Diode based passive RF frequency doublers - the basics

Introduction

Passive RF frequency doublers have always piqued my curiosity, particularly those capable of operating across a wide bandwidth. At first glance, the concept seemed to have to be deceptively simple—but I couldn’t initially grasp how it would be possible to generate higher frequencies purely through passive means, without resorting to phase-locked loops or active frequency synthesis techniques. These circuits manage to double the frequency of an RF signal without active components, and apparently their elegance lies in the clever use of nonlinearities.

With some time on my hands recently, I decided to explore the fundamental principles behind these devices more deeply. In this article, I aim to walk through the core concepts that make diode-based passive frequency doublers work, especially in broadband applications. Whether you’re approaching this topic from a theoretical interest or practical design perspective, I hope this summary offers a useful foundation.

For the purposes of this article, I will omit the discussion of impedance matching on the input and output sides. It’s worth noting, however, that any circuit employing biasing requires AC coupling at both the input and output ports, along with appropriate RF choke-based bias injection, as is standard practice in RF design.

• A recap on the Shockley equation for diodes
• The Anti-Parallel Diode pair
• Odd order single diode harmonics generation
• Conclusion
• Appendix: Exponentials of sine functions
• Appendix: The Taylor expansion of the cosh

A recap on the Shockley equation for diodes

To begin, let us consider the fundamental behavior of diodes. These devices are constructed by forming pn-junctions—typically realized as a p-well embedded within an n-well on a p-type substrate, following standard semiconductor fabrication techniques. The pn-junction inherently supports unidirectional current flow, thereby exhibiting rectifying properties.

A typical diode datasheet delineates three primary operating regions:

• The conductive region arises when the diode is forward biased beyond a specific threshold voltage. Should this voltage become excessive, the associated high current can lead to thermal damage.
• The isolation region corresponds to reverse bias conditions. Ideally, the diode does not conduct in this region; however, a small leakage current—typically in the nanoampere (nA) or picoampere (pA) range—is present in practical scenarios.
• The breakdown region occurs when the reverse voltage exceeds a critical limit. For conventional diodes, this results in irreversible damage. In contrast, Zener diodes are intentionally designed to operate in this regime and are commonly employed for voltage regulation and overvoltage protection.

Once the forward bias surpasses the threshold voltage, the diode enters its conducting state. In large-signal scenarios, the resulting current is most accurately modeled using the Shockley equation, which incorporates the effects of carrier drift, thermal diffusion, and recombination-generation mechanisms. The current-voltage (I-V) characteristics found in datasheets often appear approximately linear due to logarithmic scaling of the current axis:

[ \begin{aligned} I &\approx I_S(T) \left(e^{\frac{U}{nU_t}} - 1\right) \end{aligned} ]

Where

• $I$ is the current flowing through the diode
• $U$ is the applied bias voltage
• $I_S(T)$ is the temperature dependent saturation current. The saturation current is the leakage current flowing through a diode while being operated in reverse biased (blocking) mode.
• U_t is the thermal voltage given by $U_t = \frac{k_B T}{q}$ with the elementary charge q. This accounts for the thermal diffusion.
• The empirical emission coefficient $n$ that takes into account deviations of the real diode from the optimal model. This is mostly caused by recombination effects that reduce the Saturation current. An ideal diode would have a emission coefficient of $1$.

The relevant constants used are (in SI system):

• The Boltzmann constant $k_B \approx 1.38 * 10^{-23} \frac{Ws}{K}$
• The elementary charge $q = e = 1.6 * 10^{-19} As$

The Anti-Parallel Diode pair

Let’s now take a look at the current flowing over the anti-parallel connected diode pair put in series into the signal path (not as a shunt) when applying a drive voltage of $U$ - we utilize the odd symmetry flip to cancel the fundamental and odd harmonics:

[ \begin{aligned} I_{total}(U) &= I_{diode}(U) + I_{diode}(-U) \\ &= I_S(T) \left(e^{\frac{U}{n U_t}} - 1\right) + I_S(T) \left(e^{\frac{-U}{n U_t}} - 1\right) \\ &= 2 I_S(T) \left(\cosh \left(\frac{U}{n U_t}\right) - 1\right) \end{aligned} ]

We can now apply a Taylor expansion of the $\cosh$:

[ \begin{aligned} \cosh \left(\frac{U}{nU_t}\right) &= 1 + \frac{1}{2!}\left(\frac{U}{nU_t}\right)^2 + \frac{1}{4!}\left(\frac{U}{nU_t}\right)^4 + \ldots \end{aligned} ]

