Log-Antilog Circuit Applications

The RMS value of a signal is the square root of the mean of its square:

Vrms = √( (1/T) ∫₀^T v² dt )

For a sine wave v = Vm sin ωt, Vrms = Vm/√2 = 0.707 Vm. Log and antilog amplifiers can compute this directly. Squaring and square-rooting become multiplication of a log by 2 and by ½.

Block diagram (explicit method):

 v(t) -> [Precision rectifier |v|]
          -> [LOG] -> [x(-2)] -> [ANTILOG] = v^2
          -> [LPF / averager]              = mean(v^2)
          -> [LOG] -> [x(-1/2)] -> [ANTILOG]
          -> Vrms

Steps:

1. A precision full-wave rectifier gives |v|, because a log amp accepts only one polarity.
2. The log amp gives −K ln|v|. An inverting amplifier of gain −2 makes it 2K ln|v|. The antilog amp then gives v².
3. A low-pass filter or averager, with a time constant much longer than the signal period, gives the mean of v². For a sine wave this is Vm²/2.
4. A second log amp, a gain of −½ and an antilog amp take the square root:

Vo = √( mean(v²) ) = √(Vm²/2) = Vm/√2 = 0.707 Vm

Implicit method (fewer stages): the output is fed back into the log–antilog block to form Vo = mean(v²/Vo). At steady state Vo² = mean(v²), so Vo = Vrms directly. This gives a wider dynamic range.

The result is a true-RMS reading, correct for any waveform and not only sine waves. It is used in true-RMS voltmeters and power measurement. Example IC: AD536.

Why log–antilog circuits are required

Logarithms turn multiplication into addition and powers into multiplication by a constant:

ln(xy)  = ln x + ln y
ln(x/y) = ln x − ln y
ln(xⁿ)  = n ln x

Adders, subtractors and gain stages are easy to build with op-amps. So log and antilog amplifiers let simple linear circuits perform non-linear operations:

• Analog multiplication and division of signals
• Powers and roots (square, square root, xⁿ), RMS and vector magnitude
• Compression of signals with a very wide dynamic range (5–6 decades) into a small voltage range, for example in sonar, radar, audio and dB meters
• Linearizing sensors with exponential responses, and pH or light measurement

Antilog (analog) multiplier

Building blocks used (all voltages in volts, Vref = 1 V, K = scale factor such as VT or 1 V/decade):

• Log amp (L): output = −K ln(Vin), for Vin > 0
• Antilog amp (E): output = exp(Vin/K)
• Inverting amp of gain −m: output = −m × input

 V1 -> [LOG A1] --Va--[R]--+
                           |       R
 V2 -> [LOG A2] --Vb--[R]--+--+--/\/\/--+
                              |         |
                            (-)\  A3    |
                       GND--(+)  >------+--> Vs
 Vs -> [ANTILOG A4] -> Vo

Working:

1. Log stages: A1 and A2 (each with a transistor in feedback) give:

Va = −K ln V1
Vb = −K ln V2

2. Inverting summer A3 with all resistors R:

Vs = −(Va + Vb) = K (ln V1 + ln V2) = K ln(V1·V2)

3. Antilog stage A4:

Vo = exp(Vs/K) = exp( ln(V1·V2) ) = V1·V2

With a general reference voltage the scaled result is Vo = V1·V2/Vref. For example, V1 = 2 V and V2 = 3 V give Vo = 6 V (Vref = 1 V).

Division: replace the summer with a difference amplifier. Then Vs = K ln(V1/V2) and Vo = V1/V2.

Limitation: the inputs must be positive (one-quadrant operation). Matched transistors and temperature-compensating resistors are needed for accuracy.

Square-root circuit

Taking a square root means multiplying the logarithm by ½: √Vi = exp(½ ln Vi).

