Hello,In the datasheet below we have PAE value(formula below) and saturation point.
What are the maximal Pin and Power at the maximal P1dB saturation point?
https://www.qorvo.com/products/d/da004195

 

What are the maximal Pin and Power at the maximal P1dB saturation point?

Pin depends upon what power added efficiency you choose or want. Unclear what you mean by "maximal P1dB saturation point". For a given frequency there is a single 1dB compression point (P1dB) and output saturation (Psat) usually occurs above P1dB. Amplifier P1dB is not given but page 6 from the datasheet has everything you need.

As an example, for a PAE = 25% @ 10GHz and 100mA quiescent drain current, Pin should be 9dBm. Pout would then be 30.5dBm which results in Pdc ~ 4.47W. This translates to a DC current of around 178mA from a 25V supply voltage.

 

Hello 0rian, why do you use quicent current in your mathematical logic.
Quicent current only when there is no RF pin present in system.
All our power math is when teh Pin is present in the calculations.

 

The plots shown in the datasheet are only valid for a specific quiescent current (100mA @ 25V supply). Changing the current would alter the plots and lead to a slightly different outcome.

 

quicent current is the Ids current that is applied before we put the RF signal.
after we put rf signal to the gate the drain cuurent will rise by the transimpedance formula.
I cant understand why its important if the current will rise at the moment we put RF input?
Thanks.

 

I cant understand why its important if the current will rise at the moment we put RF input?

I don't get where you are stuck at the moment. What is the importance of the rising current, who said it is important and what has that to do with your original question?

 

Hello Orion , so the quecent current is irrelevant in PAE calculation ?
We only take the Id when RFin is being powered on ?
Correct ?
Thanks .

 

When you turn on the amplifier, it draws a quiescent current. This is your bias current. This is the amount of power the device draws just being on.

When the amplifier receives a signal on the input, the gate voltage is modulated, so the drain current will also be modulated. The average (RMS) voltage of the input signal will lead to an average (RMS) increase in current drawn on the drain. This is why class-B, F/F inverse, J/J inverse, and many other amplifier types are more efficient that class-A amplifiers (up to about 75%, roughly, compared to an absolute max of 50%, though some class-D amps can get higher at a severe cost to linearity), because the default "on" current is zero, so the only DC power draw comes when there's a signal being applied to the transistor's gate.

This total current drawn at the drain when signal is applied is what corresponds to the total DC power being drawn by the supply. The DC power is therefore the supply voltage multiplied by the total drain current drawn for a given input signal level.

PAE is therefore the difference between the output signal power and the input signal power divided by this total DC power from when the device actually has an input power signal.

So, no: quiescent current is not what is used for PAE calculations, as that will under-estimate your total DC power drawn (sometimes significantly).

 

...so the quecent current is irrelevant in PAE calculation ?

Idq is not directly needed for the actual PAE calculation but is certainly not irrelevant.

The calculation I've done at the beginning of the thread are based on the PAE vs. Pin and Pout vs. Pin (@ F = 10GHz) plots from the datasheet on page 6. Those plots were aquired at and are only valid under certain test conditions: CW output signal, Vd=25V, Idq=100mA and T=25°C. If you change any of these test conditions, you would have to make your own measurements (PAE vs. Pin and Pin vs. Pout) at the frequency of interest, redo the calculations and get new results.

 

So why quecent current is important ? why we need to start with this amount of current threw the drain before apllying the RF signal?
Thanks.

 

So why quecent current is important ? why we need to start with this amount of current threw the drain before apllying the RF signal?
Thanks.

Because quiescent current is what sets the operating point for the transistors - what defines the small-signal gain (as transconductance is dependent on drain voltage and current) and is partially responsible (among other factors) for saturation power, so therefore also dictates the large-signal gain. It also is the baseline of DC power you'll consume before applying RF.