Mathematical Beauty of Modulation with Diode

The equation that describes the current-voltage (I-V) relationship of a diode is known as the Shockley diode equation (or the ideal diode equation). It is given by:

I_{d} = I_{s} (e^{\frac{V _{d}}{n V _{T}}} - 1)

Where:

$I_{d}$ : Diode current (the current flowing through the diode).
$V_{d}$ : Voltage across the diode.
$I_{s}$ : Saturation current (also called the reverse saturation current), which is a very small current that flows when the diode is reverse-biased.
$V_{T}$ : Thermal voltage, approximately equal to $V_{T} = \frac{k T}{q}$ , where:
- $k$ is Boltzmann's constant ( $1.38 \times 1 0^{- 23} J/K$ ),
- $T$ is the absolute temperature in kelvin,
- $q$ is the charge of an electron ( $1.6 \times 1 0^{- 19} C$ ). At room temperature ( $T \approx 300 K$ ), $V_{T} \approx 25.85 mV$ .
$n$ : Ideality factor (or emission coefficient), which accounts for non-idealities in the diode. For an ideal diode, $n = 1$ , but for real diodes, $n$ typically ranges from 1 to 2.

Key Points:

When $V_{d}$ is positive (forward bias), the exponential term dominates, and the diode conducts significant current.
When $V_{d}$ is negative (reverse bias), the $- 1$ term becomes negligible, and the current is approximately $- I_{s}$ , which is very small.
The ideality factor $n$ adjusts the slope of the I-V curve, accounting for recombination and other non-ideal effects in real diodes.

This equation is fundamental in understanding diode behavior in electronic circuits.

The Shockley diode equation can be approximated as a polynomial expansion of the form:

I_{d} = a_{0} + a_{1} V_{d} + a_{2} V_{d}^{2} + a_{3} V_{d}^{3} + \dots

This is achieved by expanding the exponential term $e^{\frac{V _{d}}{n V _{T}}}$ in the Shockley diode equation into a Taylor series around $V_{d} = 0$ . Let’s break this down step by step.

Step 1: Rewrite the Shockley diode equation

The Shockley diode equation is:

I_{d} = I_{s} (e^{\frac{V _{d}}{n V _{T}}} - 1)

We focus on the exponential term $e^{\frac{V _{d}}{n V _{T}}}$ and expand it using the Taylor series.

Step 2: Taylor series expansion of $e^{x}$

The Taylor series expansion for $e^{x}$ around $x = 0$ is:

e^{x} = 1 + x + \frac{x ^{2}}{2 !} + \frac{x ^{3}}{3 !} + \frac{x ^{4}}{4 !} + \dots

Substitute $x = \frac{V _{d}}{n V _{T}}$ into the series:

e^{\frac{V _{d}}{n V _{T}}} = 1 + \frac{V _{d}}{n V _{T}} + \frac{( \frac{V _{d}}{n V _{T}} ) ^{2}}{2 !} + \frac{( \frac{V _{d}}{n V _{T}} ) ^{3}}{3 !} + \dots

Simplify each term:

e^{\frac{V _{d}}{n V _{T}}} = 1 + \frac{V _{d}}{n V _{T}} + \frac{V _{d}^{2}}{2 ( n V _{T} ) ^{2}} + \frac{V _{d}^{3}}{6 ( n V _{T} ) ^{3}} + \dots

Step 3: Subtract 1 from the expansion

The Shockley diode equation includes $e^{\frac{V _{d}}{n V _{T}}} - 1$ , so subtract 1 from the expanded series:

e^{\frac{V _{d}}{n V _{T}}} - 1 = \frac{V _{d}}{n V _{T}} + \frac{V _{d}^{2}}{2 ( n V _{T} ) ^{2}} + \frac{V _{d}^{3}}{6 ( n V _{T} ) ^{3}} + \dots

Step 4: Multiply by $I_{s}$

Now multiply the entire series by $I_{s}$ to get $I_{d}$ :

