How A Diode Detectors Produce A Voltage Proportional To Incident Power - By David Alexander Straight

Consider the following circuit.

V₁ and R₁ represent an RF source, where V₁ is an ideal voltage source and R₁ is the impedance of the source. R₂ is used to match the detector circuit so that the source terminates in its character istic impedance (R₁). C₁ is large so that V(t) appears across the diode (AC shunted to ground by C₁ ).

If C₁ has zero volts across its terminals at t=0⁺ and at this same instant V(t) is applied (which is the same as when V₁ is applied) the capacitor will begin to charge. Note that V₁ magnitude is on the order of milli-volts.

The current in the diode is described by the Ebers-Moll equation

where is the diode current

is the saturation current of the diode

is charge in Coulombs

is Boltzmann’s constant

is the temperature in Kelvins.

The current can be approximated about the origin using a Taylor Series Expansion

and , the voltage across the diode, was replaced by

Simplification of Equation 4 shows three terms, a DC term and two AC terms consisting of the original frequency and a second harmonic.

The average of this current is , where Tn is a multiple of the period of i(t).
When is substituted for Equation 5 the average current for the pure sinusoid input, , is . Note that it contains a voltage square d term indicating that the DC component of current is proportional to power. The three sources I’ve used in writing this derivation stop at this point (or stop with the derivation of Equation 3). My goal is to prove that the voltage, Vout, that d evelops across C₁ in Figure 4 is proportional to the voltage input (i.e. power). My motivation stemmed from the fact that if the voltage, , across the diode remained constant then a con stant current (DC) would be entering the capacitor, which according to would yield an unbounded Vout across the capacitor, the circuit would be unstable and accumulating endless energy, if this were the cas e I’m sure the electric company would have utilized this circuit. Realizing that the voltage across the terminals of the diode had to change

with time as the capacitor built up a stored charged, my first step was to find a function for the voltage across the diode. Then, I would be able to solve for the voltage across the capacitor.

The circuit I used following this pursuit is shown in Figure 5 below.

Where V_TH and R_TH make up the Thévenin equivalent of the source of Figure 4 The voltage loop around this circuit is which is the same as . Substituting Equation 3 into this gives

At first, I was eager to solve this for a general solution. However, this will have to wait for the time being. My experiences from school and my Co-Op have pointed out the value of knowing when to approximate something. This is especially valuable w hen time is of the essence, which I’ve found to be true both at work and school. Rather than learning how I might go about solving this differential equation I decided to take a more intuitive approach. I started with a basic I-V curve of a diode and a sked myself questions about how this circuit could work.

The obvious fact was the voltage developing across the capacitor is not unbounded, it reaches some steady state. The reason the capacitor reaches a steady state voltage is that as it charges up the bias voltage across the diode changes. When a small AC signal is applied to the diode circuit, the diode will conduct in both directions see Equation 3. However, it does favor the forward bias current a little more than the reverse bias current for small AC signals, enabling it to rectify even at very low voltage values. Incidentally, the reason why the average current, when solved for, contains a DC component, is that it is rectifying. At some point the capacitor will charge up to the exact voltage required that causes the forward bias conduct ion current to exactly match the reverse bias conduction current in such a way as to maintain the same average number of coulombs in the capacitor, thus there is an average voltage, or DC voltage across the capacitor.

Rearranging the voltage loop equation gives of Figure 5 gives which is approximately the same as

Provided the voltage across R_TH is small compared to the voltage across V_d. Equation 7 is a better description of what the diode sees over time, when the circuit reaches a steady state value. I used this to calculate the diode’s current, which is

If the circuit is to reach a steady state value then the average current (which is the same as the DC component of the current) will be zero, therefore is set to zero and A is solved for giving

This is the value is very close the voltage built up on the capacitor (because very little voltage develops across the Thévenin resistance) and it can be shown that it is proportional to the voltage squared. If Equation 10 is rearranged algebrai cally then some constant, B, will equal the voltage squared.

Equation 11