Does resonance in a MOT actually boost power output?
ScottH, Sun Mar 12 2017, 08:10PMI read somewhere that resonance only makes the relationship between voltage and current change (in or out of sync), but not really boost power, which somehow makes the arcs longer. I also notice that my MOTS get hot real fast and draw 1.5 to 2x the current on primary without resonant caps. I don't know if all that extra current draw is just the power factor or if there is another reason why.
The resonant caps make the output arcs much brighter and longer, so the voltage must be much higher?
Does resonance increase the current output also?
How can current and voltage be out of phase if voltage pushes a current?
Re: Does resonance in a MOT actually boost power output?
Martin Slinning, Sun Mar 12 2017, 09:16PM
Hi ScottH
You are woundering about the magic of electricity i think..
Faraday found out that a current want to stay in its state, and oppose any change.. This is natures "conservatisme", any change is unwanted.
Martin Slinning, Sun Mar 12 2017, 09:16PM
Hi ScottH
You are woundering about the magic of electricity i think..
Faraday found out that a current want to stay in its state, and oppose any change.. This is natures "conservatisme", any change is unwanted.
Re: Does resonance in a MOT actually boost power output?
Inducktion, Sun Mar 12 2017, 09:31PM
I don't know the entire story, but my understanding is that it cancels out the reactance (or, AC resistance) of components in your circuit.
A microwave oven primary has quite a bit of inductance; A random measurement I found online showed 63 mH of inductance.
At 60 hz that calculates out to be about 23.75 ohms of reactance, on top of the resistance of the wire, and the resistance + reactance of the secondary.
However if you run the transformer in resonance that reactance drops to almost zero, and then you're just fighting the resistance of the wire.
That's how DRSSTC's can get such incredible arcs; both the primary AND secondary are driven in resonance with eachother, so the maximum power outputs you can get are limited only by the resistance of the wires, and your power supply.
Please correct me if I'm wrong anyone by the way. I'm PRETTY sure this is how it works, but... I have been wrong before.
Inducktion, Sun Mar 12 2017, 09:31PM
I don't know the entire story, but my understanding is that it cancels out the reactance (or, AC resistance) of components in your circuit.
A microwave oven primary has quite a bit of inductance; A random measurement I found online showed 63 mH of inductance.
At 60 hz that calculates out to be about 23.75 ohms of reactance, on top of the resistance of the wire, and the resistance + reactance of the secondary.
However if you run the transformer in resonance that reactance drops to almost zero, and then you're just fighting the resistance of the wire.
That's how DRSSTC's can get such incredible arcs; both the primary AND secondary are driven in resonance with eachother, so the maximum power outputs you can get are limited only by the resistance of the wires, and your power supply.
Please correct me if I'm wrong anyone by the way. I'm PRETTY sure this is how it works, but... I have been wrong before.
Re: Does resonance in a MOT actually boost power output?
Dr. Slack, Mon Mar 13 2017, 07:05AM
Yes, resonance can cancel out series impedances, resulting in higher power output. Of course this comes from higher power input, there's nothing magical going on.
In the case of a MOT, the shunts create a leakage inductance which limits short circuit current. Putting a capacitor in series with either primary or secondary can cancel some or all of this, resulting in higher short-circuit current flowing.
Voltage and current with be in phase at a resistor. At a capacitor, current will lead voltage by 90 degrees. At an inductor, voltage leads current by 90 degrees.
You might combine this last two to see that if a current flows through both an inductor *and* a capacitor in series, the voltage across each will be in anti-phase, so reducing the impedance of the pair. When the individual voltages have the same magnitude, the total voltage will be zero, the pair are resonant, and the remaining series impedance is due to only the resistive losses in the components.
If you combine an inductor and capacitor in parallel, so they see the same voltage, then their currents are in anti-phase, and the terminal current of the combination will be less than the terminal current of either of them. At resonance, no input current is required to support any circulating current between the two, until losses get in the way.
Dr. Slack, Mon Mar 13 2017, 07:05AM
Yes, resonance can cancel out series impedances, resulting in higher power output. Of course this comes from higher power input, there's nothing magical going on.
