I am trying to design a circuit that gives me an underdamped sine wave.

I am controlling the SCR from an external transformer by microcontroller.

When I simulate this circuit, V3 has no effect on the circuit. I mean the capacitor is not getting charged to lets say 50V to oscillate. Instead, it seems it is oscillating with V2 and the transformer. I don't know if it is the proper way to make an oscillator circuit.

  1. How can I improve this design and solve this problem?
  2. Since I need to switch to have underdamped oscillating several times, I need switching. Can you advise me another topology or circuit example to achieve this if there is a better way to do this?

(Sine wave frequency and switching frequency is not important for now.)


simulate this circuit – Schematic created using CircuitLab

enter image description here

  • V(3) : Output
  • V(4) : Pulse
  • 2
    \$\begingroup\$ Do you need this circuit in practice, or just for simulation? I ask because I am puzzled by the arrangement of your circuit, but also because I see you went through the trouble of making a schematic in the built-in schematic editor, yet you show a screenshot of MicroCap. \$\endgroup\$ Commented Oct 1, 2020 at 20:13
  • \$\begingroup\$ Yes. It was because some parts in microcap was not clearly visible. Some expression as well. It took 2 min. Wasn't problem for me. People are doing me favor so I need to simplify something. \$\endgroup\$
    – Ismail
    Commented Oct 1, 2020 at 21:01
  • 1
    \$\begingroup\$ You say V1 has no effect but I don't see V1 in your schematic. \$\endgroup\$
    – Andy aka
    Commented Oct 3, 2020 at 10:15
  • \$\begingroup\$ I edited sorry. It was V3. I think microcap has same bugs. Because even if I don't send pulse to SCR, it conducts. \$\endgroup\$
    – Ismail
    Commented Oct 3, 2020 at 19:43

1 Answer 1


You should have nothing lossy connected in series and in parallel with the resonant circuit parts during the oscillation. The switch should stay ON during the oscillation. Your thyristor turns OFF in the first half cycle when the current drops under the hold limit.

The charging source for the capacitor should be floating during the expected oscillation. The series switch unfortunately must be there. It's losses kill the oscillation very soon except if there's enough charged energy available.

To make a working circuit I increased the inductance radically from your 50uH to keep the max current reasonably small for low cost parts. The capacitance is also increased for more charged energy.

enter image description here

The voltage drop is hundreds of millivolts in semiconductor switches. You can see how fast the oscillation decays due that drop. Reducing inductance increases the current and the charged energy can be used before low cost switches turn their state fully. Simulation is your way to check the situation if you avoid calculations.

About the circuit: Q1 charges C1 to 12V during the 0V-state of pulse V1.

At t=100ms V1 jumps to +5V. Q2 stops conducting. 7,5 volt zener D2 is as level translator between +5V and +12V.

Q2 starts to conduct and L1C1 oscillates. The recharging of C1 happens when V1 is again 0V.

The inductor is ideal in this simulation. The resistance and possible iron core losses of a practical inductor would make the decay of the oscillation substantially faster.

You wrote you are going to put a 50 Ohm load. That's not impossible, actually I guess you have already done some calculations which show that 50uH & 6uF resonator can in theory oscillate substantially a long time with that load and 50V intial charging voltage:

enter image description here

NOTE: Battery V1 is the marker for the initial 50V without using state variable edits, do not expect you can use that idea in a real circuit because the current should be possible to both directions.

But insert some voltage loss in the switch. The current is so high that normal diode voltage drop dissipates the energy in few cycles:

enter image description here

To sustain the oscillation longer you must either have a low drop switch or work with much lower current. Current reduction keeping the voltage needs higher inductance. That affects the oscillation frequency.

Another way to sustain the oscillation is to leave the semiconductor switch out of the oscillation current path. Charge the initial energy as inductor current. Turn the switch OFF to start the oscillation. You do not need especially high voltage because the 50uH coil can be made of thick wire.

But your DC source must be able to output say 18A. You connect the parallel resonant circuit with a transistor to a DC source. The transistor must be current limited to 18A. After the inductor current has risen to 18A you turn the transistor suddenly OFF and the oscillation starts.

The current must be limited because otherwise the capacitor in parallel with the inductor would short the DC source. It cannot have a switch due the losses.

I must admit I do not know enough of your whole case. An amplified circuit would output those bursts with much less hassle.

  • \$\begingroup\$ Thanks. I will analyse your answer tomorrow. Slightly off topic but you said that I shouldn't connect series or parallel lossy component. But I need to drive 50 ohm resistor with this oscillation. I think values of inductor and capacitor can be arranged to do that. \$\endgroup\$
    – Ismail
    Commented Oct 1, 2020 at 21:15

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