Op-Amp Bipolar Transistor Logarithmic Amplifier

The first stage (A1 with Q1) is a log amplifier. Q2 with A2 is an antilog stage whose base is lifted by VR. The overall circuit multiplies Vi by an exponential function of VR, a voltage-controlled gain.

Circuit (as given)

                 Q1 (base GND)
          C +-----|<|-----+ E
 Vi-[R1]-+--(-)\         |
         GND-(+) A1>-----+---VE---+
                                  | E
                    VR ---- B --[Q2]
                                  | C        R2
                                  +----+---/\/\/--+
                                       |          |
                                     (-)\ A2      |
                                 GND-(+)  >-------+--> Vo

Assumptions

Q1 and Q2 are matched npn transistors (same Is). The op-amps are ideal, so both inverting inputs are at 0 V. The transistor law is:

IC = Is · exp(VBE / VT),   so   VBE = VT ln(IC / Is)

Stage 1 (log amplifier)

The current Vi/R1 flows into the collector of Q1 (collector at virtual ground, base grounded):

IC1 = Vi / R1
VBE1 = VB1 − VE = 0 − VE
VE = −VBE1 = −VT ln( Vi / (R1·Is) )

Stage 2 (antilog amplifier)

Q2 has its base at VR, its emitter at VE, and its collector at the virtual ground of A2:

VBE2 = VR − VE = VR + VT ln( Vi / (R1·Is) )

IC2 = Is · exp(VBE2 / VT)
    = Is · exp(VR/VT) · exp( ln( Vi/(R1·Is) ) )
    = Is · exp(VR/VT) · Vi / (R1·Is)
    = (Vi / R1) · exp(VR / VT)

The collector current IC2 is drawn from A2's inverting node. It is supplied through R2 from the output, so:

Vo = IC2 · R2

Result

Vo = (R2/R1) · Vi · exp(VR / VT)

• Is cancels because the transistors are matched. The output does not depend on the strongly temperature-dependent saturation current.
• With VR = 0, Vo = (R2/R1)Vi, a plain linear amplifier.
• The gain changes exponentially with VR: every 60 mV of VR (at 300 K, VT ≈ 26 mV) multiplies the gain by about 10. So the circuit is a voltage-controlled exponential gain amplifier.
• VT = kT/q still appears, so a temperature-compensating resistor is needed in the VR path for accuracy.

An antilog (exponential) amplifier gives an output proportional to the exponential of its input. With two matched transistors and two op-amps, the saturation current Is cancels out and a reference current sets the scale.

Circuit

 Vref-[R1]--+--(-)\
            |C     A1>------+
           Q1  GND-(+)/     |
       B=GND \E             |
              +-----node E--+
              |E
           Q2  \C                     Rf
 Vin-[R2]-+-B   +---------+--------/\/\/----+
          |               |                 |
        [R3]            (-)\                |
          |                 A2>-------------+--> Vo
         GND         GND-(+)/

Q1 and Q2 are a matched pair with their emitters tied at node E. A1 forces Q1 to carry Iref = Vref/R1. The base of Q2 receives Vb = Vin·R3/(R2 + R3) (a divider used for scaling and temperature compensation).

Derivation

The transistor law is IC = Is·exp(VBE/VT).

Q1 (base at 0 V, collector at virtual ground):

IC1 = Vref/R1,   VBE1 = 0 − VE
VE = −VT ln( Vref / (R1·Is) )

Q2 (base at Vb, emitter at VE):

VBE2 = Vb − VE = Vb + VT ln( Vref/(R1·Is) )
IC2  = Is · exp(VBE2/VT)
     = Is · exp(Vb/VT) · Vref/(R1·Is)
     = (Vref/R1) · exp(Vb/VT)

A2 works as a current-to-voltage converter. IC2 flows from the output through Rf:

Vo = Rf · IC2 = (Rf·Vref/R1) · exp(Vb/VT)

With Vb = K·Vin, where K = R3/(R2 + R3):

Vo = (Rf·Vref/R1) · exp( K·Vin / VT )
   = (Rf·Vref/R1) · 10^( K·Vin / (2.303 VT) )

Points to note

• Is (strongly temperature dependent) cancels because Q1 and Q2 are matched.
• VT = kT/q remains. Making R3 (or R2) a temperature-sensitive resistor keeps K/VT constant.
• At 300 K, 2.303VT ≈ 60 mV. With K = 1, the output rises ten times for every 60 mV increase of Vin.
• The output is always of one polarity, set by Vref.

