RF Power

What exactly is RF power?

Two transmitters operate at 2.45 GHz. Does that mean they transmit the same power?

No. Both signals complete 2.45 billion cycles per second, but their voltage amplitudes can differ. One can swing between larger positive and negative voltages than the other. Frequency tells us how fast the voltage oscillates; amplitude tells us how large that oscillation is.

To see why amplitude matters, suppose both outputs are sine waves centered on 0 V, each connected to an identical ideal 50 Ω resistor. Across the same resistance, a larger voltage drives a larger current, transferring more energy to the resistor each second. RF power is the rate at which an RF signal transfers energy.

Power is measured in watts (W), and energy in joules (J). One watt means one joule transferred each second. A milliwatt (mW) is one thousandth of a watt, so 10 mW means 0.010 joules transferred per second.

Doubling the voltage across the same resistor doubles the current too. Power depends on their product, so the average power becomes four times larger. For the same waveform and resistance, average power is proportional to voltage amplitude squared.

For our two 2.45 GHz sine waves, a peak voltage of 1 V across the 50 Ω load gives 10 mW average power; a peak voltage of 2 V gives 40 mW. These averages cover complete cycles. The second signal transfers four times as much energy each second, even though its frequency is unchanged. The waveform and load matter: voltage amplitude alone does not specify power.

Leaving either transmitter on for longer delivers more energy without changing its average power. Over a stated interval, average power is the transferred energy divided by elapsed time:

Milli also applies to energy: 1 millijoule (mJ) is 0.001 J. Micro (µ) means one millionth; pico (p) means one trillionth. We will use these prefixes for smaller powers and energies throughout the lesson.

Where does the power go?

In our resistor example, the delivered energy becomes heat. A transmitter can instead send energy through a cable to an antenna, which releases energy as electromagnetic waves. RF power carried along the cable is called conducted power; the part that leaves the antenna as waves is radiated power.

The antenna may not radiate all the power arriving from the cable. Some can be reflected back along the cable at the connection to the antenna, called the antenna feed.

The power arriving at the feed is incident power; the power returning along the cable is reflected power. Subtract that returned power from the arriving power to find the accepted power: the net amount entering the antenna.

Some accepted power heats the antenna; the rest is radiated.

The antenna's radiation efficiency is radiated power divided by accepted power: in the figure, 7 mW / 8 mW = 87.5%.

Before power reaches the antenna, the cable can also convert some of it to heat. A generator's output rating therefore does not tell us how much power the antenna will radiate. That rating also assumes a particular load; the source-and-load example later in this lesson shows why the connected load matters.

Stating that a transmitter delivers 10 mW average power at its output connector tells us both how much power is delivered and where. Return now to the 1 V peak sine across our 50 Ω resistor. Its voltage alternates between positive and negative, yet it keeps delivering energy. Next, we will follow the voltage and current through a cycle to see how that produces the 10 mW average.

How a sinusoid delivers energy

A sine-wave voltage averages to zero. Why can it still heat a resistor?

RF Signals introduced the sine wave's changing voltage. Here, follow a 2.45 GHz sine with a 1 V peak across an ideal 50 Ω resistor. Its voltage averages to zero; the new question is what happens when we multiply it by the current.

Current heats the resistor whichever way it flows. When the voltage changes polarity, the current reverses with it. Voltage and current therefore have the same sign at each instant, so their product is positive in both halves of the cycle.

Ohm's law gives the current at each instant: voltage divided by resistance. At the positive peak, $\frac{1\,\mathrm V}{50\,\Omega}=0.020\,\mathrm A=20\,\mathrm{mA}$. Multiplying 1 V by 0.020 A gives 0.020 W, or 20 mW. At the negative peak, $(-1\,\mathrm V)(-0.020\,\mathrm A)$ gives the same power. At a zero crossing, both readings and their product are zero.

