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I have a circuit (see pic) that includes the primary winding of a 1:20 self wound transformer.

Without some form of additional resistance this circuit will draw too much current from the power supply and damage the FET I’m using so I plan to put a rheostat inline with the primary winding to control the maximum current flowing in it.
 
Given the basic equation for power expended in a resistor of P=I^2R, I’m trying to work out the maximum power rating for the rheostat.
 
If I set the input voltage to 50V and set a 100R rheostat to 50Ohms then my current should be 1.0A. In that case the power rating of the pot will need to be 50W x2 (for de-rating 50% of the track)= 100W i.e. a very beefy rheostat!
 
My query then is this - is it normal practice in circuit design to insert a fixed value resistor, in this case say 100R (as shown), so that the maximum current with the pot at zero is still 0.5A and the max power dissipation 12.5W? On the other hand, if the rheostat is set to zero then will it have no power dissipation capacity at all and be damaged - or are they built differently to potentiometers?
 
Any other thoughts on how I can safely reduce the current flow welcomed.
 
Thank you
 

HV Supply.jpeg

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If I set the input voltage to 50V and set a 100R rheostat to 50Ohms then my current should be 1.0A. In that case the power rating of the pot will need to be 50W x2 (for de-rating 50% of the track)= 100W i.e. a very beefy rheostat!

I believe you are correct. Yes they are beefy - see: https://www.ebay.com/itm/OHMITE-RHEOSTAT-100-OHMS-100-WATT-MODEL-K-STOCK-NO-0451/223494235238?hash=item34094b6466:g:MHYAAOSwJHhcwbwy

You could consider using a 1 ampere fast blow fuse.  Ideally one could put a second power mosfet at the output of the power supply and use that to control the current and control it via a potentiometer.

What is 1 ampere when using a pulse amplifier?  Is it the peak pulse current or  would you have to integrate the sums of the currents?  I would think it relates to the power dissipated by the power transistors in the power supply.

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What are you using for input pulses?

Any idea what the inductance of the primary and secondary windings of the transformer are? Else does anyone have a good guess what they may be? I have never used inductors enough to get any concept of what  transformer inductance like that would be.

I can run the circuit in the simulator if you like.

 

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I'm putting in gated square wave pulse where the HF part is 100-500Hz and the gating frequency 10-20Hz. I've tried to measure the primary and secondary inductances and got a reading of 720mH for the secondary but just a resistance reading of 155 Ohms for the primary. Based on the turns ratio I'm guessing the primary has an inductance of around 5mH?

That would be great if you can quickly simulate the circuit. 😀

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As a first try  the circuit below  has a transformer with the primary of 1.8mh and 0.1 ohm while the secondary is 720mh and 150 ohms. The inductance is related to the number of turns squared ratio: 25^2 to 500^2 =  625/250000 =    0.0025 thus ( 0.0025  *  720mh) = 1.8mh  So your guess is a lot better than mine would be.

circuit.png.359464a9807ecbb964492442e612838c.png

For input the pulses are 2ms on and 2ms off for a period of 4ms. Amplitude is 8 volts.

input.png.8b43f3b9e4261b82a1df1edc9c0a26a3.png

 

The out put would make a good TENs device for an elephant.

output.png.049f1fb5f647a56fef02f8bcf1c166bc.png

 

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Hi Harry,

I have now built the circuit using an additional FET to regulate the current as shown in pic 1 and it does give me fine control.
 
I’ve used a diode in two different orientations, and also without it to, see the effect on the output but in all cases I get a waveform on the transformer secondary (no load) as in pics 2-4 which shows + and - voltage swings instead of just a + as in your simulation, which is what I need. Pic 2 has no diode, Pic 3 has cathode towards the IRF530 Source and Pic 4 has the diode cathode towards the IRF840 Drain (i.e. reversed).
 
What do you think might be causing this and can I ‘rectify’ them with one or more diodes being strategically placed?
 
Thanks
 
Julian

Pic 1.jpeg

Pic 2.JPEG

Pic 3.JPEG

Pic 4.JPEG

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What happens when you connect the load? You may try to connect a small capacitor across the output if you have one that has the  required voltage rating. The last trace maybe as good as you can get. I will try adding distributed capacitance to the windings to see if that produces similar results in the simulator.

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Hi Harry,

I tried using a 0.22uF, 2kV capacitor across the secondary windings and Pic 1 shows the result without it and Pic 2 with the capacitor.

