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Create 50-60Hz positive pulses from mains, using only transformers?

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neazoi

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Hi, this is a kind of weird question. I was wondering if I can take only the positive pulses out of the AC mains (after a step down transformer) but using only transformers, capacitors, coils and resistors. No semiconductor or tube.
A way is the synchronous rectifier using either relays or motors, but I was wondering if this can be done my some king of saturated reactor or anything else comes to your mind so that mechanical movement can be avoided?

Are there historically such attempts?
 

KlausST

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Hi,

For a short try you could use a low capacitance electrolytic capacitor, but the capacitor will soon refuse to work, because AC is not healthy.

I can't imagine a really good solution.

Klaus
 

FvM

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The passive components you mention have a symmetrical characteristic and don't show any rectifying effect. A principle way to get some kind of rectifying effect would be a saturated reactor with DC premagnetization, using a permanent magnet.
 
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neazoi

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The passive components you mention have a symmetrical characteristic and don't show any rectifying effect. A principle way to get some kind of rectifying effect would be a saturated reactor with DC premagnetization, using a permanent magnet.
Right, so that an asymetric ac is produced. The positive cycle to be half AC and the negative more like a squared ac.

I am thinking of using this circuit to supply pulses to an RF LC, so that damped wave oscillation is produced. I have successfully used a relay for producing these pulses to the LC, but I am thinking how the noisy and unreliable relay could be replaced.

Do I really need half cycle of the AC in the case of the transformer? I mean, what if I just supply to the LC the whole cycle (positive and negative)? It will be like positive and negative pulses.

I have no idea if a sharp pulse is required for the LC to produce damped wave or not though? Maybe the 50-60Hz AC could not do the trick and a square wave is needed instead?
 

BradtheRad

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This sounds as though it's similar to the magnetic amplifier, a method of feeding small DC current through an auxiliary winding, to control power through the main winding.

I have yet to experiment with the concept, but you may find a way to saturate the flux to one polarity or the other. The result could be partial or total rectification of AC.
 

neazoi

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This sounds as though it's similar to the magnetic amplifier, a method of feeding small DC current through an auxiliary winding, to control power through the main winding.

I have yet to experiment with the concept, but you may find a way to saturate the flux to one polarity or the other. The result could be partial or total rectification of AC.
The saturable reactor can do this to an extent. What I am trying to do is different.
Feed pulses to an LC to produce damped RF.

My question is how sharp should the pulse be and also if I can feed different polarity pulses (from both cycles of the AC) so that no rectification would be actually needed.
 

Anna Conda

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Hi, this is a kind of weird question. I was wondering if I can take only the positive pulses out of the AC mains (after a step down transformer) but using only transformers, capacitors, coils and resistors. No semiconductor or tube.
A way is the synchronous rectifier using either relays or motors, but I was wondering if this can be done my some king of saturated reactor or anything else comes to your mind so that mechanical movement can be avoided?

Are there historically such attempts?
The passive components you mention have a symmetrical characteristic and don't show any rectifying effect. A principle way to get some kind of rectifying effect would be a saturated reactor with DC premagnetization, using a permanent magnet.
This thread is tending towards the "magnetic rectifier" which mirage was chased by a good many engineers from the 30's to the 50's... in short there is no way to produce pulses with a net average DC without a diode or similar (even diodes etc used to precharge magnetic elements) magnets are no help either...

Mag amps are very useful for producing very sharp edged pulses...

If you draw net DC from a transformer with a single output diode for example this will be reflected in the mains input current, there is a limit of 5mA average you may draw from the mains supply for reasons of earth stake corrosion. Also if you draw enough DC from the mains you will cause asymmetry on the mains waveform and this can lead to local transformers drawing excess current (flux stair casing) blowing their fuses and breakers....
 

The Electrician

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If you draw net DC from a transformer with a single output diode for example this will be reflected in the mains input current, there is a limit of 5mA average you may draw from the mains supply for reasons of earth stake corrosion. Also if you draw enough DC from the mains you will cause asymmetry on the mains waveform and this can lead to local transformers drawing excess current (flux stair casing) blowing their fuses and breakers....
Reflected in what way? Will there be DC in the mains input of a transformer with a single output diode?
 

