Strange energy loss in pulsing circuit

Discussion in 'The Projects Forum' started by renginear, Feb 22, 2007.

  1. renginear

    Thread Starter Member

    Feb 22, 2007
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    Hello All

    I am designing a ultrasound pulsing circuit using an H-bridge configuration with around 500V DC bus at a frequency of 100 kHz. The current required for the design is a pulsed peak of 12A. However, since this is a pulsed application, my average power is less than 5W.

    I have my power source as a low-current (mA) high-voltage power supply trickle-charging a capacitor of 200 uF, which supplies the energy required during the ON time.

    I am using the IR2110 as the driver in standard bootstrap configuration with a 12V Vccand my N-ch mosfets have been: IRFS9N60A, IRF830AS, FQB3N60C, and similar ST mosfets.

    Every time, I see the same scenario: With my H-bridge disabled, I see the full 500V on the DC bus. Once the circuit starts switching, I see only around 300-325V across the load. Since ultrasonic excitation depends a lot on peak voltage, this is causing a huge deficiency on ultrasonic signal. However, I do not see any big tell-tale signs of trouble, like over-heating of mosfets, or pronounced droop in the voltage waveform or any HF oscillations. My dead-time is as high as 400 ns to prevent shoot-thru and my inputs are opto-isolated -- there is no RF noise to speak of. So, I am really confused here.

    The things I thought could be wrong were:

    1) ESR of my storage caps~ they are around 500 mOhm
    2) Insufficient gate drive causing high drop across mofet -- not verified yet, but from IRF tech support, I should have no problems driving small mosfets like IRF820 with the IR2110. Also, I do not see any heat from the mosfets and the performance does not degrade with increase in operating frequency.
    3) Bootstrap cap - again, I have low ESR tantalum cap in parallel with series ceramic cap -which meets teh guidelines.


    I am sure the has to be something I have missed here, so I would appreciate any comments regarding my circuit and any tests i could try to narrow my problem area. I have compared my design to a legacy design by another manufacturer operating at 500V and it does not exhibit any voltage loss across the very same load.

    Thanks in advance.
     
  2. hgmjr

    Moderator

    Jan 28, 2005
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    Can you sketch the circuit and post it so that we may see how you have hooked everything up? Please include part numbers and component values.

    My first guess is that your power source is not up to the task of powering the circuit.

    Even if the power supply meets the average power requirements there is still the issue of the supply's ability to meet the instantaneous power demand. The large capacitor will help but that will only go so far in smoothing the power available to the load. The fact that you are switching the load at ultrasonic frequencies will tend to prevent the capacitor from storing enough energy between pulsed power demand. That is most likely the cause of the sag in the power supplies output.

    A schematic will help facilitate a more definitive diagnosis. Without one, all we can do is throw darts at a list of possible root causes for your problem.

    hgmjr
     
  3. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Keep in mind that if you are pulsing at 12 amps with a 50% duty cycle (6 amps average), your 200uF cap will discharge at the rate of dv/dt=I/C, which in your case equals 30V per millisecond. Keep in mind also that you have to replace all that charge during the time you are not pulsing.
    As hgmjr said, a schematic will help immensely.
     
  4. renginear

    Thread Starter Member

    Feb 22, 2007
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    Thank you for the comments. As requested, I have posted a schematic. I have included part numbers where I can. I am also working on a design based on the IXYSRF DE275-501 mosfets, but it is not ready yet.

    Basically, since my duty cycle is less than 5 % and the highest pulse duration is approximately 200 us ( 10 pulses of 10 us at 50 % duty cycle) with a 'dead time' of at least 4-5 ms....usually keep it at 15 or 20 ms at least.

    So, I presumed that the cap could provide current to my load and get trickle charged by the power supply.

    I tried to raise the cap value by putting in two 330 uFs instead of a 330 and a 20. The result was what I thought was a slow burn b/w the HV pin on the 330 uf cap next to the h-bridge to ground. It also burnt through the ground plane (inner 1 layer). but while it worked, I saw no change. I am now also wondering if my h-bridge pcb had some fault.

    I guess I am confused because I expected to see a high spike followed by a decreasing exponential sag. However I am seeing a nearly flat square wave (droop b/w 5 and 10%) always at a voltage which is approximately 70% of the bus voltage.

    Again, I am mainly an embedded person and some of this HV stuff is new to me, so I am sure I have made a simplistic assumption somewhere
     
  5. hgmjr

    Moderator

    Jan 28, 2005
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    If I read your diagram correctly, it appears that the power source and the 330 uFd cap are separated from the Half H-bridge that drives your load.

    What gauge wire are you using for this 4 to 6 inch connection? If the gauge is too small the connection wire may have some inductance that is acting as a significant resistance during the rise times of your switching signal.

    What do you think RonH?

    hgmjr
     
  6. renginear

    Thread Starter Member

    Feb 22, 2007
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    I am using a 12 Gauge wire, but for ease of connection, I connected them with molex pin and receptacle in between. This makes it around 4-6 inches long.

    I am getting a 2nd board built and in this case, I will try to put a direct short (<1 inch) 12 gauge wire connection between the two boards. Also, I am thinking of moving my 20 uf cap to next to the 330 and leaving only the ceramic caps there. I am wondering if the ground plane on the h-bridge board causes a larger area for current to flow and thus increases losses.
     
  7. hgmjr

    Moderator

    Jan 28, 2005
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    12 gauge should carry the current with little if any inductance so that thought is a bust.

