Underdamped Sine Wave Generator With Thyristors

Thread Starter

iraquois1

Joined Feb 27, 2020
16
I am trying to design a circuit to generate underdamped sine wave (actually it is a basic lc tank with switching pulse) to drive 50ohm load. I tried to switch a simple RLC circuit with mosfet switch, but Unfortunately I got so much distortions.).
I have a circuit which people say that it was working once . So I tried to simulate it but it is not working in simulation.
Problems I got:

1) Big problem is that I see oscillation wave even If I don't send pulse from V2. (Breakdown voltage of SCR is 600V)
2) And if I connect 50ohm load to output there is no ringing.

I am not sure if this is the proper way for this purpose. If someone has any idea about this circuit or another circuit idea please share. Thank you for reading.
 

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DickCappels

Joined Aug 21, 2008
6,659
Here is how I did it with a MOSFET. It was for measuring the Q of an inductor.
1601628243447.png
Note the low voltage and tiny MOSFET -this is a small signal circuit.

1601628316275.png

One trick was to only "ping" the tuned circuit with a long interval compared to the ringing frequency so that the MOSFET is off most of the time.

1601628384062.png

An enlarged view of the damped ringing. Distortion? I don't think there is much.

Here is a quickly designed circuit to demonstrate a circuit in which a Thyristor should work:
1601630261212.png

I couldn't locate a thyristor in my LTspice Library so I made something that is very close, though low power using a 2N3906 and 2N3904.

The damped waveform may be observed across C1 and L1.

1601630324125.png
This LTspice demonstration is attached as Thyristor ring 201002A .asc.
 

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Thread Starter

iraquois1

Joined Feb 27, 2020
16
Thanks a lot sir. I will try.
Do you think I can Drive a 50 ohm with this ringing by paraleling the load across the inductor and ground. Or should I change whole calculations after adding 50 ohm resistor.
 

DickCappels

Joined Aug 21, 2008
6,659
I am only beginning to understand what you want to do. If you are calculating the resonant circuit and plan to put a 50 ohm resistor across it then yes, if it were me, I would recalculate the L and C based on having the resistor in place. Remember to take the resistance of the inductor into account.
 

jeffl_2

Joined Sep 17, 2013
30
Without doing a "formal analysis" this is pretty reminiscent of how designers of amateur radio gear (typically, could be any small transmitter) would design the final stage of a transmitter, they would have a "class C" stage (your switch) drive into a "pi network" (just a resonant tank with two series capacitors of different values), the larger of the two capacitors drove the lower impedance which typically was the output which by convention was also usually at 50 ohms, the switch itself coupled its energy across the smaller capacitor at the higher impedance, and the tank circuit then did double duty as an impedance converter. There are probably dozens of other reasonable ways to design this, you could put a coupling tap on the inductor, it could be a separate link winding turning the inductor into a transformer. There are loads of "sample circuits" on the web and even tools to do the calculation for you! I might have misunderstood some feature requirement so this may not apply but this is usually how this sort of thing is done in my experience.
 

Thread Starter

iraquois1

Joined Feb 27, 2020
16
Without doing a "formal analysis" this is pretty reminiscent of how designers of amateur radio gear (typically, could be any small transmitter) would design the final stage of a transmitter, they would have a "class C" stage (your switch) drive into a "pi network" (just a resonant tank with two series capacitors of different values), the larger of the two capacitors drove the lower impedance which typically was the output which by convention was also usually at 50 ohms, the switch itself coupled its energy across the smaller capacitor at the higher impedance, and the tank circuit then did double duty as an impedance converter. There are probably dozens of other reasonable ways to design this, you could put a coupling tap on the inductor, it could be a separate link winding turning the inductor into a transformer. There are loads of "sample circuits" on the web and even tools to do the calculation for you! I might have misunderstood some feature requirement so this may not apply but this is usually how this sort of thing is done in my experience.
Sorry but I actually didn't understand what kind of circuit are you talking about sorry. Can you share me with what key word should I search it to understand what is it? I have sort of success with this circuit lately. The purpose is that : first capacitor gets charged by power supply V3 in the picture and then V3 closes. Right after that V2 triggers the thyristor. So all I have left after last step is "Ron of thyristor, L and C" But I observe some distortions in the signal like sometimes it does not act like pure underdamped. For example, one stage might generate peak larger than peak of previous stage.
 

Papabravo

Joined Feb 24, 2006
14,701
A class C stage is an amplifier that conducts for less than one-half of a cycle. The bias point is chosen so that the active device (tube, FET, BJT) spends most of the cycle in the cutoff state.

https://www.circuitstoday.com/class-c-power-amplifier#:~:text=Class C power amplifier is,80° to 120°.

A "pi" filter has a certain resemblance to the greek letter pi. It is composed of two capacitors and an inductor. It is often used in RF amplifiers.

https://www.digikey.com/en/articles...ically one,inductors and a coupling capacitor.
 

Thread Starter

iraquois1

Joined Feb 27, 2020
16
A class C stage is an amplifier that conducts for less than one-half of a cycle. The bias point is chosen so that the active device (tube, FET, BJT) spends most of the cycle in the cutoff state.

https://www.circuitstoday.com/class-c-power-amplifier#:~:text=Class C power amplifier is,80° to 120°.

A "pi" filter has a certain resemblance to the greek letter pi. It is composed of two capacitors and an inductor. It is often used in RF amplifiers.

https://www.digikey.com/en/articles/pi-t-filters-match-rf-impedances#:~:text=Pi filters are basically one,inductors and a coupling capacitor.
Actually I know both but this part:
the larger of the two capacitors drove the lower impedance which typically was the output which by convention was also usually at 50 ohms, the switch itself coupled its energy across the smaller capacitor at the higher impedance, and the tank circuit then did double duty as an impedance converter.
Which is confusing me :).
The problem with Class C is that: If you turn on the switching componnent there will be a glitch in the circuit (high current due to charge). Because of that I plan to connect load across inductor so I can charge just capacitor first(whithout causing any distortion on the load) and after closing a switching element between L and C.
 

Papabravo

Joined Feb 24, 2006
14,701
Actually I know both but this part:

Which is confusing me :).
The problem with Class C is that: If you turn on the switching componnent there will be a glitch in the circuit (high current due to charge). Because of that I plan to connect load across inductor so I can charge just capacitor first(whithout causing any distortion on the load) and after closing a switching element between L and C.
I think the statement is fairly clear. The two capacitors in the pi filter are not equal. The bigger capacitor has the lower impedance and the smaller capacitor has the higher impedance so you get both an impedance match and a filter at the frequency of interest. You know the output of the pi will be matched to 50 + j0 and the input will be matched to the source impedance of the Class C amplifier output. It will actually be a complex conjugate match.
 
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