My colpitts oscillator I made on LTSpice oscillated at 5 MHz at a peak-to-peak voltage of 260V. When I put all the components down on my breadboard and measured it with a real oscilloscope, the frequency was only 2MHz with less than a peak-to-peak of 400mV.

How can I fix both the voltage output and frequency? To be clear, my goal is to make an oscillator at 5MHz with a peak voltage of 100V to 500V.

The oscillator in LTSpice

The LTSPice simulation - 5MHz - 260 Vpp

  • 1
    \$\begingroup\$ You should set your maximum time step to be smaller than what it currently is. \$\endgroup\$
    – Andy aka
    Apr 11 at 15:16
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    \$\begingroup\$ Don't forget that your scope probe will be a significant load on the output. Put the probe in your simulation. Also put in the actual series resistance and shunt capacitance of the output inductor. \$\endgroup\$ Apr 11 at 15:52
  • \$\begingroup\$ What is the impedance of your probe? Specifically, R + jX at the test frequency. You also need loss elements of L2 and C3, and probably L1. \$\endgroup\$ Apr 11 at 16:51
  • \$\begingroup\$ You can obtain +/- 50 Vpeak with a load 10 Meg + 7 pF. \$\endgroup\$
    – Antonio51
    Apr 11 at 18:04
  • \$\begingroup\$ @Antonio51 Thank you for your response. I added 7pF + 10 Meg in series with L2 and got a positive peak of 16V and a negative peak of -1V. I must misunderstand you, where precisely is the load placed? \$\endgroup\$ Apr 12 at 19:29

2 Answers 2


A breadboard will have capacitance between the connection rows. The jumper wires will have inductance. Real components will have parasitic resistance, inductance and capacitance that a simulation may or may not model well. These will tend to bring the frequency of an oscillator down.

To get the frequency up where you want it you'll have to adjust the inductance and capacitances in your circuit to account for the stray R/L/C.

As for the voltage, do you really think you're going to get 260 V output from a 9 V source? What you're seeing there is most likely because you have no load on the output of your simulation. Try adding a resistor across L2, something like the input impedance of your scope, maybe 1 Meg., see if you still get 260 V.

  • \$\begingroup\$ Yes, I believe that was the real question. I was confused as to why a 9V would even output 260V in the first place. I added a resistor across L2, got around 12V - makes more sense that way. \$\endgroup\$ Apr 11 at 14:51
  • \$\begingroup\$ What is the peak-to-peak voltage one can expect from a real implementation? \$\endgroup\$
    – PMF
    Apr 11 at 18:57
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    \$\begingroup\$ @george_2roz - you have a high-Q resonator after the oscillator (C3 and L2) that can magnify the voltage at the output of Q1 to very large values if the losses are low enough. Real-world losses will significantly reduce that. \$\endgroup\$ Apr 12 at 1:56

I changed some components and I retain ultimately this behavior.

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