Logarithmic Amplifier using Diode and Transistor

Engineeering Projects

Logarithmic Amplifier using Diode and Transistor

  • It produces output that is proportional to the logarithmic input
  • It is a non-linear amplifier used for amplification or compression of a wide range of input signals for better resolution.
  • It can be used direct DB display on a spectrum analyzer.

You may also like: Antilogarithmic Amplifier

Log Amplifier using a Single Diode and Op-Amp

The circuit arrangement for the logarithmic amplifier/converter is illustrated in figure 1. Here a silicon diode D is connected in the feedback path and the current via the diode is dependent upon the output voltage.

Now from the diagram,

I_f = I_s\times (e^{\dfrac{V_D}{\eta V_T}})

Where Is = Saturation Current

VT = Thermal Voltage

\eta = material constant

If \dfrac{V_D}{\eta V_T}>1

Then, e^{\dfrac{V_D}{\eta V_T}}>>1

logarithmic amplifier using single diode

Now, from the above condition we can write

I_f = I_s \times e^{\dfrac{V_D}{\eta V_T}}

 

V_D = \eta V_T ln(\dfrac{I_f}{I_s})

 

V_o = -V_D

 

= -\eta V_T\times ln(\dfrac{I_f}{I_s})

Now,

I_f = I = \dfrac{V_s}{R}

V_o = -\eta V_T\times ln(\dfrac{V_s}{I_s\times R})      …..(1)

V_o = k_1\times ln(k_2\times V_s)       …..(2)

Where,

k_1 = -\eta V_T and k_2 = \dfrac{1}{R\times I_s}

Hence, V0  is the logarithmic function of input voltage Vin.

It has offset term \eta V_T ln(\dfrac{1}{R T_s}) and scale factor \eta V_T

Log Amplifier using a transistor and op-amp

The circuit arrangement for the logarithmic amplifier is illustrated in figure 2. Here the NPN transistor is connected in the feedback path.

logarithmic amplifier using single transistor

I_c = I_s \times (e^{\dfrac{V_{BE}}{V_T}}-1)

 

If \dfrac{V_{BE}}{V_T}> 1

Then, e^{\dfrac{V_{BE}}{V_T}}>>1

Thus, we can write,

I_c = I_s\times e^{\dfrac{V_{BE}}{V_T}}

 

V_{BE} = V_T\times ln(\dfrac{I_c}{I_s})

 

V_o = -V_{BE} = -V_T\times ln(\dfrac{I_c}{I_s})

 

I_c = I = \dfrac{V_s}{R}

 

Therefore, V_o = -V_T\times ln (\dfrac{V_s}{R\times I_s})         ……(3)

Log amplifier using two op-amps and two matched diodes

 

logarithmic amplifier using two matched diodes

RT = temperature dependent resistor

  • This circuit eliminates the saturation term by giving a constant current source.
  • Assuming two diodes are matched

For diode

I_f = I_s \times (e^{\dfrac{V_D}{\eta times V_T}}-1)

 

If \dfrac{V_{BE}}{V_T}> 1

Then, e^{\dfrac{V_{BE}}{V_T}}>>1

Thus, we can write

I_f = i_s \times (e^{\dfrac{V_d}{\eta \times V_T}})

Output diode voltage

V_D = \eta \times V_T ln \times (\dfrac{I_f}{I_s})         ….(4)

Now, form figure 3

V^+-V_{D2}+V_{D1} = 0

 

V^+ = V_{D2} -V_{D1}

 

V_{D2} = \eta_2 \times V_{T2} \times ln(\dfrac{I_R}{I_{s2}})          …..(5)

Similarly,

V_{D1} = \eta_1 \times V_{T1} \times ln(\dfrac{I_R}{I_{s1}})           ……(6)

We know that, I_R = \dfrac{V_s}{R}

Now, putting the value of IR in the above equation of voltage output of diode.

\eta_1 \times V_{T1} \times ln (\dfrac{V_s}{R\times I_{s1}})

Assuming both diodes are matched

i.e. \eta_1 = \eta_2 = \eta

 

V_{T1} = V_{T2} = V_T

 

I_{s1} = I_{s2} = I_s

Subtracting equation 5 from equation 6

V_{D2}-V_{D1} = \eta \times V_T \times ln(\dfrac{I_R}{I_s})-\eta \times V_T \times ln(\dfrac{V_s}{R\times I_s})

 

= \eta \times V_T [ln(\dfrac{I_R}{I_s})-ln(\dfrac{V_s}{R \times I_s})]

 

\eta \times V_T [ln (\dfrac{I_R \times R \times I_s}{I_s \times V_s})]

 

V^+ = -\eta \times V_T \times ln(\dfrac{V_s}{R\times I_R})

 

V_0 = (1 + \dfrac{R_f}{R_1 + R_3}) \times V^+

 

= [1 + \dfrac{R_f}{R_1 + R_T}] \times [-\eta \times V_T \times ln (\dfrac{V_s}{R \times I_R})]

 

V_0 = -\eta \times V_T \times (1 + \dfrac{R_f}{R_1 + R_T}) \times ln(\dfrac{V_s}{R\times I_s})            …….(7)

  • Temperature-sensitive scale factor \eta \times V_T can be compensated by making the gain of Op-Amp (A2) temperature-sensitive.
  • Using temperature-dependent resistor RT in such a way that \eta \times V_T and RT has an opposite effect and the circuit is independent of temperature.

Log amplifier using two matched transistors and two op-amps

  • It provides wide dynamic range amplification for input voltage then the diode

logarithmic amplifier using two matched transistors

From figure, we can write

V^+ - V_{BE2} + V_{BE1} = 0

 

V^+ = V_{BE2}-V_{BE1}          ……(8)

For transistor,

I_c = I_s \times (e^{\dfrac{V_{BE}}{V_T}}-1)

 

If \dfrac{V_{BE}}{V_T}> 1

 

Then, e^{\dfrac{V_{BE}}{V_T}}>>1

Thus, we can write

I_c = I_s \times e^{\dfrac{V_{BE}}{V_T}}

 

V_{BE} = V_T \times ln(\dfrac{I_c}{I_s})

Assuming two transistors are matched.

I_{s1} = I_{s2} = I_s

 

V_{T1} = V_{T2} = V_T

For transistor T1

V_{BE1} = V_{T1} \times ln(\dfrac{I_C1}{I_{s1}})

For transistor T2

V_{BE2} = V_{T2} \times ln (\dfrac{I_{c2}}{I_{s2}})

From equation 8

V^+ = V_{T2} \times ln(\dfrac{I_{c2}}{I_{s2}})-V_{T1} \times ln(\dfrac{T_{c1}}{I_{s1}})

 

V^+ = V_T \times ln(\dfrac{I_{c2}}{I_{S1}})-V_T \times ln(\dfrac{I_{c1}}{I_{s2}})

 

I_{c1} = \dfrac{V_s}{R_1}

 

I_{c2} = \dfrac{V_R}{R_2}

 

V^+ = -V_T \times ln(\dfrac{V_s}{R_1}\times \dfrac{R_2}{V_R})

 

As, V^+ = V_{BE2}-V_{BE1} is very small and I_{B2}<<I_{c2}

 

V_{0} = (1+ \dfrac{R_4}{R_3})\times V^+

 

V_0 = -V_T\times (1 + \dfrac{R_4}{R_3})\times ln(\dfrac{V_s}{V_R}\times \dfrac{R_2}{R_1})

 

  • Temperature-sensitive factors can be compensated by making the gain of op-amp A2 also temperature-sensitive by introducing the temperature-dependent resistor so as to make the circuit independent of temperature.

 

Leave a Comment