# Microstrip Loss Calculations

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Here's a page on the effect of surface roughness on conductor loss!

Here's a separate page on microstrip loss due to substrate conductivity.

**Updated July 2022** to correct the '"final" equation, thanks to Hayrat. On this page we will review the math for calculating the RF loss of microstrip transmission lines over frequency. We may (or may not) refer to a paper written by Goldfarb and Platzker, entitled *Losses in GaAs Microstrip*, IEEE Transactions on Microwave Theory and Techniques, Volume 38, No. 12, pp 1957-1962, for accurate data to compare the calculations to.

### Loss due to metal

The metal loss calculation for microstrip is a bit messy if you want high accuracy, beyond what we want to get into right now. However, we can always make a good approximation that's much easier to deal with. As with any attenuation-due-to-metal calculation, microstrip attenuation can be expressed as a simple function of radio frequency resistance per meter R' and the line's characteristic impedance Z0, in either Nepers/meter or dB/meter:

Here's the devil in the details: the resistance/meter cannot be found with a simple equation like it can in coax. There are actually five surfaces that are conducting the wave, and their contributions are all geometry-dependant! The primary surfaces are the bottom of the strip and the ground plane below it, but there is conduction on the top surface, and even some on the edges, especially for thick conductors such as one ounce copper. We'll offer just an approximate solution, if you really need to know, use an expensive simulator, or look up some more accurate equations!

Just considering the bottom of the microstrip, the resistance here is:

R'(bottom of strip)=Rsh/W

Where Rsh is the RF sheet resistance (function of resistivity, permeability and frequency) and W is the width of the conductor. Let's assume that the edges of the strip are much smaller than the width of the strip and therefore can be neglected. Let's *postulate* (the less scientific word for this is *pidooma*) that the top of the strip reduces the strip resistance overall by 20% (because it is in parallel with the bottom resistance):

R'(entire strip)=0.8*Rsh/W

Later we intend to prove that the top of the strip is actually doing some of the conduction. This can be done using Sonnet, and adding a thin layer of nickel to the top of the conductor, and watching the loss increase! Come back later and we'll have a page on that, we promise!

If we postulate that the fields spread out uniformly by a factor five under the conductor (this obviously is going to be geometry dependent, fat lines will have less spreading than thin lines!), and that the same metal is used for the strip and the groundplane (same Rsh), then:

R'(groundplane)=0.2*Rsh/W

Now we *add* the strip and groundplane resistances together to get the total resistance/length:

R'=0.8*Rsh/W+0.2Rsh/W=Rsh/W

Voila, we get a very simple expression that is used in some textbooks. But please understand that our simple expression could just as easily come out 1.05*Rsh/W, for example, if we picked a spreading factor of four in the groundplane, etc.). This is just an approximation, don't use it in cases where you really need an accurate prediction of loss

**Update July 2022,** thank to Hayrat. Previously, this equation was Loss/length=8.686(Rsh/W*2xZ0), which is incorrect. The 2xZ0 term should be in the denominator, which we corrected below. Thanks! Finally, we see that the loss/length in dB is:

Loss/length=8.686(Rsh/W)/(2xZ0)

The length units will be whatever you used for the width units (W), typically meters are used.

If anyone has any comments on this approximation and whether it is useful or not, please contact us! We'll postulate that the accuracy of this expression is within +/-50% of the real answer.

Now it's time for a Microwaves101 rule of thumb!

For microstrip, you can (approximately) cut metal losses in half by doubling the dielectric thickness and keeping the same system impedance. For example, going from 10 mil to 20 mil alumina, or two-mil to four-mil GaAs. The reason for this is that the conductor width scales with substrate height, so thicker substrates mean fatter conductors.

### ADS microstrip loss trade study

For grins we used ADS to compare the losses of some different microstrip media, we chose some of our favorite hard substrate materials. If you want us to add to the list, make a suggestion! We separately calculated metal loss and dielectric loss tangent loss. The trick is to use infinite conductivity in one case, and tandD=0 in the other case. We'll summarize the data in tables below. Note that the calculation does not take into account surface roughness, or underplating. Also please don't send us arguments that ADS is not as accurate as HFSS for computing losses, that might be true but the error in this trade space is slight. BTW, some of the dielectric properties came from an old MIC Technologies catalog in case you were wondering about references...

Here's two tables on alumina (Al_{2}O_{3}). The first table is for 10 mil, the second one is for five mil. You'd only use five mil on millimeterwave jobs, (greater than 30 GHz). Check out the loss tangent loss, it is very nearly proportional to frequency. The metal loss closely follows the square root of frequency.

