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How to findout Power Cosnumption of LNA using ADS

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hari_preetha

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I have designed LNA using ATF54143 transistor of biasing Vds = 3V and Ids =60 mA and used Keysight Technology ADS for design of LNA. I would like to know Power Consumption of designed LNA.
How to know power consumption of designed LNA using ADS.

Please find attached of designed LNA, Data values of ATF54143
 

Attachments

  • ADS_DC_Consumption_Finding_Ckt.jpg
    ADS_DC_Consumption_Finding_Ckt.jpg
    468.9 KB · Views: 186
  • ATF54143_Data_Values_3V_60mA.jpg
    ATF54143_Data_Values_3V_60mA.jpg
    125.3 KB · Views: 131
  • Power_Consumption_Example.jpg
    Power_Consumption_Example.jpg
    282.4 KB · Views: 162

pancho_hideboo said:
180mwatts.
That’s all.
Click to expand...
How did you find out 180 mW, I didn't understand it.
Is there any extra circuit to connect for finding DC power consumption using ADS. In ADS Cookbook, they have given example for finding out DC power consumption (Below screenshot is attached from ADS example for finding DC consumption). What method we should follow for finding out DC power consumption of any LNA circuit.
Below attachment, there is one more LNA design that is stacking of two transistors, how to find out the DC power consumption this attached LNA circuit.
 

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  • ADS_Cook_Example_Finding_DC_Power_Consumption.jpg
    ADS_Cook_Example_Finding_DC_Power_Consumption.jpg
    59.8 KB · Views: 128
  • DC_Power_Consumption_Measuring_Two_Stacing_Transistor.jpg
    DC_Power_Consumption_Measuring_Two_Stacing_Transistor.jpg
    484.1 KB · Views: 85

pancho_hideboo said:
Can you understand very common basics regarding circuit ?

3V * 60mA = 180mWatts for previous circuit.

6V * 108mA = 672mWatts.
Click to expand...

Dear sir,
There is lot in confusion in calculating total power consumption of LNA ckt. In the first post, you can notice there current is negative -60.9 mA, and later post two stacked transistors ckt there also you can notice current is negative -108 mA. Then we should get get 3 X -60.9 mA = -182.7 mW and 6 x -108mA =- 648 mW. which one is correct, please clarify.
Instead of manually calculating Power Consumption of LNA circuit, is there any option or technique to calculate total power consumption of LNA ckt while simulating using Keysight Technology ADS. Please guide me.
 

It is amazing what people learn at engineering schools.

Why do you think a negative power consumption can be correct?
Why do you think that you need to add all current values at circuit nodes?

Don't you think that a single current probe at the (only) 3V power supply should be used?
Don't you think that power consumption would be a positive value?
 

hari_preetha said:
There is lot in confusion in calculating total power consumption of LNA ckt.
Click to expand...
Simply, you can not understand very basics regarding circuit at all.

hari_preetha said:
Then we should get get 3 X -60.9 mA = -182.7 mW
and 6 x -108mA =- 648 mW.
which one is correct, please clarify.
Click to expand...
Both are correct.

Surely see and understand all branch currents in your circuit.
40.8mA is flowing in resistor for biasing in latter circuit.
That't all.

hari_preetha said:
Instead of manually calculating Power Consumption of LNA circuit,
is there any option or technique to calculate total power consumption of LNA ckt
while simulating using Keysight Technology ADS. Please guide me.
Click to expand...
Simply, save Vdd*Idd in dataset.
Or calcilate Vdd*Idd in DataDisplay Window.
It is too easy thing.
 
Last edited:

Hi,

I agree with the others.

Just to clarify:
To calculate power consumption on a DC supply you need to use
* average (DC) current
Not RMS current, not instantaneous current

Power = V_DC x I_AVG

Klaus
 

KlausST said:
To calculate power consumption on a DC supply you need to use
* average (DC) current
Not RMS current, not instantaneous current
Power = V_DC x I_AVG
Click to expand...
hari_preetha uses DC Analysis.

So RMS current, instantaneous current, I_AVG all have no relation.
 

Hi,

hari_preetha uses DC Analysis.

So RMS current, instantaneous current, I_AVG all have no relation.
Click to expand...
True.
But it seems there is a lack of general power calculation understanding.

Klaus
 

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