Absorbing constant factors in constants for each power we thus arrive at:

[ \begin{aligned} I_{total}(U) &\propto a_2 U^2 + a_4 U^4 + \ldots \end{aligned} ]

When we now assume a sine driving signal $U(t) = A \sin(\omega t)$ this yields:

[ \begin{aligned} I_{total}(U) &\propto a_2 A^2 \sin^2(\omega t) + a_4 A^4 \sin^4(\omega t) + \ldots \end{aligned} ]

When taking a look into product and exponential rules for sines (see Appendix) we can expand the exponents:

[ \begin{aligned} I_{total}(U) &\propto a_2 A^2 \frac{1}{2} \left(1 - \cos(2\omega t) \right) + a_4 A^4 \left(\frac{3}{4} + \frac{1}{4} \cos(2\omega t) + \frac{1}{8} \cos(4\omega t)\right) + \ldots \\ \end{aligned} ]

When we now drop all terms above 4th order expansion:

[ \begin{aligned} I_{total}(U) &\propto a_2 A^2 \frac{1}{2} + a_4 A^4 \frac{3}{4} - a_2 A^2 \frac{1}{2} \cos(2 \omega t) + \frac{1}{4} a_4 A^4 \cos(2 \omega t) + \frac{1}{8} a_4 A^4 \cos(4\omega t) \\ &= \left(\frac{1}{2} a_2 A^2 + \frac{3}{4} a_4 A^4 \right) + \left(-\frac{1}{2} a_2 A^2 + \frac{1}{4} a^4 A^4 \right) \cos(2 \omega t) + \frac{1}{8} a_4 A^4 \cos(4 \omega t) \\ \end{aligned} ]

We can see the total current contains:

• A constant DC component
• A second harmonic at $2 \omega$
• A fourth harmonic at $4 \omega$
• Higher even harmonics will also be present when we perform the expansion to higher levels (those terms from the expansion will also contribute to the prefactors of lower frequency components)

We can utilize this configuration without any biasing (i.e. fully passive). Each diode conducts during one of the half cycles. The fundamental cancels out naturally, the second harmonic dominates. Biasing actually would shift the symmetry.

Odd order single diode harmonics generation

So then the next question that arises - how could we generate odd harmonics? “Simply” by using only a single diode in the signal path. Let’s again start with the Shockley equation:

[ \begin{aligned} I_{total} = I_S(T) \left(e^{\frac{U}{n U_t}} - 1\right) \end{aligned} ]

Let’s again assume the driving signal $U(t) = A \sin(\omega t)$ and perform the Taylor expansion we get:

[ \begin{aligned} I(t) &= c_1 \sin(\omega t) + c_2 \sin^2(\omega t) + c_3 \sin^3(\omega t) + \ldots \\ &= c_1 \sin(\omega t) + c_2 \frac{1}{2} \left(1 - \cos(2\omega t) \right) + c_2 \left( \frac{3}{4} \sin(\omega t) - \frac{1}{4} \sin(3\omega t) \right) + \ldots \\ &= c_1 \sin(\omega t) + \frac{1}{2} c_2 - \frac{1}{2} c_2 \cos(2 \omega t) + \frac{3}{4} c_2 \sin(\omega t) - \frac{1}{4} c_2 \sin(3 \omega t) + \ldots \\ &= \frac{1}{2} c_2 + \left(c_1 + \frac{3}{4} c_2 \right) \sin(\omega t) - \frac{1}{2} c_2 \cos(2 \omega t) - \frac{1}{4} c_2 \sin(3 \omega t) + \ldots \\ c_m &= I_s \frac{A^m}{m! (n U_T)^m} \end{aligned} ]

As one can see this term includes a DC term, the fundamental as well as odd and even harmonics. This configuration is usually used with biasing to reach the exponential conductive region of the diode instead of the quasi-linear mode (and blocking during the second half wave). It is possible to operate triplers without bias for large input signals, for small signals a large asymmetric waveform and thus odd harmonics are generated - this clipping produces odd harmonics. Alternatively one can operate the diode in its exponential conductive mode - the nonlinearity again produces even and odd harmonics.

