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UCC28701 use with a current sensing transformer

Other Parts Discussed in Thread: UCC28701

Hello,

I'm designing a "simmer" supply for flashlamps, with a 24V input, 300V open-circuit output and 300mA constant-current regulation. I'm confident the overall principle will work, having already prototyped the circuit.

I would like to reduce the power dissipation in the primary current sense resistor, which is currently in the region of 1.5 to 2W, and was wondering if it is possible to use a current-sense transformer instead? For example, this one from Mouser - 

www.mouser.com/.../-13316.pdf

I'd presumably place it in the drain of the MOSFET rather than the source (to avoid any effect on the gate drive). Are there any problems which a current transformer might introduce?

Best regards,

Lindsay Wilson

  • Theoretically I don't see why a current sense transformer would not work but the nagging issue that comes to mind would be the frequency range of operation; the device you link to shows a minimum range of 50kHz but the UCC28701 will operate much lower than this over its normal range of operation.

  • Another thought is to make sure the CS pin doesn't go below ground or otherwise violate the Abs Max rating (noise on the drain during switching...)
  • Hi,

    Thanks for the reply. I did wonder about that, but then again, it's not as if the chip will have the same maximum duty cycle at lower frequencies - that would surely be limited by the various thresholds on the CS pin. For example, if it takes say 5us for the current through the main transformer primary to reach the CS threshold, will the lower frequency operation then not consist of a string of 5us pulses at (e.g.) a few kHz? It can't increase the duty cycle and give a longer pulse, because that would go over the current threshold.

    If that's the case, then I think the current transformer would still be operating within its limits - I found a datasheet for Epcos' transformers (www.farnell.com/.../1872311.pdf) which lists the various maximum Volts*Time product. If the current transformer + burden resistor were arranged for say a 500mV output over 5us, that's only 2.5V*us which should be well below the maximum ratings.

    Am I understanding that right or missing something obvious?!?

    Best regards,

    Lindsay
  • I thought of that after I hit the "reply" so I don't think it will saturate...can you keep me posted as to how successful you are? I am very curious!
  • Lindsay, Lisa,

    I think that the effect of the CS traffo on the Rlc resistor should be considered.

    Assume the CS shunt in the source was say 1-ohm, with Rlc of 1 kohm to get the peak current adjustment with line to adjust for turn-off delay.

    If it's changed to use a CT with ratio of say 100:1, the 1-ohm shunt becomes a 100-ohm burden on the CT output. This is more significant wrt the Rlc value, and will produce 10% more CS adjustment than needed (CS offset current into 1.1 kohm instead of 1-kohm). In this case, Rlc should be decreased to 900 ohm.

    If the CT ratio was 1000:1, Rlc would have to drop to zero.


    I think it this should be taken into consideration when calculating the value of Rlc.

    Thanks,
    Bernard
  • Bernard, Lisa,

    Many thanks for the replies - what I'm probably going to try is make up a board with both a normal current-sense resistor and a CT, so I can switch between them once I'm certain it's operating reliably.

    Good info about the correct choice of Rlc - I was wondering how to determine that given the CT burden!

    I'm afraid I'm still rather unclear about the exact purpose of the Rlc resistor. As I understand it, this is to compensate for variations in the input supply voltage since, at a lower voltage, the dI/dt in the transformer primary will be less, and if Rlc wasn't used, that would introduce an error into the calculation of secondary current. Is that about right? If I wasn't concerned about variations in input supply voltage (e.g. it's going to be operated from a known 24V supply), could I ignore Rlc?

    Things are slightly different in my design because I'm choosing not to use a separate auxiliary winding, instead taking the voltage feedback directly from the secondary (there is no need in this design for an isolated output, and it simplifies the transformer construction). The component values from the equations seem to work out ok using Npa=Nps (and Nas=1), but might there be any caveats I'm missing? 

    If it's any use, here's the circuit I'm currently working on - imajeenyus.com/.../full_schematic.pdf

  • Seemed to miss the last bit of the message - open circuit voltage is 300V, and I'm aiming for a constant-current regulation of 300mA.

    Apologies for the questions getting a little off the original topic.

    Best regards,

    Lindsay
  • Lindsay,

    This controller was designed for universal AC input range from 85-265 Vrms. Over this wide range, the slope of the inductor current will vary a lot. Due to the delay in turning off the external FET, the peak current will overshoot slightly compared to the target level. At higher input voltage with higher di/dt, this overshoot can be very pronounced. This higher peak current will then cause a big variation in the CC-mode limit, since that function regulates the diode duty cycle in order to indirectly regulate the output load or average current - on the assumption that peak current is also regulated.

    In order to counteract the effect of prop-delay current overshoot, during the on-time interval, the current flowing out of the VS pin (which will be proportional to the input voltage, turns ratio and upper RS1 resister value) is internally scaled down and mirrored, and sourced out of the CS pin. This current will produce a small offset at the CS pin, depending on the value of Rlc and the shunt/burden. And since the current out of the VS pin increases as line increases, so the CS offset also increases. If Rlc value is correctly chosen, the CS offset can exactly cancel the effect of the current overshoot with line, using eqn 17 in the datasheet.