Building blocks used (all voltages in volts, Vref = 1 V, K = scale factor such as VT or 1 V/decade):

• Log amp (L): output = −K ln(Vin), for Vin > 0
• Antilog amp (E): output = exp(Vin/K)
• Inverting amp of gain −m: output = −m × input

                                R/2
                          +---/\/\/---+
                          |           |
 Vi -> [LOG] --V1'--[R]---+--(-)\     |
                                  >---+--> V2'
                         GND--(+)/
 V2' -> [ANTILOG] -> Vo

Derivation:

1. Log amp: V1' = −K ln Vi
2. Inverting amplifier with gain −(R/2)/R = −½:

V2' = −½ × (−K ln Vi) = (K/2) ln Vi

3. Antilog amp:

Vo = exp(V2'/K) = exp(½ ln Vi) = Vi^(1/2)

Result: Vo = √Vi. With a general reference, Vo = √(Vref·Vi). For example, Vi = 9 V gives Vo = 3 V (Vref = 1 V). The input must be positive.

Circuit for V0 = √((V1·V2)² + (V3·V4)²)

Plan: form (V1V2)² and (V3V4)² with log–antilog stages, add them, then take the square root.

 V1->[LOG]-+
           +->[summer, gain -2]->[ANTILOG]->(V1V2)^2 -+
 V2->[LOG]-+                                         |
                                       [SUMMER] <----+
 V3->[LOG]-+                              ^    |
           +->[summer, gain -2]->[ANTILOG]+    | S
 V4->[LOG]-+              (V3V4)^2             v
          S -> [LOG] -> [gain -1/2] -> [ANTILOG] -> V0

Step by step:

1. Log amps give −K ln V1, −K ln V2, −K ln V3 and −K ln V4.
2. Inverting summer with gain 2 (input resistors R, feedback 2R):

Va = −2(−K ln V1 − K ln V2) = 2K ln(V1V2)

The antilog gives exp(Va/K) = (V1V2)². In the same way the second branch gives (V3V4)².

3. A non-inverting summer (or an inverting summer followed by an inverter) gives:

S = (V1V2)² + (V3V4)²

4. The square-root block (log, gain −½, antilog) gives:

V0 = √S = √( (V1V2)² + (V3V4)² )

All inputs must be positive (or first passed through precision rectifiers, since the result depends only on squares). The circuit uses 8 log or antilog amps (5 log, 3 antilog) plus the summing and scaling op-amps.

The idea is to work in the logarithmic domain:

ln V0 = α (ln Vx − ln Vy) + ln Vz
V0 = exp[ α ln(Vx/Vy) + ln Vz ] = (Vx/Vy)^α · Vz

So the circuit must find logs, subtract, multiply by α, add, and take the antilog.

Building blocks used (all voltages in volts, Vref = 1 V, K = scale factor such as VT or 1 V/decade):

• Log amp (L): output = −K ln(Vin), for Vin > 0
• Antilog amp (E): output = exp(Vin/K)
• Inverting amp of gain −m: output = −m × input

Block diagram

 Vx->[LOG A1]--Vx'-->(-)[A4: diff amp]
 Vy->[LOG A2]--Vy'-->(+)[ gain 1     ]--Vd
                                          |
             [A5: non-inverting, gain α]<-+
                          |
                          Ve
                          v
                    (+)[A6: diff amp]--Vs
 Vz->[LOG A3]--Vz'->(-)[ gain 1     ]   |
                                       v
                          [ANTILOG A7] --> V0

Design steps

1. Log amplifiers (transistor in feedback, matched, all with the same K):

Vx' = −K ln Vx,  Vy' = −K ln Vy,  Vz' = −K ln Vz

2. Difference amplifier A4 (all four resistors R = 10 kΩ), with Vy' on the (+) input and Vx' on the (−) input:

Vd = Vy' − Vx' = K (ln Vx − ln Vy) = K ln(Vx/Vy)

3. Non-inverting amplifier A5 with gain α = 1 + Rf/R1:

Ve = α K ln(Vx/Vy) = K ln( (Vx/Vy)^α )

Since α > 1, a non-inverting amplifier can give it directly:

Rf = (α − 1) R1

For example, α = 2.5 with R1 = 10 kΩ gives Rf = 15 kΩ. Make Rf variable to set any α.

