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The short answer is yes.
when you inhibit count you effectively stop the clock signal to the flip-flops. Once there is no clock the outputs of flip-flops remain unchanged and so the output of the chip remains unchanged (page 3 of the attached datasheet).
It is worth having a look at the actual paper describing the experiment.
The setup can be seen in the Fig1 and consists of a microwave source (i.e. magnetron) followed by a directional coupler to measure forward and reflected power a waveguide with a sample in it and the termination.
The...
The correct relation between return loss and reflection coefficient is RL = -20log10|Gamma|
so in your example Gamma=02 -> RL = 14
I always thought VNAs show Snm parameters instead of Return/insertion losses.
Because it is a lumped element design there should be no traces between balun components. The next best thing would be 50Ohms as short as possible.
Since you cannot physically have 50ohm traces make them as short as possible.
One more thing - the DC power line to L1 should have much higher...
The vias should stay. it is not as bad in the bottom layer as I thought. But if you can run reset line in the top layer, then do it.
73mil with 63mil board gives you 65Ohm! (Epsilon r around 4.5).
You need to rearrange the balun and keep the non-50Ohm rf traces length to absolute minimum. It...
I wouldn't worry about the antenna just yet. I see a bit of a problem with your layout in the RF section. First of all all the RF traces should be 50 Ohms. Second of all the placement of the inductors is not ideal - I'd put L2 symmetrically between RFN and RFP, similarly L3 and arrange all caps...
I'd be worried. According to the datasheet the saturation voltage is around 1.2V ~ 300uGy. If you are getting 430 it means there is another Xray ON somewhere... or you are reading the ADC incorrectly..
RL is only negative for active devices
as per the definition RL=10 *log10 (Pin/Pr )
for a passive device Pin > Pr therefore Pin/Pr > 1 -> log10(Pin/Pr) > 0
simples :)
ps
Wikipedia also explains why is return loss often expressed as a negative number.
I would venture a guess that the 1N4148 diode is there to prevent damage to the LED when inserted the wrong way round. 1N4148 has the reverse voltage of 20V and a typical white LED will have the reverse voltage of 5-10V.
The upshot of that arrangement is that whatever the voltage is in an off...
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