Determine Mazilli Output

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  • #2
Baluncore
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The AC output voltage and current will be determined by the turns ratio of the transformer.
I see 5+5 = 10 turns centre tapped for the transformer primary, but no turns count for the secondary.
Building circuits like that is a challenge for beginners because power semiconductors, switching transients and high output voltages do not mix well with beginners. Take care.
Do you have access to a free simulation program such as LTspice?
That will allow you to optimise the design for your purpose and study the performance.
 
  • #3
Thanks a bunch, I am learning to use the software right now,
 
  • #4
Baluncore
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See attached file. Remove the .txt to make it a .asc file for LTspice.
Modelling transformers can be tricky. Inductance is proportional to the square of the turns count.
Note the imbalance of the two 470R gate bias resistors that is needed to start oscillation of the numerical model.
 

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  • #5
Thanks a bunch, very cool software, seems on "Vout" that it is at 30khz and by 10uh + 10uh, a 20uh to 4000uh ratio is equal to a 200:1 ratio for the inductance, and the square root of 200 is about 14, so a 14:1 ratio for the turns, with the turns at 5+5=10 turns for the primary, the secondary coil should be 10*14=140 turns. I take this as I should get 30khz if I put 140 turns on it, but that I am free to add them for a different frequency and of course a higher voltage output. I'll be tinkering with the inductance and turns ratios to find a combination of highest frequency and highest voltage, limiting it at about 10kv so that I can put it into a full wave voltage multiplier with 20kv capacitors, the high frequency desired for the smoothing of the output.
 
  • #6
Baluncore
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seems on "Vout" that it is at 30khz and by 10uh + 10uh, a 20uh to 4000uh ratio is equal to a 200:1 ratio for the inductance, and the square root of 200 is about 14, so a 14:1 ratio for the turns, with the turns at 5+5=10 turns for the primary, the secondary coil should be 10*14=140 turns.
You lost me there in your first leap of beginners reasoning. But I did warn you with …
Modelling transformers can be tricky. Inductance is proportional to the square of the turns count.
The resonant frequency of the converter comes from the capacitance C1, in parallel with the primary winding inductance L2 & L3 combined. But L2 and L3 are wound on the same core, each with the same turns count, with coupling, Kt = 100%.
So the combination is not 10uH + 10 uH = 20 uH.
If L = 10 uH for n turns, then for 2n turns, L = 40 uH.
It got you there when you forgot to apply the square of 2 = 4, because they are combined into the same inductor with a centre tap.
The converter resonance will therefore be at frequency = 1 / ( 2 * Pi * Sqrt( 40uH * 0u68F ) ) = 30.51657 kHz
Which is very close to what the model gives you.

Next we consider output voltage. The converter operates by alternatively grounding each end of the primary. So for my previous attached diagram, the square wave voltage across the whole primary is therefore twice supply voltage, 2 * 30V = 60V. L1 provides some isolation from the supply so the sine wave ringing has an amplitude of about Sqrt(2) of that. 60V * 1.4142 = 84.85V.

The secondary to primary inductance ratio is 4000 uH / 40 uH = 100.
The turns ratio will therefore be only Sqrt(100) = 10.
The secondary voltage will therefore be 10 * 84.85V = 848.5V. Once it settles, the model gives us close to 93.5V for the primary and 935V for the secondary. I believe the difference is due to the effects of L1 inductance and it's internal Rs on the conversion from square to sinewave.
The kt value of 100% for the transformer is also probably closer to 99%.
 
  • #7
Thank you, that is extremely helpful, so with a turns "ratio" of 10, the number of turns on the secondary coil is 100?
 
  • #8
Baluncore
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so with a turns "ratio" of 10, the number of turns on the secondary coil is 100?
That will be true if there are a total of 10 turns on the primary.
 
  • #9
Can you explain the amplitude of the "sine wave ringing" sqrt(2)? Looking for formulas to that online and can't find any that match the sqrt of 2
 
  • #10
Baluncore
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Sorry, I was asleep at the time of that post. I'm glad you picked it up as I felt uneasy then and am more uneasy now. Simply put, I was converting the RMS power of the DC supply to peak amplitude of the sinewave. It seemed like a good idea at the time.

But looking at it differently, a square wave has a fourier transform y = Sin(x) + 1/3 Sin(3x) + 1/5 Sin(5x) + … So the only energy available at the fundamental frequency is the Sin(x) term. So where does the extra amplitude come from? I now think it must be somehow via L1.

The primary sinewave is seated on the ground reference voltage by virtue of either one of the two conducting MOSFETs. The instantaneous voltage of that sinewave varies sinusoidally with time. Half that instantaneous voltage appears at the centre tap of the transformer and is there tied to the fixed DC supply through inductor L1. Since the arc, transformer and power supply are all solid, the series L1 will drop voltage and so limit current when an arc occurs. But what about when there is no arc?

Don't expect a quick explanation. I will need to think about it today and investigate it on my return.
P.S. Where does the term "Mazilli" in the title come from?
 
  • #11
jim mcnamara
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Vladimir Mazilli? In the attachment to post #1.
 
  • #12
berkeman
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So when supplying this circuit, with a 200 uh inductor,
http://cdn.instructables.com/FIY/EAZN/G9NH8XOG/FIYEAZNG9NH8XOG.LARGE.jpg

with the volts and amps values, how can I calculate the output without having to resort to measuring the distance between starting arcs? And does anybody know the amps capacity of the circuit?
So, what do you want to do with the output? I'll temporarily lock this thread to wait for your PM response...
 
  • #13
berkeman
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Based on my PM conversation with the OP, the thread will remain closed.
 

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