4. Difference amplifier A6 (all resistors R), with Ve on the (+) input and Vz' on the (−) input:

Vs = Ve − Vz' = K ln( (Vx/Vy)^α ) + K ln Vz
   = K ln( (Vx/Vy)^α · Vz )

5. Antilog amplifier A7:

V0 = exp(Vs/K) = (Vx/Vy)^α · Vz

Check

Take Vx = 4 V, Vy = 2 V, Vz = 1.5 V, α = 2. Then V0 = (2)² × 1.5 = 6 V.

Notes: Vx, Vy and Vz must be positive. The log amps use matched transistors and temperature-compensating resistors, so Is and VT cancel. The output is scaled by the reference voltage (1 V here).

A log amplifier gives an output proportional to the logarithm of its input, Vo = −V_T ln(Vi/(R·Is)), and an antilog amplifier gives Vo = −R·Is·e^(Vi/V_T). Used together, they turn multiplication, division and powers into simple addition, subtraction and scaling of voltages.

Main applications:

1. Analog multiplier: log V1 + log V2, then antilog → Vo ∝ V1·V2.
2. Analog divider: log V1 − log V2, then antilog → Vo ∝ V1/V2.
3. Raising to a power / taking roots: scale the log output by m (resistor ratio), then antilog → Vo ∝ Vi^m (m = 2 for squaring, m = 0.5 for square root).
4. Multifunction converter: Vo = Vy·(Vz/Vx)^m, one circuit giving multiply, divide, power and root.
5. True RMS converter: squaring, averaging and square-rooting a signal using log-antilog blocks.
6. Signal compression: a log amp compresses a wide dynamic range (e.g. 1 mV to 10 V) into a small output range, used in audio, sonar, radar and pH/light measurement.
7. Decibel (dB) meters: output directly proportional to log of signal level.
8. Linearising sensors that have exponential response (e.g. photodiodes, thermistors).

Limitation: simple log amps work only for one polarity of input (one-quadrant) and need temperature compensation because V_T and Is depend on temperature.

A four-quadrant multiplier gives Vo = K·Vx·Vy for both positive and negative values of Vx and Vy. The standard circuit is the Gilbert cell (variable-transconductance multiplier), used in ICs such as AD633 and MC1496.

Circuit

           +Vcc        +Vcc
            |            |
           RL           RL
            |  Io1   Io2 |
            +---+    +---+---> Vo (differential)
            |   |    |   |
           Q3  Q4   Q5  Q6     upper pairs: input Vx
            \  /     \  /       (Q3,Q6 bases at +Vx/2,
             \/       \/         Q4,Q5 bases at -Vx/2)
             I1        I2
             |         |
            Q1 ------ Q2       lower pair: input Vy
              \      /
               \    /
                IEE (tail current source)
                 |
               -VEE

Collectors of Q3 and Q5 are joined (current Io1), and collectors of Q4 and Q6 are joined (current Io2).

Derivation

For a BJT differential pair with tail current I and differential input V, the collector current difference is

ΔI = I · tanh(V / 2V_T)

Lower pair (Q1, Q2), input Vy:

I1 − I2 = IEE · tanh(Vy / 2V_T)

Upper pairs: Q3–Q4 is fed by I1 and Q5–Q6 by I2, both driven by Vx but cross-connected:

(IC3 − IC4) =  I1 · tanh(Vx / 2V_T)
(IC6 − IC5) =  I2 · tanh(Vx / 2V_T)

Output current difference:

ΔIo = Io1 − Io2 = (IC3 + IC5) − (IC4 + IC6)
    = (IC3 − IC4) − (IC6 − IC5)
    = (I1 − I2) · tanh(Vx / 2V_T)
    = IEE · tanh(Vx/2V_T) · tanh(Vy/2V_T)

For small inputs (|V| << 2V_T), tanh(u) ≈ u:

ΔIo ≈ IEE · Vx · Vy / (4 V_T²)
Vo  = RL · ΔIo = (IEE · RL / 4V_T²) · Vx · Vy
Vo  = K · Vx · Vy,   K = IEE·RL / (4 V_T²)

Since tanh is an odd function, the sign of ΔIo follows the signs of both Vx and Vy, so the circuit works in all four quadrants. In practical ICs, emitter-degeneration resistors and pre-distortion (log) circuits extend the linear range so that K becomes a fixed constant, usually 1/10 V⁻¹ (Vo = Vx·Vy/10).

