Electric Fields and Resonance

  • Thread starter Samson4
  • Start date
  • #26
sophiecentaur
Science Advisor
Gold Member
24,829
4,627
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..
 
  • #27
243
15
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?
 
  • #28
34,482
10,606
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.
 
  • #29
sophiecentaur
Science Advisor
Gold Member
24,829
4,627
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.
 
  • #30
243
15
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. Im 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 im wrong in thinking this then this whole circuit is doomed from the get go.
question 5.png
 
  • #31
sophiecentaur
Science Advisor
Gold Member
24,829
4,627
I altered a simple Crystal AM transmitter circuit to meet my needs. Im 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 im wrong in thinking this then this whole circuit is doomed from the get go.
View attachment 75171
I altered a simple Crystal AM transmitter circuit to meet my needs. Im 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 im wrong in thinking this then this whole circuit is doomed from the get go.
View attachment 75171
I'm afraid that circuit will not work. The ground of the audio input signal is not shown but we assume it is the same ground as the rest of the circuit.
AM stands for Amplitude Modulation. Modulation is a non-linear process and needs a nonlinear element in any circuit. You are connecting the audio input to the battery input - which is low impedance - so the supply volts to the 1MHz unit won't alter and neither will the level of the carrier signal produced. If your 1MHz source produces a carrier with a level proportional to the supply volts (it may or may not) then you need to be altering the supply volts with the audio signal. For low power, you can change the DC supply to the 1MHz oscillator, using a transistor that's driven by the audio and produces a supply voltage that varies in about a mean DC value.
You really should Google Amplitude Modulation Circuits and see the variety of ways in which it is achieved.

PS The best your circuit can do is to Add the two signals together - which is not Modulation
PPS To protect their circuits, most pieces of audio equipment have AC coupled outputs.
 
Last edited:

Related Threads on Electric Fields and Resonance

  • Last Post
Replies
1
Views
1K
Replies
9
Views
898
Replies
1
Views
462
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
1
Views
567
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
5
Views
8K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
12
Views
2K
Top