Conclusion

In summary, we have seen that diodes can be effectively employed to generate harmonic content from RF signals. Even harmonics arise naturally from configurations using anti-parallel diode pairs, which operate without the need for any external bias. In contrast, a single diode in the signal path can generate both even and odd harmonics, particularly when driven into its exponential conduction region. This approach often benefits from biasing, especially when working with low-level signals.

It is important to note, however, that harmonic generation is only part of the complete design. Practical implementations must also consider how to efficiently transfer power into and out of the nonlinear circuit. Impedance matching—whether narrowband (e.g., using Pi or L networks) or broadband (e.g., via transformers or resistive divider networks)—is crucial to minimize reflections and maximize power conversion at the desired harmonic frequencies.

Appendix: Exponentials of sine functions

Since they are often used here a short overview of multiplications and exponentials of sine and cosine functions. The relations that we are going to look in are:

Term	Equation
$\sin(x)$	$\frac{e^{ix} - e^{-ix}}{2i}$
$\cos(x)$	$\frac{e^{ix} + e^{-ix}}{2}$
$\sin(\alpha) * \sin(\beta)$	$\frac{1}{2} \left( \cos(\alpha - \beta) - \cos(\alpha + \beta) \right)$
$\cos(\alpha) * \cos(\beta)$	$\frac{1}{2} \left( \cos(\alpha + \beta) + \cos(\alpha - \beta) \right)$
$\sin(\alpha) * \cos(\beta)$	$\frac{1}{2} \left( \sin(\alpha + \beta) + \sin(\alpha - \beta) \right)$
$\sin^2(x)$	$\frac{1}{2} \left(1 - \cos(2x) \right)$
$\cos^2(x)$	$\frac{1}{2} \left(\cos(2x) + 1 \right)$
$\sin^3(x)$	$\frac{3}{4} \sin(x) - \frac{1}{4} \sin(3x)$
$\sin^4(x)$	$\frac{3}{4} + \frac{1}{4} \cos(2x) + \frac{1}{8} \cos(4x)$

So now let’s go into detail. First let’s take a look at the Euler formulation of sine and cosine and the implications for multiplications of sines and cosines:

[ \begin{aligned} \sin(x) &= \frac{e^{ix} - e^{-ix}}{2i} \\ \cos(x) &= \frac{e^{ix} + e^{-ix}}{2} \\ \sin(\alpha) * \sin(\beta) &= \frac{e^{i\alpha} - e^{-i\alpha}}{2i} * \frac{e^{i\beta} - e^{-i\beta}}{2i} \\ &= \frac{1}{4 i^2} \left(e^{i(\alpha + \beta)} - e^{i(\alpha - \beta)} - e^{-i(\alpha - \beta)} + e^{-i(\alpha + \beta)}\right) \\ &= -\frac{1}{4} \left(e^{i(\alpha + \beta)} + e^{-i(\alpha + \beta)} - \left(e^{i(\alpha - \beta)} + e^{-i(\alpha - \beta)}\right)\right) \\ &= -\frac{1}{2} \left( \frac{e^{i(\alpha + \beta)} + e^{-i(\alpha + \beta)}}{2} - \frac{e^{i(\alpha - \beta)} + e^{-i(\alpha - \beta)}}{2} \right) \\ &= \frac{1}{2} \left( \underbrace{\frac{e^{i(\alpha - \beta)} + e^{-i(\alpha - \beta)}}{2}}_{\cos(\alpha - \beta)} - \underbrace{\frac{e^{i(\alpha + \beta)} + e^{-i(\alpha + \beta)}}{2}}_{\cos(\alpha + \beta)} \right) \\ &= \frac{1}{2} \left( \cos(\alpha - \beta) - \cos(\alpha + \beta) \right) \\ \cos(\alpha) * \cos(\beta) &= \frac{e^{i\alpha} + e^{-i\alpha}}{2} * \frac{e^{i\beta} + e^{-i\beta}}{2} \\ &= \frac{1}{4} \left(e^{i(\alpha + \beta)} + e^{i(\alpha - \beta)} + e^{-i(\alpha - \beta)} + e^{-i(\alpha + \beta)}\right) \\ &= \frac{1}{2} \left(\underbrace{\frac{e^{i(\alpha + \beta)} + e^{-i(\alpha + \beta)}}{2}}_{\cos(\alpha + \beta)} + \underbrace{\frac{e^{i(\alpha - \beta)} + e^{-i(\alpha - \beta)}}{2}}_{\cos(\alpha - \beta)} \right) \\ &= \frac{1}{2} \left( \cos(\alpha + \beta) + \cos(\alpha - \beta) \right) \\ \sin(\alpha) * \cos(\beta) &= \frac{e^{i\alpha} - e^{-i \alpha}}{2i} \frac{e^{i\beta} + e^{-i \beta}}{2} \\ &= \frac{1}{4i} \left( e^{i(\alpha + \beta)} + e^{i(\alpha - \beta)} - e^{-i(\alpha - \beta)} - e^{-i(\alpha + \beta)}\right) \\ &= \frac{1}{2} \left( \underbrace{\frac{e^{i(\alpha + \beta)} - e^{-i(\alpha + \beta)}}{2i}}_{\sin(\alpha + \beta)} + \underbrace{\frac{e^{i(\alpha - \beta)} - e^{-i(\alpha - \beta)}}{2i}}_{\sin(\alpha - \beta)}\right) \\ &= \frac{1}{2} \left( \sin(\alpha + \beta) + \sin(\alpha - \beta) \right) \end{aligned} ]