For this signal, $v(t)=1\sin(2\pi ft)\,\mathrm V$ and $i(t)=20\sin(2\pi ft)\,\mathrm{mA}$. Here $f=2.45\,\mathrm{GHz}$. Their product is $p(t)=20\sin^2(2\pi ft)\,\mathrm{mW}$. Instantaneous power moves from 0 to 20 mW; its average is exactly 10 mW.

The power trace has two humps during every voltage cycle. Squaring removes the sign, so the positive and negative half-cycles have the same power shape. Power therefore repeats at twice the 2.45 GHz carrier frequency, even though the carrier frequency itself did not change.

Would a higher carrier frequency deliver more energy by fitting more cycles into each second? With the same peak voltage and ideal resistance, each cycle becomes shorter and delivers less energy. Doubling the frequency gives twice as many cycles, each carrying half as much energy. The average power stays at 10 mW.

RMS: the amplitude that predicts heating

What DC voltage would heat the same resistor by the same amount?

RF Signals connected root mean square (RMS) voltage with equal heating in a resistor. Now examine how waveform shape and the measurement interval determine it. Square the changing voltage so both polarities contribute, average the squared values over a stated measurement interval, then take the square root. The result is the DC voltage that produces the same average heating in the same resistor.

For a repeating waveform, averaging over one or several complete periods gives the same RMS value. A signal that does not repeat still has an RMS value over a chosen measurement interval.

First try a simple four-step voltage that holds 0 V, +1 V, 0 V, then −1 V for equal amounts of time. Its average voltage is zero. Squaring gives 0, 1, 0, and 1 V²; their average is $\frac{0+1+0+1}{4}\,\mathrm{V^2}=0.5\,\mathrm{V^2}$. The square root is about 0.707 V.

Now apply the same process to a continuously changing sine wave, including every part of the cycle.

For a zero-offset sine wave, the average squared voltage over complete cycles is half the squared peak. Taking the square root gives $V_{\mathrm{rms}}=\frac{V_{\mathrm{pk}}}{\sqrt2}$. The sine wave shown above has a 1 V peak, so its RMS voltage is about 0.707 V.

Recall that peak-to-peak voltage is the full swing: for symmetric peaks about zero, $V_{\mathrm{pp}}=2V_{\mathrm{pk}}$. Peak-to-peak voltage alone does not determine RMS voltage; the waveform shape matters too.

Voltage crest factor compares the largest voltage magnitude with RMS voltage. For this zero-offset sine, $V_{\mathrm{pk}}/V_{\mathrm{rms}}=\sqrt2\approx1.41$. It describes the voltage peaks relative to the amplitude that predicts heating.

For the stated ideal resistive load, RMS voltage and current lead directly to average power.

Across the same resistor, equal RMS voltages give equal average powers—even when the waveforms have different shapes.

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From voltage to power: the load and measurement point

Does “1 V RMS” tell us how much power is being delivered?

When the load voltage is known

Not by itself. The same RMS voltage produces different current and heating in different resistances. For an ideal resistor, divide the squared RMS voltage by the stated resistance.

RF generators commonly specify their output voltage for a 50 Ω load. Other loads are possible; 75 Ω is common in cable-TV systems. When using the resistor-power formula, use the actual load resistance and the RMS voltage across it. The generator's voltage setting may assume a different load.

When the generator setting is fixed

To understand why the voltage measured at a generator's output depends on its load, represent the generator as an ideal voltage source followed by an internal 50 Ω resistance. With an external 50 Ω load connected, the same current passes through both equal resistances, so they share the voltage equally. To provide 1 V RMS across the load, the internal source must therefore supply 2 V RMS. The generator's 1 V setting refers to the expected load voltage.

The diagram's nearly 2 V RMS oscilloscope reading assumes the input draws very little current, leaving almost no voltage drop across the generator's internal resistance. A real oscilloscope and its probe can draw more current at RF because they also store and release charge as the voltage changes. This changes the voltage at the generator's output. Impedance & Reactance explains how this happens.

The earlier 50 Ω and 75 Ω examples both assume 1 V RMS across the load. Changing the load on the generator in this diagram does not keep that voltage fixed: it changes how the internal 2 V RMS is shared between the source resistance and the load.

