How to prohibit negative pulses on a transformer?

Hello, I have a small toroidal transformer with a primary and a secondary. I am driving it with a square waveform on the primary.
This generates a positive pulse at the secondary during the rising edge of the square wave and a negative pulse during the falling edge of the square wave.

I am looking for a way to somehow prohibit the negative pulse from being generated at the secondary, or at least make it much lower amplitude.
No work must be done after the primary, all the work should be done in the clock or the primary.

A first thought that I had is this:
Well why not drive the transformer with an inverted sawtooth clock? http://sparkbangbuzz.com/cds-fet/ns0749-cr-50.jpg
As the fast rising edge of the sawtooth pulse comes in the primary, a positive pulse will be generated at the secondary. But then, the falling edge of the inverted sawtooth will be much smoother than that of the square wave, causing a much less amplitude (or/and more width?) on the output pulse.

Any thoughts about it, or any other ways i can achieve this?

Quite generally, if you drive a transformer with a square wave voltage, you get a square wave at the secondary. Getting positive/negative pulses presumes unsaid conditions where the transformer isn't working as a voltage transformer any more.

FvM said:

Quite generally, if you drive a transformer with a square wave voltage, you get a square wave at the secondary. Getting positive/negative pulses presumes unsaid conditions where the transformer isn't working as a voltage transformer any more.

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Are you sure about this? I thought transformers were AC devices.
A square wave is like 0v-5v-0v. When a pulse 0v->5v rising edge comes in the primary, then is that the transformer produces positive output. The next thing is the falling edge of the pulse where 5v->0v falling edge comes in the primary. Then the transformer produces negative output. At the time interval where the pulse at the primary stays at 5v or 0v, there is no output at the transformer secondary.

ya its going to be a impulse signal in the output when giving square wave..........

for a transformer the output will be the differentiated function of input..

so for a every increase in input you will get positive output and for every decrease in input you will get negative signal, so whatever the AC input you will get the negative part..

for generating a purely positive in the output,
you sould give a ramp which never saturates or stops which is practicaly impossible...

for generating a limited negative output,
you can use a inverted ramp as you say, you can reduce the negative component by reducing the decaying rate..

Firstly the transformer output is DC-free, you get a level shifted square wave if the input coltage contains a DC component.

Secondly, the transformer has a L/R time constant, where R is the primary winding resistance and L the primary inductance. The flat top will decay exponentially with L/R at the secondary. So if the pulse period is large compared the time constant, you get in fact positive and negative pulses.

Finally, in most cases, you don't drive the transformer with a voltage rather than a source with internal resistance Rs. The time constant reduces to L/(Rs+R).

ya its going to be a impulse signal in the output when giving square wave..........

for a transformer the output will be the differentiated function of input..

Click to expand...

You are continuing the fault of the original post, assuming transformer and signal parameters that haven't been specified.

FvM said:

Firstly the transformer output is DC-free, you get a level shifted square wave if the input coltage contains a DC component.

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I believe, We cant say its a level shifted square wave at all......

Of course you can get a nearly level shifted square wave. Just a matter of transformer L, R and pulse period for a real transformer.

An "ideal" transformer blocks DC and passes AC, so you get surely a square wave.

I am saying That is just a deformed impulse signal due to the inability of transformers ideal characteristics and it is not going to happen in all transformers.....

Put a diode in series with the output so the negative transient is suppressed or put a shunt diode across the primary of the transformer.
Frank

FvM said:

An "ideal" transformer blocks DC and passes AC, so you get surely a square wave.

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Hi Its totally unacceptable......... It is exactly block the DC COMPONENT and allows the AC COMPONENT.........

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chuckey said:

Put a diode in series with the output so the negative transient is suppressed or put a shunt diode across the primary of the transformer.
Frank

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according OP we shd nt touch the secondary and how could a shunt diode will avoid negative output??

Venkadesh_M said:

I am saying That is just a deformed impulse signal due to the inability of transformers ideal characteristics and it is not going to happen in all transformers.....