In the case of a MOT, the shunts create a leakage inductance which limits short circuit current. Putting a capacitor in series with either primary or secondary can cancel some or all of this, resulting in higher short-circuit current flowing.
Voltage and current with be in phase at a resistor. At a capacitor, current will lead voltage by 90 degrees. At an inductor, voltage leads current by 90 degrees.
You might combine this last two to see that if a current flows through both an inductor *and* a capacitor in series, the voltage across each will be in anti-phase, so reducing the impedance of the pair. When the individual voltages have the same magnitude, the total voltage will be zero, the pair are resonant, and the remaining series impedance is due to only the resistive losses in the components.
If you combine an inductor and capacitor in parallel, so they see the same voltage, then their currents are in anti-phase, and the terminal current of the combination will be less than the terminal current of either of them. At resonance, no input current is required to support any circulating current between the two, until losses get in the way.
Re: Does resonance in a MOT actually boost power output?
ScottH, Mon Mar 13 2017, 09:45AM
Does resonance work with the MOTS wired in parallel as well? By terminal current, do you mean the current drawn by the primary? Can you explain how current and voltage can be out of phase?
I thought a voltage drives the current, making power. When the current is out of phase with voltage, does VxA=W if they aren't working together at the same time?
If possible, do you know roughly how many Henries of inductance the secondary of a large MOT has?
ScottH, Mon Mar 13 2017, 09:45AM
Dr. Slack wrote ...
Yes, resonance can cancel out series impedances, resulting in higher power output. Of course this comes from higher power input, there's nothing magical going on.
In the case of a MOT, the shunts create a leakage inductance which limits short circuit current. Putting a capacitor in series with either primary or secondary can cancel some or all of this, resulting in higher short-circuit current flowing.
Voltage and current with be in phase at a resistor. At a capacitor, current will lead voltage by 90 degrees. At an inductor, voltage leads current by 90 degrees.
You might combine this last two to see that if a current flows through both an inductor *and* a capacitor in series, the voltage across each will be in anti-phase, so reducing the impedance of the pair. When the individual voltages have the same magnitude, the total voltage will be zero, the pair are resonant, and the remaining series impedance is due to only the resistive losses in the components.
If you combine an inductor and capacitor in parallel, so they see the same voltage, then their currents are in anti-phase, and the terminal current of the combination will be less than the terminal current of either of them. At resonance, no input current is required to support any circulating current between the two, until losses get in the way.
Yes, resonance can cancel out series impedances, resulting in higher power output. Of course this comes from higher power input, there's nothing magical going on.
In the case of a MOT, the shunts create a leakage inductance which limits short circuit current. Putting a capacitor in series with either primary or secondary can cancel some or all of this, resulting in higher short-circuit current flowing.
Voltage and current with be in phase at a resistor. At a capacitor, current will lead voltage by 90 degrees. At an inductor, voltage leads current by 90 degrees.
You might combine this last two to see that if a current flows through both an inductor *and* a capacitor in series, the voltage across each will be in anti-phase, so reducing the impedance of the pair. When the individual voltages have the same magnitude, the total voltage will be zero, the pair are resonant, and the remaining series impedance is due to only the resistive losses in the components.
If you combine an inductor and capacitor in parallel, so they see the same voltage, then their currents are in anti-phase, and the terminal current of the combination will be less than the terminal current of either of them. At resonance, no input current is required to support any circulating current between the two, until losses get in the way.
Does resonance work with the MOTS wired in parallel as well? By terminal current, do you mean the current drawn by the primary? Can you explain how current and voltage can be out of phase?
I thought a voltage drives the current, making power. When the current is out of phase with voltage, does VxA=W if they aren't working together at the same time?
If possible, do you know roughly how many Henries of inductance the secondary of a large MOT has?
Re: Does resonance in a MOT actually boost power output?
Dr. Slack, Mon Mar 13 2017, 03:11PM
The other important inductance in a MOT is the leakage inductance. That's the inductance due to the shunts, that appears in series with the MOT. This is the inductance that's resonated out with a series capacitor. I don't have an estimate for this inductance.
