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let's consider AD8318, a chip which contains a logarithmic amplifier, often used for power measurements.

I have read its datasheet and there is a thing that I do not understand: no maximum input RF voltage is declared. We may see from page 6 that only maximum input power is declared (12 dBm). Why? How can I know which is the maximum input RF voltage?

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Becasue there isn't a max voltage, only a max power. You can put in whatever voltage you want, as long as the power referenced to 50Ω is kept below 12dBm. The requirement forces you to look at the frequency content of your signal and ensure that the total power of that signal does not exceed the rating. So measure or calculate the total power of whatever is upstream and keep it below that requirement.

At DC you should limit the voltage to this requirement.

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  • \$\begingroup\$ But how can this be physically possible? How can the isolation of the input terminals resist to any voltage? \$\endgroup\$ – Kinka-Byo Sep 23 at 18:09
  • \$\begingroup\$ I think your missing the point, you won't be able to put in a voltage large enough to damage the part if you keep the power low. If you did calculate the max voltage for the power requirement (which would be difficult to do for all frequencies) then you would find it would be pretty low. They are saying that you don't need to calculate the voltage and worry about it, just worry about the power. \$\endgroup\$ – Voltage Spike Sep 23 at 18:14
  • \$\begingroup\$ Ok perfect, thank you very much \$\endgroup\$ – Kinka-Byo Sep 23 at 18:25
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Using P = V^2/R, at Power = 0.001 watt, and R = 50 ohm, you can compute Vrms = 0.223 volts.

Then scale up by 2 * 1.414, to compute Vpeakpeak = 0.632 volts.

The +12dBm input is 4 factors of 2X power (3dB) stronger than 0dBm.

The +12dBm input is 2 factors of 2X voltage (6.01dB) stronger than 0dBm.

Thus we know the +12dBm is 2 * 2 * 0.223 voltRMS.

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