[SOLVED] Parallel Resonant Circuit with 50 Ohm source Math

[chiques]

I went through this tutorial which seems fairly straight forward: Parallel Resonant Circuit Math

My question is, how does this change when I switch over to a source that has a 50 ohm impedance?

Zin = 50 Ohms
Zout = 36.4 Ohms

 

[D.A.(Tony)Stewart]

What would you expect when you add resistance to a 0 ohm source greater than the reactive load absolute value and almost but less than 2x the load?

Less than 6 dB loss and almost flat response or a bandpass with < 6dB loss near resonance and tapering down at -12 dB/octave on either side?

 

[KerimF]

I went through this tutorial which seems fairly straight forward: Parallel Resonant Circuit Math

Just to be sure, is the voltage source, on the circuit at the top side, a real one and has, therefore, an internal resistance/impedance close to zero ohm?
If this is the case, the parallel RLC circuit is shorted by the voltage source (in AC analysis).

For instance, an RLC tank is usually driven by a current source like the collector of a bjt transistor.

But if a voltage source (V) is used and has a relatively high resistance as 50 ohm (Rin), it is better to transform it to a Norton configuration first which is a current source (V/Rin) in parallel with the resistance (Rin). By doing this, it becomes obvious that the load of the LC tank is, 60//50 (about 27.3 ohm) in your case.

 

[FvM]

I understand the question as simple exercise problem rather than practical tank circuit question.

You want to calculate circuit currents with non-zero source impedance. You can analyze the circuit as voltage divider, but the LC load has a complex impedance. It's not sufficient to know a Z number, the phase angle respectively separate R and X values must calculated.

 

[BigBoss]

Use complex notation or phasor diagram.

 

[chiques]

Use complex notation or phasor diagram.

Something like this? I’m not sure about the way I calculated the voltage coming from the signal generators 50 Ohm impedance though...

Are the current magnitudes the Irms currents or are they peak currents?

 

[D.A.(Tony)Stewart]

Normally the units of measurements should not change, but were oddly not given. Normally we say 60 Line is 120V RMS so 12Vac 60 Hz could also be RMS and thus all the currents in RMS. But then if we use trig. for sine voltages we usually express it as Peak*sin() so I suspect they assumed 12V peak.


But then I doubt this was done and the results look wrong. I calculated and attenuated output at 60 Hz at -6.6dB @ 31 deg

if the current thru 60 ohms is 0.094A @ 30.651 deg then the voltage is 5.64 and 5.64/12 = attenuation ratio= 0.47 or -6.55 dB which agrees with my calc.

When choosing transformer voltages we assume RMS, but then use ratios with same units in and out.

 

[FvM]

Are the current magnitudes the Irms currents or are they peak currents?

The same unit type as the source voltage, whatever it is. What do you prefer?

Can confirm calculated source current, didn't continue further. Not sure if it's of much practical interest to calculate current and voltage phase relative to source voltage, but it can be done of course.

 

[BigBoss]

When you use complex/phasor convention, you find the right value.
So your math is correct.

 

[chiques]

When you use complex/phasor convention, you find the right value.
So your math is correct.
View attachment 182427
View attachment 182428

This is reassuring.
I'm curious as to why the phases are so off though??

 

[BigBoss]

This is reassuring.
I'm curious as to why the phases are so off though??

No, it was a weirdness of AWR. Voltage Source has a particular state. There is 90 degree lagging anyway..
Keysight ADS gives the right results. There is single error in your MathCad space on The Current of the Capacitor. If you correct that error, you'll find the same results.

 

[chiques]

No, it was a weirdness of AWR. Voltage Source has a particular state. There is 90 degree lagging anyway..
Keysight ADS gives the right results. There is single error in your MathCad space on The Current of the Capacitor. If you correct that error, you'll find the same results.
View attachment 182460

I also had some typos on my imaginary signs for the resistor and inductor so all those match your simulation results.

I reviewed the current across the capacitor section and I can't seem to find anything wrong. Anything that stands out to you other than the phase is different? Maybe ADS calculated 180-59.04=120.96????

 

[BigBoss]

Z=-0.33+j*0.055 Rect-->Polar = 0.0641405 \[ \angle \]120.96376
according to my hand calculator.

 

[chiques]

Z=-0.33+j*0.055 Rect-->Polar = 0.0641405 \[ \angle \]120.96376
according to my hand calculator.

Here is my hand calculator version in radians and degrees. Still scratching my head

 

[KerimF]

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 quadrant and close to its end (close to 180 deg).

 

[chiques]

View attachment 182465
Here is my hand calculator version in radians and degrees. Still scratching my head

I used the polar solver in my TI and got the same results you did.

I found the problem. I totally forgot there are conditions when converting to polar form (it's been a while....)

I get what you get now

 

[KerimF]

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 (close to 180 deg).

--- Updated Apr 25, 2023 ---

And
Z=-0.033+j*0.055 Rect --> Polar = 0.064140471∠120.9637565

--- Updated Apr 25, 2023 ---

Z=-0.33+j*0.055 Rect-->Polar = 0.0641405 \[ \angle \]120.96376
according to my hand calculator.

I was confused because of the mistyping in the real number which, I noticed now, is -0.033. Sorry

 

[chiques]

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 (close to 180 deg).

--- Updated Apr 25, 2023 ---

And
Z=-0.033+j*0.055 Rect --> Polar = 0.064140471∠120.9637565

--- Updated Apr 25, 2023 ---

I was confused because of the mistyping in the real number which, I noticed now, is -0.033. Sorry

 

[KerimF]

BigBoss said:
Z=-0.33+j*0.055 Rect-->Polar = 0.0641405 \[ \angle \]120.96376
according to my hand calculator.

I was confused because of the mistyping in the real number which, I noticed now, is -0.033. Sorry

 

[BigBoss]