Dr. Slack, Mon Mar 13 2017, 03:11PM
Does resonance work with the MOTS wired in parallel as well?Yes
By terminal current, do you mean the current drawn by the primary?I mean the current drawn at the terminal of whichever component I'm talking about, so the inductor, the capacitor, or the combination of the two
Can you explain how current and voltage can be out of phase?Current and voltage are independent. It's only when you are driving a resistor that they're in phase. If you're driving a current into a capacitor, then the voltage is the time integral of the current/C. If you're putting a voltage across an inductor, then the current is the time integral of the voltage/L.
I thought a voltage drives the current, making power.For a resistor
When the current is out of phase with voltage, does VxA=W if they aren't working together at the same time?No. That's where you get power factor coming into the equation. For a capacitor or inductor across a supply, the voltage and current are in quadrature, no power is dissipated, the power factor is zero, you have a finite VA, but zero W.
If possible, do you know roughly how many Henries of inductance the secondary of a large MOT has?Never measured one. Here's a rough estimate though for one I have. Core area 70x35mm, approx magnetic length 260mm, assume iron has ur of 2000, guess at 2200 turns. H field for 1 amp one turn = current/length = 1/0.26, B field = u*Hfield = 4pie-7*2000/0.26, flux = area*Bfield = 0.07*0.035*4pie-7*2000/0.26, inductance = n^2.flux = 114H. Of course, should the iron saturate, the ur and hence the inductance will plummet.
The other important inductance in a MOT is the leakage inductance. That's the inductance due to the shunts, that appears in series with the MOT. This is the inductance that's resonated out with a series capacitor. I don't have an estimate for this inductance.
Re: Does resonance in a MOT actually boost power output?
ScottH, Mon Mar 13 2017, 05:48PM
So why is the primary current draw much less when the secondary is in resonance, if resonance makes it act like the shunts are not there? Also, why don't my transformers get hot when resonating and putting out much more power? When I run the Mots with no caps, they overheat fast and blow the fuse.
Other YouTube videos measuring draw current with vs without caps show that the primary current draws more from the wall with resonant caps, but the opposite is true in my case as my setup probably pulls 20a at 120v (caps) vs 32a (no caps). Can you explain how that works? Thanks.
ScottH, Mon Mar 13 2017, 05:48PM
Dr. Slack wrote ...
The other important inductance in a MOT is the leakage inductance. That's the inductance due to the shunts, that appears in series with the MOT. This is the inductance that's resonated out with a series capacitor. I don't have an estimate for this inductance.
Does resonance work with the MOTS wired in parallel as well?Yes
By terminal current, do you mean the current drawn by the primary?I mean the current drawn at the terminal of whichever component I'm talking about, so the inductor, the capacitor, or the combination of the two
Can you explain how current and voltage can be out of phase?Current and voltage are independent. It's only when you are driving a resistor that they're in phase. If you're driving a current into a capacitor, then the voltage is the time integral of the current/C. If you're putting a voltage across an inductor, then the current is the time integral of the voltage/L.
I thought a voltage drives the current, making power.For a resistor
When the current is out of phase with voltage, does VxA=W if they aren't working together at the same time?No. That's where you get power factor coming into the equation. For a capacitor or inductor across a supply, the voltage and current are in quadrature, no power is dissipated, the power factor is zero, you have a finite VA, but zero W.
If possible, do you know roughly how many Henries of inductance the secondary of a large MOT has?Never measured one. Here's a rough estimate though for one I have. Core area 70x35mm, approx magnetic length 260mm, assume iron has ur of 2000, guess at 2200 turns. H field for 1 amp one turn = current/length = 1/0.26, B field = u*Hfield = 4pie-7*2000/0.26, flux = area*Bfield = 0.07*0.035*4pie-7*2000/0.26, inductance = n^2.flux = 114H. Of course, should the iron saturate, the ur and hence the inductance will plummet.
The other important inductance in a MOT is the leakage inductance. That's the inductance due to the shunts, that appears in series with the MOT. This is the inductance that's resonated out with a series capacitor. I don't have an estimate for this inductance.