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I think you are both right, although I cannot explain it as you do.

In my experiments, if the continuous pulse frequency is quite high, the transformer indeed can show a sine wave like waveform at the output (decaying gradually) and this is probably because of the effects FvM specified earlier. In fact when a single (monostable) high frequency rising edge pulse comes in, a positive pulse (positive part of sine) comes out, which seems to decay, probably as the energy "stored" in the winding is lost.
This is a simple very practical explanation, I do not know if I am wrong and I cannot explain it scientifically.

The inverse sawtooth I have thought causes a very fast rising edge on the positive portion, whereas is much more slowly decays the falling edge. I want to experiment with this but I think this will cause only the positive pulse to appear at the secondary, plys some much less amount of the negative.
If you see here **broken link removed** this is what is happenning, but the input pulse (top line) is a capacitor discharging, so the negative pulses are passing through it from the square wave generator. But if you focus only on the positive pulse which "simulates" a single pulse of inverted sawtooth, you can notice that I am talking about.
I hope my explanation is valid...

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Venkadesh_M said:

Hi Its totally unacceptable......... It is exactly block the DC COMPONENT and allows the AC COMPONENT.........

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according OP we shd nt touch the secondary and how could a shunt diode will avoid negative output??

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Yes, a diode does not avoid the negative portion. Remember, when speaking about the negative here, we are talking about the FALLING EDGE of the positive pulse.
And Yes, any work must be done on the primary or before it in my case.

Hi this figure shows the inputs and corresponding outputs,

**broken link removed**

for every change in flux only there will be a voltage in the output

If you change the frequency the waves will be affected by the L and R components as FvM said

but for the designed frequency inputs these are the outputs by this you can understand and give the correct signal

In all cases the core saturation is not mentioned, If you reduce the frequecy that will also happen in the output.......

Hi this figure shows the inputs and corresponding outputs

Click to expand...

The diagram is correct, if the transformer is driven by a current waveform and you observe the output (or input) voltage. That's surely not the general case of transformer operation.

You can however get similar waveforms under specific conditions. If the input is a voltage source, characteristic waveform times must be very large compared to transformer L/R.

Hi how do you genenrate the input signal?? with 555??

Venkadesh_M said:

Hi how do you genenrate the input signal?? with 555??

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Yes I am using a 555 generating a square wave (not 50% duty cucle). The top line shows this signal passing through a capacitor to the transformer primary.

What is the value of series capacitance ?? If you increase it or adding a small resistor will increase the decaying time and reduce the negative amp........

Venkadesh_M said:

What is the value of series capacitance ?? If you increase it or adding a small resistor will increase the decaying time and reduce the negative amp........

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It is 100nF and the generator frequency is 20KHz. I have not experimented with altering the values of the capacitor yet.
A negative pulse out of the capacitor will still occur I think. That is why I have thought of the inverse sawtooth. I do not focus on the current setup, I focus more of "different ways to do it", I am not limited by the setup.

Ya Whatever the values of components It is not going to be zero negative because the Area of positive side will be equal to the negative side.........
but you can control the Amplitude by slowly decaying....
Don change the frequency just put a 400nF using two parallel caps and look at the waveform.......

Venkadesh_M said:

Ya Whatever the values of components It is not going to be zero negative because the Area of positive side will be equal to the negative side.........
but you can control the Amplitude by slowly decaying....
Don change the frequency just put a 400nF using two parallel caps and look at the waveform.......

Click to expand...

Maybe a combination of this and the inverse sawtooth will give even better results.
The only advantage of the inverse sawtooth alone, is that is is more frequency independent.

So, you cannot think of a better way to achieve such a result?
I wanted to make sure that an inverse sawtooth is a good way to do go.

Youu know you are already giving the inverted saw tooth !!! with exponential decay... I am just saying to extend it... I think you are meaning the huge negative pulse??

that can be easily arrested by putting parallel diode in the primary as chuckey said.....