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I don't understand your math (possibly you are considering a different circuit from mine). In any case If you have:
Ic costant current drawn
V voltage across the ideal capacitor
Vc voltage across the series C+ESR
and we suppose negligible the current from the power supply (i.e.: considering...
Not sure to have correctly understood your comment, however I'm referring to a one shot discharge of a fully charged capacitor. I didn't read about repetitive pulses.
There is something strange in your simulation. The resistor R1 can just sink some of the current from C1 subtracting it from that going to the MOS. Could you post you LT project ?
In order to calculate the capacitor you shoud write down the equations of the circuit, in any case you could...
I think is not so easy to check the impedances of the bias path especially at the higher frequencies. This is mainly due to the difficulty to accurately calibrate the launchers.
The circuit can oscillate (self-oscillation) with no RF at the input.
You said that in the simulation the circuit...
It seems your circuit is not stable and the transistor starts to oscillate. Possibly your idea to eliminates the need of bypass capacitor is not working. However in order to help you should attach the schematic diagram.
In the video you posted I can see an amplitde modulate sinewave with a superimposed DC (from the oscilloscope measurement). The DMM is set on "DC" so it only measures the DC components of the signal that is constant. You coud set the DMM on "AC" to see the RMS (if it is a true RMS voltmeter) but...
The summer is composed by 10 resistor of values [R, 2R, 3R...2^(N-1)R] connected between MSB, MSB-1, MSB-1...LSB] bit and inverting pin of the opamp. Let's call the feedback resitor Rf.
When only the LSB bit is connected to Vi and the others left open, the output voltage will be...
Here is my analysis. It should be valid as far as the op-amp can be considered (almost) ideal. If the frequency increases and the input capacitance (between inverting and non-inverting pin) starts to play a role this quite simple math loose validity. How much can be high the frequency depends...
Probably there is a LDO (low drop out voltage regulator) insider the antenna, then the lna is always supplied with the same voltage, thus the current consumption doesn't change on the whole input voltage range
No, the (low level) RF signal will not affect the DC measurement. However the insertion of the measurement system, if not properly RF decoupled will affect the RF performances of the antenna.
If you want just measure once the current consumption you can use the multimeter that has to be removed...
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