Do Oscillations Occur in Inductors When Modifying AM Transmitter Circuits?

[Samson4]

Discharging a capacitor through an inductor creates oscillations. This is about as much as I understand about it. I'm having a hard time explaining my question so I attached a pic.
My questions:

1. In circuit A, are there oscillations in the inductor as the battery charges the first capacitor?

2. In circuit B, does the pulsed dc signal allow for oscillations in the inductor? Does the frequency of the pulses effect the resonant frequency of the lc circuit?

 

[mfb]

In (A), what is the other end of the voltage source connected to? Assuming ground, you will get oscillations if you suddenly connect it.

In circuit B, does the pulsed dc signal allow for oscillations in the inductor?

Yes.

Does the frequency of the pulses effect the resonant frequency of the lc circuit?

The resonance frequency is a property of the LC part only.
The actual frequency of voltage and current you'll see in this circuit can be different from the resonance frequency.

 

[Samson4]

In (A), what is the other end of the voltage source connected to? Assuming ground, you will get oscillations if you suddenly connect it.
Yes.The resonance frequency is a property of the LC part only.
The actual frequency of voltage and current you'll see in this circuit can be different from the resonance frequency.

What about in this circuit? If the pulsed dc doesn't charge the capacitor completely, would this allow for amplitute modulation at the frequency of the lc circuit?

 

[mfb]

If your frequency matches and you can vary the DC pulse length or height, maybe.

 

[Samson4]

If your frequency matches and you can vary the DC pulse length or height, maybe.

Which frequency has to match?

Im thinking the dc pulses would be a dc biased audio input. Does that mean the lc circuit would have to match the audio frequency?

I really appreciate your time. I wish I had someone around here to go to for questions but I'm not in school and I have no knowledgeable friends.

 

[Samson4]

Also, does a capacitor have a similar resistance to change that inductors have?

 

[zoki85]

Also, does a capacitor have a similar resistance to change that inductors have?

Capacitor and Inductor have same magnitudes of impedance at resonant freq

 

[Samson4]

Because the charge is moving through the conductor and onto the capacitor plates at the same rate?

 

[zoki85]

Yes, but the same rate must be the resonant rate

 

[Samson4]

Understood, thank you for clearing that up.

What about that circuit in post 3. Can that possibly produce an amplitude modulated signal?

 

[zoki85]

Understood, thank you for clearing that up.

What about that circuit in post 3. Can that possibly produce an amplitude modulated signal?

Possibly

 

[mfb]

Which frequency has to match?

The pulse frequency and the AC generation.

 

[Samson4]

The pulse frequency and the AC generation.

Wait, now I am confused. I thought you could excite LC circuits at any harmonic of the pulse base frequency. Can the exciting frequency be 1/4 the frequency of the lc circuits resonant frequency?

 

[nasu]

You can excite it at any frequency you want. But you have a resonance condition only at the specific resonance frequency.
You don't have resonance at 1/4 of the resonant frequency. Nor at multiples of it. If you increase frequency, the reactance of the capacitor decreases and that of the inductor increases. They won't be equal again, as they are at resonance.

 

[Samson4]

This is good stuff, thank you guys.

Does every inductor oscillate until the energy is depleted? For example; if I quickly switched on/off a dc supply to an inductor, would it resist the magnetic field, then resist the collapsing magnetic field etc? What if I just turn the dc power on? Will it still try to oppose the collapsing field?

Happy Halloween

 

[mfb]

Wait, now I am confused. I thought you could excite LC circuits at any harmonic of the pulse base frequency. Can the exciting frequency be 1/4 the frequency of the lc circuits resonant frequency?

Sending a DC pulse every N cycles will work as well, or even in some weird pattern but synchronized to the oscillation, but sending it randomly won't work.

 

[Samson4]

Sending a DC pulse every N cycles will work as well, or even in some weird pattern but synchronized to the oscillation, but sending it randomly won't work.

Wow, we must have posted at the exact same time. I think you just answered my question; but, just to clarify. If there is a battery and an inductor; upon closing the circuit, will the inductor initially oscillate because of it's own changing magnetic field?

 

[mfb]

Without capacitor? It will just increase its current until it looks like a short circuit. With a capacitor, you get oscillations.

 

[Samson4]

Then what would keep this from sustaining oscillations in itself?
Capacitor A is asymmetric, with the electrode on the left having twice the capacitance as the electrode on the right. Using the change of capacitance when the system oscillates, wouldn't the oscillations sustain themselves?

 

[mfb]

There is no such thing as an asymmetric capacitance. Capacitance is a property of the capacitor, not of its electrodes.

You can put some active, amplifying element there, but then the drive source is obvious.

 

[Samson4]

But what if we are talking about two separate plates? Can we have two separate plates that we are treating as two separate capacitors?

 

[mfb]

Two plates far away from each other without anything else nearby? They will act like two independent, tiny capacitors against ground. No coupling in between...

 

[Samson4]

Ok, I've been thinking on this for a couple days now. I still don't see the solution in this circuit.

L1 initially is a changing magnetic field, inducing a field into L2. L2 is wired so that the induced current charges C2a negative.
The battery charges C1a positive.

Now this is where I am lost.
The magnetic field at L2 will now collapse, and charge C2a positive.
L3 will also collapse and charge C1b positive and C2b negative.
But; the battery supply is still charging C1a positive, while L1 will be sustained.

