Logarithmic Converter with Diode

A logarithmic converter circuit is an analog electronic circuit that produces an output voltage proportional to the logarithm of the input voltage or current. This circuit is widely used in applications such as signal compression, dynamic range compression, and measurement systems where a wide range of input values needs to be processed linearly. The core component of a logarithmic converter is often a diode or a bipolar transistor, which inherently exhibits a logarithmic relationship between its current and voltage in the forward-biased region.

Diode Equation

The current-voltage relationship for an ideal diode is given by the Shockley diode equation:

I_{D} = I_{S} (e^{\frac{V_{D}}{n V_{T}}} - 1)

Where:

$I_{D}$ : Diode current
$I_{S}$ : Reverse saturation current (constant for the diode)
$V_{D}$ : Voltage across the diode
$n$ : Ideality factor (typically between 1 and 2)
$V_{T}$ : Thermal voltage ( $V_{T} = \frac{k T}{q}$ , where $k$ is Boltzmann's constant, $T$ is temperature in Kelvin, and $q$ is the charge of an electron)

At room temperature ( $T \approx 300 K$ ), $V_{T} \approx 26 mV$ .

By leveraging the exponential characteristic of these semiconductor devices, the circuit can accurately convert a linear input signal into a logarithmic output. For a deeper dive into the design and working principles of logarithmic amplifiers, check out this detailed guide on Logarithmic Amplifier Design.

The following is a simple one diode logarithmic converter circuit diagram.

Deriving $V_{out} = k \log (V_{in})$

We assume:

The input voltage $V_{in}$ is accurately proportional to the input current $I_{in}$ , i.e., $I_{in} = k_{1} V_{in}$ , where $k_{1}$ is a proportionality constant.
The diode is operating in the forward-biased region, where $e^{\frac{V_{D}}{n V_{T}}} ≫ 1$ , so the Shockley diode equation simplifies to:
$I_{D} \approx I_{S} e^{\frac{V_{D}}{n V_{T}}}$
The output voltage $V_{out}$ is taken across the diode, so $V_{out} = V_{D}$ .

Step 1: Express $V_{D}$ in terms of $I_{D}$

From the simplified diode equation:

I_{D} \approx I_{S} e^{\frac{V_{D}}{n V_{T}}}

Take the natural logarithm of both sides:

\ln (I_{D}) = \ln (I_{S}) + \frac{V_{D}}{n V_{T}}

Solve for $V_{D}$ :

V_{D} = n V_{T} (\ln (I_{D}) - \ln (I_{S}))

Step 2: Substitute $I_{D} = I_{in}$

Since $I_{D} = I_{in}$ and $I_{in} = k_{1} V_{in}$ :

V_{D} = n V_{T} (\ln (k_{1} V_{in}) - \ln (I_{S}))

Simplify using logarithm properties:

V_{D} = n V_{T} (\ln (V_{in}) + \ln (k_{1}) - \ln (I_{S}))

Step 3: Combine constants

Let $k = n V_{T}$ and $C = n V_{T} (\ln (k_{1}) - \ln (I_{S}))$ . Then:

V_{D} = k \ln (V_{in}) + C

If we ignore the constant term $C$ (or assume it is compensated in the circuit), we get:

V_{out} = k \ln (V_{in})

Final Result

The output voltage $V_{out}$ is proportional to the natural logarithm of the input voltage $V_{in}$ :

V_{out} = k \ln (V_{in})

This logarithmic relationship is commonly used in analog circuits, such as logarithmic amplifiers.

Logarithmic Converter with Diode

Diode Equation

Deriving $V_{out} = k \log (V_{in})$

Step 1: Express $V_{D}$ in terms of $I_{D}$

Step 2: Substitute $I_{D} = I_{in}$

Step 3: Combine constants

Final Result

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Logarithmic Converter with Diode

Diode Equation

Deriving Vout=klog⁡(Vin)

Step 1: Express VD​ in terms of ID​

Step 2: Substitute ID=IinID​=Iin​

Step 3: Combine constants

Final Result

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Deriving $V_{out} = k \log (V_{in})$

Step 1: Express $V_{D}$ in terms of $I_{D}$

Step 2: Substitute $I_{D} = I_{in}$