Alumina, H=254 um (10 mils), W=247 um, Er=9.9, TanD=0.0002, 3um gold, conductivity 3.5E7 mhos/meter | |||

Frequency (GHz) | loss tangent loss (dB/cm) |
metal loss (dB/cm) |
total loss (dB/cm) |
---|---|---|---|

0 | -0 | -0.033 | -0.033 |

10 | -0.005 | -0.119 | -0.124 |

20 | -0.009 | -0.175 | -0.184 |

30 | -0.014 | -0.217 | -0.231 |

40 | -0.019 | -0.249 | -0.268 |

Now let's reduce the dielectric thickness by half, from 10 to 5 mils. See what happens to the dielectric loss tangent loss? Almost nothing! But the metal loss is nearly *doubled* (thanks for pointing this out, Les!) We agree that loss should technically be shown as positive values in dB in this case, but this is a convention that is not always followed, perhaps out of mental laziness. No one that matters is going to interpret "-0.5 dB loss" as "+0.5 dB gain" for a passive network! It's like pointing out that "February" isn't pronounced "February", the speaker isn't going to suddenly change his ways because you pointed out that he's an idiot. Hmm, maybe that wasn't a good example!

Alumina, H=127 um (5 mils), W=120 um, ER=9.9, TanD=0.0002, 3um gold, conductivity 3.5E7 mhos/meter | |||

Frequency (GHz) | loss tangent loss (dB/cm) |
metal loss (dB/cm) |
total loss (dB/cm) |
---|---|---|---|

0 | 0 | -0.069 | -0.069 |

10 | -0.004 | -0.227 | -0.231 |

20 | -0.009 | -0.333 | -0.342 |

30 | -0.013 | -0.415 | -0.429 |

40 | -0.018 | -0.483 | -0.501 |

It's time for a Microwaves101 Rule of Thumb!

For fixed line impedance on microstrip (such as fifty ohms), dielectric loss is almost constant for any thickness substrate. Metal loss is reduced by thicker substrates, and is almost inversely proportional. Therefore doubling the thickness cuts the metal loss in half, but does nothing to loss tangent loss.

Here's microstrip on 10 mil quartz. Quartz is the lowest loss substrate there is. Why is it so low in attenuation? Two reasons. It has extremely low loss tangent (0.0001), and because it has low dielectric constant the line widths for 50 ohms are much fatter than other substrates, which reduces metal loss.

Quartz, H=254 um (10 mils), W=540 um, ER=3.8, TanD=0.0001, 3um gold, conductivity 3.5E7 mhos/meter | |||

Frequency (GHz) | loss tangent loss (dB/cm) |
metal loss (dB/cm) |
total loss (dB/cm) |
---|---|---|---|

0 | 0 | -0.015 | -0.015 |

10 | -0.001 | -0.061 | -0.063 |

20 | -0.003 | -0.091 | -0.094 |

30 | -0.004 | -0.114 | -0.118 |

40 | -0.005 | -0.132 | -0.138 |

Here's four-mil (100um) GaAs is used on most MMICs, up to Ka-band.

GaAs, H=100 um (4 mils), W=70 um, ER=12.88, TanD=0.0004, 3um gold, conductivity 3.5E7 mhos/meter | |||

Frequency (GHz) | loss tangent loss (dB/cm) |
metal loss (dB/cm) |
total loss (dB/cm) |
---|---|---|---|

0 | 0 | -0.117 | -0.117 |

10 | -0.010 | -0.356 | -0.366 |

20 | -0.020 | -0.520 | -0.540 |

30 | -0.030 | -0.647 | -0.677 |

40 | -0.040 | -0.752 | -0.793 |

Here's two-mil (50um) GaAs is used on higher frequency MMICs, and sporty power amps. The advantage of two-mil GaAs is that via hole inductance is reduced, and the smaller size of vias also allows individual source vias for large power FETs which is like getting free gain! Again, this example illustrates that dielectric loss tangent loss is pretty much constant with substrate height, but metal loss is nearly doubled when the substrate height is cut in half (following our rule of thumb).

GaAs, H=50 um (2 mils), W=34 um, ER=12.88, TanD=0.0004, 3um gold, conductivity 3.5E7 mhos/meter | |||

Frequency (GHz) | loss tangent loss (dB/cm) |
metal loss (dB/cm) |
total loss (dB/cm) |
---|---|---|---|

0 | 0 | -0.240 | -0.240 |

10 | -0.010 | -0.663 | -0.673 |

20 | -0.019 | -0.957 | -0.977 |

30 | -0.029 | -1.187 | -1.217 |

40 | -0.039 | -1.382 | -1.421 |

Here's beryllium oxide (BeO). This substrate is used as a heat spreader for high power devices. What sucks about it is that it is totally toxic. We recommend you use BeO as a last resort, and don't be surprised if some TV lawyer comes after you! Here we used 10 mil BeO, which is readily available. This medium has lower metal loss than alumina owing to wider line widths (its dielectric constant is less than alumina), but dielectric loss is higher than GaAs or alumina.

Beryllium oxide (BeO) 99.5%, H=254 um (10 mils), W=350 um, ER=6.7, TanD=0.003, 3um gold, conductivity 3.5E7 mhos/meter | |||

Frequency (GHz) | loss tangent loss (dB/cm) |
metal loss (dB/cm) |
total loss (dB/cm) |
---|---|---|---|

0 | 0 | -0.024 | -0.024 |

10 | -0.056 | -0.089 | -0.145 |

20 | -0.112 | -0.132 | -0.244 |

30 | -0.171 | -0.164 | -0.335 |

40 | -0.232 | -0.189 | -0.421 |

### Loss due to substrate conductivity

This topic is on a separate page.

**Author : **Unknown Editor