A log amplifier gives an output proportional to the logarithm of its input. In the matched-transistor version, a second identical transistor carrying a reference current cancels the saturation current Is, which depends strongly on temperature.

Circuit

 Vin-[R1]-+-(-)\   Q1: C at (-), E at Vo1
          |  A1 >--Vo1--[R3]--+--[R4]--+
 GND-----(+)/                 |        |
                             (-)\      |
                                 A3>---+--> Vo
                             (+)/
 Vref-[R2]-+-(-)\   Q2        |
           |  A2 >--Vo2--[R3]-+--[R4]--GND
 GND------(+)/
 (bases of Q1, Q2 grounded; Q1, Q2 matched)

A1 and A2 are log stages. A3 is a difference amplifier of gain R4/R3.

Derivation

The transistor law gives VBE = VT ln(IC/Is).

Stage A1:

IC1 = Vin/R1
Vo1 = −VBE1 = −VT ln( Vin / (R1·Is) )

Stage A2:

IC2 = Vref/R2
Vo2 = −VBE2 = −VT ln( Vref / (R2·Is) )

Difference amplifier A3:

Vo = (R4/R3)(Vo2 − Vo1)
   = (R4/R3)·VT [ ln(Vin/(R1·Is)) − ln(Vref/(R2·Is)) ]
   = (R4/R3) · VT · ln( (Vin·R2) / (Vref·R1) )

Transfer function

With R1 = R2:

Vo = (R4/R3) · VT · ln( Vin / Vref )
   = 2.303 (R4/R3) VT · log10( Vin / Vref )

• Is has cancelled, because Q1 and Q2 are matched and at the same temperature.
• At 300 K, VT ≈ 25.9 mV. With R4/R3 = 1, the output changes by about 60 mV per decade of Vin.
• When Vin = Vref, the output is 0.
• Vin and Vref must be positive (for npn transistors).
• VT = kT/q still varies with temperature. It is compensated by making R3 a temperature-sensitive resistor.

A log amplifier produces Vo ∝ ln(Vin). The basic one-transistor log amp gives Vo = −VT ln(Vin/(R1·Is)). This has two temperature-dependent terms: Is (which roughly doubles every 10 °C) and VT = kT/q. The matched-transistor log amp removes Is and a temperature-sensitive resistor removes the VT effect.

Circuit

 Vin-[R1]-+-(-)\  Q1 (C at -, E at out, B GND)
          |  A1 >---Vo1---[R3]--+--[R4]--+
 GND-----(+)/                   |        |
                               (-)\      |
                                   A3>---+--> Vo
                               (+)/
 Vref-[R2]-+-(-)\  Q2            |
           |  A2 >--Vo2--[R3]----+--[R4]--GND
 GND------(+)/
   Q1, Q2: matched pair on one chip (same temperature)
   R3: temperature-sensitive resistor (RTC)

Operation

1. A1 with Q1 (log of input). The current Vin/R1 flows through Q1, and A1's output settles at −VBE1:

Vo1 = −VT ln( Vin / (R1·Is) )

2. A2 with Q2 (log of reference). It is driven by a fixed current Vref/R2:

Vo2 = −VT ln( Vref / (R2·Is) )

3. A3 subtracts the two outputs with gain K = R4/R3:

Vo = K (Vo2 − Vo1) = K·VT·ln( Vin·R2 / (Vref·R1) )

With R1 = R2: Vo = K·VT·ln(Vin/Vref).

The output follows the logarithm of the input ratio. It is about 60 mV per decade per unit K at room temperature.

How the temperature effect is minimized

1. Is is removed by matching. Q1 and Q2 are made on the same chip (a matched pair), so they have the same Is and the same temperature. In the subtraction, ln(1/Is) − ln(1/Is) = 0, so the large drift of Is disappears.
2. VT is compensated by a temperature-sensitive gain. The output is still proportional to VT = kT/q, which rises by about 0.33 %/°C at 25 °C. So the gain K is made to fall at the same rate. R3 (or RTC in a non-inverting gain stage, K = 1 + R2/RTC) is a resistor with a positive temperature coefficient of about +0.33 %/°C. As T rises, RTC rises and K falls, so K·VT stays constant.
3. Using a reference current (Vref/R2) also fixes the zero point of the log curve (Vo = 0 at Vin = Vref).
4. Low-bias-current, low-offset op-amps reduce errors at small input currents.

