Insertion loss indicates how much energy gets lost while it is going through a component and Insertion Gain indicates how much energy is obtained while it is going through a component. It is same as |S21| in s-parameter terms or transmission coefficient.

Every cable, connector, filter and switch in an RF path has an insertion loss, so this one number appears on almost every datasheet. Let's start with how it is calculated. Then we will see where the lost power goes, how losses add up in a chain, and why a small loss in the wrong place can be expensive.

• How do you calculate insertion loss ?
• Where does the lost power go ?
• How do insertion losses add up in a chain ?
• Why does insertion loss grow with frequency ?
• Why does the location of a loss matter ?

How do you calculate insertion loss ?

Insertion loss compares two powers. The first is the power that reaches the load when the component is not there. The second is the power that reaches the load after you insert the component. The ratio in dB is the insertion loss. When the source and the load are both matched to the reference impedance, usually 50 ohm, this ratio is set by the transmission coefficient S21 alone.

So the working formula is IL = -20 log₁₀|S21| in dB. The factor is 20 rather than 10 because S21 is a ratio of wave amplitudes, and power goes with the square of amplitude. In other words, the "same as |S21|" in the opening means the same information as |S21|, written in dB with the sign flipped. The table below shows the scale with a few values.

For an amplifier, |S21| is larger than 1. Then -20 log₁₀|S21| is negative, and we usually quote the positive number as the insertion gain. For example, |S21| = 10 is an insertion gain of 20 dB.

• Insertion loss is -20 log₁₀|S21| : this holds when the source and the load are matched to the reference impedance.
• A 3 dB loss delivers half the power : |S21| is then 0.707, and |S21|² is 0.5.
• Gain is negative loss : an amplifier with |S21| greater than 1 has an insertion gain instead.

Where does the lost power go ?

A passive component cannot create power, so the power that does not come out of port 2 must go somewhere. It goes to only two places. Part of it reflects back toward the source, and the rest turns into heat inside the component. Knowing which part dominates tells you how to fix a high insertion loss.

Figure 1 splits the incident power for a component with |S11| = 0.2 and |S21| = 0.9. The reflected power is |S11|² = 4 %, the transmitted power is |S21|² = 81 %, and the remaining 15 % is dissipated.

Figure 1. Where the incident power goes in a passive two-port with |S11| = 0.2 and |S21| = 0.9. Most of the 0.92 dB insertion loss here is heat, not reflection.

• Reflection gives the mismatch loss : -10 log₁₀(1 - |S11|²) = 0.18 dB. A better match at port 1 removes this part.
• Heat gives the dissipative loss : -10 log₁₀(|S21|²/(1 - |S11|²)) = 0.74 dB. Only better conductors, a better dielectric or a shorter path reduce this part.
• The two parts add in dB : 0.18 dB + 0.74 dB = 0.92 dB, which is the insertion loss from the table above.
• A lossless component has |S11|² + |S21|² = 1 : all of its insertion loss is mismatch loss. See S-Parameter and Return Loss.

How do insertion losses add up in a chain ?

A real RF path is a chain of parts, and you usually need the total loss from the first part to the last. In linear terms each part multiplies the power by its own factor. In dB, multiplication becomes addition, so the total insertion loss is the sum of the individual losses. This is the same bookkeeping as a Link Budget.

Let's add up a typical path from an antenna connector to a receiver input. The values in the table below are examples chosen for the calculation, not data for a specific product.

The dB column adds to 3.4 dB. The linear column multiplies to 0.457, which is 10^-3.4/10, so the two columns agree. Less than half of the power survives five small losses. This is why RF engineers count every tenth of a dB.

One warning about the simple sum: it assumes that every part is matched. When two mismatched parts sit next to each other, the reflections between them can add or cancel, depending on the electrical length between them. The real total can then differ from the sum, and the difference changes with frequency.

• Losses in dB add : the linear factors multiply, and the log turns that into a sum.
• Small losses accumulate quickly : five parts of 0.1 dB to 1.5 dB leave only 46 % of the power.
• The sum assumes matched parts : reflections between mismatched parts make the real total ripple around the sum.

Why does insertion loss grow with frequency ?

Datasheets for cables and connectors give insertion loss at several frequencies, and the number always rises with frequency. Two physical effects cause this, and they grow at different rates. Knowing both helps you estimate the loss at a frequency the datasheet does not list.

The first effect is conductor loss. At RF, the current flows only in a thin layer near the surface of the conductor. The thickness of this layer is the skin depth, δ = √(ρ/(π f μ)). For copper it is about 6.5 micrometers at 100 MHz, 2.1 micrometers at 1 GHz and 0.39 micrometers at 28 GHz. A thinner layer means a higher resistance, and the conductor loss in dB grows roughly with √f. So a cable with 1 dB of conductor loss at 1 GHz has about 2 dB at 4 GHz.

The second effect is dielectric loss. The insulating material absorbs a small part of the field in every cycle, and the fraction is given by its loss tangent. This loss in dB grows roughly in proportion to f, so it takes over at high frequency. See Loss Tangent for the details.

• Conductor loss grows with √f : the skin depth shrinks as frequency rises, so the resistance rises.
• Dielectric loss grows with f : it is small at low frequency and dominant at millimeter wave.
• Always read the loss at your own frequency : a value quoted at a lower frequency underestimates the loss.

Why does the location of a loss matter ?

The same 1 dB of insertion loss can be harmless in one place and expensive in another. On the receive side, the cost is noise figure. On the transmit side, the cost is output power. Let's check both with numbers.

A passive loss at room temperature has a noise figure equal to its loss. Suppose a 1 dB cable sits in front of an LNA with a 1.5 dB noise figure. The noise figures then add in dB, and the receiver noise figure becomes 2.5 dB. Now move the same cable after the LNA, which has 20 dB of gain. With the Friis formula for noise, F = F₁ + (F₂ - 1)/G₁, the total becomes 1.51 dB. The cable almost disappears from the result, because the LNA gain has already lifted the signal above the noise that the cable adds.

On the transmit side, every dB between the power amplifier and the antenna is lost output power. A 20 W amplifier followed by 1 dB of loss delivers 15.9 W to the antenna, and with 3 dB of loss it delivers only 10.0 W. The lost power becomes heat in the cable and the filter.

• Loss before the LNA adds directly to the noise figure : 1 dB of cable in front of a 1.5 dB LNA gives 2.5 dB. See Noise Figure.
• Loss after a high gain stage matters little for noise : the same cable after 20 dB of gain adds only 0.01 dB.
• Loss after the PA costs output power : 3 dB turns 20 W into 10 W. This is one reason for mounting radios close to the antenna.