Hi guys I am trying to implement galvanic isolation for my lm555 square wave generator. I am designing a marx generator so this is essential. I am new with transformers, at the moment all I want to get is the same square wave at the secondary side of the transformer but LTSpice doesn't even want to run it. Can anyone give me some advice? Am I missing any components? Cheers. Here is the circuit i am trying. The square wave is at around 8kHz enter image description here

This is the output across the resistor after grounding the secondary side. enter image description here

  • \$\begingroup\$ What error are you getting? What happens if you connect a ground symbol to one of the nodes on the secondary side?(SPICE doesn't much like fully isolated circuits) \$\endgroup\$
    – The Photon
    Mar 28, 2018 at 3:57
  • \$\begingroup\$ @ThePhoton, the problem is that i want to have that side of the transformer isolated, so connecting the same ground there wouldn't be an option for me. I tried it and it does give a square wave it only gives positive and negative peaks \$\endgroup\$
    – David
    Mar 28, 2018 at 4:05
  • 5
    \$\begingroup\$ It can be isolated in real life but not in the simulation. SPICE is simply not capable of modeling true isolation, because there is no single solution to the circuit when there is isolation since the secondary could be offset by an arbitrary DC voltage. \$\endgroup\$
    – The Photon
    Mar 28, 2018 at 4:07
  • \$\begingroup\$ Anyway if you want an answer you will need to tell us what error you got, or how the results were different from what you expected (and why you think they're incorrect). We can't read your mind so you need to share all the relevant information about your problem with us in writing. \$\endgroup\$
    – The Photon
    Mar 28, 2018 at 4:08
  • \$\begingroup\$ Maybe a few femtofarad capacitor that will simulate the capacitive coupling between the two sides of the transformer could help Spice? \$\endgroup\$ Mar 28, 2018 at 5:10

1 Answer 1


I grounded the secondary side and it lets me simulate now. But the waveform at the output is not the square wave i am expecting

No it won't be a square wave with primary inductance at 100 nH. Irrespective of the secondary winding, just think how low an impedance this presents to the top and bottom (flat surfaces) of a square wave.

It looks like your square wave is about 5 kHz and so at 5 kHz, the 100 nH primary has an impedance of \$2\pi\times 5000 \times 0.1\text{E-6}\$ = 0.00314 ohms.

You need to make the primary inductance many times higher. For instance, the primary magnetization inductance for a regular 230 V 50 Hz transformer is in the region of 10 henries. If you accept a degredation of this to 1 henry then divide it by 100 to put it in the 5kHz range you are looking at an inductance of 10 mH (and not 100 nH). You are about 100,000 times too small in my off-the-cuff calculation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.