Hi

What is the difference between resonantly pulsing a parallel LC circuit with DC vs. connecting it to an AC tuned to resonance frequency?
I am not able to find a link that would discuss this.

Will it resonant at the same frequency?
What effect will it have on the circuit or its components?

Many Thanks

 

With AC, the capacitor has a low impedance path for discharge on the negative cycle. With pulsed DC, the cap can only discharge through the inductor.

You can build yourself an LC simulator in software - even Excel - and answer this for yourself. The formulas are not that bad. I would recommend including R, since there is no real circuit without it anyway.

Of course you can likely find a simulator online as well.

 

It will resonant at the primary frequency of the pulse edges but the odd harmonics of a square wave pulse will have a good fraction of energy that will be wasted as heat in circuit components. Then you have the DC offset (that varies according to the duty cycle of the pulses) of the signal that carries no frequency domain energy.

 

That means a pure sine wave "not sampled/composed" is needed to avoid DC pulsing. Am I right?

 

isnt that what transistors and tubes do when connected to a resonant circuit, pulse the resonant to keep up the oscilations? if you correctly pulse the resonant lc \circuit, it will keep ringing at its natural resonant frequency. do not pull the dc to ground between pulses. it spoils the "Q" of the tank and kills the ringing.
resonance of paralell lc circuit is when the xC = Xl. the impedance is maximum at resonance.

 

Back in the vacuum tube days the class C amplifier only conducted for about
90° of the resonant frequency, so the tank circuit was feed a pulse to excite the circuit. All the energy is in the circulating current, which is very high.

 

It will resonant at the primary frequency of the pulse edges but the odd harmonics of a square wave pulse will have a good fraction of energy that will be wasted as heat in circuit components. Then you have the DC offset (that varies according to the duty cycle of the pulses) of the signal that carries no frequency domain energy.

If you pulse instead with a very narrow pulse width, you can time it to feed energy into the resonance at the best time (phase position).

That will be highly efficient, compared to a 50:50 squre wave. Provided you are only feeding energy one direction, on the HI duty (ie, not a push pull driver).

 

The RB:

Could you please show this concept in a LTspice file that I can run?
When you say best phase position, would that be when the energy transfer is the fastest? If so, What would the circuit looks like?

Thank you.

 

If you pulse instead with a very narrow pulse width, you can time it to feed energy into the resonance at the best time (phase position).

That will be highly efficient, compared to a 50:50 squre wave. Provided you are only feeding energy one direction, on the HI duty (ie, not a push pull driver).

Sure, but there is a compromise between output power and efficiency. In the case of the class C amplifier the conduction angle (~same as duty cycle) controls the possible output power per pulse at a given voltage and current per pulse. As the angle changes we can see what happens below. To increase the avg power with small angles we need to increase the peak instantaneous power of the pulses and design the circuit to handle possibility much higher voltages and/or currents for short periods of time.

Figure 8.5: (a) RF power and efficiency as a function of the conduction angle; (b) Fourier analysis of the drain current As the conduction angle of the drain current decreases, the harmonic content of the current signal increases. The magnitude of the n th harmonic of the output drain current is given by [3]
...
By examining Figure 8.5b, it is clear that the DC component decreases monotonically as the conduction angle is reduced.

http://flylib.com/books/en/3.253.1.54/1/

 

The RB:

Could you please show this concept in a LTspice file that I can run?
When you say best phase position, would that be when the energy transfer is the fastest? If so, What would the circuit looks like?
...

Sorry I'm not a LT spice user. If LT spice can't make a posiitve pulse then use a narrow duty PWM pulse through a diode. The diode means energy is only fed forward.

Re the timing, the pulse needs to be fed into the resonance at the time when it increases the resonance. Play with the phase angle in LT spice and you will work it out.

Sure, but there is a compromise between output power and efficiency. In the case of the class C amplifier the conduction angle (~same as duty cycle) controls the possible output power per pulse at a given voltage and current per pulse. As the angle changes we can see what happens below. To increase the avg power with small angles we need to increase the peak instantaneous power of the pulses and design the circuit to handle possibility much higher voltages and/or currents for short periods of time.

I agree completely. My point was that since the period (angle) of the pulse is short it is efficient as all it's energy feeds into the correct phase of the resonance to increase the resonant energy.

It wasn't about getting the maximum possible power into the resonance, it was about showing that you can use a digital pulse to transfer the energy efficiently. Just not a square wave, and not push/pull.

 

I agree completely. My point was that since the period (angle) of the pulse is short it is efficient as all it's energy feeds into the correct phase of the resonance to increase the resonant energy.

Yep, the efficiency approaches 100% as Pout approaches 0. It's just like a spinning flywheel, if you kick (phase) the edge for a short time you add little power but if you grab the edge to add power you have to follow (phase and frequency) it around for a bit while giving a tug.