Now let us apply those rules in conjunction with the fact that $\cos(0) = 1$ and $\sin(0) = 0$ to evaluate exponentials of sines:

[ \begin{aligned} \sin^2(x) &= \sin(x) \sin(x) \\ &= \frac{1}{2} \left(\cos(x-x) - \cos(x+x) \right) \\ &= \frac{1}{2} \left(\cos(0) - \cos(2x) \right) \\ &= \frac{1}{2} \left(1 - \cos(2x) \right) \\ \cos^2(x) &= \cos(x) \cos(x) \\ &= \frac{1}{2} \left(\cos(x+x) + cos(x-x)\right) \\ &= \frac{1}{2} \left(\cos(2x) + cos(0) \right) \\ &= \frac{1}{2} \left(\cos(2x) + 1 \right) \\ \sin^3(x) &= \sin^2(x) \sin(x) \\ &= \frac{1}{2} \left(1 - \cos(2x) \right) \sin(x) \\ &= \frac{1}{2} \left( \sin(x) - \sin(x) \cos(2x) \right) \\ &= \frac{1}{2} \left( \sin(x) - \frac{1}{2} \left(\sin(x+2x) + \sin(x - 2x) \right) \right) \\ &= \frac{1}{2} \left( \sin(x) - \frac{1}{2} \left(\sin(3x) + \sin(-x) \right) \right) \\ &= \frac{1}{2} \left( \sin(x) - \frac{1}{2} \left(\sin(3x) - \sin(x) \right) \right) \\ &= \frac{1}{2} \sin(x) - \frac{1}{4} \left(\sin(3x) - \sin(x) \right) \\ &= \frac{1}{2} \sin(x) - \frac{1}{4} \sin(3x) + \frac{1}{4} \sin(x) \\ &= \frac{3}{4} \sin(x) - \frac{1}{4} \sin(3x) \\ \sin^4(x) &= \sin^3(x) \sin(x) \\ &= \left( \frac{3}{4} \sin(x) - \frac{1}{4} \sin(3x) \right) \sin(x) \\ &= \frac{3}{4} \sin(x) \sin(x) - \frac{1}{4} \sin(3x) \sin(x) \\ &= \frac{3}{4} \frac{1}{2} \left(\cos(2x) + 1 \right) - \frac{1}{4} \frac{1}{2} \left( \cos(3x - x) - \cos(3x + x) \right) \\ &= \frac{3}{8} \left(\cos(2x) + 1 \right) - \frac{1}{8} \cos(2x) + \frac{1}{8} \cos(4x) \\ &= \frac{3}{4} + \frac{1}{4} \cos(2x) + \frac{1}{8} \cos(4x) \\ \end{aligned} ]

Appendix: The Taylor expansion of the cosh

Since this is something also not commonly known out of someones basic math knowledge here is a short recap on the Taylor expansion of the $\cosh$ function:

[ \begin{aligned} \cosh(x) &= \frac{e^x + e^{-x}}{2} \\ e^{x} &= \sum_{n=0}^{\infty} \frac{x^n}{n!} \\ \to e^x + e^{-x} &= \sum_{n=0}^{\infty} \frac{x^n}{n!} + \sum_{n=0}^{\infty} \frac{(-x)^n}{n!} \\ &= \sum_{n=0}^{\infty} \frac{x^n + (-x)^n}{n!} \\ &= 2 \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} \\ \to \cosh(x) &= \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!} \\ &= 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \ldots \end{aligned} ]

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