Feeding a square pulsed DC into a transformer's primary winding.

Discussion in 'General Electronics Chat' started by BlackMelon, Apr 18, 2016.

  1. BlackMelon

    Thread Starter Member

    Mar 19, 2015
    32
    0
    Hi there!

    The attached photo is a circuit with an input waveform. Do I get it correctly? I'm so worried about the increasing current very much. (As you know, the inductor current is an integration of its voltage. Because the voltage is always greater or equal to zero, this current will keep going up)

    Background story:
    I'm trying to control a forward converter in this link: http://www.onsemi.com/pub_link/Collateral/TND378-D.PDF
    The circuit is on page 47and48. Unfortunately, the PWM controller is not with a high-side MOSFET driver, so the author did it in page 48. When Q301 is on and the other is off, the voltage across the primary winding is Vcc. Subsequently, Q302 operation gives 0V to the winding. This will be a square pulsed DC like in the attached photo.

    PS. Sometimes it's funny that the high side driver like HIP4082 does not come with a feedback pin and a pin for a setpoint resistor. while the ONSEMI has a feedback and a setpoint resistor, but no high side driver.

    Thank you
    BlackMelon
     
  2. ronv

    AAC Fanatic!

    Nov 12, 2008
    3,293
    1,262
    I think you need two switches so you can reverse the current thru the transformer.
     
  3. wayneh

    Expert

    Sep 9, 2010
    12,156
    3,063
    The current will not increase as you have plotted it. When there is no voltage applied, any stored energy will dissipate quickly and the current will return to zero during each cycle.

    Oh, never mind. I just looked at your link and realized things are more complicated.
     
    Last edited: Apr 18, 2016
  4. Brownout

    Well-Known Member

    Jan 10, 2012
    2,375
    998
    You didn't show the circuit driving the primary. Your drawing looks like the one on the link, so if I can assume your driver is the same, then you have it correct. But the answer depends upon how the input signal is driven.
     
  5. MaxHeadRoom

    Expert

    Jul 18, 2013
    10,571
    2,381
    The old tube car radios used a mechanical vibrator that switched the 12vdc polarity on the primary, there was also a synchronous type that switched both primary and secondary in synchronism reducing the need for a rectifier.
    This was not pulsed DC but alternated the polarity.
    Max.
     
  6. BlackMelon

    Thread Starter Member

    Mar 19, 2015
    32
    0
    The previous diagram is just checking my understanding, because I want to understand the circuit here: http://www.onsemi.com/pub_link/Collateral/TND378-D.PDF (in page 48). It is a high and low side driver, which is used to drive a forward converter in page 47. The transformer that I'm analyzing is XFMR2. In the HLS driver, when Q301 turns on, it's just like C302 and the transformer primary winding do a resonance stuff. I don't know whether the primary and C302 reactance are equal or not, since they are not shown. On the other hand, when Q302 turns on, we just feed 0V to the primary winding.

    This circuit supposed to give two separated ground to applied two signals to high and low side MOSFETs' gate-source, but I can't draw its waveforms now. It's very strange to get an analysis like that

    Thank you
    BlackMelon
     
  7. BlackMelon

    Thread Starter Member

    Mar 19, 2015
    32
    0
    The drive signal (DRV) is an output of IC NCP1252 in page 47. Is this what you guys are looking for?
     
  8. DickCappels

    Moderator

    Aug 21, 2008
    2,664
    634
    With pulsed DC the flux in your transformer core will "walk" towards saturation. That would mean that the pulse source's drive capability will likely limit the current.

    Would you like to put a capacitor in series with that pulse?
     
    BlackMelon likes this.
  9. MrAl

    Well-Known Member

    Jun 17, 2014
    2,440
    492
    Hello there,

    According to that drawing you are driving the primary of a transformer with a seemingly square wave voltage that ranges from 0 to 12v.

    With no load on the secondary, this is equivalent to driving a single inductor in series with a resistance with that same waveform.
    What happens is the current builds up after some cycles determined by the values of L and R, and the average current reaches a level:
    Iavg=Vpk/(2*R)

    Now this is no problem unless the winding resistance is small, but if it is small then the current ratchets up higher than the switch transistor can handle and so the transistor blows out. If at any time before that the current goes up higher than the core construction can handle then the inductance falls and the current goes up even higher, a condition known as core saturation, and the transistor blows out.

    For a couple quick examples, say we have 12 ohms winding resistance in the primary. The average current will reach:
    Iavg=12/24=1/2 amp.
    If the transistor can handle 1/2 amp then it's ok. On the other hand, if the core/winding construction can not handle a DC current of 1/2 amp then the core would saturate and the inductance would fall and then the most important calculation becomes the peak current calculation which would be:
    Ipeak=12/12=1 amp peak.
    So here we see the most important current calculation rise from 1/2 to 1 amp. If the transistor can handle 1 amp then it wont blow, but during the time the core is saturated there will be little energy transfer to the secondary.

    The above would be a very lucky setup too, because normally a transformer is designed to have low winding resistance. If the winding resistance was only 1 Ohm for example, then we'd see a big rise in current either way. That would most likely blow the transistor.