I_{d} = I_{s} (\frac{V _{d}}{n V _{T}} + \frac{V _{d}^{2}}{2 ( n V _{T} ) ^{2}} + \frac{V _{d}^{3}}{6 ( n V _{T} ) ^{3}} + \dots)

Distribute $I_{s}$ across the terms:

I_{d} = \frac{I _{s}}{n V _{T}} V_{d} + \frac{I _{s}}{2 ( n V _{T} ) ^{2}} V_{d}^{2} + \frac{I _{s}}{6 ( n V _{T} ) ^{3}} V_{d}^{3} + \dots

Step 5: Identify the coefficients

Comparing this with the polynomial form $I_{d} = a_{0} + a_{1} V_{d} + a_{2} V_{d}^{2} + a_{3} V_{d}^{3} + \dots$ , we identify the coefficients:

$a_{0} = 0$ (since there is no constant term),
$a_{1} = \frac{I _{s}}{n V _{T}}$ ,
$a_{2} = \frac{I _{s}}{2 ( n V _{T} ) ^{2}}$ ,
$a_{3} = \frac{I _{s}}{6 ( n V _{T} ) ^{3}}$ ,
and so on.

Thus, the polynomial approximation of the Shockley diode equation is:

I_{d} = \frac{I _{s}}{n V _{T}} V_{d} + \frac{I _{s}}{2 ( n V _{T} ) ^{2}} V_{d}^{2} + \frac{I _{s}}{6 ( n V _{T} ) ^{3}} V_{d}^{3} + \dots

Final Answer:

I_{d} = k = 1 \sum \infty \frac{I _{s}}{k ! ( n V _{T} ) ^{k}} V_{d}^{k}

Or explicitly:

I_{d} = \frac{I _{s}}{n V _{T}} V_{d} + \frac{I _{s}}{2 ( n V _{T} ) ^{2}} V_{d}^{2} + \frac{I _{s}}{6 ( n V _{T} ) ^{3}} V_{d}^{3} + \dots

This is the polynomial expansion of the Shockley diode equation. Note that this approximation is valid only for small values of $V_{d}$ , where the higher-order terms ( $V_{d}^{2}, V_{d}^{3}, \dots$ ) become negligible.

Amplitude Modulation

An Amplitude Modulation (AM) signal is a type of modulated signal where the amplitude of a high-frequency carrier wave is varied in proportion to the instantaneous value of a lower-frequency message (or baseband) signal. The equation for an AM signal can be expressed as follows:

General Form of an AM Signal

The AM signal $s (t)$ is given by:

s (t) = [A_{c} + m (t)] cos (2 π f_{c} t)

Where:

$A_{c}$ : Amplitude of the carrier wave (a constant).
$m (t)$ : Message (baseband) signal, which contains the information to be transmitted.
$f_{c}$ : Frequency of the carrier wave (in Hz).
$cos (2 π f_{c} t)$ : Carrier wave.

Simplified AM Signal Equation

If the message signal $m (t)$ is sinusoidal, say $m (t) = A_{m} cos (2 π f_{m} t)$ , where:

$A_{m}$ : Amplitude of the message signal,
$f_{m}$ : Frequency of the message signal,

then the AM signal becomes:

s (t) = [A_{c} + A_{m} cos (2 π f_{m} t)] cos (2 π f_{c} t)

Expanding this using trigonometric identities, we get:

s (t) = A_{c} cos (2 π f_{c} t) + \frac{A _{m}}{2} cos [2 π (f_{c} + f_{m}) t] + \frac{A _{m}}{2} cos [2 π (f_{c} - f_{m}) t]

Key Components of the AM Signal

Carrier Component : $A_{c} cos (2 π f_{c} t)$
- This is the unmodulated carrier wave with frequency $f_{c}$ and amplitude $A_{c}$ .
Upper Sideband (USB) : $\frac{A _{m}}{2} cos [2 π (f_{c} + f_{m}) t]$
- This component has a frequency $f_{c} + f_{m}$ , which is higher than the carrier frequency.
Lower Sideband (LSB) : $\frac{A _{m}}{2} cos [2 π (f_{c} - f_{m}) t]$
- This component has a frequency $f_{c} - f_{m}$ , which is lower than the carrier frequency.