So why is the primary current draw much less when the secondary is in resonance, if resonance makes it act like the shunts are not there? Also, why don't my transformers get hot when resonating and putting out much more power? When I run the Mots with no caps, they overheat fast and blow the fuse.
Other YouTube videos measuring draw current with vs without caps show that the primary current draws more from the wall with resonant caps, but the opposite is true in my case as my setup probably pulls 20a at 120v (caps) vs 32a (no caps). Can you explain how that works? Thanks.
Re: Does resonance in a MOT actually boost power output?
Dr. Slack, Mon Mar 13 2017, 06:49PM
Post schematic diagrams of how you have caps connected to the MOTs. As I said, you can resonate different inductances, it's not clear what you are doing.
Dr. Slack, Mon Mar 13 2017, 06:49PM
Post schematic diagrams of how you have caps connected to the MOTs. As I said, you can resonate different inductances, it's not clear what you are doing.
Re: Does resonance in a MOT actually boost power output?
ScottH, Mon Mar 13 2017, 09:57PM
I don't have one, but I'll explain. I have 2 Mots with primaries wired in parallel. I have 2 sets of 2 Mocs each wired in parallel connected to each HV output. Since it is wired in phase, I wired the HV output wires from the 2 Moc sets to one wire. There is the Neutral HV terminal, which is wired to the Earthed body of the Mot. Both Mot transformer bodies (cores) are sitting on a foil base so they make contact with each other.
I experience the same things when they are wired in anti-phase/ anti- parallel (Moc HV outputs are not wired together in this configuration, they arc to each other).
The caps are .75uf each. Overall capacitance of the 2X2 cap setup totals .75uf
ScottH, Mon Mar 13 2017, 09:57PM
Dr. Slack wrote ...
Post schematic diagrams of how you have caps connected to the MOTs. As I said, you can resonate different inductances, it's not clear what you are doing.
Post schematic diagrams of how you have caps connected to the MOTs. As I said, you can resonate different inductances, it's not clear what you are doing.
I don't have one, but I'll explain. I have 2 Mots with primaries wired in parallel. I have 2 sets of 2 Mocs each wired in parallel connected to each HV output. Since it is wired in phase, I wired the HV output wires from the 2 Moc sets to one wire. There is the Neutral HV terminal, which is wired to the Earthed body of the Mot. Both Mot transformer bodies (cores) are sitting on a foil base so they make contact with each other.
I experience the same things when they are wired in anti-phase/ anti- parallel (Moc HV outputs are not wired together in this configuration, they arc to each other).
The caps are .75uf each. Overall capacitance of the 2X2 cap setup totals .75uf
Re: Does resonance in a MOT actually boost power output?
Dr. Slack, Tue Mar 14 2017, 06:28AM
You've left out some connections. You draw the output arc (a) in series with the capacitor, to ground or (b) the capacitor is to ground, and you draw the arc across the capacitor?
you can use ascii art to draw a schematic if you *really* can't be bothered to install LTSPICE and do a screen cap
but SPICE is kinda handy to have on your computer anyway
Dr. Slack, Tue Mar 14 2017, 06:28AM
You've left out some connections. You draw the output arc (a) in series with the capacitor, to ground or (b) the capacitor is to ground, and you draw the arc across the capacitor?
you can use ascii art to draw a schematic if you *really* can't be bothered to install LTSPICE and do a screen cap
___________ | | | MOT == z __|_____|_____|____
but SPICE is kinda handy to have on your computer anyway
Re: Does resonance in a MOT actually boost power output?
Hazmatt_(The Underdog), Tue Mar 14 2017, 04:33PM
Check out Power Factor on Wikipedia.
As mentioned before, voltage and current are out of phase without the capacitor in parallel with the MOT secondary.
Theoretically you could measure that phase angle between the current and voltage on your scope, but it's dangerous.
The Power Factor is the cosine of the phase angle between voltage and current, PF = cos(phase).
Then Power in Watts = VA (for DC) and P = VA cos(phase) (for AC)
When we have a purely resistive circuit, power factor becomes 1 because there is no reactive component (no phase component)
For your transformer without the capacitor, voltage and current are out of phase by some amount (probably a power factor of .86, or cos(30 degrees) )
So the power in Watts delivered to your load becomes P = VA cos (phase)
But you are wondering WHY is there more power drawn without the cap then with?