    Are the leads to the 330 uFd capacitor trimmed very short?

    hgmjr
     
  8. hgmjr

    Moderator

    Jan 28, 2005
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    You may want to look into the capacitor's specification for its ESL (Effective Series Inductance).

    hgmjr
     
  9. hgmjr

    Moderator

    Jan 28, 2005
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    Voltage drop across an inductor is equal to L times di/dt.

    You are switching 12A in say 1 microsecond. That equates to 12 x 10^6 Amps/sec. If you take a voltage drop of 100 volts and divide that by 12 x 10^6, you get an inductance in the neighborhood of 8.3 microHenries.

    hgmjr
     
  10. renginear

    Thread Starter Member

    Feb 22, 2007
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    The leads to the 330 uF cap are the standard leads that 'snap' into the slot.

    One thing I noticed (before my mishap) that in case of a multi burst excitation (multi burst means that instead of one pulse at 5% duty cycle - you send 5 pulses at 50% duty cycle and then follow 1% duty cycle- this helps in raising the output of the transducer) , on my 3rd to 5th bursts, I see a steep roll-off instead of a flat peak...thus indicating that yes, I do not have enough energy.

    If I add many capacitors in parallel, will that bank be a good buffer to prevent my droop? What is a good way to calculate the energy required to supply a steady 12A, at 400V and design a capacitor bank for it. I have been looking around for a pulsed power circuit design guide, but all I found so far are kV/kA circuits used for accelerators etc...
     
  11. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Inductance of 12 inches of wire is probably considerably less than 500nH, and diameter is a second-order effect. In any case, it shouldn't cause droop. I ran a simple sim on the circuit (including inductance), and I can't see where the droop is coming from.
     
  12. hgmjr

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    Jan 28, 2005
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    Is it possible that the 1 ohm power resistor you are using in-line with the power supply is a wirewound type? Such a resistor could have some inductance.

    hgmjr
     
  13. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    Is the power supply adequate? 400V*12A*5%=240W. Hope this is the right way to calculate the power.
     
  14. renginear

    Thread Starter Member

    Feb 22, 2007
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    Thanks hgmjr

    I looked up the resistors, they are wirewound SMT. This reminded me of another fact I forgot to mention. I tried a test without the resistors just jumpering the pads and that setup caused instantaneous high-side mosfet blowouts above 50V supply voltage. I wonder if the lack of the resistor 'allowed' the extra energy to get to the probes

    (warning: all links below are pdfs)

    The resistors I have used are the WSC series ( 2 in parallel)
    http://www.vishay.com/docs/30102/wscwsn.pdf

    So, it appears I need a replacement. There is a company I found called TT electronics which has high power resistors with nH inductance. One of their app note mentions low inductance values:

    http://www.irctt.com/file.aspx?product_id=110&file_type=application_note
    (page 16)

    The CHP series is available on newark:
    http://www.irctt.com/pdf_files/CHP.pdf

    There is also a set of resistors (non-inductive) from a company linked from their website:
    http://www.bitechnologies.com/power/index.htm

    I am thinking of trying them. What do you think?
    -----

    Also regarding kubeek's calculation, yes at 5% duty cycle we would need 240W, however we try to stay below 1% for high-current probes. For high-frequency probes, 5% duty cycle is used, but their current consumption is less than 2A. Hope this helps. I am looking at a 250W version of the power supply as well - they are pin compatible.
     
  15. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    To charge and discharge a capacitor, P=f*C*V^2. For a 5nF cap, 400V p-p, 100kHz,

    P=1e5*5e-9*400^2
    P=80 watts.
    At 5% duty cycle, the average power is only 4 watts.

    The instantaneous current may be 12 amps, but it only takes
    t=V*C/I = 400*5e-9/12=167nS to fully charge or discharge the cap (this is a simplification, since the current will not be constant during that time). Current has nowhere to go after the cap is fully charged or discharged.
     
  16. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    You have to charge the capacitors with the same energy you need for the output, so if the output needs 240W then the supply must provide 240W.

    But that all depends on the load. Can you tell us what exactly it is?
     
  17. renginear

    Thread Starter Member

    Feb 22, 2007
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    The load is a probe, which has approximately between 100 and 1kOhm resistance and 1 to 5 nF capacitance.
     
  18. hgmjr

    Moderator

    Jan 28, 2005
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    Since the circuit is switching a fairly high current at fairly rapid switching rates, then it would be a good practice to look for any inductance that appears in series with the high current signal lines.

    The use of a low inductance power resistor should improve the performance of the circuit overall. That is if the wire-wound power resistor was the only source of inductance. The circuit can tolerate nano-Henries but micro-Henries and larger are going to result in losses that could be significant.

    hgmjr
     
  19. Ron H

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    Apr 14, 2005
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    OK, I was thinking the load was purely capacitive. I looked back at the schematic, and it also has a significant resistive component. I'll rethink my calculations. :(
    It would be good to get better definition of the load. 1-5nF and 50-300 ohms is a huge range. Renginear, can it be 50 ohms and 5nF simultaneously?
     
  20. Ron H

    AAC Fanatic!

    Apr 14, 2005
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    Here is an article that suggests that maybe you should add an inductor in parallel with your transducer to reduce peak currents.

    i ran a sim on your circuit with 50 ohm||5nF as the load, and it appears that the problem is the 2k resistor.
     
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