yuri Panarin, DT021/4, Electronics
Notes
Temperature Compensation
The input voltage is applied to a voltage divider that includes a temperature sensor. If R3
ic1 ≈ I s1 exp(vBE / VT ) = Vr / R5
» R4, vBC1 ~ 0V and applying
to Q1
yields
I c = I s (exp(vBE / VT ) − 1)
where Vr is a reference voltage and we have assumed VBE1>>VT (25 mV).
In Q2 VBC2 = 0V and hence : ic 2 ≈ I s 2 exp( vBE 2 / VT ) = Vo / R 5
Also:
R4
vi
= vBE1 − vBE 2
R 4 + R3
Substituting vBE1 and vBE2, and solving for vo,
if Ql and Q2 are matched yields
v
R1
R4
vo ≈ Vr
exp( − i
)
R5
VT R3 + R 4
Therefore, if the temperature
coefficient of R4 is such that
dR4/R4 = dVT/VT = l/T the
voltage divider will compensate
for the temperature dependence
of VT. At T = 298 K, the
temperature coefficient of R4
must be 3390 x 10-6K.
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Log-Antilog
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Notes
Log and antilog amp circuits include the same elements but
arranged in different feedback configurations.
Some integrated log amps have uncommitted elements allowing us
to implement antilog amps.
Some IC (like ICL8049) are a committed only antilog amp.
Some so-called multifunction converters (AD538, LH0094, 4302)
include op amps and transistors to simultaneously implement log
and antilog functions, or functions derived thereof, such as
•
•
•
•
multiplication,
division,
raising to a power,
or taking a root
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5.13
Dr. Yuri Panarin, DT021/4, Electronics
Notes
Basic Multiplier
Multipliers are based on the fundamental logarithmic relationship
that states that the product of two terms equals the sum of the
logarithms of each term.
This relationship is shown in the following formula:
ln(a x b) = ln a + ln b
This formula shows that two signal voltages are effectively
multiplied if the logarithms of the signal voltages are added.
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Multiplication Stages
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Notes
The multiplication procedure take three steps:
1. 1. To get the logarithm of a signal voltage use a Log amplifier.
V1 = ln(V1 ) and V2 = ln(V2 )
*
*
2. 2. By summing the outputs of two log amplifiers, you get the
logarithm of the product of the two original input voltages.
VO* = V2* + V2* = ln(V1 ) + ln(V2 ) = ln(V1 ⋅V2 )
3. 3. Then, by taking the antilogarithm, you get the product of the
two input voltages as indicated in the following equations:
VO = exp(VO* ) = exp[ln(V1 ⋅V 2) ] = V1 ⋅ V2
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5.14
Dr. Yuri Panarin, DT021/4, Electronics
block diagram of an analog
multiplier
Notes
The block diagram shows how the functions are connected to multiply two input
voltages.
Constant terms are omitted for simplicity.
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Basic Multiplier Circuitry
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Notes
The outputs of the log amplifier are
stated as follows:
V 
Vout (log1) = − K1 ⋅ ln in1 
 K2 
V 
Vout (log 2) = − K1 ⋅ ln in 2 
 K2 
where K1 = 0.025 V, K2 = R⋅Iebo and R =
R1 = R2= R6.
The two output voltages from the log
amplifiers are added and inverted by the
unity-gain summing amplifier to produce
the following result:
 V 
 V 
Vout ( sum ) = K1 ⋅ ln in1  + ln in 2  =
 K 2 
  K2 
 Vin1 ⋅ Vin 2 
= K1 ⋅ ln
2

 K2 
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5.15
Dr. Yuri Panarin, DT021/4, Electronics
Notes
This expression is then applied to the antilog amplifier; the expression for the
multiplier output voltage is as follows:
1
 Vout ( sum ) 
 V ⋅ V 
 = − K 2 ⋅ exp K1 ⋅ ln in1 2 in 2  =
Vout (exp) = − K 2 ⋅ exp
K
K
1
1


 K 2 

 V ⋅V 
V ⋅V
= − K 2  in1 2 in 2  = − in1 in 2
K2
 K2 
The output of the antilog (exp) amplifier is a constant (1/K2) times the product
of the input voltages.
The final output is developed by an inverting amplifier with a voltage gain of —
K2.
 V ⋅V 
Vout = − K 2  − in1 in 2  = Vin1 ⋅ Vin 2
K2 

