A simple question about LC parallel circuit

Thread Starter

AlwaysNumber1

Joined Dec 4, 2016
52
Hi AN1,
There is a an excellent tutorial here on the AAC site.
https://www.allaboutcircuits.com/te...rrent/chpt-6/parallel-tank-circuit-resonance/

If you look though the Tutorials Index you should find answers to many of your questions.
Eric
View attachment 120456
I do look in those tutorials
But this one does not answer my question :(

I will change it a little bit: when their inductance are equal, MAXIMUM impedance happens. Then why the resonant frequency passes through ?
 

Colin55

Joined Aug 27, 2015
519
I have covered this down to the finest detail on my website in the FM bug articles and projects on Talking Electronics.com.
We commonly call this type of circuit a TANK CIRCUIT because the very first radio pioneers used hundreds of watts to get any sort of transmission distance.
But when they combined a capacitor across a coil in the output circuit, they found the output improved enormously and they could reduce the power. They thought the combination was similar to a tank (of water) that stored energy and released it at the right time. That's how it got the name TANK CIRCUIT.
No-one has ever described it so the beginner can understand. And you have to go into "fairyland" to believe how it works because it is so amazing.
 

crutschow

Joined Mar 14, 2008
25,669
in a parallel LC, how does resonance occur ?
When the frequency is such that the inductive reactance equals the capacitive reactance, then the two reactances completely cancel each other, and the only circuit impedance is any stray resistance in the circuit.
The oscillating current then reaches its maximum value.
This is resonance.

The simulation of a simple LC parallel resonant is shown below.
A 1 mA, 0.1ms pulse is applied to the LC circuit which causes it to oscillate at its resonant frequency.

At the end of the pulse, all the pulse energy is stored in the inductor, with the current at its peak and the capacitor (Sig) voltage is zero.

This inductive current now starts to charge the capacitor, and when all the inductive energy is transferred to the capacitor, the inductor current becomes zero and the capacitor voltage is at its maximum

Now the capacitor voltage starts the inductor current in the opposite direction until all the capacitive energy is stored back in the inductor with the current again at its maximum and thc capacitor voltage at zero.

Thus the stored circuit (tank) energy is transferred back and forth between the inductance and the capacitance twice per cycle, until the energy is dissipated by the stray circuit resistance (here showing the amplitude decay due to a 2Ω inductor resistance).

upload_2017-2-17_0-50-7.png
 

BR-549

Joined Sep 22, 2013
4,938
Resonance is not a mathematical condition. It's a physical force, mechanical, balanced condition.

What makes it so hard to understand is that ANY current amplitude is composed of many different rates of amplitude.

Current is the average of many rates. It's full of rate potentials.

It's the resistance that sets up and causes the multiple rates, because of different distance collisions.

The resistance selects the average amount of current and the reactance selects what rate (frequency) the current is.

Reactance filters, (selects) the rate or frequency of current. This is what tunes the radio station in.
 

Colin55

Joined Aug 27, 2015
519
A 1 mA, 0.1ms pulse is applied to the LC circuit which causes it to oscillate at its resonant frequency.
At the end of the pulse, all the pulse energy is stored in the inductor, with the current at its peak and the capacitor (Sig) voltage is zero.
This inductive current now starts to charge the capacitor, and when all the inductive energy is transferred to the capacitor, the inductor current becomes zero and the capacitor voltage is at its maximum
Now the capacitor voltage starts the inductor current in the opposite direction until all the capacitive energy is stored back in the inductor with the current again at its maximum and the capacitor voltage at zero.

The above is quite untrue and garbled.
 

Colin55

Joined Aug 27, 2015
519
You cannot apply a pulse to a parallel circuit and talk about what is happening because the pulse has to be the exact duration and this timing is determined by the circuit.
That's the first mistake in the discussion above.

Now I will tell you what actually happens.

I have included a circuit containing a parallel cap and inductor and it is called a TUNED CIRCUIT. The value of the inductor creates the timing of the first half of the wave and the value of the capacitor creates the second half of the wave.
To make these two parts equal, the time taken to charge the capacitor has to be equal to the time taken for the magnetic energy in the inductor to be released and passed to the capacitor.
 