Does C1a react like a change in capacitance and discharge? If it does, will it change the polarity of the magnetic field at L1?

or

Does the electric field between C1a and C1b forbid current from being induced by the collapsing magnetic fields at L3 and L2?

 

[mfb]

You can combine the two capacitors to one, it will not change the setup.
For a realistic circuit you'll need a resistor in it, otherwise it will probably oscillate forever.

It is probably easier to analyze the setup with calculations instead of qualitative descriptions.

 

[Samson4]

You can combine the two capacitors to one, it will not change the setup.
For a realistic circuit you'll need a resistor in it, otherwise it will probably oscillate forever.

It is probably easier to analyze the setup with calculations instead of qualitative descriptions.

So this would work and oscillate until the battery lost it's charge? Is there a circuit like this I can read on. That's the main problem, I don't have a resource for this type of setup.

 

[sophiecentaur]

So this would work and oscillate until the battery lost it's charge? Is there a circuit like this I can read on. That's the main problem, I don't have a resource for this type of setup.

That circuit is a linear one so there is no possibility of modulation unless you want to include some strange behaviour inside the battery (not altogether possible but . . . ). As you haven't included any switches then I am not sure how you want this circuit to do AM.
If the battery is a true voltage source then it has no internal resistance and it can dissipate no energy and the solution is indeterminate. If, as you should do, you introduce some resistance in the circuit you can combine the battery resistance, wire resistance etc. as a small series resistance on the output connection of the battery and you can consider it as a normal RLC circuit. (It is always best to reduce complicated networks as much as possible before analysis)
No net charge can leave the battery after the first cycle of the resonance because the Capacitors are open circuits. It will not 'go flat'. There will be an oscillation with a natural frequency of 1/2π√(LC) where C is the equivalent Capacitance to the two Capacitances in series and the L is the equivalent to the network of self and mutual inductances (I assume that L1 and L2 are coupled in some way?). The oscillations will die down exponentially and the final voltage across the Series LC part of the circuit will be equal to the battery voltage. The only energy lost will be the initial energy that flowed into the circuit at switch on (CV²/2) and the number of cycles to drop the energy to 1/e will be the Q of the circuit..

 

[Samson4]

That circuit is a linear one so there is no possibility of modulation unless you want to include some strange behaviour inside the battery (not altogether possible but . . . ). As you haven't included any switches then I am not sure how you want this circuit to do AM.
If the battery is a true voltage source then it has no internal resistance and it can dissipate no energy and the solution is indeterminate. If, as you should do, you introduce some resistance in the circuit you can combine the battery resistance, wire resistance etc. as a small series resistance on the output connection of the battery and you can consider it as a normal RLC circuit. (It is always best to reduce complicated networks as much as possible before analysis)
No net charge can leave the battery after the first cycle of the resonance because the Capacitors are open circuits. It will not 'go flat'. There will be an oscillation with a natural frequency of 1/2π√(LC) where C is the equivalent Capacitance to the two Capacitances in series and the L is the equivalent to the network of self and mutual inductances (I assume that L1 and L2 are coupled in some way?). The oscillations will die down exponentially and the final voltage across the Series LC part of the circuit will be equal to the battery voltage. The only energy lost will be the initial energy that flowed into the circuit at switch on (CV²/2) and the number of cycles to drop the energy to 1/e will be the Q of the circuit..

This circuit isn't for Am it's for producing the frequency that will be AM. With this in mind, how would you keep the circuit oscillating?

L1 and L2 are coupled. Adding an induced load and resistor at L1, would that keep it oscillating?

 

[mfb]

So this would work and oscillate until the battery lost it's charge?

No, as there are no ideal electronic components. Even with superconductors you would still get some damping from the emission of electromagnetic waves, and the system would reach an equilibrium after a while.

how would you keep the circuit oscillating?

You'll need some active elements like transistors. Oscillators are common electronic components.

 

[sophiecentaur]

This circuit isn't for Am it's for producing the frequency that will be AM. With this in mind, how would you keep the circuit oscillating?

L1 and L2 are coupled. Adding an induced load and resistor at L1, would that keep it oscillating?

There are many forms of oscillator, based around resonant circuits. They all involve using feedback from an amplifying device (like the howl round in speaker systems) and a frequency selective circuit (something like your suggested circuit) to make sure than only one frequency gets fed back and re-amplified. Your suggested "switch" would need to be an 'active' component and it would need to be triggered to switch in step with the tuned circuit so that it kept the energy up to the required level as signal power is passed to the rest of the circuit.
Google electronic oscillator circuits - you will see hundreds of different versions.

 

[Samson4]

There are many forms of oscillator, based around resonant circuits. They all involve using feedback from an amplifying device (like the howl round in speaker systems) and a frequency selective circuit (something like your suggested circuit) to make sure than only one frequency gets fed back and re-amplified. Your suggested "switch" would need to be an 'active' component and it would need to be triggered to switch in step with the tuned circuit so that it kept the energy up to the required level as signal power is passed to the rest of the circuit.
Google electronic oscillator circuits - you will see hundreds of different versions.

I altered a simple Crystal AM transmitter circuit to meet my needs. I am basically trying to get am modulation without the use of a diode. Does this meet the criteria? Also, the output from audio sources such as iPods are dc biased correct? If I am wrong in thinking this then this whole circuit is doomed from the get go.