For example, replace the external 50 Ω load connected to the generator with a 25 Ω load. The generator's internal resistance remains 50 Ω. The smaller load draws more current through both itself and the generator's internal 50 Ω resistance. The RMS current is now $\frac{2\,\mathrm V}{50\,\Omega+25\,\Omega}\approx26.7\,\mathrm{mA}$. More voltage therefore drops across the internal resistance, leaving only about 0.667 V RMS across the load. The load now receives about 17.8 mW, compared with 20 mW for the 50 Ω load.

If we adjusted the generator to maintain 1 V RMS across the 25 Ω load, it would receive 40 mW. In the example above, we kept the generator unchanged and replaced the 50 Ω load with a 25 Ω load. The voltage across the load fell from 1 V RMS to about 0.667 V RMS.

With this generator's internal voltage held fixed, a 50 Ω load receives the most power: 20 mW. A higher or lower load resistance receives less. This maximum deliverable power is the source's available power. It describes the source under the stated load condition; cable losses, reflection and antenna heating can still reduce the power that is eventually radiated.

Where is the measurement made?

Now consider a separate cause of a changed power reading. A cable turns some signal energy into heat, so less power reaches its output than enters its input. A reading therefore needs a location. That location, such as the generator connector or cable output, is called the reference plane.

Watts now have physical meaning, but RF systems often span factors of millions or billions. A logarithmic language makes those ratios easier to see and combine.

Why RF engineers use decibels

What is the easiest way to follow a signal through several stages of gain and loss?

Suppose a cable reduces 10 mW to 5 mW, then an amplifier increases that 5 mW to 50 mW. The cable multiplies power by 0.5; the amplifier multiplies it by 10. Together they multiply power by five. Decibels let us calculate the same combined change by adding instead of multiplying.

If a power amplifier increases signal power by a factor of 100, its output-to-input power ratio is $\frac{P_{\mathrm{out}}}{P_{\mathrm{in}}}=100$. A base-ten logarithm asks: what power of ten gives this ratio? Since $10^2=100$, $\log_{10}(100)=2$. For power ratios, this logarithm expresses the ratio in bels. One bel equals 10 decibels (dB), so multiplying 2 by 10 gives the amplifier's power gain of +20 dB. The power has increased 100 times; the factor of 10 converts the logarithmic value to decibels.

An attenuator that reduces signal power to one tenth has $\frac{P_{\mathrm{out}}}{P_{\mathrm{in}}}=0.1$. Since $10^{-1}=0.1$, $\log_{10}(0.1)=-1$. Multiplying by 10 gives a change of −10 dB, also called a 10 dB loss. Equal dB steps therefore mean equal multiplying factors, not equal extra watts.

Engineers often round the doubling or halving change to 3 dB; at that precision, say “approximately double” or “approximately half.”

To compare 20 mW with 5 mW, divide output by input: $\frac{20\,\mathrm{mW}}{5\,\mathrm{mW}}=4$. On a calculator, $\log_{10}(4)$ is about 0.602; multiply by ten to get +6.02 dB. To convert +6.02 dB back to a power ratio, divide 6.02 by ten, then calculate 10 raised to that result: $10^{0.602}\approx4$.

Before dividing the two powers, express both in watts or both in milliwatts. For an amplifier with a 10 mW input and a 1 W output, convert 1 W to 1,000 mW before dividing: $\frac{1{,}000\,\mathrm{mW}}{10\,\mathrm{mW}}=100$. The units cancel, and $10\log_{10}(100)=20$ gives +20 dB gain. Expressing both powers in watts would give the same ratio.

If one stage multiplies power by $r_1$ and the next by $r_2$, the chain ratio is $r_1r_2$. Because $\log_{10}(r_1r_2)=\log_{10}r_1+\log_{10}r_2$, their decibel values add. This is a property of logarithms, not a special behavior invented for RF.