Difference between four-quadrant and single-quadrant multiplier

A log amplifier cannot accept negative input (ln of a negative number is undefined), so a log-antilog multiplier is only single-quadrant. A four-quadrant multiplier works in all four quadrants of the Vx–Vy plane.

Output voltage of four-quadrant multiplier (Gilbert cell)

         +Vcc            +Vcc
          RL              RL
          |  Io1     Io2  |
          +--+--  ...  +--+----> Vo
         Q3  Q4       Q5  Q6    <- Vx (cross-coupled)
           \/           \/
           I1           I2
           Q1 ---------- Q2     <- Vy
                  |
                 IEE
                  |
                -VEE

Q3, Q5 collectors join to give Io1; Q4, Q6 collectors join to give Io2.

For a differential pair: ΔI = I·tanh(V/2V_T).

1. Lower pair:

I1 − I2 = IEE·tanh(Vy/2V_T)

2. Upper pairs (both driven by Vx, outputs cross-coupled):

IC3 − IC4 = I1·tanh(Vx/2V_T)
IC6 − IC5 = I2·tanh(Vx/2V_T)

3. Output:

ΔIo = (IC3 + IC5) − (IC4 + IC6)
    = (I1 − I2)·tanh(Vx/2V_T)
    = IEE·tanh(Vx/2V_T)·tanh(Vy/2V_T)

4. For small inputs, tanh(u) ≈ u:

ΔIo ≈ IEE·Vx·Vy / (4V_T²)
Vo  = RL·ΔIo = K·Vx·Vy,  K = IEE·RL/(4V_T²)

Because tanh is odd, a change of sign of Vx or Vy changes the sign of Vo, so all four quadrants are covered. Practical ICs (AD633) linearise the cell and give Vo = Vx·Vy/10 V.

A multiplier can be built by taking the log of each input, adding the logs and then taking the antilog, because ln V1 + ln V2 = ln(V1·V2). A divider uses subtraction instead: ln V1 − ln V2 = ln(V1/V2).

Building blocks

• Log amp (transistor in feedback): Vo = −V_T ln(Vi / (R·Is))
• Antilog amp (transistor at input): Vo = −R·Is·e^(Vi/V_T)

Multiplier circuit

V1 -->[LOG AMP 1]--Va--R--+
                          |
                     [INVERTING   ]
                     [SUMMER, Rf=R]--Vc-->[ANTILOG]--> Vo
                          |
V2 -->[LOG AMP 2]--Vb--R--+

1. Log amp outputs:

Va = −V_T ln(V1/(R·Is))
Vb = −V_T ln(V2/(R·Is))

2. Inverting summer (Rf = R):

Vc = −(Va + Vb) = V_T [ln(V1/RIs) + ln(V2/RIs)]
   = V_T ln( V1·V2 / (R·Is)² )

3. Antilog amp (transistor polarity chosen to accept this sign; output written in magnitude):

Vo = R·Is·e^(Vc/V_T) = R·Is · V1·V2/(R·Is)²
Vo = V1·V2 / (R·Is) = K·V1·V2,   K = 1/(R·Is)

So the output is proportional to the product of the two inputs.

