555 Pulse Transformer

Thread Starter

p75213

Joined May 24, 2011
70
I am attempting to build a NE555 timer controlled step up transformer (3:250). R2 and R3 will be variable resistors when it is completed. Input is 12V and I'm stepping that up to about 1000V. The image shows the voltage across L1. I'm not sure why spice is showing a declining voltage wave. Is it the "ringing" of L1? In any case I need a 12V pulse. Any help would be appreciated.NE555Pulser.PNG
 

MisterBill2

Joined Jan 23, 2018
3,885
You should check the voltage across the capacitors, also the supply voltage. Note that the claimed voltage across the inductor goes below ground, implying that the circuit being measured is not quite the same as the one drawn. In addition, that IRF9130 is heating up quite a bit, which raises it's resistance a bit.
 

ebp

Joined Feb 8, 2018
2,332
I do wish people would label nodes so we aren't left guessing what plot belongs where.

I'm assuming the transformer is modeled as being air core with perfect coupling between the primary and secondary, so there is no issue with core saturation. Is something specifying the turns ratio of the transformer? It looks to me like it is 1:1, in spite of the different inductances (which is of course impossible in reality if the coupling coefficient is 1, but does the simulation software know that? - it would seem not since the measured voltage is (presumably) equal to the supply rail).

Show us the current in the FET, L1 and D1.

Here's what I think is happening:
FET turns on, storing energy in the primary (L1) inductance
The FET turns off.
The schottky diode recirculates the current in L1, removing only a small amount of energy (forward voltage of diode times current times off time)
FET turns on and adds more to the stored energy. Each subsequent cycle increases the stored energy and current level in L1. The FET transconductance limits how much current can be delivered, so more and more voltage is dropped across the FET instead of appearing across L1.

Any energy stored in the transformer must be removed every cycle.
 
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Thread Starter

p75213

Joined May 24, 2011
70
I do wish people would label nodes so we aren't left guessing what plot belongs where.

I'm assuming the transformer is modeled as being air core with perfect coupling between the primary and secondary, so there is no issue with core saturation. Is something specifying the turns ratio of the transformer? It looks to me like it is 1:1, in spite of the different inductances (which is of course impossible in reality if the coupling coefficient is 1, but does the simulation software know that? - it would seem not since the measured voltage is (presumably) equal to the supply rail).

Show us the current in the FET, L1 and D1.

Here's what I think is happening:
FET turns on, but because of the effective 1:1 ratio of the transformer excess energy is stored in the transformer primary.
The FET turns off.
The schottky diode recirculates the current in L1, removing only a small amount of energy (forward voltage of diode times current times off time)
FET turns on and adds more to the stored energy. Each subsequent cycle increases the stored energy and current level in L1. The FET transconductance limits how much current can be delivered, so more and more voltage is dropped across the FET instead of appearing across L1.

Any energy stored in the transformer must be removed every cycle.
The turns ratio is mentioned in my post - 3:250
 

ebp

Joined Feb 8, 2018
2,332
Yes, I know you mentioned the turns ratio, but if the waveform you showed us is measured across the load resistor, the simulator does not know about it. The first pulse should be stepped up by a factor of about 83, so about 1000 V.

You cannot get DC from the secondary of a transformer.
 

Thread Starter

p75213

Joined May 24, 2011
70
Yes, I know you mentioned the turns ratio, but if the waveform you showed us is measured across the load resistor, the simulator does not know about it. The first pulse should be stepped up by a factor of about 83, so about 1000 V.

You cannot get DC from the secondary of a transformer.
Pulsed dc? The same as the primary.
 

ebp

Joined Feb 8, 2018
2,332
Pulsed DC? Nope! You have to use a rectifier for that.

After thinking about it a bit more, I suspect that the circuit is just running out of power supply headroom to cram more current into the primary.

I'm done here until the OP shows us some current waveforms.
 

Thread Starter

p75213

Joined May 24, 2011
70
Pulsed DC? Nope! You have to use a rectifier for that.

After thinking about it a bit more, I suspect that the circuit is just running out of power supply headroom to cram more current into the primary.

I'm done here until the OP shows us some current waveforms.
How is the secondary AC when the primary is DC? The only thing I can think of is that it is resonating with itself.
 

MisterBill2

Joined Jan 23, 2018
3,885
Yes, I know you mentioned the turns ratio, but if the waveform you showed us is measured across the load resistor, the simulator does not know about it. The first pulse should be stepped up by a factor of about 83, so about 1000 V.

You cannot get DC from the secondary of a transformer.
You know that and I know that BUT does the SPICE simulator know that? Simulation programs only give good answers when they have perfectly accurate models. What size heat sink does your model have on that power FET? Also, try displaying the secondary voltage of that transformer. My point is that while what you entered is probably right, what you did not include is what is needed to get the right answer.
 

ebp

Joined Feb 8, 2018
2,332
OK - the current waveforms confirm most of what I suspected - very substantial energy is being stored in L1 because there is no place for it to go. The circuit quickly reaches the point where no further energy can be added because there isn't enough voltage to drive any higher current through the ON resistance of the FET. I suspect the model does not include the temperature rise in the FET, which as MisterBill says, increases the ON resistance.

So next step: show us the voltage across L1 and across L2. I'm still pretty sure the turns ratio isn't making it into the model. The first pulse, while the FET is ON, should produce about 1000 volts across the load. [edit - this simulation only needs to run for two or three cycles of the 555]

There are fairly simple ways to make this work the way you want it to, but we need to be sure the transformer model is basically correct before we proceed.

A real transformer needs a magnetic core, and that introduces a bunch of new problems, but they too can be solved.

Try a simulation with D1 removed, just to see what happens [edit - try this with everything else just as it is now. In real life it would avalanche the body diode of the FET, but what happens with the simulation depends on the FET model]
 
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Bordodynov

Joined May 20, 2015
2,402
Without a diode, the voltage at the drain of the transistor will reach a breakdown and the transistor will break through. The transistor will certainly not fail instantly, but quickly.
 

dendad

Joined Feb 20, 2016
3,028
How is the secondary AC when the primary is DC? The only thing I can think of is that it is resonating with itself.
Transformer secondaries can only be AC. They cannot be DC, pulsed or otherwise. To get DC of any sort you will need to add a rectifier circuit and even then, only the output is "DC", the secondary still is AC.
That is a function of the transformer. It can only pass power when the magnetic field is changing. A steady field has no power transfer, so you get an output during the magnetic field rising, then the output falls as the field stabilizes. Next, if the primary current is removed, the field collapses so you get a secondary output, but in the opposite direction as the rising, and when the field drops to zero, the secondary output also returns to zero.

You can test this if you have an oscilloscope to connect to the output and put a 1.5V battery on the input.
If no oscilloscope, try it with a red and green LED tied in parallel, back to back, include a series resistor to limit the current. Maybe 2.2Kohms, (depends on the transformer) and you will see one LED flash as you put the battery on, and the other as you take the battery off. This shows you are getting AC as current flows both ways. If it was pulsed DC, only one LED will flash.
 

Bordodynov

Joined May 20, 2015
2,402
The original circuit is bad, both with a diode and without. In both cases, everything will end with the death of the transistor!
Display the power of the transistor (use the ALT key).
 
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