The result is a log amplifier that is accurate over several decades of input with little drift.

A log amplifier gives an output voltage proportional to the logarithm of the input voltage. The basic circuit places a diode or a transistor (transdiode connection) in the feedback path of an inverting op-amp. It uses the exponential current–voltage law of the PN junction.

Circuit

              Q1 (npn)
        C +----|<|----+ E
          |   B=GND   |
 Vin-[R]--+--(-)\     |
             A   >----+--> Vo
     GND--(+)/

(A diode with its anode at the inverting input and its cathode at the output can replace Q1.)

Working and derivation

1. The inverting input is at virtual ground, so the input current is I = Vin/R. All of it flows into the collector of Q1.
2. The base and collector are both at 0 V, so VBE = −Vo. The transistor law gives:

IC = Is · exp(VBE/VT)

3. Equating the two currents:

Vin/R = Is · exp(−Vo/VT)
Vo = −VT ln( Vin / (R·Is) )
   = −2.303 VT log10( Vin / (R·Is) )

At 300 K, VT = kT/q ≈ 26 mV. So the output changes by about −60 mV for each tenfold (decade) increase in Vin.

Features and limitations

• It works only for one input polarity (positive Vin for an npn transistor). A protective diode is often added.
• The output is small (hundreds of mV). A following amplifier scales it up.
• Is and VT both depend on temperature, so the basic circuit drifts. Matched transistor pairs and temperature-compensating resistors are used in practical log amps.
• The transistor version works over a wider current range (about 7 decades) than the diode version.

Uses: compressing wide-range signals, dB meters, and analog multiplication and division.

An antilog (exponential) amplifier produces an output proportional to the exponential of its input. Using two matched transistors with a reference current cancels the saturation current Is. A temperature-sensitive resistor then compensates VT.

Circuit

 Vref-[R1]--+--(-)\
            |C     A1>------+
           Q1  GND-(+)/     |
       B=GND \E             |
              +-----node E--+
              |E
           Q2  \C                     Rf
 Vin-[R2]-+-B   +---------+--------/\/\/----+
          |               |                 |
        [RTC]           (-)\                |
          |                 A2>-------------+--> Vo
         GND         GND-(+)/

A1 forces Q1's collector current to Iref = Vref/R1 by driving the common emitter node E. The base of Q2 gets Vb = Vin·RTC/(R2 + RTC).

Input–output relation

Using VBE = VT ln(IC/Is):

Q1:

IC1 = Vref/R1,  base at 0 V
VE  = −VT ln( Vref/(R1·Is) )

Q2:

VBE2 = Vb − VE
IC2  = Is · exp(VBE2/VT)
     = Is · exp(Vb/VT) · Vref/(R1·Is)
     = (Vref/R1) · exp(Vb/VT)

A2 (current-to-voltage):

Vo = Rf · IC2

Result:

Vo = (Rf·Vref/R1) · exp[ (RTC/(R2 + RTC)) · Vin / VT ]

For example, with K = RTC/(R2 + RTC) and 2.303VT/K = 1 V:

Vo = (Rf·Vref/R1) · 10^(Vin)

Minimizing the effect of temperature

1. Matched pair. Q1 and Q2 are made on the same chip and kept at the same temperature. Their Is values are equal and cancel, as shown above. Is is the most temperature-sensitive term, roughly doubling every 10 °C.
2. Compensating VT = kT/q. The exponent is K·Vin/VT, and VT rises about 0.33 %/°C. RTC is a resistor with a positive temperature coefficient (about +0.3 %/°C, with R2 ≫ RTC). Then K ≈ RTC/R2 rises with T at the same rate as VT, so K/VT, and therefore the exponent, stays constant.
3. A stable reference Vref and low-bias-current op-amps keep the scale factor Rf·Vref/R1 constant.

With these measures the antilog output stays accurate over a wide temperature range.

This is the temperature-compensated log amplifier. Q1 carries the signal current, while Q2 with A2 carries a fixed reference current. The divider R2–RTC sets the scale and cancels the effect of VT.

Assumptions

Q1 and Q2 are matched (same Is). The op-amps are ideal. The reference resistor (marked R2 at A2's input) is called R2' here to keep it separate from the divider R2. VB is small compared with VRef.