Divider from the multiplier

Replace the summer by a difference amplifier (subtractor) so that the log of the divisor is subtracted:

V1 -->[LOG AMP 1]--Va--->(+)
                         [SUBTRACTOR]--Vc-->[ANTILOG]--> Vo
V2 -->[LOG AMP 2]--Vb--->(−)

Vc = Vb − Va = V_T [ln(V1/RIs) − ln(V2/RIs)]
   = V_T ln(V1/V2)
Vo = R·Is · e^(Vc/V_T) = R·Is · (V1/V2)
Vo = K'·V1/V2,   K' = R·Is

Another simple way: put the multiplier in the feedback path of an op-amp. With Vz at the input resistor and the multiplier output K·Vo·Vx fed back, the virtual-ground condition gives Vz = −K·Vo·Vx, so Vo = −Vz/(K·Vx), a divider.

Notes:

• Inputs must be positive (one-quadrant operation).
• Matched transistors on the same chip cancel Is; temperature compensation is needed for V_T.

Each input stage is a log amplifier (transistor in the feedback path), and the last stage is an inverting summer with gain −2R/R = −2.

Log amplifier outputs

For a log amp with input resistor R and grounded-base transistor in feedback:

V01 = −V_T ln( V1 / (R·Is) )
V02 = −V_T ln( V2 / (R·Is) )
V03 = −V_T ln( V3 / (R·Is) )

where V_T = kT/q ≈ 26 mV at room temperature and Is is the reverse saturation current (transistors assumed matched).

Summing amplifier

Inputs through R, feedback 2R:

V0 = −(2R/R)(V01 + V02 + V03)
   = −2 (V01 + V02 + V03)

Substituting:

V0 = 2V_T [ ln(V1/RIs) + ln(V2/RIs) + ln(V3/RIs) ]
   = 2V_T ln( V1·V2·V3 / (R·Is)³ )

Answer: V0 = 2V_T · ln[ V1·V2·V3 / (R·Is)³ ]

This can also be written as V0 = V_T ln[ (V1·V2·V3)² / (R·Is)⁶ ], i.e. the output is proportional to the log of the square of the product of the three inputs. The circuit works only for positive V1, V2, V3.

A multifunction converter is a log-antilog circuit whose output is

Vo = Vy · (Vz / Vx)^m

By choosing the inputs and the exponent m, the same circuit performs multiplication, division, squaring, square-root and other powers (e.g. the Burr-Brown 4302 IC).

Circuit realization

Vz -->[LOG AMP]--V1-->(+)
                      [SUBTRACTOR]--V3-->[GAIN m]--V4
Vx -->[LOG AMP]--V2-->(−)                (pot/R ratio)
                                              |
Vy -->[LOG AMP]--V5---------------->[SUMMER]<-+
                                       |
                                       V6
                                       |
                                  [ANTILOG AMP]--> Vo

Working and derivation

Take each log amp as V = V_T ln(Vin/Vref), with Vref = R·Is (signs handled by inverting stages).

1. Log of Vz and Vx:

V1 = V_T ln(Vz/Vref)
V2 = V_T ln(Vx/Vref)

2. Subtractor:

V3 = V1 − V2 = V_T ln(Vz/Vx)

3. Gain stage (voltage divider or amplifier, gain m):

V4 = m·V_T ln(Vz/Vx) = V_T ln[(Vz/Vx)^m]

4. Add log of Vy:

V6 = V4 + V_T ln(Vy/Vref)
   = V_T ln[ (Vy/Vref)·(Vz/Vx)^m ]

5. Antilog amp:

Vo = Vref · e^(V6/V_T)
Vo = Vy · (Vz/Vx)^m

Functions obtained

m < 1 is set with a resistor divider; m > 1 with a non-inverting amplifier. Matched transistors cancel Is, and all inputs must be positive.

Four-quadrant multiplier

A multiplier gives Vo = K·Vx·Vy. If both inputs can be positive or negative and the output takes the correct sign, the multiplier works in all four quadrants of the Vx–Vy plane and is called a four-quadrant multiplier.