Orson Cart

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Will there be DC in the mains input of a transformer with a single output diode?
Correct, every time, ever plugged in a hair dryer on half power (achieved with a diode in series with the element) and noticed a local transformer buzzing when you do this? standards allow for short term use items like this but 24 hr average DC into the neutral must be < 5mA.
 

KlausST

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Hi,

While you can't transmit DC voltage in both directions with a transformer,
But a pulsed current (maybe caused by a half wave rectifier at transfirmer secondary) can be transmitted.

The secondary current will not change the magnetic flux in the iron core.( only in a very small value caused by primary sray inductance and DC resistance).

Klaus
 

FvM

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The apparently simple problem reveals as rather complex at second sight. Besides windings resistance converting part of the asymmetrical distorted AC current into DC, you also have the nonlinear core characteristic.

The secondary current will not change the magnetic flux in the iron core.
The secondary DC current will cause a strong DC flux component which in turn enforces an asymmetrical magnetizing current.
 

The Electrician

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Correct, every time, ever plugged in a hair dryer on half power (achieved with a diode in series with the element) and noticed a local transformer buzzing when you do this?
Of course, but this is not what we were talking about; the hair dryer is directly connected to the distribution transformer, not through another small transformer.

Give me a feel for the magnitude of the effect. Suppose I have a suitably large 1 to 1 isolation transformer and I connect the secondary to a 50Ω resistor in series with a diode. I would expect a current of about 2.4 amps (in the U.S., with 120 VAC at the outlets in the home) without a diode, so about 1.2 amps (RMS, AC+DC) with the diode.

What would be the typical approximate magnitude of the DC current in the primary?
 

FvM

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What would be the typical approximate magnitude of the DC current in the primary?
I guess less than 1 mA if the transformer is suffciently far from saturation.
 

The Electrician

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I guess less than 1 mA if the transformer is suffciently far from saturation.
My question was directed to Orson Cart.

What do you say, Orson Cart?

Also, what can you tell me, Anna Conda?
 
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Orson Cart

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It does depend on the design of the Tx and how far towards Bsat it goes normally. Also the type of Tx, e.g. toroidal is much worse than EI. For a 3kVA isolation Tx, with Bpk = 1.8T say, with normal transformer steel material, drawing 1.2 amps rms via a diode, could easily lead to 10 amps peaks in the primary as the resistance of the wires is low. Try it your self...!

Note for supply transformers, they are specified to withstand 2 x half cycles of the same polarity in a row without fuse blowing saturation levels to keep the supply network intact during faulty tap change operation and other events, thus they operate at about 1.0T peak under normal operation.

Most commercial transformers operate at higher Bpk and hence you often get that characteristic Whuummm! at turn on where the flux in the Tx geats very near saturation and the primary current is very high for a few cycles, half second or so, at turn on - if the Tx happens to be turned on near the zero volt crossing, giving a whole half cycle of volt-seconds in one direction...
 

KlausST

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Hi,

You may get saturation at power_on caused by remanence and/or the time where the input voltage starts.
And you may get saturation when input "voltage" is unsymmetric = has DC components.

But this is not the case when secondary current is unsymmetric = DC current components.

Imagine a flyback transformer has a catch diode at its secondary. Here the current is only in one direction. It can't flow in reverse.

Also you may use a diode at a standard transformer secondary output. Whithout (big) core flux change.

The core flux is proportional to the integral of voltage over time. Not current.

But in detail an unsymmetric current on the secondary side will cause an unsymmetric current on the primary side. Now any power source has series impedance Lets say 0.5 ohms with a 230V system. And if the unsymmetric current is 5A, this causes a 2.5V DC component at the primary side. And thus the core sees a small DC component and therefore thre will be increased current in the opposite side (increasing slowly over a lot of fullwaves). It will compensate the unsymmetric current a little, but not completely.

Klaus
 

FvM

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Give me a feel for the magnitude of the effect. Suppose I have a suitably large 1 to 1 isolation transformer and I connect the secondary to a 50Ω resistor in series with a diode. I would expect a current of about 2.4 amps (in the U.S., with 120 VAC at the outlets in the home) without a diode, so about 1.2 amps (RMS, AC+DC) with the diode.
The question about the expectable primary DC current hasn't been answered yet. My < 1 mA guess was extrapolated from a measurement with a small (20 VA) transformer and a half-wave rectifier DC load of about 50% Irated. Obviously this transformer is operating at a low core flux.