Colin55

Joined Aug 27, 2015
519
The circuit starts with the 4k7 turning on the transistor.
At the same time the 4p7 will get fully charged and will have about 3v across it.
The resistance between the collector and emitter leads will reduce as the transistor turns ON and this will cause the voltage across the air timmer to increase. At the same time the voltage across the coil (inductor) will increase and current will flow through the inductor and produce magnetic flux.
This magnetic flux will cut all the turns of the coil and produce a "back voltage" that will oppose the incoming voltage. This will prevent a high current flowing through the inductor and in fact almost no current will flow through it.
At the same time the voltage on the top lead of the 4p7 will decrease and since the 4p7 is charged, the lower lead will also drop in value.
 

Colin55

Joined Aug 27, 2015
519
This will lower the voltage on the emitter of the transistor.
There are two ways to turn a transistor ON. Either increase the voltage on the base and keep the emitter from moving or lower the voltage on the emitter and keep the base fixed.
That's what happens in this case. The 1n capacitor on the base keeps the base fixed at the frequency at which this circuit operates and as the collector voltage drops, the emitter is pushed towards the 0v rail.
At the same time the 4p7 is discharging and while the transistor is turning on more and more, the above conditions apply.
But a point comes when the 4p7 is fully discharged and the transistor is turned on a fair bit but it is not turning on any more. We don't know the voltage across the air trimmer but the inductor will have the same voltage across it and it will be producing the maximum amount of EXPANDING FLUX. But the rate at which this flux expands is getting less and less because the transistor is turning on more and more but at a reduced rate.
Eventually this rate drops to a point where the circuit cannot deliver expanding flux and the flux starts to collapse. The collapsing flux cuts the turns of the inductor and produces a voltage in the OPPOSITE DIRECTION. This voltage discharges the capacitor and it keeps delivering energy to the capacitor to charge it in the opposite direction.
The voltage on the collector of the transistor rises and the top lead of the 4p7 rises and the cap starts to charge. The pulls the bottom lead of the 4p7 up and the emitter rises too. The reduces the voltage between the base and emitter leads and it turns the transistor OFF. The transistor effectively disappears from the circuit and the voltage on the lower lead of the air-trimmer rises and can rise to a voltage higher than the 3v supply.
 

Colin55

Joined Aug 27, 2015
519
The magnetic flux (energy) charges the air trimmer and 4p7 and eventually the 4p7 is charged and it no longer keeps pulling the emitter lead "up."
The voltage on the emitter lead reduces and the voltage between the base and emitter increases and this starts to turn ON the transistor to create the next cycle.

See . . . the actual operation of the circuit is nothing like anyone has described before.
Creating flowery technical language doesn't explain anything.
The timing of the circuit NOT just the two components, but includes the timing of the pulse delivered by the transistor at the beginning of each cycle.
 
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BR-549

Joined Sep 22, 2013
4,938
Let's put a coil across a 120VAC source. We will wind the coil until we get 1 amp going thru it. That means the coil has about 120 ohms of reactance.
So we have 120V source with coil and ammeter showing 1 amp. What happens when I put a capacitor with 120 ohms reactance across the coil?

a. current increases.

b. current decreases.

c. current remains the same.

Try it. Any frequency will work. Use a signal generator. Not many people have 60 Hz chokes.

Just make sure the reactants are equal at frequency.
 

crutschow

Joined Mar 14, 2008
25,669
At the end of the pulse, all the pulse energy is stored in the inductor,


This is incorrect.
No it's not.
Look at the simulation graphs.
At the end of the pulse the inductor current is at a maximum and the capacitor voltage is zero thus, at that instant, all the energy is stored in the inductor current.
My description exactly describes what happens in the simulated circuit.

It's your description that seems rather garbled. :rolleyes:
 
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BR-549

Joined Sep 22, 2013
4,938
In any resonant circuit.......The voltage and current of the cap and coil is always equal and inverted.

In a series circuit, the voltage may be higher than the source.......in a parallel circuit........the current may be higher than the source.
 

Colin55

Joined Aug 27, 2015
519
"the inductor current is at a maximum and the capacitor voltage is zero"

How can the current in the inductor be a maximum when the voltage across the inductor is zero??????

Do you understand what you are saying ?????
 
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