            Vy
     II      |      I
  Vx<0,Vy>0  |  Vx>0,Vy>0
   Vo < 0    |   Vo > 0
 ------------+------------ Vx
  Vx<0,Vy<0  |  Vx>0,Vy<0
   Vo > 0    |   Vo < 0
     III     |      IV

A log-antilog multiplier works only in quadrant I (one-quadrant). A four-quadrant multiplier is needed for AC signals, e.g. in modulators, phase detectors and power measurement.

Circuit (Gilbert cell)

         +Vcc            +Vcc
          RL              RL
          |  Io1     Io2  |
          +--+--  ...  +--+----> Vo
         Q3  Q4       Q5  Q6    <- Vx (cross-coupled)
           \/           \/
           I1           I2
           Q1 ---------- Q2     <- Vy
                  |
                 IEE
                  |
                -VEE

Collectors of Q3, Q5 give Io1; collectors of Q4, Q6 give Io2.

Derivation

Differential pair with tail current I and input V: ΔI = I·tanh(V/2V_T).

Lower pair:   I1 − I2   = IEE·tanh(Vy/2V_T)
Upper pairs:  IC3 − IC4 = I1·tanh(Vx/2V_T)
              IC6 − IC5 = I2·tanh(Vx/2V_T)

ΔIo = (IC3 + IC5) − (IC4 + IC6)
    = (IC3 − IC4) − (IC6 − IC5)
    = (I1 − I2)·tanh(Vx/2V_T)
    = IEE·tanh(Vx/2V_T)·tanh(Vy/2V_T)

For small signals, tanh(u) ≈ u:

ΔIo ≈ IEE·Vx·Vy/(4V_T²)
Vo  = RL·ΔIo = (IEE·RL/4V_T²)·Vx·Vy = K·Vx·Vy

The product of two odd functions keeps the correct sign for every combination of input polarities, so the circuit is four-quadrant. IC versions (AD633, MC1495) give Vo = Vx·Vy/10.

The RMS value is Vrms = √( average of Vi² ). A log-antilog circuit computes this implicitly: it forms Vi²/Vo, averages it and feeds the result back as Vo. This avoids a separate squarer whose output range would be very large.

Principle

Vo = avg( Vi² / Vo )
Vo is DC, so Vo = avg(Vi²)/Vo
Vo² = avg(Vi²)
Vo  = √avg(Vi²) = Vrms

Circuit (block form)

Vi -->[ABSOLUTE ]--|Vi|-->[LOG]--V1--x2--+
      [VALUE CKT]                         |
                                     [SUBTRACT]--V3
         +-->[LOG]--V2------------->(−)   |
         |                           [ANTILOG]--V4
         |                                |
         |                         [LOW-PASS RC]
         |                         [ (averager) ]
         |                                |
         +--------------------------------+--> Vo

Derivation

1. Precision full-wave rectifier gives |Vi| (log amp needs a positive input; |Vi|² = Vi²).
2. Log amp and gain of 2:

V1 = 2V_T ln(|Vi|/Vref) = V_T ln(Vi²/Vref²)

3. Log of the output (feedback):

V2 = V_T ln(Vo/Vref)

4. Subtractor:

V3 = V1 − V2 = V_T ln( Vi² / (Vref·Vo) )

5. Antilog:

V4 = Vref·e^(V3/V_T) = Vi² / Vo

6. Low-pass filter (time constant much larger than signal period) takes the average:

Vo = avg(Vi²/Vo) = avg(Vi²)/Vo
Vo = √avg(Vi²)

For a sinusoid Vi = Vm sin ωt

avg(Vi²) = (1/T)∫₀ᵀ Vm² sin²ωt dt
         = (Vm²/T)∫₀ᵀ (1 − cos2ωt)/2 dt
         = Vm²/2
Vo = √(Vm²/2) = Vm/√2 = 0.707·Vm

Example: for Vi = 10 sin ωt V, the circuit gives Vo = 7.07 V DC.

Advantages: it is a true RMS converter (correct for any waveform, not only sine), and its dynamic range is wide because the internal signal Vi²/Vo stays close to Vi in size.