That's because there is your Real Power (cosine) and your Reactive power VAR (Volt Amps Reactive) (sine).
So real power = VA cos(phase), and Reactive power = VA sine(phase), and the actual power being delivered is the vector sum.
The reactive power is power consumed but not being transferred to the load.
Power = VA cos(30) = VA * .8660
Reactive = VA sin(30) = VA * .5 < SEE! this is not insignificant loss!
With the capacitor correcting the phase angle to 0 degrees, you have the current and voltage now in phase with each other
Real power W = VA cos (0) = VA, and Reactive power VAR = VA sin(0) = 0
Hazmatt_(The Underdog), Tue Mar 14 2017, 04:33PM
Check out Power Factor on Wikipedia.
As mentioned before, voltage and current are out of phase without the capacitor in parallel with the MOT secondary.
Theoretically you could measure that phase angle between the current and voltage on your scope, but it's dangerous.
The Power Factor is the cosine of the phase angle between voltage and current, PF = cos(phase).
Then Power in Watts = VA (for DC) and P = VA cos(phase) (for AC)
When we have a purely resistive circuit, power factor becomes 1 because there is no reactive component (no phase component)
For your transformer without the capacitor, voltage and current are out of phase by some amount (probably a power factor of .86, or cos(30 degrees) )
So the power in Watts delivered to your load becomes P = VA cos (phase)
But you are wondering WHY is there more power drawn without the cap then with?
That's because there is your Real Power (cosine) and your Reactive power VAR (Volt Amps Reactive) (sine).
So real power = VA cos(phase), and Reactive power = VA sine(phase), and the actual power being delivered is the vector sum.
The reactive power is power consumed but not being transferred to the load.
Power = VA cos(30) = VA * .8660
Reactive = VA sin(30) = VA * .5 < SEE! this is not insignificant loss!
With the capacitor correcting the phase angle to 0 degrees, you have the current and voltage now in phase with each other
Real power W = VA cos (0) = VA, and Reactive power VAR = VA sin(0) = 0
Re: Does resonance in a MOT actually boost power output?
ScottH, Wed Mar 15 2017, 01:37PM
So if I just include a good PF correction cap parallel to the primarys input wires, that would give me the results of the resonant caps on the secondary, since I'm just making the current and voltage in phase?
Here's a link
Its wired just like this Schematic, except I have 2 Mots (each wired the same way) and no VARIAC.
The 2 HV outputs after the caps are wired together into a single hot, then arced to ground for the parallel setup. For the Anti-parallel setup, the 2 HV outputs after the caps are arced together with primaries fed 180 deg out of phase.
Is it bad that I have a 80uf PF cap in parallel with the wires going into the primary, since the Mocs correct PF? Would the PF cap make the PF better, or worse in this scenario?
ScottH, Wed Mar 15 2017, 01:37PM
Hazmatt_(The Underdog) wrote ...
Check out Power Factor on Wikipedia.
As mentioned before, voltage and current are out of phase without the capacitor in parallel with the MOT secondary.
Theoretically you could measure that phase angle between the current and voltage on your scope, but it's dangerous.
The Power Factor is the cosine of the phase angle between voltage and current, PF = cos(phase).
Then Power in Watts = VA (for DC) and P = VA cos(phase) (for AC)
When we have a purely resistive circuit, power factor becomes 1 because there is no reactive component (no phase component)
For your transformer without the capacitor, voltage and current are out of phase by some amount (probably a power factor of .86, or cos(30 degrees) )
So the power in Watts delivered to your load becomes P = VA cos (phase)
But you are wondering WHY is there more power drawn without the cap then with?
That's because there is your Real Power (cosine) and your Reactive power VAR (Volt Amps Reactive) (sine).
So real power = VA cos(phase), and Reactive power = VA sine(phase), and the actual power being delivered is the vector sum.
The reactive power is power consumed but not being transferred to the load.
Power = VA cos(30) = VA * .8660
Reactive = VA sin(30) = VA * .5 < SEE! this is not insignificant loss!
