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In general terms, the PCB industry considers an RF circuit board to be any high frequency PCB that operates above 100MHz.

IRF510 - The device dissipates 43 watts so as long as the kit doesn't draw more that about 3.5 amps max at 12 volts, it won't fail. At 5 watts and 50% efficiency it won't draw more than 1 amp and can handle full reflected SWR.

https://www.rapidonline.com/Catalogue/Search?Query=CDIL

LED Wemos D1 Mini - The built-in led is on pin D4, and it is inverted.

Z=Vcc2 / 2PO
In [7]: (22 * 22) / (2 * 5)
Out[7]: 48.4

[dhiru@zippy DS1054Z_screen_capture]$ python2 OscScreenGrabLAN.py png 192.168.1.16
Instrument ID: RIGOL TECHNOLOGIES,DS1054Z,DS1ZA203916662,00.04.04.SP3

Receiving screen capture...
Saved file: 'captures/DS1054Z_DS1ZA203916662_2021-09-21_18.23.17.png'

From the internet,

7805 -> Without an extra heatsink, you can burn off up to 2W. If you need more current heatsinks are cheap and effective. Every 11 degrees Celsius cooler you run your semiconductors, reliability doubles.

  • Get the max die temperature from the regulator datasheet

  • Get the Theta(j-a) number from the datasheet in degrees C per watt.

  • Calculate the power dissipated by your device. Example: 9 volts in, 5 volts out at 1 amp would be 4 watts.

  • Multiply the watts dissipated by Theta(j-a) to get the temperature rise.

  • Add the temperature rise to the maximum ambient temperature the device will be exposed to.

  • If the number computed in #5 is greater than the max die temperature, then you need to use a heatsink or some other way to actively cool the device.

  • If you use a heatsink, you will use a slightly different set of calculations involving the regulator's Theta(j-c) (Not Theta(j-a).

For a 7805 it's 65 C/W junction to air.

This means that if the regulator produces 1 W of heat, the die will be 65 degrees C hotter than the ambient air. The maximum operating temperature is 125 C. Pune temperature (inside the case) -> 70 c.

Can dissipate -> (125 - 70) / 65 -> 0.84 watts. P = VI. This allows around 100mA of current when input voltage is 14v.

Note: Try the Small mark option instead of the regular Real drill option in KiCad.

Note 2: Set track width to >= 2.2mm in KiCad - try to work with this. For the copper pour, set clearance to 0.850.

...

LTspice notes:

  • Download LTspice (Google "Download LTspice"). It's free.

  • ~/Documents/LTspiceXVII/lib

    Extract external components to this path.

  • Extract the contents of LTspiceXVII_2019Jan29.zip (Google for this file) properly to ~/Documents/LTspiceXVII path.

...

From Daniel Alonso,

A unnecessarily large L doesn't do harm by itself, but the 'series' self resonant frequencies decrease when you increase the turns count. A rule-of-thumb is choosing an inductive reactance of 4 times the load resistance, at the lower frequency. Supposing the circuit is to drive a 50 ohm load, to get a reactance of 200 ohms at 3.5MHz you will need L = 200/(2pi3.5) = 9.1uH. Or if you are more conservative with 10 times the R, it should be 22.7uH.

From Sandeep Lohia,

Inductive reactance on primary side is critical and the magic number is 36.

36 ÷ lowest operating frequency.
36 ÷ 7 = 5 uh

Inductance on secondary not critical but turns ratio is...

...

From Farhan,

At 25 watts, the RF peak voltage is 50v. This is easily handled by the regular disc ceramics rated for 60v. Higher power will require expensive capacitors.