Pic 3 is with no capacitor but instead with a diode across the secondary windings, as shown in the circuit in Pic 5 (but without the bifilar coil that I have yet to add in).

Pic 4 is with the load attached, the load being a simple cylindrical electrolysis cell in distilled water.

It seems that with the load attached, and only about 50mA being drawn from the Variable PSU, the voltage across the secondary windings and cell drops dramatically to around 3 V which is disappointing. I'm hoping the when the bifilar coil is added that I will get some decent voltage spikes to compensate. Is this voltage drop caused purely by whatever current is flowing through the electrolysis cell and is there a simple way to boost it back up?

Julian

Pic 1.JPG

Pic 2.JPG

Pic 3.JPG

Pic 4.JPG

Pic 5.JPG

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Can you measure the resistance of the cell? Possible it got comtaminated. Also knowing the size of the electrodes and their separation  in water one could get a first order  approximation of the cell's capacitance.

 

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Hi Harry,

The cell measure 1.164 M Ohms and a capacitance measurement using the same meter showed 1.058mF. That last reading is surprising since I did a calculation last year based on plate area, separation and the dielectric of water which came to 5.18nF so either the meter reading is unreliable in that context or the distilled water surrounding the cell is contributing in some way.

With the resistance measurement that would suggest that the average current flow through the cell was 50/1.164 E6 = 42uA where 50 is a typical rounded voltage peak I'm getting on the scope. I'm unclear how such a small current (about 30mA from the variable PSU) would result in such a big voltage drop.

Is the way to keep the voltage across the cell up a beefier transformer and variable PSU?

Thanks for your time.

Julian

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In the simulator (with 100 ohms between the PSU and the coil and 1 ohm resistor in the source lead of the ITF840) the output is about 1.2kv peak with or without the 1.164 meg  resistor. With the 1.058nF capacitor it drops to about 750v and with the 5.18nF it drops to about 450v. The current through the secondary coil is +29ma to -42ma peaks.

The current through the primary is 0 to 1.8 amperes peak.  This is with a 100 ohm resistor between the PSU and the coil.

At the 1 ohm resistor in the source lead  of the ITF840 it shows  1 ampere peaks. I found an ITF840 spice model.

If you put a small resistor in the source lead of the IFR840 you could measure the voltage across it and see if the current is what you expect.

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Hi Harry,

I inserted a 1.5 Ohm resistor into the Source line of the IRF840 as suggested (Pic 1) and calculated typical readings of the primary current of 30mA when the secondary was unloaded and about 160mA when the cell was connected (Pic 2), The value under load could have gone higher but not as high as 1A. This was with a square wave input of 200Hz from the pulse circuit and 12V from the PSU as before but with my bifilar coils in place (Pic 3).

Once again the voltage across the cell drops to 3-4V under load and reads about 220V when the cell is not connected. I know to expect a voltage drop when more current is flowing but I can’t explain why it drops so much when the load is attached. Since my aim is to put 1000+V across the electrodes of the cell (with distilled water) how can I achieve this if the moment the load is connected the voltage drops to almost nothing? I don't think it's a weakness in my variable power supply that feeds the transformer and it's the same with or without the bifilar coils that are designed to limit the current in the cell.

Julian

 

 

 

Pic 1.jpg

Pic 2.jpg

Pic 3.jpeg

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I would think you are not hitting it with enough current. Using a resistor instead of the IRF530 as it is faster to use rather then dorking around with two resistors to simulate a pot. I am using a input pulses that 2.5 ms  on and 2.5 ms off at 6.8 volts peak

No bifilar coil nor diode in the output:

  • resistor      current    output capacitor  output resistor   output voltage
  • 100              1A          1.058 nf               1.16 meg            720 volts
  • 100              1A            0                        1.16                  1.05 kv
  •  50               2A           1.058 nf.              1.16                 1.0 kv
  •  50               2A            0                         1.16                  1.2 kv

The current being a  symmetrical pulse(?) would be  only be half of a continuous draw from the power supply. 

 

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Hi Harry,

In your first reply at the start of this thread you suggested using a second MOSFET to control the current instead of a variable resistor, which is why I modified the circuit to include the IRF530. While it works to allow fine control of the current are you now suggesting I put a 50 Ohm resistor instead, based on your simulation? If so then this would surely need to be a 200W resistor (massive?) to be able to cope with 2A through it and I certainly don't have one of those - which is why I presume you suggested the FET in the first place.

J

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