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# Rheostat power rating

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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

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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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Thanks Harry,

That Rheostat is far too big to incorporate and your idea of a MOSFET (or BJT?) to control the current is far more elegant and practical. I have revised my circuit in the attached so would that work ok?

Jules

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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.

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

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

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Thanks for doing that. The sharp pulses are fine for what I want which is to feed an electrolyser. You used an IRF530 and, as I have quite a few, I hope the IRF840 will work just as well.

Thank you

J

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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

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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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Harry, I will set it up with the load in the next day or so and also add a 1uF, 2kV capacitor across the secondary coil and report back. I'd be interested to see what the simulations shows.

Thanks

Julian

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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

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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?

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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Thanks. I will see what current is flowing in the primary and report back.

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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

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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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I was using the fixed resistors as the simulator dos not have a concept of a potentiometer. Putting the 530 back in an using a combination of two resistors  between 30k - 17k to 46k -  1k (where the lower value is to ground) I get good results with 40k - 7k  and 30k - 17k but the peak current is 2 amperes for  some reason.  At 44k - 3k the current  is 900ma but the output is down to 50v. I will continue looking at it.

If you have a symmetrical  pulse waveform say of 1 ampere I gather the rms value of the current is 1 * 0.707   and the rms voltage would be 100 * 0.707 so the equivalent power would be: 0.707 * 70.7 = 49.98 watts  not the 100v * 1a = 100 watts of a continuous drain. So one may be able to run it above 1 ampere.

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Hi Harry,

I'm not clear where your two variable resistors are and in the attached Pic 1 I have put where I think one is. My supply is set at 50V and the gate pulses at 200Hz, 50% so 'hoping' to be able to get 750-1kV out at the secondary when connected to the electrolysis cell.

As I come from a Physics background and not electronics I'm trying to understand why increasing the current in the primary will increase the voltage across the cell. Is the problem that the 530 is throttling both the voltage and the current and hence the multiplication by the secondary?

From my limited experience of Spice I think one can have a pot in the design together with a simple statement that increments the value of a resistor R. I attach a grab of an early attempt (Pic 2) at a regulator using this approach that someone else advised me on together with the relevant files (not sure which is essential so I've provided three). Perhaps you can incorporate some bits of it into your simulation?

I appreciate your efforts, thank you,

Regards,

Julian

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Don't laugh that is an emulation of your circuit. I was thinking that there is not enough load in the circuit
with only the coil (the mosfets are acting like switches only off or on) so I started out with 7 watt incandescent
lamps but could not get two mosfet to work so I switched to only one; that works well.
In the photo at the lower right is an automobile ignition coil and the big black thing is an 80 volt battery (borrowed from my chainsaw).

Working up to 13 bulbs (normally 7 watt bulbs) I got 0.571 ma through a 7 ohm resistor. Switching to a 100 watt
bulb got 0.594 ma, the IRF530 got up to 32 degrees C.  I do not have any IRF580's. The input signal is 6 volts peak. The 50k resistor is a pot. max'ed out.

I turn the input up from low to high to prevent any inrush of current into the lamp(s), I do not know if that is necessary or not.

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Harry, that's very industrious of you to build a physical version of the circuit.

Have you have substituted my upper 530 for a light bulb because it is just acting as a variable resistor?

Also could you clarify if you got a high voltage on the secondary of the ignition coil? If you did, even with the small currents you measured in the primary, what does this all mean for my current design 'HV Supply 1B' shown a few posts back?

Julian

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Using the 100 watt bulb and two different ignition coils this is what I got:

AC Output(VOM) 1 meg load        Input sig. 204 Hz   Peak voltage  across 7ohms      coil type
67.4 v                                             6.0v                          1.84  v                                    std auto ignt. coil
159.8 v                                           9.68 v                        5.44 v

17.4  v                                            8.0                            4.0 - 4.2 v                             Capacitor discharge coil

The std auto coil is a 12 volt and a few amperes coil. While for the CDI coil I am running it via discharging a 1 mfd capacitor charged to  350 volts in my experimental chain saw ignition system. So neither are designed for what  you are doing. Speaking of coils/transformers Scherz and Monk have information on transformers in their "Practical Electronics for Inventors" you may check it out the next time you are in a book store.

I think using a incandescent lamp (as a ballast resistor) and controlling the current via the input signal is useful but  using a mosfet as  current regulator  maybe useful also. This regulator is from the IRF530 pdf sheet. I use a similar circuit in my ignition system floating a 9 volt battery at 350 volts.

The 12 volts at the battery got cut off.

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Harry , I need to unpick what you have found. The highest output across the secondary of your coil was 159.8V, when I presume one would expect several kV? (I don't know what the turns ratio is in such a coil but ignition systems are designed to produce 5-15kV)

Am I correct in saying that to get a higher secondary output one will need a bigger primary current - to store more energy in the primary coil? Also are you saying that the current in the primary is a function of the Gate frequency - perhaps a lower frequency will allow more current into the primary?

What's DUT?

Regards,

Julian

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We know the output voltage is related to the ratio of the number of turns. And the output current related to the inverse of the number of turns. The more voltage gives one more current at the primary so in an ideal transformer the power is the same on both sides of the transformer. But in a pulse transformer is it not the rate of change of current in the primary that gives rise to the change  in magnetic flux that produces the voltage?  I need to learn more about pulse transformers.

I had in mind trying out different frequencies on the automobile  coil/transformer  but had a problem when I connect the second probe to the current monitoring resistor while the first probe  was floating on the output of the coil. The  100 watt lamp when full brightness and it when down  hill from there. I will  continue later.

Anyway I had put a voltage divider across the output of the auto. coil (a 1094k and a 9.84k) and got + and - 32.8  volts peaks on the scope at the 9.84k. I calculate the output voltage at +3679 volts  and -3679 volts peaks.

DUT  device under test?

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