Modulation Index ( $μ$ )

The modulation index $μ$ is defined as the ratio of the amplitude of the message signal to the amplitude of the carrier signal:

μ = \frac{A _{m}}{A _{c}}

For proper AM modulation, $μ$ should satisfy $0 \leq μ \leq 1$ . If $μ > 1$ , overmodulation occurs, leading to distortion in the demodulated signal.

Final AM Signal Equation

In terms of the modulation index $μ$ , the AM signal can also be written as:

s (t) = A_{c} [1 + μ cos (2 π f_{m} t)] cos (2 π f_{c} t)

Or equivalently:

s (t) = A_{c} cos (2 π f_{c} t) + \frac{μ A _{c}}{2} cos [2 π (f_{c} + f_{m}) t] + \frac{μ A _{c}}{2} cos [2 π (f_{c} - f_{m}) t]

Summary

The general equation for an AM signal is:

s (t) = A_{c} [1 + μ cos (2 π f_{m} t)] cos (2 π f_{c} t)

Where:

$A_{c}$ : Carrier amplitude,
$μ$ : Modulation index ( $μ = \frac{A _{m}}{A _{c}}$ ),
$f_{m}$ : Message signal frequency,
$f_{c}$ : Carrier signal frequency.

AM Modulation

To derive the equation for modulation in the context of AM diode modulaton, we start with the Shockley diode equation and incorporate the modulation signal

V_{m}

(audio signal) and the carrier signal

V_{c}

The total voltage across the diode is the sum of these two signals:

V_{d} = V_{m} + V_{c}

Here:

$V_{m}$ : Modulating signal (e.g., audio input).
$V_{c}$ : Carrier signal (e.g., high-frequency sinusoidal wave).

The goal is to analyze how the diode's current $I_{d}$ responds to this combined voltage $V_{d}$ , which represents amplitude modulation (AM). Let’s break this down step by step.

Step 1: Shockley Diode Equation

The Shockley diode equation is:

I_{d} = I_{s} (e^{\frac{V _{d}}{n V _{T}}} - 1)

Substitute $V_{d} = V_{m} + V_{c}$ :

I_{d} = I_{s} (e^{\frac{V _{m} + V _{c}}{n V _{T}}} - 1)

Step 2: Expand the Exponential Term

Expand $e^{\frac{V _{m} + V _{c}}{n V _{T}}}$ using the Taylor series expansion around $V_{d} = 0$ :

e^{\frac{V _{m} + V _{c}}{n V _{T}}} = e^{\frac{V _{c}}{n V _{T}}} \cdot e^{\frac{V _{m}}{n V _{T}}}

Now expand each term separately. For small values of $V_{m}$ (compared to $n V_{T}$ ), we can approximate $e^{\frac{V _{m}}{n V _{T}}}$ as:

e^{\frac{V _{m}}{n V _{T}}} \approx 1 + \frac{V _{m}}{n V _{T}} + \frac{( \frac{V _{m}}{n V _{T}} ) ^{2}}{2 !} + \dots

Thus:

e^{\frac{V _{m} + V _{c}}{n V _{T}}} \approx e^{\frac{V _{c}}{n V _{T}}}

1 + \frac{V _{m}}{n V _{T}} + \frac{( \frac{V _{m}}{n V _{T}} ) ^{2}}{2 !} + \dots

Step 3: Substitute Back into the Diode Equation

Substitute the expanded exponential term back into the diode equation:

I_{d} = I_{s}

e^{\frac{V _{c}}{n V _{T}}}

1 + \frac{V _{m}}{n V _{T}} + \frac{( \frac{V _{m}}{n V _{T}} ) ^{2}}{2 !} + \dots

