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The LTspice simulator, I know, doesn't expect an electrically isolated circuit as the one on the secondary side. To simulate a circuit as yours, I had to also add a ground node on the secondary side, directly or via a certain element likely a resistor (perhaps having a rather very high...
This is what I also understood so far. So, I wonder if this topic is simply an academic one that has no practical application (no useful output power).
Or perhaps, its purpose is to maximize the cooling of the input generator by extracting from it the maximum power (when its load is also 50R...
Perhaps I miss something because I couldn't see yet a clear definition of the output 'load' other than it is related to the resonant coil. Does it have a value? Or it is strictly the usual losses of a coil (which is not a useful power, I guess). But there is also the possibility that the power...
The full bridge rectifier has two sides AC and DC. The node 'GND' can exist on one of these two sides only.
For example, on your schematic post (#7), you need to connect the negative node of the bridge (between DIODE1 and DIODE3) with the lower node of the load (R1) after removing the two GNDs...
You may be right, but I can't relate the measuring of the RMS of a varied DC current and the resetting of ADC zero current reference which is usually close to 512 (10-bit ADC).
1682765776
I am afraid that the reading of ADC at zero current is affected, in the long run, by many factors (the...
I also couldn't get you well :(
For instance, the current is the charging one of an acid battery (sinusoidal tips) or the discharging one of a square wave inverter (with a varying duty cycle to regulate the RMS output voltage in case the battery voltage >12V).
Hi,
I attached a schematic of a DC current measurement. Its output is read by an ADC pin of ATmega8. The ADC maximum voltage is 2.6V.
Since the average input current could be positive or negative, the opamp output is adjusted to be close to 1.3V when there is no current.
The current sensor is a...
The right answer is:
Z=-0.33+j*0.055 Rect --> Polar = 0.334551939∠170.5376778
The magnitude should be close to the real number because its imaginary one is much smaller (this can be seen by drawing the 2 vectors).
Also, it is clear that the vector is in the second quadrant and close to its end...
The right answer is:
Z=-0.33+j*0.055 Rect --> Polar = 0.334551939 ∠" tabindex="0" id="isPasted">∠170.5376778
The magnitude should be close to the real number because its imaginary one is much smaller (this can be seen by drawing the 2 vectors).
Also, it is clear that the vector is in the second...
Perhaps you saw and opened it before I deleted the unlisted one.
When I delete the unlisted one (to save space), the TWO blogs, the old one (unlisted) and the new one (listed) were both deleted completely. And this retuned me to zero square.
I hope what happened to me will lead soon to fix (by...
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