This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

TIDA-070002: Reference Design

Part Number: TIDA-070002
Other Parts Discussed in Thread: LM139, UC1843B-SP, LM117, UC1843, TPS50601A-SP, LM137

Hello, 

I have some questions about the TIDA-070002 reference design:

1. why isolate the feedback loop, what advantage does this give?

2. would you agree that the over-current facility isn't strictly required which means that INA901 and LM139 could be removed from the design?

3. What functions are L1 and D11 performing?

4. why are there so many output capacitors after the secondary output from T1?

5. why are there so many input capacitors before T1?

thank you very much.

Kind Regards,

Rajan.

  • Hey Rajan,

    1. The benefit of the isolated feedback is to keep it as an isolated converter. The 28 V bus for satellites often have very noisy grounds which can interfere with signals. This creates the isolation requirement.

    2. Agreed

    3. D11 isn't really doing much, its there mostly to ensure the line stays below 6 VDC. L1 is reducing the voltage ripple.

    4. If you are talking about C21 through C28 that has to do with the output filter which is briefly gone over in 2.4.5. If you are just asking about why there is so many, I like to have 2 A per cap maximum. L1 means the capacitors before it need to be able to handle the total Iout.

    5. The input current isn't too high on this design so I likely could have gone with less here. This design ended up having small changes and being used for the EVM for the UC1843B-SP as well and our EVMs generally err on the side of caution.

    Thanks,
    Daniel
  • Hello Daniel (are you the same Daniel I spoke to about the LM117?),

    Thank you for your replies which are very helpful.

    Given that the input voltage to the primary winding can range between 20 and 40V, will the secondary always
    output +5V or will this also scale based on turns ratio and the exact input bus voltage?

    Is the large output capacitance related to the use of the UC1843?

    How would T1 change if I wanted to generate secondary voltages of +/-5V instead of just +5V?

    Kind Regards,
    Rajan.
  • Daniel,

    Is D8 a snubber?

    What's the purpose of D4, D5 and D9?

    Kind Regards,
    Rajan.
  • Hey Rajan,

    I am the same Daniel that you spoke to about the LM117. I support all power products (more or less, sometimes I get help) related to space.

    The secondary output will always be 5 V, the duty cycle changes to keep it this way.

    Is the output capacitance that large? I believe section 2.4.5 in the user's guide also talks about how I got to the total output capacitance value.

    If you wanted to generate +/- 5 V instead of just 5 V, what you would likely have to do is add another secondary winding that is similar to the current one and then simply flip the circuit (the output diode mostly). This winding would not be regulated like the 5 V rail.

    Depending on your input voltage a flybuck using the TPS50601A-SP might work for this as well (although this is something I have been working on testing.)

    D8 is a schottky diode to prevent the output voltage from going negative. This is recommended but not required.

    D9 is to rectify the feedback signal coming from T2.

    D4/5 is a Zener snubber circuit (Also called a flyback clamp). D4 can be replaced with an RC, but this circuitry is required in some fashion.

    Thanks,
    Daniel
  • Hi Daniel,

    Hope you are well.

    If one assumes Vin = 28V, efficiency is 80% and the maximum output current is 10A, can we infer that Pin = 50/0.8 =
    62.5 W and Iin = 62.5 / 28 = 2.23 A.

    Is T1 a coupled inductor to allow energy storage during the on-time and energy transfer during the off-time?

    Is the turns ratio of T1 playing a role in setting the output to +5V?

    Kind Regards,
    Rajan.
  • Hey Rajan,

    T1 is in fact a coupled inductor. It can be confusing as it is almost universally referred to as a transformer.

    You are correct on the functionality.

    The turns ratio of T1 plays a role in determining the duty cycle which in turn sets the output to 5 V.

    Thanks,

    Daniel

  • Daniel,

    What is the purpose of D2? This doesn't appear to be the normal diode one would find at the secondary side of a fly-back.

    I've asked the manufacturer of T1 to send me its datasheet as I'm trying to understand how the turns ratio is impacting the design.

    If Vin = +20V, is Vsec = +5V?
    If Vin = +40V, is Vsec = +5V?