With the capacitor correcting the phase angle to 0 degrees, you have the current and voltage now in phase with each other
Real power W = VA cos (0) = VA, and Reactive power VAR = VA sin(0) = 0
Check out Power Factor on Wikipedia.
As mentioned before, voltage and current are out of phase without the capacitor in parallel with the MOT secondary.
Theoretically you could measure that phase angle between the current and voltage on your scope, but it's dangerous.
The Power Factor is the cosine of the phase angle between voltage and current, PF = cos(phase).
Then Power in Watts = VA (for DC) and P = VA cos(phase) (for AC)
When we have a purely resistive circuit, power factor becomes 1 because there is no reactive component (no phase component)
For your transformer without the capacitor, voltage and current are out of phase by some amount (probably a power factor of .86, or cos(30 degrees) )
So the power in Watts delivered to your load becomes P = VA cos (phase)
But you are wondering WHY is there more power drawn without the cap then with?
That's because there is your Real Power (cosine) and your Reactive power VAR (Volt Amps Reactive) (sine).
So real power = VA cos(phase), and Reactive power = VA sine(phase), and the actual power being delivered is the vector sum.
The reactive power is power consumed but not being transferred to the load.
Power = VA cos(30) = VA * .8660
Reactive = VA sin(30) = VA * .5 < SEE! this is not insignificant loss!
With the capacitor correcting the phase angle to 0 degrees, you have the current and voltage now in phase with each other
Real power W = VA cos (0) = VA, and Reactive power VAR = VA sin(0) = 0
So if I just include a good PF correction cap parallel to the primarys input wires, that would give me the results of the resonant caps on the secondary, since I'm just making the current and voltage in phase?
Dr. Slack wrote ...
You've left out some connections. You draw the output arc (a) in series with the capacitor, to ground or (b) the capacitor is to ground, and you draw the arc across the capacitor?
You've left out some connections. You draw the output arc (a) in series with the capacitor, to ground or (b) the capacitor is to ground, and you draw the arc across the capacitor?
Here's a link
Its wired just like this Schematic, except I have 2 Mots (each wired the same way) and no VARIAC. The 2 HV outputs after the caps are wired together into a single hot, then arced to ground for the parallel setup. For the Anti-parallel setup, the 2 HV outputs after the caps are arced together with primaries fed 180 deg out of phase.
Is it bad that I have a 80uf PF cap in parallel with the wires going into the primary, since the Mocs correct PF? Would the PF cap make the PF better, or worse in this scenario?
Re: Does resonance in a MOT actually boost power output?
Dr. Slack, Wed Mar 15 2017, 05:41PM
In that case, you're resonating out the leakage inductance, not resonating with the secondary inductance. I don't have an estimate for the leakage inductance, so don't know how near those caps are to resonance.
You could make some measurements from which you could estimate the the leakage inductance. The safest way to do this would be to drive the HT side from the mains, putting your test capacitors in series with the HT, and measuring the AC current output by simply shorting the LV output across with an AC ammeter. You can use both caps in series, or parallel, or one of them, to get three distinct values, and no cap will give you a 4th.
Dr. Slack, Wed Mar 15 2017, 05:41PM
In that case, you're resonating out the leakage inductance, not resonating with the secondary inductance. I don't have an estimate for the leakage inductance, so don't know how near those caps are to resonance.
You could make some measurements from which you could estimate the the leakage inductance. The safest way to do this would be to drive the HT side from the mains, putting your test capacitors in series with the HT, and measuring the AC current output by simply shorting the LV output across with an AC ammeter. You can use both caps in series, or parallel, or one of them, to get three distinct values, and no cap will give you a 4th.
Re: Does resonance in a MOT actually boost power output?
ScottH, Wed Mar 15 2017, 11:22PM
I could set the resonance at different frequencies (for the secondary) and still get max power output, right? I was tinkering with some resonant online calcs, and the cap values are based on the resonant frequencies you set. Should I set the resonance of the secondary to 60hz if that is the primary input frequency and secondary frequency before caps?
