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Transmission lines on printed circuit boards must be impedance controlled lines since we want to transmit the full power and avoid the reflections.

Microstrip lines have three parameters that we can change to achieve 50 ohm impedance:

  • copper thickness (T)
  • substrate thickness (T)
  • track width (W)

Due to design constraints, there is only one parameter (track width,) left to achieve 50 ohm.

For example on 0.76mm Rogers4350B substrate at a certain frequency and with 35 micron copper thickness, line width becomes 0.37mm for a 50 ohm transmission line. But this relatively thin line must support transmitting 45dBm RF power.

My question is:

  • How much RF power can a microstrip line handle with a specified width?
  • How can we calculate this?
  • What parameters should we consider on the basis of this problem?
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  • \$\begingroup\$ You have more design freedom than you imply. You can pick a substrate with lower Dk to get a wider trace, you can pick a substrate with a higher thermal conductivity, you can pick a substrate that can operate at higher temperatures. You can pick a different substrate thickness (higher to give you a wider 50-ohm trace, or lower to reduce the thermal resistance between the trace and ground plane). Etc. \$\endgroup\$
    – The Photon
    Feb 21, 2020 at 17:29

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A simple method might be: -

  • Convert power to watts: 45 dBm is 32 watts
  • Divide by Z0 to get current squared
  • Take the square root to get current (0.8 amps RMS)

Then use an on-line calculator to reveal the minimum thickness required for a given temperature rise of the copper. This on-line calculator suggests that the minimum trace width (1 oz copper = 35 micron) for a 40 degC rise is 0.35 mm.

I personally think no more than 20 degC is something I'd be comfortable with (generalism alert) so, the trace width has to be at least 0.53 mm.

Conversely, for a 0.37 mm trace width, the current should be no more than 0.77 amps RMS.

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    \$\begingroup\$ worth considering 2oz copper (or asking the mfg for additional plating on one layer) perhaps? \$\endgroup\$
    – user16324
    Feb 21, 2020 at 11:47
  • \$\begingroup\$ I doubt the online calc takes skin effect into account \$\endgroup\$
    – Mike
    Feb 21, 2020 at 13:37
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    \$\begingroup\$ @Mike I agree, anything much above about 14 MHz would need to account for the fact that the usable copper is going to reduce. See this calculator. At (say) 50 MHz about half the track thickness is being used. Needs to be considered but, I did say (right at the start) that it was a simple method. \$\endgroup\$
    – Andy aka
    Feb 21, 2020 at 13:48

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