    In your case, if the input voltage is a fixed 24-V supply, then you won't need to worry so much about Rlc value.

    Thanks,
    Bernard
  • Bernard,

    Thank you so much for the clear explanation, that makes perfect sense now!

    I'm also wondering if the following would be possible (feature creep!). Although my initial aim is to achieve a regulated 300mA output at up to 300V, most commercial simmer supplies have an open-circuit voltage of around 1200V, although their current regulation range is much lower. For example, suppose the maximum power output is 90W. The supply will regulate current to 300mA up to 300V, then it will "top-out" and be limited to 90W power at up to 1200V (i.e. delivering 75mA at 1200V).

    So, suppose I initially designed the various values to suit a 300mA, 300V supply, but then decrease the value of Rs2 (the lower resistor in the feedback divider) to give a regulated voltage of 1200V, whilst keeping everything else (transformer inductance, turns ratio, current-sense resistor) exactly the same. My think is that, in current-regulated operation, the voltage feedback is below the threshold, so the supply will work exactly as before, but when the required output voltage rises above 300V, the supply delivers its maximum power of 90W, until it tops out at 1200V.

    Do you think this is likely to work? I might not need to go as high as 1200V, but some flashlamps do require that.

    Best regards,

    Lindsay
  • Lindsay,

    I am not familiar with these simmer power supplies, or why the open cct voltage is normally so high - is this consequence of how they are designed/controlled, or is it actually a system requirement/benefit.

    It may be possible to design the power stage to regulate high in CV mode at ~1 kV, and set the CC limit @ 300 mA. If the power stage is designed for 90 W, then it will deliver up to 90 mA in CV mode with regulated 1 kV output. Thereafter it will power limit, output voltage will fall as output current increases (at about 90 W product), and once the current gets to 300 mA it will go into CC mode, where the output voltage will fall to a level that depends on the load impedance (Vo = Rload * Iocc).

    From looking at your sch, the pri peak current is about 23 A (I see why you want to use a CS Traffo), with 5 uH inductance, at 70 kHz that's about 90 W of power. The 300-V output reflects back to the primary at about 23 V, so that's about right for 50% max duty cycle at 24 V input.

    At no-load, the controller will drop down in freq to a min of 1 kHz, so R3 needs to be able to absorb 0.5*Lpri*Ipk_min^2*Fmin, which is about 150 mW - so the 2k2 value for R3 is way too low, at 300 V out, R2 should be more like 560 k-ohm.

    When Q1 is on, the dot-end of the secondary will be at about -312 V, while the cathode of D1 is tied to the 300-V output. So the reverse voltage across D1 is (300 - (-312)) = 612 V. So I think D1 needs to be at least 800-V rated rather than the 400 V shown.


    If you want to design for higher 1000-1200 V at no load, you will need much higher turns on the secondary. Which means even higher rating for D1 (prob need 2-3 diodes in series). To keep the CC limit at 300 mA, the Rcs has to get much lower to compensate for the turns ratio, so the primary peak currents will be much higher than now. That then forces the inductance value to be very small to keep the power handling capability to 90 W. I am not sure that the approach will be practical.

    If you follow eqn's 21-25 in the datasheet, choose Nps to get your desired CV output voltage (1 kV or whatever). Choose Rcs to get your 300 mA CC limit using this Nps. That then dictates the required pri peak current. Then in eqn 25, replace the term (Vocv + Vf+ Vocbc)*Iocc with the max power limit (90 W or whatever). This will produce a design that regulates at high output 1 kV up to 90 W; thereafter it will power limit at 90 W as load current is increased, Vout will roll off; once the current hits CC limit of 300 mA, it will clamp the current and then Vout will fall to a level that will depend on the load impedance.

    I am not sure if this is what your system requires - you cannot guarantee that Vout will be 300 V at 300 mA, it will depend on the load impedance - it will only be at 300 V if the load is 1 k-ohm.

    I hope this helps you out.

    Thanks,
    Bernard
  • Bernard,

    Many thanks for the reply. Most simmer supplies, although they'll only provide the specified current up to maybe a few hundred volts, will still provide some current up to 1200V or so. After the main discharge through the flashlamp occurs, the resistance can increase for a short period, so it's good to have a bit of headroom in the output voltage, even if the output current can't be maintained. Some supplies actually provide a "boost" in current immediately after the main discharge, just to ensure the lamp stays lit.

    (If you're interested in the specs of these sort of supplies, Analog Modules make probably the most widely-used units. More info on their site at www.analogmodules.com/.../simmer-supplies.php)

    The actual impedance of the flashlamp is highly nonlinear and usually negative - it has pretty much a constant voltage drop over a wide range of currents, although it decreases slightly at higher currents. Once I have a suitable constant-current supply (that can work with a normal resistive load), I'll try it with a series output inductor and the flashlamp load, which is what seems to be done in the Analog Modules supplies (flyback supply + series output inductor)

    Thanks for the pointers on the component design - once I get a 300V supply working reliably, I'll have a go at altering for a higher output.

    Best regards,

    Lindsay