- 1

Distribute $e^{\frac{V _{c}}{n V _{T}}}$ :

I_{d} = I_{s}

e^{\frac{V _{c}}{n V _{T}}} + e^{\frac{V _{c}}{n V _{T}}} \frac{V _{m}}{n V _{T}} + e^{\frac{V _{c}}{n V _{T}}} \frac{( \frac{V _{m}}{n V _{T}} ) ^{2}}{2 !} + \dots - 1

Step 4: Simplify the Terms

Carrier Term : The first term, $e^{\frac{V _{c}}{n V _{T}}}$ , represents the unmodulated carrier signal.
Modulation Term : The second term, $e^{\frac{V _{c}}{n V _{T}}} \frac{V _{m}}{n V _{T}}$ , represents the interaction between the carrier and the modulating signal. This term produces sidebands (upper and lower sidebands) around the carrier frequency.
Higher-Order Terms : Higher-order terms like $e^{\frac{V _{c}}{n V _{T}}} \frac{( \frac{V _{m}}{n V _{T}} ) ^{2}}{2 !}$ represent distortion and harmonic components. These are typically small for small-signal modulation ( $V_{m} ≪ n V_{T}$ ).

Step 5: Approximation for Small-Signal Modulation

For small-signal modulation ( $V_{m} ≪ n V_{T}$ ), higher-order terms can be neglected. The diode current simplifies to:

I_{d} \approx I_{s} (e^{\frac{V _{c}}{n V _{T}}} - 1) + I_{s} e^{\frac{V _{c}}{n V _{T}}} \frac{V _{m}}{n V _{T}}

Let’s define:

$I_{d c} = I_{s} (e^{\frac{V _{c}}{n V _{T}}} - 1)$ : DC component of the diode current.
$I_{a c} = I_{s} e^{\frac{V _{c}}{n V _{T}}} \frac{V _{m}}{n V _{T}}$ : AC component due to modulation.

Thus:

I_{d} \approx I_{d c} + I_{a c}

Step 6: Frequency Domain Representation

In the frequency domain:

The carrier signal $V_{c}$ generates a central frequency component at $f_{c}$ .
The modulating signal $V_{m}$ introduces sidebands at $f_{c} + f_{m}$ (upper sideband) and $f_{c} - f_{m}$ (lower sideband).

The resulting spectrum of $I_{d}$ contains:

A strong carrier at $f_{c}$ ,
Two sidebands at $f_{c} + f_{m}$ and $f_{c} - f_{m}$ .

This corresponds to amplitude modulation (AM) .

Final Modulation Equation

The diode current $I_{d}$ during modulation can be expressed as:

I_{d} = I_{s} (e^{\frac{V _{c}}{n V _{T}}} - 1) + I_{s} e^{\frac{V _{c}}{n V _{T}}} \frac{V _{m}}{n V _{T}}

Where:

$I_{s} (e^{\frac{V _{c}}{n V _{T}}} - 1)$ : DC component (carrier).
$I_{s} e^{\frac{V _{c}}{n V _{T}}} \frac{V _{m}}{n V _{T}}$ : AC component (sidebands).

This equation describes how the diode generates an AM signal when subjected to a combined voltage $V_{d} = V_{m} + V_{c}$ .

AM Demodulation

Now we will see the AM demodulation process with equation. Substitute the AM signal

s (t)

V_{d}

in the Shockley diode equation and calculate the coefficients

a_{0}, a_{1}, a_{2}, \dots

, we need to expand the diode current

I_{d}

in terms of the Taylor series expansion.