Return to the opening cable and amplifier: reducing 10 mW to 5 mW is approximately −3 dB, and increasing 5 mW to 50 mW is +10 dB. Adding gives approximately +7 dB, corresponding to five times the starting power.

dBm and dBW: absolute levels with fixed references

How can decibels describe a power level rather than just a change in power?

A gain of 10 dB tells us that power increased tenfold, but we still need the starting power to know the result. dBm compares power with a fixed 1 mW reference. A power level of +10 dBm therefore means 10 mW, and 0 dBm means 1 mW. dBW uses 1 W as its reference, so 0 dBW means 1 W.

In our earlier example, the sine wave delivers 10 mW of average power to the 50 Ω resistor. That is +10 dBm or −20 dBW. Both readings describe the same power.

In dBm, we compare that power with 1 mW; in dBW, with 1 W (1,000 mW). Dividing by this larger reference makes the power ratio 1,000 times smaller. Since $10\log_{10}(1000)=30$, we subtract 30 from the dBm value to express the same power in dBW.

Gain and loss through an RF chain

What reaches the load after an attenuator, amplifier, and cable?

Follow one signal from the source through an attenuator, an amplifier, and a cable. The attenuator deliberately reduces signal power, often to protect the next device from too much input power. In this example there are no reflections: the removed signal power becomes heat. The cable also turns some signal power into heat along its length.

The amplifier increases signal power by drawing energy from its DC supply. Its power gain is output signal power divided by input signal power. The gain describes how much the amplifier changes the signal; finding its output power also requires knowing its input power.

When we connect the source to the attenuator, the attenuator's input becomes the source's load. Its input circuitry receives the signal power. The amplifier's input then acts as the load for the attenuator. Each input or output connection is called a port.

At 2.45 GHz, each output in this example is connected to the 50 Ω load it is designed to drive. These connections are matched: signal power enters the next device without being reflected back at the connection in this ideal example.

Begin with the source level in dBm. Add each amplifier gain in dB and subtract each loss in dB. For example, “3 dB loss” means subtract 3 dB once. Follow the result to the next connector before applying the next block.

Before moving the slider below, predict the gain needed to reach the target from the displayed source level after both losses. Then use the slider to check your calculation.

These calculations assume the stated 50 Ω connections and constant amplifier gain at 2.45 GHz. With constant gain, doubling the input power in watts doubles the output power in watts. Near the amplifier's limit, the output increases by less than this. Its gain has fallen—a behavior called compression. The amplifier can also distort the waveform as it approaches this limit.

A data sheet might specify 20 dB gain at 2.45 GHz, with a 3.3 V supply and a small input signal. That value applies to those conditions: changing the supply voltage or frequency can change the gain, and a larger input may cause compression. Use gain specified near your operating conditions when calculating the output power.

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Why voltage ratios use 20 log

If voltage doubles, does power merely double too?

Across the same ideal resistance, power is proportional to RMS voltage squared. Doubling voltage therefore makes four times the power. That square is where the factor of 20 comes from—it is not a separate definition to memorize.

For example, 1 V RMS across 50 Ω gives $\frac{(1\,\mathrm V)^2}{50\,\Omega}=0.020\,\mathrm W$. With 2 V RMS across the same resistance, $\frac{(2\,\mathrm V)^2}{50\,\Omega}=0.080\,\mathrm W$. The power ratio is four, so the change is $10\log_{10}(4)\approx+6.02\,\mathrm{dB}$. Since four is the voltage ratio squared, the same result is $20\log_{10}(2)$. The rule follows from $\log_{10}(x^2)=2\log_{10}(x)$ for positive ratios.

When the two resistances differ

The examples above use the same resistance for both voltages. With different resistances, the voltage ratio alone does not tell us the power ratio. One volt RMS across 50 Ω gives 20 mW; the same voltage across 100 Ω gives 10 mW. The voltage is unchanged, but doubling the resistance halves the power. To compare powers across different resistances, multiply the squared voltage ratio by $R_1/R_2$. Here that resistance ratio is $\frac{50\,\Omega}{100\,\Omega}=\frac{1}{2}$:

Impedance & Reactance extends these resistor examples to loads that also store and return energy. Two-Port Networks & S-Parameters (in preparation) develops power comparisons for waves travelling toward and away from RF ports.

dBc: relative to a carrier

A spur is measured at −40 dBc. What does that tell us—and what is still missing?