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Four-Quadrant Multipliers
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Notes
Four-Quadrant Multiplier is a device with two inputs and one
output.
V1
Vout
Vout = k ⋅ V1 ⋅ V2
V2
Typically k = 0.1 to reduce the possibility of output overload.
It is called four-quadrant since inputs and output can be positive or
negative.
An example device is Motorola MC1494, powered by ± 15 V
power supply
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5.16
Dr. Yuri Panarin, DT021/4, Electronics
Multiplier Applications
Notes
Alongside the multiplication Multipliers have many uses
such as:
•
•
•
•
•
Squaring
Dividing
Modulation / demodulation
Frequency and amplitude modulation
Automatic gain control
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Notes
AM & Squaring
Amplitude Modulation
VLF
Vout
VRF
Squaring circuit
Vin
Vout
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Notes
Divider by feedback
Divider
Vm = K ⋅ Vx ⋅ Vout
i1 =
Vin
R1
i2 =
Vm
R2
Vin = −Vm = − K ⋅ Vx ⋅Vout
Vout =
Vm
Vin
=−
K ⋅ Vx
K ⋅ Vx
Square root: If Vx = Vout
Vout = −
Vin
K ⋅ Vout
Vout =
− Vin
K
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Test problems
Sketch the diagram for transdiode log amplifier and define its gain.
Describe the stability problem of this circuit.
Suggest the model to improve stability range. Use the BJT common –base
small-signal model shown on the Figure.
4.
In this circuit let R=10 kΩ, 1 mV < Vi < 10 V, Cµ + Cn = 20 pF, VA = 100 V,
rd = 2 MΩ, and f1= 1 MHz. Find suitable values for Cf and RE.
For this circuit, find the time needed for output voltage to come within 1 % of
its final value (in worst case).
Discuss the factors limiting the dynamic range of transdiode log amplifier.
Suggest the methods to increase dynamic range
For LM394 rs ==0.5 Ω, and IS =0.25pA (at room temperature). Estimate the
log conformity error at IC=1mA, 100 µA and 10 µA.
Estimate the max dynamic range with-in log conformity 1%
6.
7.
8.
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Notes
1.
2.
3.
5.
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Notes
Transistor parameters re and ro depend on the operating current Ic,
ro = V A / I C
re = αVT / I C ≈ VT / I C
where VA is called the Early voltage (typically ~ 100 V). Cµ is the basecollector junction capacitance. Both Cµ and Cn are typically ~10 pF range.
R1 = R || ro || (rd + R ) and R 2 = re + RE
KCL at the summing junction yields
Eliminating ie and rearranging yields
vn (1 / R1 + jω (Cn + C µ ) ) + α ⋅ ie + jωC F (vn + vo )
vn
1 + jωR1C1
1 + jωR 2C F
= vo
R1
R2
where ie=-vo/R2 and C1=Cn+Cµ+CF
1 vo R 2 1 + jωR 2C F
≡ =
β vn R1 1 + jωR1C1
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Analogue Multipliers
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Notes
In analog-signal processing the need often arises for a circuit that
takes two analog inputs and produces an output proportional to
their product.
Such circuits are termed analog multipliers.
There are two different approaches to analog multipliers
One of them is based on log/antilog amplifiers
Another utilizes the exponential transfer function of bipolar
transistors (Gilbert cell) .
In following sections we consider applications of IC multipliers
based on log/antilog amplifiers
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Notes
Log/Antilog Converter
The log and antilog functions can be combined in slide rule fashion to
perform such operations as
• multiplication,
• division,
• exponentiation, and
• root computation.
With the help of simple op amp circuitry it can be configured for
additional operations, such as
• multifunction conversion and
• non-integer exponent approximations,
• coordinate conversion, and
• true rms-to-dc conversion.
Although now the tendency is to implement these functions digitally,
considerations of cost and speed often require their implementation in
analog hardware.
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Multifunction Converters
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Notes
A multifunction converter (4302) is a circuit that accepts three inputs, Vx, Vy, and Vz
and yields an output Vo of the type:
m
Vo = KV y Vz 
 Vx 
where K is a suitable scale factor (typically K = 1),
and m is a user-programmable exponent, in the
range 0.2 < m < 5
where K is a suitable scale
factor (typically K = 1),
and m is a userprogrammable exponent, in
the range 0.2 < m < 5
By proper selection of
input configuration and
exponent, the circuit can be
programmed for a variety
of operations:
Vo = VxV y , Vx / Vz , Vzm ,
n
Vz , 1 / Vx etc.
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Notes
4302 block diagram
The circuit diagram of 4302 is shown with frequency compensation and reversepolarity protection omitted for simplicity.
Vy
V
V
By op amp action, we have
V
I z = z Io = o
Ix = x I y =
Rz
Ro
Ry
Rx
The voltages at pins 6
and 12 are proportional
to the log ratios of the
corresponding currents:
I 
V6 = VT ln z 
 Ix 
I 
V12 = VT ln o 
 Iy 
 