Here's how we proceed step by step:

Step 1: Recall the Shockley Diode Equation

The Shockley diode equation is:

I_{d} = I_{s} (e^{\frac{V _{d}}{n V _{T}}} - 1)

Substitute $V_{d} = s (t)$ , where $s (t)$ is the AM signal:

s (t) = A_{c} [1 + μ cos (2 π f_{m} t)] cos (2 π f_{c} t)

Thus:

I_{d} = I_{s} (e^{\frac{s ( t )}{n V _{T}}} - 1)

Step 2: Expand $e^{\frac{s ( t )}{n V _{T}}}$ Using the Taylor Series

The exponential term $e^{\frac{s ( t )}{n V _{T}}}$ can be expanded as:

e^{\frac{s ( t )}{n V _{T}}} = 1 + \frac{s ( t )}{n V _{T}} + \frac{( \frac{s ( t )}{n V _{T}} ) ^{2}}{2 !} + \frac{( \frac{s ( t )}{n V _{T}} ) ^{3}}{3 !} + \dots

Subtract 1 (as per the Shockley equation):

e^{\frac{s ( t )}{n V _{T}}} - 1 = \frac{s ( t )}{n V _{T}} + \frac{( \frac{s ( t )}{n V _{T}} ) ^{2}}{2 !} + \frac{( \frac{s ( t )}{n V _{T}} ) ^{3}}{3 !} + \dots

Multiply by $I_{s}$ :

I_{d} = I_{s}

\frac{s ( t )}{n V _{T}} + \frac{( \frac{s ( t )}{n V _{T}} ) ^{2}}{2 !} + \frac{( \frac{s ( t )}{n V _{T}} ) ^{3}}{3 !} + \dots

Step 3: Substitute $s (t)$

Substitute $s (t) = A_{c} [1 + μ cos (2 π f_{m} t)] cos (2 π f_{c} t)$ into the equation. This gives:

I_{d} = I_{s}

\frac{A _{c} [ 1 + μ cos ( 2 π f _{m} t ) ] cos ( 2 π f _{c} t )}{n V _{T}} + \frac{( \frac{A _{c} [ 1 + μ c o s ( 2 π f _{m} t ) ] c o s ( 2 π f _{c} t )}{n V _{T}} ) ^{2}}{2 !} + \dots

Step 4: Expand Each Term

First-Order Term ( $a_{1}$ ):

The first-order term is:

\frac{s ( t )}{n V _{T}} = \frac{A _{c} [ 1 + μ cos ( 2 π f _{m} t ) ] cos ( 2 π f _{c} t )}{n V _{T}}

This term contributes to the linear relationship between $I_{d}$ and $s (t)$ . The coefficient $a_{1}$ is:

a_{1} = \frac{I _{s}}{n V _{T}}

Second-Order Term ( $a_{2}$ ):

The second-order term involves $(\frac{s ( t )}{n V _{T}})^{2}$ . Substituting $s (t)$ :

(\frac{s ( t )}{n V _{T}})^{2} = (\frac{A _{c} [ 1 + μ cos ( 2 π f _{m} t ) ] cos ( 2 π f _{c} t )}{n V _{T}})^{2}

Expanding this term will produce components at frequencies $2 f_{c}$ , $2 f_{c} \pm f_{m}$ , etc. The coefficient $a_{2}$ is:

a_{2} = \frac{I _{s}}{2 ( n V _{T} ) ^{2}}

Higher-Order Terms ( $a_{3}, a_{4}, \dots$ ):

Similarly, higher-order terms involve powers of $\frac{s ( t )}{n V _{T}}$ . For example:

The third-order term $a_{3}$ is proportional to $\frac{I _{s}}{6 ( n V _{T} ) ^{3}}$ ,
The fourth-order term $a_{4}$ is proportional to $\frac{I _{s}}{24 ( n V _{T} ) ^{4}}$ , and so on.