A spur is an unwanted frequency component. The carrier is the intended RF tone in this example. The spectrum below shows a peak for the carrier and another for the spur. Each peak's horizontal position shows its frequency, and its height shows its power level in dBm.

The suffix “c” names the reference: the carrier. A component at −40 dBc has one ten-thousandth of the carrier power under the stated measurement conditions. Its absolute power remains unknown until the carrier level is known at the same reference plane.

Record the carrier power, measurement point, and instrument settings with the dBc result. The measurement section explains how to choose the frequency range and averaging interval.

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How separate powers combine

When can separate powers be added, and when does phase matter?

Even with no incoming transmission, components in a receiver generate random electrical fluctuations called noise. Suppose two parts of the receiver contribute noise to the same output. Their fluctuations can be uncorrelated: knowing one contribution does not tell us which way the other is fluctuating.

Their instantaneous voltages still add, as in RF Signals. Sometimes the fluctuations reinforce and sometimes they oppose. For uncorrelated noise with no DC offset, the extra power from reinforcement and the reduction from opposition balance out in a sufficiently long average. Each contribution's own average heating remains, so their average powers add.

Cascaded dB values can be added because they describe ratios along one path. Separate uncorrelated contributions require another calculation: convert to milliwatts, add the powers, then convert back to dBm. Watts and milliwatts express linear power, meaning twice the number represents twice the power.

Compare the three cases below: how much does a second contribution add as it becomes weaker than the first?

These examples add uncorrelated contributions. Two copies of the same tone can keep a stable phase relationship; such signals are called coherent. Their voltages can reinforce or cancel, so phase must be included before calculating their combined power.

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Apply it: calculate and report RF power

Can you choose the calculation at each step, then describe exactly what the results mean?

Use this new case to connect voltage, gain and loss, relative power, and timing. During transmission, a 915 MHz zero-offset sine measures 0.4 V peak-to-peak across the attenuator's matched 50 Ω input at P0. This is the actual load voltage, not an unloaded generator setting.

• The signal passes through a 3 dB attenuator, an amplifier with 20 dB gain, and a cable with 7 dB loss.
• All connections are matched to 50 Ω, and the amplifier maintains the stated gain at 915 MHz.
• At the final load, P3, a measured spur is −30 dBc during the on-time, relative to the carrier there.
• The carrier transmits for 3 ms every 12 ms, with zero carrier power during the 9 ms idle time.

Calculate the following before opening the solution. Keep full precision through each calculation and round only the displayed results. Report the carrier and spur separately.

1. Find the source peak and RMS voltages, then its on-time power at P0 in milliwatts and dBm.
2. Find the on-time carrier power at P1, P2, and P3 in milliwatts and dBm.
3. Find the on-time spur power at P3 in microwatts and dBm.
4. Find the duty cycle and full-period carrier average at P3 in milliwatts and dBm.
5. Write a measurement report. Which conditions are given, and which instrument details still need recording?

Choose the calculation you need

• Voltage to power: establish RMS voltage across the actual resistive load, then use $P=V_{\mathrm{rms}}^2/R$. A peak-voltage shortcut also needs the waveform shape.
• A chain of gains and losses: start with a power level in dBm, add gains in dB, and subtract losses in dB under the stated matching and gain conditions.
• Separate uncorrelated contributions: convert to milliwatts, add, then convert back to dBm. Coherent signals also require phase and circuit information.
• A changing power over time: add the energy delivered in each interval and divide by total time. With zero idle power, multiply duty cycle by the average power during the on-time; for a flat burst, this is its constant on-state power.
• A measurement report: name the location and load, defined power quantity, included frequencies, time interval, and instrument mode and settings.

Apply this with your own values: convert between watts, dBm, and sinusoidal voltage, keeping the load resistance explicit.