I want to correct a point from the previous discussion. A linear transformer can't transmit any DC component to the primary side in steady state. This can be easily derived from network theory. The half-wave rectified secondary current has to be decomposed in harmonic components of the fundamental and a DC component. In a linear system, the transmission of each component can be calculated independently, the DC component is blocked by the transformer. After re-composing the AC components to a primary current, the DC component must be zero.

Only the non-linear B=f(H) core characteristic gives a "chance" to get DC primary current.
 

The Electrician

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It does depend on the design of the Tx and how far towards Bsat it goes normally. Also the type of Tx, e.g. toroidal is much worse than EI. For a 3kVA isolation Tx, with Bpk = 1.8T say, with normal transformer steel material, drawing 1.2 amps rms via a diode, could easily lead to 10 amps peaks in the primary as the resistance of the wires is low. Try it your self...!
Ok. Suppose the transformer is a 1 kW isolation transformer with E-I laminations, loaded as I described in post #12. Typically, what DC current would one expect in the primary?
 

Orson Cart

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HTML:
Ok. Suppose the transformer is a 1 kW isolation transformer with E-I laminations, loaded as I described in post #12. Typically, what DC current would one expect in the primary?
It depends on the nature & size of the DC load on the sec.

- - - Updated - - -

A linear transformer can't transmit any DC component to the primary side in steady state
unfortunately we have steel cored transformers which are non linear...

- - - Updated - - -

And you may get saturation when input "voltage" is unsymmetric = has DC components.

But this is not the case when secondary current is unsymmetric = DC current components.

Imagine a flyback transformer has a catch diode at its secondary. Here the current is only in one direction. It can't flow in reverse.

Also you may use a diode at a standard transformer secondary output. Whithout (big) core flux change.

The core flux is proportional to the integral of voltage over time. Not current.

But in detail an unsymmetric current on the secondary side will cause an unsymmetric current on the primary side. Now any power source has series impedance Lets say 0.5 ohms with a 230V system. And if the unsymmetric current is 5A, this causes a 2.5V DC component at the primary side. And thus the core sees a small DC component and therefore thre will be increased current in the opposite side (increasing slowly over a lot of fullwaves). It will compensate the unsymmetric current a little, but not completely.
Klaus states " core flux is proportional to the integral of voltage over time. Not current." but then goes on to show that sec currents will change the primary volts symmetry leading to DC drawn on the primary.
Such self contradictions are not helpful to newbies.
The mains impedance is often in the order of 0.05 ohm, so a net DC offset of 50mV gives 1 amp DC average in the primary.
Until people have done the test on a bench with a 1 - 3kW Tx and a non linear load (hair dryer on 1/2 power) either on the sec or on the mains nearby, then contributions are about the same as the stuff exiting the hair dryer....
 

KlausST

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Hi,

It is no contradiction.
Please reread my post carefully and take care about the difference between voltage and current.

*****

And in my description "in detail" i described the differenc between a theoretical simulation, (where a source is considered to have zero impedance) and a real circuit. In a real circuit you have source impedance.

Imagine an AC SOURCE with a 1 Ohms series resistor. And a LOAD consisting of a diode connected with a 10 ohms load resistor in series.

Now measure the voltage across the load with a scope. You will see that one halfwave has full size, but the other has about 90% of fullsize. The difference between both results in a DC voltage component caused by the unsymmetric load current.

If you now assume the same "unsymmetric" voltage at a primary of a transformer you will see that the iron core will slowly saturate.

****
In my eyes the core flux depends on voltage, but the core flux does not depend on current. An unloaded transformer saturates at X voltage. But a transformer with full (symmetric) load saturates at maybe 0.95 * X voltage. The 0.95 comes from series impedance in windings.
If the core flus would depend on current, then a unloaded transformer would almost never saturate, and it would saturate with increasing current. But the opposite is the case.
****

I'm sorry if i confused you.

Klaus
 

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