    For a fly-back, Vout = [Vin(reflected) * D] / [1-D]
    where Vin(reflected) = Vin/turns ratio

    Kind Regards,
    Rajan.
  • Hey Rajan,

    If you would like something to get an intuitive sense of how a flyback converter works and what changes with voltage ect. I would suggest downloading something like powerstage designer in which you can change the output current ect. and look at how the signals change.
    It is fairly basic, but gives a good starting point.

    What the turns ratio of the transformer divides the switch node down.
    Thus if the switch node is switching between 0 to 20 Volts and the turns ratio is 2, this would cause the secondary side to switch between 0 and 10 V.

    If the switch node is switching from 0 to 40 Volts and the turns ratio is 2, this would cause the secondary side to switch between 0 and 20 V.

    If I want a 5 Volt output the duty cycle would look at the secondary side switch node and figure out the duty cycle from that. Thus in the case its 20 Vin they duty cycle would look at the 0 to 10 secondary side and be 50%.

    If the input was 40 V than the secondary side would see 0 to 20 V and be 25%. This corresponds with the equation you mentioned.

    You are correct the output will always be 5 Volts to be clear.
    Note the duty cycle equation oyu are using doesn't take efficiency into account and efficiency can change the duty cycle.

    Thanks,
    Daniel
  • Daniel,

    What other, cheaper options exist for the PWM control and the generation of the feedback signal? For the latter, could several op-amps be used to create the error amplifier?

    I am exploring option, voltage-mode PWM may require less overall parts.

    Kind Regards,

    Rajan.

  • Hey Rajan,

    As far as generation of the feedback signal, you can use optocouplers, but those have issues with the CTR (Current Transfer Ratio) degrading over radiation.

    The cheapest and easiest method would be simply making the topology non-isolated, in which you can simply use feedback resistors

    I don't believe voltage mode will really use less parts, because most voltage mode controllers still have an ILIMIT pin which require the same current sensing resistor current mode control uses.

    It doesn't help that most of our space grade controllers are current mode control, which can be switched to voltage mode control, but that requires generating a ramp signal to put into the CS pin.

    Voltage mode will certainly be the simplest, however unless you are using voltage feedforward mode, this will cause heavy changes to the frequency response due to changes in the input voltage which is less prevalent in current mode control.

    Thanks,

    Daniel

  • Hello Daniel,

    Hope you are well.

    Can you check the part number of the coupled inductor from Wurth please or email its datasheet.

    I asked Wurth  for a datasheet of part 750317433 but they cannot find it.

    Kind Regards,

    Rajan.

  • Hey Rajan,

    Here you go.

    I believe the reason for them not being able to find it is that I asked for it custom750317433r00 Prelim Spec Sheet.pdf

    Thanks,

    Daniel

  • Thank you Daniel, so you got one custom made based on 5V at 10A

    Have you considered Coilcraft's Hexa-Path parts which contains six windings to allow you to create a 'semi-custom' magnetic
    www.coilcraft.com/.../hexa-path.pdf
  • Hey Rajan,

    Thanks for the suggestion, I haven't heard of them before. I will consider it in the future.

    Thanks,
    Daniel
  • H Daniel,

    Hope you are well.

    The datasheet of the LM137 does not specify clearly how to preset the output magnitude, e.g. within figure 14 on page 13 of the datasheet, what is V1? Is this mean't to be -Vin?

    Should Vout = -1.25(Vin + R2/120) + (-Iadj * R2)

    If the above formula is correct, then the output magnitude is sensitive to all input voltage changes.

    Kind Regards,

    Rajan

  • Hey Rajan,

    Next time you have a question on a separate part I would ask that you start a new question.
    This helps both our tracking of issues as well as allows people who have the same question to search for it.

    I believe the datasheet means to give -Vout = -1.25*(1+R2/120)+(-IADJ*R2) as the equation.
    Note that the -1.25 is the reference voltage and 120 was R1 in the equation.

    This is EQ1 on page 8 of the LM137 datasheet.

    Thanks,
    Daniel
  • thank you Daniel,

    apologies for asking a question about lm137 in this thread, I only looked at the QML datasheet and the equation is listed in the front page of the commercial datasheet