And you're saying my PF cap only affects the leakage inductance part, and the resonant caps are to resonate my secondary?
ScottH, Wed Mar 15 2017, 11:22PM
Dr. Slack wrote ...
In that case, you're resonating out the leakage inductance, not resonating with the secondary inductance. I don't have an estimate for the leakage inductance, so don't know how near those caps are to resonance.
You could make some measurements from which you could estimate the the leakage inductance. The safest way to do this would be to drive the HT side from the mains, putting your test capacitors in series with the HT, and measuring the AC current output by simply shorting the LV output across with an AC ammeter. You can use both caps in series, or parallel, or one of them, to get three distinct values, and no cap will give you a 4th.
In that case, you're resonating out the leakage inductance, not resonating with the secondary inductance. I don't have an estimate for the leakage inductance, so don't know how near those caps are to resonance.
You could make some measurements from which you could estimate the the leakage inductance. The safest way to do this would be to drive the HT side from the mains, putting your test capacitors in series with the HT, and measuring the AC current output by simply shorting the LV output across with an AC ammeter. You can use both caps in series, or parallel, or one of them, to get three distinct values, and no cap will give you a 4th.
I could set the resonance at different frequencies (for the secondary) and still get max power output, right? I was tinkering with some resonant online calcs, and the cap values are based on the resonant frequencies you set. Should I set the resonance of the secondary to 60hz if that is the primary input frequency and secondary frequency before caps?
And you're saying my PF cap only affects the leakage inductance part, and the resonant caps are to resonate my secondary?
Re: Does resonance in a MOT actually boost power output?
Dr. Slack, Thu Mar 16 2017, 07:10AM
No, the PF cap resonates the primary inductance, the output caps resonate the leakage inductance.
'Resonance' only has a signaifiant effect when the resonant freuqency is equal tot he driving frequency. If the driving frequency is off resonance, then the effect is reduced.
So a small PF cap will reduce the magnetising current that the mains has to supply, as it becomes larger to resonance, all of the mag current is supplied by the cap, and it it gets too large, the mains supply has to supply excess capacitive current. This affects only the supply side current, and doesn't affect the output at all.
The leakage inductance limits the output current (to some extent, the winding resistance does as well). Using a large series output C reduces the effective leakage inductance, increasing the output current. As the output C becomes smaller, to resonance, the outpu current is maximum and determined only the by the winding resistance. As it becomes yet smaller, the outout current is limited by the capacitor.
The output winding inductance is not resonanted, but as it's big and in parallel with the output voltage, does not significantly affect the output.
Dr. Slack, Thu Mar 16 2017, 07:10AM
No, the PF cap resonates the primary inductance, the output caps resonate the leakage inductance.
'Resonance' only has a signaifiant effect when the resonant freuqency is equal tot he driving frequency. If the driving frequency is off resonance, then the effect is reduced.
So a small PF cap will reduce the magnetising current that the mains has to supply, as it becomes larger to resonance, all of the mag current is supplied by the cap, and it it gets too large, the mains supply has to supply excess capacitive current. This affects only the supply side current, and doesn't affect the output at all.
The leakage inductance limits the output current (to some extent, the winding resistance does as well). Using a large series output C reduces the effective leakage inductance, increasing the output current. As the output C becomes smaller, to resonance, the outpu current is maximum and determined only the by the winding resistance. As it becomes yet smaller, the outout current is limited by the capacitor.
The output winding inductance is not resonanted, but as it's big and in parallel with the output voltage, does not significantly affect the output.
Re: Does resonance in a MOT actually boost power output?
Kolas, Thu May 11 2017, 10:16PM
I have experimented with this using a much more robust and electrically insulated device. I found the output to be greatly increased. This did; however, come at the cost of much higher stresses on the secondary of the transformer ultimately lead to it's demise.
it also caused the pole pig to draw significantly more power when loaded.
Kolas, Thu May 11 2017, 10:16PM
I have experimented with this using a much more robust and electrically insulated device. I found the output to be greatly increased. This did; however, come at the cost of much higher stresses on the secondary of the transformer ultimately lead to it's demise.
it also caused the pole pig to draw significantly more power when loaded.
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