Currents

Q1: its collector is at A1's virtual ground (0 V), its base is grounded and its emitter is at VA.

IC1 = vi / R1
VBE1 = 0 − VA   →   VA = −VT ln( vi / (R1·Is) )

Q2: A2's inverting input is held at VB (virtual short with its + input). Q2's base is also at VB, and its emitter is at VA (driven by A2's output).

Iref = IC2 = (VRef − VB)/R2' ≈ VRef/R2'
VBE2 = VB − VA = VT ln( Iref / Is )

Solving for VB

VB = VA + VT ln(Iref/Is)
   = −VT ln( vi/(R1·Is) ) + VT ln( Iref/Is )
   = −VT ln( vi / (R1·Iref) )

Is has cancelled.

Output voltage

A1's output V0 feeds the divider, so VB = V0·RTC/(R2 + RTC). Therefore:

V0 = (1 + R2/RTC) · VB

V0 = −(1 + R2/RTC) · VT · ln( vi / (R1·Iref) )
   = −(1 + R2/RTC) · VT · ln( vi·R2' / (R1·VRef) )

In log10 form:

V0 = −2.303 (1 + R2/RTC) VT · log10( vi·R2' / (R1·VRef) )

Why it is temperature compensated

• Is cancels because Q1 and Q2 are matched.
• VT = kT/q rises with temperature (+0.33 %/°C). RTC has a positive temperature coefficient, so as T rises RTC rises and (1 + R2/RTC) falls. With R2 ≫ RTC the factor is about R2/RTC, so the product (R2/RTC)·VT stays nearly constant.

For example, with (1 + R2/RTC)·2.303VT = 1 V, the output falls by 1 V per decade of vi.

The circuit has two log amplifiers (A1, A2), a unity-gain difference amplifier (A3) and a non-inverting amplifier (A4) with a temperature-compensating resistor RT. Its output is proportional to the log of the ratio V1/VR.

Step 1: log amplifiers

Q1 and Q2 have their bases grounded and their collectors at virtual ground. Using VBE = VT ln(IC/Is):

Vo1 = −VBE1 = −VT ln( V1 / (R1·Is) )
Vo2 = −VBE2 = −VT ln( VR / (R1·Is) )

Step 2: difference amplifier A3 (all resistors R)

Vo1 goes to the inverting input through R (feedback R). Vo2 goes to the non-inverting input through R, which has R to ground, so V+ = Vo2/2.

Vo3 = (1 + R/R)(Vo2/2) − (R/R)·Vo1
    = Vo2 − Vo1
    = −VT ln( VR/(R1·Is) ) + VT ln( V1/(R1·Is) )
    = VT ln( V1 / VR )

Is cancels because Q1 and Q2 are matched.

Step 3: non-inverting amplifier A4

V0 = (1 + R2/RT) · Vo3

Result

V0 = (1 + R2/RT) · VT · ln( V1 / VR )
   = 2.303 (1 + R2/RT) · VT · log10( V1 / VR )

where VT = kT/q (about 25.9 mV at 300 K).

Notes

• V0 = 0 when V1 = VR. It rises by 2.303(1 + R2/RT)VT for every decade of V1/VR.
• Is has been removed by the matched pair. The remaining temperature dependence is in VT. RT is a positive-temperature-coefficient resistor, so (1 + R2/RT) falls as T rises, keeping (1 + R2/RT)·VT nearly constant.
• Example: to get 1 V per decade at 300 K, 2.303 × 0.0259 × (1 + R2/RT) = 1, so 1 + R2/RT ≈ 16.8.

Advantage of the BJT log amplifier over the diode log amplifier

In a BJT (transdiode) log amplifier, only the collector current flows through the feedback path. The collector current follows the ideal exponential law IC = Is·exp(VBE/VT) very closely over a wide range, about 7 to 9 decades of current. A diode current includes recombination and leakage components. Its ideality factor η varies between 1 and 2, so the diode follows the log law accurately over only about 3 to 4 decades. The BJT version therefore has a wider dynamic range and better log accuracy.

Applications of log–antilog amplifiers

• Analog multiplication and division (log, add or subtract, then antilog)
• Raising to a power, square and square root, and RMS computation
• Compressing signals with a wide dynamic range (audio, sonar, radar)
• dB and decibel meters, light (photodiode) and pH measurement
• Analog computation, function generation and true-RMS meters