    To find out if a given construction is going to saturate, you can raise the pulse voltage little by little and check for a sharp rise in current at the end of the switch cycle. Normally you see what looks like a triangular waveform, but if the core starts to saturate the wave shoots up sharply near the end of what used to be the triangle wave. That immediately tells you it's going to blow if something is not changed.
    To help the situation an air gap is usually introduced into transformer core. The gap has the effect of stretching out the BH curve along H and thus makes it require more current to saturate. The down side is the inductance falls, so it's a tradeoff. Adding turns helps to get the inductance back up a little more while also making it saturate more easily again, but luckily adding more turns does less to make it saturate than the air gap helps to stop it from saturating so if there is enough window area we eventually find a workable design :)
     
    BlackMelon likes this.
  10. BlackMelon

    Thread Starter Member

    Mar 19, 2015
    32
    0
    A capacitor in series with that pulse is also in series with the primary coil. The total impedance of the cap and the primary winding is less than the winding alone (Ztotal = jwL-j(1/wC)). I think this will worsen the situation, because as you said, "the pulse source's drive capability will likely limit the current" means that the pulse source gives out highest current it can does. Reducing this impedance is dangerous in my opinion. So, I still don't understand why the app note put a capacitor in that way. (And don't know why they put a diode in that direction too)


    MrAl, that was a clear and informative explanation. I now know all the important parts that come to play this game. Thank you very much. Could you please analysis the driver in the app note of my previous reply also?

    Thank you everyone
    BlackMelon
     
  11. BlackMelon

    Thread Starter Member

    Mar 19, 2015
    32
    0
    I meant I just want to know more about the diode in parallel with the capacitor in page 48 of the app note (High side driver). Will this only block current from flowing into the primary winding?
     
  12. DickCappels

    Moderator

    Aug 21, 2008
    2,664
    634
    As I said earlier but in different words, the DC component of the waveform (without a series capacitor) will push the core towards saturation unless the transformer is designed to handle DC. Approaching saturation will result in increased magnetizing current which will lead to higher copper and conduction losses in the transformer and switching device and in some cases leads to failure.
     
  13. BlackMelon

    Thread Starter Member

    Mar 19, 2015
    32
    0
    Were you trying to say that the capacitor filters the DC component out? The position of it is similar to that of a high pass filter. I should have think it this way before considering about resonance stuff. :(
     
  14. DickCappels

    Moderator

    Aug 21, 2008
    2,664
    634
    Yes. That is what capacitors do.
     
  15. BlackMelon

    Thread Starter Member

    Mar 19, 2015
    32
    0
    http://www.onsemi.com/pub_link/Collateral/TND378-D.PDF
    Thank you, I now understand how the capacitor removes the dc offset. However, in page 48, I still can't get a complete waveform of U301 primary winding by my analysis.

    As in the attached picture, I was trying to use a superposition technique to think of a signal as a DC component plus an AC component. In the DC part, everything is clarified as we discussed before. However on the AC part, I was hoping to apply a filter's equation directly, but the only roadblock is the diode, especially when it is off. Is the only way to analysis it doing a second order C-L circuit? If so, I think I might need to find the primary winding's inductance, even I just want to understand its operation.
     
  16. BlackMelon

    Thread Starter Member

    Mar 19, 2015
    32
    0
    Oops! sorry for the last reply, I forgot that the superposition is not valid for a non-linear component like a diode. Completely messed it up LOL

    So, is there any way to completely analyze this?
     
  17. Roderick Young

    Member

    Feb 22, 2015
    408
    168
    XFMR2 is a pulse transformer. The idea is that C302 is very large compared to the pulse duration, so it acts like a short when Q301 turns on, and resonance with the transformer winding does not come into play. When Q301 turns off and Q302 turns on, two things are reset: the voltage on C302 goes back to (near) zero, and the current in the transformer winding goes back to near zero. You would think that it would oscillate for a while, but the frequency is so low that the energy dissipates itself in D301 and other losses in the circuit before much ringing can occur.

    Yes, the pulse transformer has no DC connection between the primary and either of the secondaries, allowing the high side drive to be referenced to the high side voltage, and the low side to be referenced to the low side (might be ground).
     
    BlackMelon likes this.
  18. DickCappels

    Moderator

    Aug 21, 2008
    2,664
    634
    Keep in mind that the circuit on page 48 is a small signal stage rather than a power stage.

    @Roderick Young I am impressed by your analysis.
     
    BlackMelon likes this.
  19. BlackMelon

    Thread Starter Member

    Mar 19, 2015
    32
    0
    @Roderick Young Oh! Thank you very much for your explanation. However, there's something left in my mind. The way to reduce inductor's current is to apply a negative voltage across it accordingly to the equation V=L*(di/dt). When Q302 turns on, the voltage across the XFMR2, V6-V1, is equal to Q302's C-E voltage (0.2) plus D301 forward voltage (0.7). The polarity of this voltage is still the same as when Q301 turns on. I mean the voltage across the XFMR2 does not go to negative (reverse its polarity). So, how can it decrease its current? Considering wire's resistance, is it so low that the current of XFMR2 latches too long?

    Thanks
     
  20. BlackMelon

    Thread Starter Member

    Mar 19, 2015
    32
    0
    @Roderick Young I consider your last explanation every words. That's clear and simple. I appreciate. However, I did this calculation because I just want to develop the idea that you've proposed before to be able to draw a waveform.

    So, I did some calculation (a 2nd order problem) about when Q301 is ON. As shown in the picture, the waveform is the voltage across the primary winding of XFMR2. The winding is shown in the attached picture as an inductor.

    Is my following understanding correct?
    In the view of frequency, due to a large capacitance value, the period of free-oscillated inductor's voltage is very long compared to the ON-time of Q301, so the actual voltage across the inductor is trimmed by the two transistors. (In the manner that Vcc is connected to this L-C circuit for a very very short time before disconnecting it) What's remained is the red waveform. If the capacitor is large enough, the voltage across an inductor at Q301's ON-time will be almost constant near Vcc, which corresponds to the idea that the large capacitor shorts the Vcc to the primary winding when Q301 is on.
     
    Last edited: Apr 24, 2016
Loading...