Calculating Pulse Transformer Frequency at Voltage

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To calculate the minimum allowable frequency for a pulse transformer with an ET value of 45 V·µs, one must understand the relationship between voltage, flux, and time. The equation Φ = volts × time indicates that the transformer can handle a maximum flux of 45 V·µs, which translates to various voltage and time combinations. For a square wave input of +/-6VAC, the half-cycle duration can be calculated to determine the frequency, with lower frequencies risking saturation. Given the bipolar nature of the square wave, operating at half the calculated frequency may be acceptable if the duty cycle remains at 50%. Understanding these principles ensures the transformer operates within its specifications.
Honk
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Hi.
I need some help in explaining in a simple way how to calculate the minimum allowable
frequency at a certain voltage when selecting a pulse transformer.
The one I'm looking at has an ET value of 45Vusec.

I'm feeding it 22.5KHz at +/-6VAC square wave. It seems to work fine but I fear it is operating out of spec.
Can someone nice her please help me (in s simple way) how to calculate the data from the ET value?

Regards / Honk
 
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Honk said:
Can someone nice her please help me (in s simple way) how to calculate the data from the ET value?

I really like that Butler link.

I'm no pulse transformer expert, but hre's the basic physics you start from.

Volts in a coil is proportional to Δflux /Δtime. dΦ/dt

so Φ = ∫volts dt
for one half cycle of your square wave volts is constant so it can move to other side of integral sign
Φ = volts X time
and they're telling you your transformer can hold enough flux to withstand 45 volt- microseconds, that is 45 volts for 1 microsecond or 1 volt for 45 microseconds
or 6 volts for 7.5 microseconds.

So what is frequency of a square wave with half cycle duration of 7.5 microseconds ?
Any lower frequency saturates it.

Note that is for unidirectional applied voltage.
Since your bipolar square wave drives the core between negative and positive Φmax not zero and max, you probably would get away with running it at half that frequency provided wave is square ie 50% duty cycle ..

See? You knew that already.

old jim
 
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