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m=1
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Notes
1. V6 and V12 are derived directly from V11 so that V6 = V12 = V11 .
By V6 = VT ln (I z / I x ) = V12 = VT ln (I o / I y ) this implies Iz/Ix = Io/Iy that is,
Vz/Vx = Vo/Vy.
Thus,
V 
Vo = V y  z 
 Vx 
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Notes
m<1 and m>1
2. m < 1: V6 is derived directly from V11 while V12 is derived from V11 via a voltage
divider, V12=mV11, where m=R2/(R1+R2).
Letting V12=mV6 yields ,
ln(I o / I y ) = m ln( I z / I x ) = ln( I z / I x ) m
that is,
(Io / I y ) = (I z / I x )m
This, in turn, yields ,
(Vo / Vy ) = (Vz / Vx ) m
that is,
m
V 
R2
Vo = V y  z  , where m =
<1
R1 + R 2
 Vx 
3. m > 1: V12 is derived directly from V11 while V6 is derived from V11 via a voltage
divider, V6=(1/m)V11, where (1/m)=R2/(R1+R2).
Letting V6=V12/m yields
m
V 
R1 + R 2
Vo = V y  z  , where m =
>1
R2
V
 x
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Multiplication and Division
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Notes
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Exponentiator - Root Extractor
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Notes
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Notes
4302 Adjustment
In each configuration the scale factor is calibrated by setting the input(s) to 10 V
and adjusting Ry for Vo = 10 V.
To maintain the accuracy of division at low signal levels, the input offset errors
of the X and Z op amps must be nulled as follows
1. With Vz = Vx = 10.0 V, adjust R1 for Vo = 10.0 V.
2. With Vz = Vx = 100 mV, adjust R2 for Vo = 10.0 V.
3. With Vx = 100 mV and Vz = 10.0 mV, adjust R3 for Vo 1.00 V.
Repeat the procedure, if necessary.
The 4302 provides the following accuracies:
• multiply, ±0.25 percent;
• divide, ±0.25 percent;
• square, ±0.03 percent;
• square root, ±0.07 percent.
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Dr. Yuri Panarin, DT021/4, Electronics
Notes
Test (2004 Suppl.)
The circuit diagram of 4302 is shown in Fig.2 with frequency compensation and reversepolarity protection omitted for simplicity. Assume that RX = RY = RZ = RO. The pins 6,
11, 12 are connected as follows where R1=R2= 15 kΩ. Find the expression for Output
Voltage VO.
[13 marks]
(c) Make appropriate changes/connections to produce expression for Output Voltage
[5 marks]
V = 5⋅ 3 V /V
o
1
2
RY
VY
iY
QY
QO
iO
RO
13
AY
VX
1
VZ
7
RX
iX
RZ
iZ
AO
QZ
2
VO
QX
AY
AO
6
11
12
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FT221/4 Electronics – 6 Log and AntiLog Amplifiers
4302 Test
V 
Vo = V y  z 
 Vx 
Vo = 5 ⋅ 3 V1 / V2
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Notes
m
1. m=1/3, pins 6 and 11 are short circuit, pin 12 – volt. divider
2. m=R2/(R1+R2)=1/3, 3 R2=R1+R2, R1=2 R2
3.
4.
5.
6.
V1 is connected to pin VZ
V2 is connected to pin VX
pin VY is connected to +5 V
R3=2R4 is voltage divider for +5V
+15V
15V
14
VY
13
Vx
1
VZ
7
-15V
10
3
2
4302
Vo
R3
6
+5V
11
12
R4
R1
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4302 Test
4302 Test
V 
Vo = V y  z 
 Vx 
Vo = 16 ⋅V1 / V 24
m
15V
-15V
14
VY
V 
Vo = V y  z 
 Vx 
Vo = 2 ⋅V1 / V2
10
15V
3
13
VY
Vx
1
VZ
7
VZ
7
6
93
12
94
4302 Test
V 
Vo = V y  z 
 Vx 
m
14
V 
Vo = V y  z 
 Vx 
Vo = V13 /V22
15V
VY
11
Vo
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4302 Test
Vo = 2 ⋅V1 V2
2
12
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3
4302
1
11
10
13
Vo
Vx
6
-15V
14
2
4302
m
-15V
10
15V
3
13
14
2
4302
VY
13
1
Vx
1
VZ
7
VZ
7
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2
6
95
-15V
10
4302
Vo
Vx
6
m
11
Vo
12
96