Step 5: Collect Coefficients

The diode current $I_{d}$ can now be expressed as a polynomial in terms of $s (t)$ :

I_{d} = a_{0} + a_{1} s (t) + a_{2} s (t)^{2} + a_{3} s (t)^{3} + \dots

Where:

$a_{0} = 0$ (no constant term),
$a_{1} = \frac{I _{s}}{n V _{T}}$ ,
$a_{2} = \frac{I _{s}}{2 ( n V _{T} ) ^{2}}$ ,
$a_{3} = \frac{I _{s}}{6 ( n V _{T} ) ^{3}}$ ,
$a_{4} = \frac{I _{s}}{24 ( n V _{T} ) ^{4}}$ ,
and so on.

Final Answer:

The coefficients for the polynomial expansion of $I_{d}$ are:

a_{0} a_{1} a_{2} a_{3} a_{4} = 0, = \frac{I _{s}}{n V _{T}}, = \frac{I _{s}}{2 ( n V _{T} ) ^{2}}, = \frac{I _{s}}{6 ( n V _{T} ) ^{3}}, = \frac{I _{s}}{24 ( n V _{T} ) ^{4}}, ⋮

These coefficients describe how the diode current $I_{d}$ depends on the AM signal $s (t)$ when expanded as a polynomial.

Mathematical Beauty of Modulation with Diode

Key Points:

Step 1: Rewrite the Shockley diode equation

Step 2: Taylor series expansion of $e^{x}$

Step 3: Subtract 1 from the expansion

Step 4: Multiply by $I_{s}$

Step 5: Identify the coefficients

Final Answer:

Amplitude Modulation

General Form of an AM Signal

Simplified AM Signal Equation

Key Components of the AM Signal

Modulation Index ( $μ$ )

Final AM Signal Equation

Summary

AM Modulation

Step 1: Shockley Diode Equation

Step 2: Expand the Exponential Term

Step 3: Substitute Back into the Diode Equation

Step 4: Simplify the Terms

Step 5: Approximation for Small-Signal Modulation

Step 6: Frequency Domain Representation

Final Modulation Equation

AM Demodulation

Step 1: Recall the Shockley Diode Equation

Step 2: Expand $e^{\frac{s ( t )}{n V _{T}}}$ Using the Taylor Series

Step 3: Substitute $s (t)$

Step 4: Expand Each Term

First-Order Term ( $a_{1}$ ):

Second-Order Term ( $a_{2}$ ):

Higher-Order Terms ( $a_{3}, a_{4}, \dots$ ):

Step 5: Collect Coefficients

Final Answer:

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Mathematical Beauty of Modulation with Diode

Key Points:

Step 1: Rewrite the Shockley diode equation

Step 2: Taylor series expansion of ex

Step 3: Subtract 1 from the expansion

Step 4: Multiply by Is​

Step 5: Identify the coefficients

Final Answer:

Amplitude Modulation

General Form of an AM Signal

Simplified AM Signal Equation

Key Components of the AM Signal

Modulation Index (μ)

Final AM Signal Equation

Summary

AM Modulation

Step 1: Shockley Diode Equation

Step 2: Expand the Exponential Term

Step 3: Substitute Back into the Diode Equation

Step 4: Simplify the Terms

Step 5: Approximation for Small-Signal Modulation

Step 6: Frequency Domain Representation

Final Modulation Equation

AM Demodulation

Step 1: Recall the Shockley Diode Equation

Step 2: Expand enVT​s(t)​ Using the Taylor Series

Step 3: Substitute s(t)

Step 4: Expand Each Term

First-Order Term (a1​):

Second-Order Term (a2​):

Higher-Order Terms (a3​,a4​,…):

Step 5: Collect Coefficients

Final Answer:

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Step 2: Taylor series expansion of $e^{x}$

Step 4: Multiply by $I_{s}$

Modulation Index ( $μ$ )

Step 2: Expand $e^{\frac{s ( t )}{n V _{T}}}$ Using the Taylor Series

Step 3: Substitute $s (t)$

First-Order Term ( $a_{1}$ ):

Second-Order Term ( $a_{2}$ ):

Higher-Order Terms ( $a_{3}, a_{4}, \dots$ ):