I have recently put together this circuit, which works basically as expected.

enter image description here

Due to my lack of certain parts, however, I had to replace the 68 Ohm resistor with a 52 Ohm resistor, and the MOSFET I'm using is an IRF3205.

I am powering the circuit with a 7.4V LiPo battery.

Despite the changes I've made, the circuit drives the transformer fairly well, and the MOSFET doesn't get too hot, especially with a decent heatsink.

However, after attaching the secondary of my transformer to a high voltage bridge rectifier, the output measured by my multimeter was above 600V! With 10 turns on the primary, and 275 turns on the secondary, I expected about 200V.

Output Voltage = Input Voltage * (Secondary Turns / Primary Turns)

Output Voltage = 7.4 * (275 / 10)

Output Voltage = 7.4 * 27.5

Output Voltage = 203.5

I figured a smoothing capacitor might help with the reading on my multimeter, so I added a 150uF 450V capacitor in parallel to the multimeter.

Still, the voltage quickly rose to the cap's maximum rated voltage (according to my multimeter), so I deactivated the circuit.

Is my multimeter wrong?

Or is my math wrong?

What is the cause behind this?

  • \$\begingroup\$ The output is AC. What did you expect a capacitor across the secondary to do for you? \$\endgroup\$
    – JRE
    Commented Jan 26, 2019 at 23:53
  • \$\begingroup\$ It's worth mentioning that your IRF3205 is rated for 55V drain-to-source, and the schematic says it should be rated at least 200V. This is not the cause of your problem, though. \$\endgroup\$
    – Hearth
    Commented Jan 26, 2019 at 23:58
  • \$\begingroup\$ Flyback boosts voltage with I*t so duty cycle affects output voltage. \$\endgroup\$ Commented Jan 26, 2019 at 23:59
  • \$\begingroup\$ electronics.stackexchange.com/search?q=flyback+voltage+ratio \$\endgroup\$ Commented Jan 27, 2019 at 0:01
  • \$\begingroup\$ Where do you think the energy stored in the magnetic field in the transformer is going to go when the MOSFET switches off? The input waveform to the transformer is not a sine wave, so your voltage equation does not hold. \$\endgroup\$ Commented Jan 27, 2019 at 1:43

5 Answers 5


It's a bit of a misnomer to call the flyback converter magnetics "a transformer" because it is being used as an inductor with a load so the inductance is chosen to support the low DCR conduction losses and same for the secondary.

Normally the inductance ratio is adjusted to give around 50% duty cycle for the desired load or less at minimum losses.

Basic math

The energy storage density has a limit to transfer about 100~200W max before the topology becomes less efficient than a forward converting transformer topology where the transformer no longer has to store the energy being transferred as it is coupled by the mutual coupling with a much lower impedance path.


You have no load on the secondary, after rectification, so the peak DC voltage is going to be the maximum negative and positive voltage swing at the transformer secondary. It is now what you estimated, let's call it 'n'. It is \$n^*2\$. You must account for transformer negative swings as well as positive swings. The rectifier happily sums them to get the value you have now.

A proper minimum load would be like 2 100K 1 watt resistors in series. With 2 in series they should not arc until you get close to 1,000 volts. But as a small load they should get the voltage down closer to what you estimated.

Your estimated DC peak voltage will not occur until you have about a 90% load on the DC outputs. In other words you can use 2 50K 1 watt resistors and see the DC peak voltage drop even more. With these resistors use the \$I^2R\$ equation so things do not overheat.

You are using a NE555 which has a noisy output due to shoot-through currents. Be sure its power pins are well bypassed with 100nF and 100uF capacitors. The TLC555 will give you much better performance.

You are lacking a snubber circuit on the transformer primary which absorbs overshoot and ringing. It is built with a resistor and capacitor in parallel in series with a fast rectifier. Many online switching power supply schematics give the values for them, fine tuned to not over-dampen which would make the MOSFET run a bit hotter. You can use values for lower voltage supplies.


Well, you unwittingly put together a flyback converter of sorts. You are storing energy in your transformer inductance, and the only way out is to raise the voltage until something breaks down or parasitic losses dissipate it.

Place another path for that stored energy, e.g., although wasteful a diode across the primary should do it, and it will behave more as the simple transformer you were expecting.


You don't give enough information to provide real help. What output do you expect ...voltage? …..current? ….power?

However, with what you have, you have basically built an ignition circuit.
The change from 60 Ohms to 52 Ohms in series with the gate makes absolutely no difference.

When the MosFet is on you store energy in the transformer core (the L value), and when you turn it off the back EMF on the MosFet side will hit whatever the V(BR)DSS value for your FET is (luckily you are suing a FET with good avalanche capability). By transformer action, the peak output will be 27.5 times greater than this value ...probably in the 1.5kV or above range.

A ten turn primary probably has a resistance of only 2-300 mOhms, so you primary current is trying to be around 40A from your 7.4V battery. You won't be getting there with the device you have I'd bet because your wiring won't support that current.

What is limiting the output voltage(due to the back EMF) is the avalanche characteristics of your FET:

enter image description here

The V(BR)DSS is stated to be above 55V, but they don't say how high it might be. If we assume an avalanche at 55V, the peak output value you'd see would be about 1.5kV, but perhaps higher.

In the forward direction I would suggest you 7.4V is falling due to the resistance of your wiring, so you don't see 200V, but much less.

End result is you'd see a forward conduction voltage of less than 200V and back EMF kick of 1.5kV. Certainly not what you wanted.

I'd suggest:

  1. You need to limit the current on the input by adding a resistor between the supply and the top of the transformer.
  2. You need to limit the excursion of the MosFet Drain by adding a Zener between Drain and ground. The value of this Zener * N give you the voltage output. So for example if you use a 33V Zener you get about 900V out.

You still have a working mosfet? You are lucky! Your circuit is an inductive voltage peak generator. It resembles a little Slayer Exciter except it has separate oscillator.

You connect a coil for a moment to DC and then suddenly turn the switch off. When one tries to cut off the current of an inductor, all coils on the same core develop just as high voltage as is needed to let at least one of the coils output current through something. Then the current and induced voltage peak die soon off because the magnetic energy is dissipated, but the voltage is a moment just as high as is needed to allow current to go. You may have a spark somewhere or your multimeter in the secondary enjoys something substantially dangerous or your mosfet leaks but for some reason it has still survived.

A transformer (it's also 2 coils on the same core) is normally connected all the time to AC source, not to DC, which is in turns connected and disconnected. Transformers can easily get damaged if one connects it to DC and then switches the DC off by removing the current path. A spark can punch the insulation irreversibly if one succeeds to remove the wire fast enough.

Your best bet is to learn how induction works and how it generates voltage when current changes in an inductor. Then you should learn how switch mode power supplies work. You can start from this older case: How does the inductor ''really'' induce voltage?


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