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# Schottky diode with maximum efficiency when input power exceeds 30 dBm.

#### True koke

##### Newbie level 6
I'm looking for a Schottky diode to use for the rectifier now.
I intend to design a rectifier with maximum efficiency when the input power exceeds 30 dBm(31, 32, 33...dBm).

What diode should be used in this case?
If you know the name of the diode I want, please let me know.
What should be considered in the datasheet of the diode?

perhaps mention the volts and current ?

... and frequency!

+30 dBm at 50 Ohm is 7.07 Vrms
+30 dBm at 75 ohm is 8.66 Vrms
+30 dBm at 600 ohm is 24.5 Vrms

I'm looking for a Schottky diode to use for the rectifier now.
I intend to design a rectifier with maximum efficiency when the input power exceeds 30 dBm(31, 32, 33...dBm).

What diode should be used in this case?
If you know the name of the diode I want, please let me know.
What should be considered in the datasheet of the diode?
Do you wonder if frequency, current and PIV specs or cost matters? Define acceptable efficiency.

Diode tables on Distributor sites do not list dBm or efficiency.

So which parameters can you specify from the following example headers as a potential selection criteria.? Then what values would you think might work best to sort by?
Quantity Available
PriceSeriesPackageProduct StatusDiode TypeVoltage - Peak Reverse (Max)Current - MaxCapacitance @ Vr, FResistance @ If, FPower Dissipation (Max)Operating TemperaturePackage / CaseSupplier
--- Updated ---

If you convert dBm to mA then choose that as a minimum then sort by Resistance (min) and capacitance (min) and not to exceed price (\$) which do you choose to achieve TBD % efficiency using I*R/ (IR+Vf(I)) * 100%.

Last edited:

Rectification efficiency is supposed to be the power ratio from DC out to AC in. Yes, Watts out / Watts in. For a single diode circuit, as in half wave rectification it is 40.6% efficiency.
So to start, instead of using a diode; using two will double efficiency.

For a single diode circuit, as in half wave rectification it is 40.6% efficiency.
May I suggest, this is incorrect as it depends on output voltage level, relative to the input unless you assume some unspecified fixed voltage.

I'd ask what is the meaning of "efficiency" here? Ohmic, Vf@If?
Or some leak-back attributes (just been through a very high
frequency charge pump design where this was a big deal and
a design compromise - "forward efficiency" vs "reverse efficiency,"
find sweet spot)?

Seems to me you'd be well off to create a test circuit and start
substituting vendor models, and figuring out what about which
ends up giving you what you wanted (or close as practical).
If you don't know a priori what makes the desired outcome.

For a single diode circuit, as in half wave rectification it is 40.6% efficiency.
Apparently you are making some assumptions that might not apply here. Problem is that the OP didn't explicitly tell the conditions. Nevertheless some assumptions are quite obvious: RF rectification, lossless LC matching circuit. Under ideal conditions, losses occur only in the diode.

This link states..."Rectifier efficiency is the ratio of output DC power to the input AC power.
For a half-wave rectifier, rectifier efficiency is 40.6%."

The initial statement is a true definition, yet the value is false. The author is somehow conflating capacity to efficiency.
A true statement is that the average resistor voltage
is Vp/ pi = 31.8% for half-wave and 63.6% for full wave assuming Vout >> Vf diode. In this case, diode power losses could be 1% if Vout was near 100V and thus efficiency power would be>> 99%.