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SN6507: OK to use with a buck regulator?

Greg
Expert 2031 points
Part Number: SN6507
Other Parts Discussed in Thread: SN6501

Tool/software:

Greetings,

I plan on using the SN6507 in a new product for a customer.  I have some questions that I would appreciate some help on.

My application requires a 24V power input and a regulated isolated 5V output.  It appears I will be able to achieve a high level of efficiency and power output by converting 24V down to about 16.5V using the Wurth 750319696 transformer.  From 16.5V I would use a buck regulator to drop down to 5V.  The datasheet discusses the use of LDOs for regulating the output, but no mention of buck regulators.  My concern is the effect a buck regulator might have on the startup of the SN6507 circuit.  I realize that the SN6507 is often used in low EMI applications, but I'm not taking advantage of that feature.  I just like the cost and performance of the SN6507.  I am open to other isolated solutions.

Per Wurth's documentation, it appears 400 kHz achieves a high efficiency, so that is the frequency I would use.  I would let the SR pin float and use the default slew rate.  I would not use Duty Cycle correction.  I would tie UVLO to VCC and use the default value instead.  But I could be convinced to add some resistors if there is justification.  I didn't see anything in the datasheet that discusses the tradeoffs of using a higher UVLO as opposed to the default value of 2.8V.

I didn't see anything in the datasheet that discusses how to choose the value for Over Current Protection.  The chosen value of resistor enters into the equation for the soft-start time and CSS capacitor value.  The value of CSS affects the choice of output capacitance (no more than 10 times the CSS value).  The design example chooses a peak current limit of 500 mA for a load current of 200 mA.  If I want to maximize the output current capacity, what reasons are there for avoiding 1.3 A as the ILIM value?  Since the SN6507 has thermal overload protection, what harm is there is setting it to the max value?  1.3 Amps is on the curve provided by Wurth in the transformer's datasheet.  If that current was sustained, the transformer would get pretty hot!  (Over a 90-degree rise).  But I assume it would survive since it is a peak value and it is transient.

The buck regulator requires an input capacitor of 10 uF.  Can the 10 uF output capacitor of the SN6507 circuit be shared with the buck regulator?  The 10 uF output capacitance dictates a CSS capacitance of at least 1 uF in order to meet the 10x requirement.  This results in a Soft-Start time of about 4 ms.  How does this time affect the Soft-Start time of the buck regulator?  Perhaps this is where UVLO settings for both the SN6507 and the buck regulator should be considered?  Should the soft-start time of the buck regulator be longer than the startup time of the SN6507 to prevent it from being starved?  On the other hand, I could make an argument for the start time of the buck regulator being shorter.  It would simply regulate an input voltage that is relatively slowly rising.

If a buck regulator is not going to work, could I use an LDO in between the SN6507 circuit and the buck regulator?

Thank you for your help.

Kind regards,

Greg

  • 0
    TI__Mastermind 32975 points

    The buck converter is in series with the SN6507 output. The total efficiency will be low, the buck converter has a pulsed input current which means a large input cap for the buck (output cap for the SN6507). I would recommend to adjust the transformer turns ratio such that the output voltage is ~0.5V-1V higher than the desired regulated output voltage then use an output LDO. 

    Steve

  • 0 in reply to Steven Mappus1
    Expert 2031 points

    Thanks Steve.  You said that the total efficiency would be low, but it seems an LDO at lower output voltages (e.g. 5V) would be less efficient than a buck.  I would want about 1 volt of headroom to sit comfortably above the dropout voltage over varying loads, temperature, and tolerances.  For a 5V LDO, 1 V out of 6 V would be wasted in the LDO.  A 90% efficient buck would be more efficient than that.  Ultimate efficiency isn't an issue since my application isn't battery-powered.

    The SN6507 datasheet doesn't have as many graphs as the SN6501 or SN6505 datasheets, so it is hard to know what the actual power outputs and efficiencies are.  I looked at the transformer datasheets from Wurth, which show typical application values for each transformer in their series that is meant to work with the SN6507.  It looks like they chose a temperature rise of about 44 degrees C as the criteria for the stated current.  The two transformers that drop from 12V or 24V down to 5V are only capable of delivering 3.5 to 4 Watts, compared to the others that are capable of 6 Watts.  Given that data, I would want to convert to a higher voltage, and then use a buck converter to regulate it at 5V.  I realize that this is based on Wurth's transformers designed for the SN6507, so other transformers might work out better.

    If I treat the SN6507 circuit and the buck circuit as independent modules, it seems this should work out.  The SN6507 requires 10 uF on its output, and the buck circuit also requires 10 uF on its input.  If those capacitances satisfy each circuit, the two combined should be enough.  The combined 10 uF capacitors would require a larger CSS capacitor.  But a slower rise time probably isn't a problem.  2 uF of CSS would result in a soft start time of about 8 ms.

    The Output Switch Current of the SN6507 is specified as 0.4 A for VCC between 3 and 6 V, and 0.5 A between 6 and 36V.  After looking at Wurth's data, perhaps the 24V to 12V transformer would be the best option if followed up by a buck regulator.  I was thinking about the UVLO question, and it seems I could set the buck regulator's UVLO to a voltage greater than 5V, for example 8V.  The buck wouldn't start up until the SN6507's output is above 8V.  The buck has a soft-start, so I wouldn't be losing a soft-start functionality for the complete power supply, though the entire startup would be a little longer.  If the SN6507's startup is cut short once the buck regular begins to draw current, would that be a problem?  It wouldn't happen until it reaches 8V, and the impact would be gradual, due to the soft-start of the buck regulator.

    I didn't see anything in the documentation that describes how to choose the peak current limit value.  What are the drawbacks to setting it to a limit that is higher than is needed?  The design example specifies a limit of 500 mA and a load current of 200 mA.  What is the peak current limit protecting?  Is it the switch's 500 mA recommended max?  Or the transformer's primary winding?  Table 8-3 in the datasheet lists recommended resistor values that extend the peak limit all of the way to 1.3 A, so it probably isn't the SN6507's output.

    Thanks,

    Greg

  • +1 in reply to Greg
    TI__Mastermind 32975 points

    Don't worry about the SN6507 start up since it is not closed loop. Once the SN6507 begins the soft start cycle it continues until the Css is fully charged. Your output diodes have a max rated current, transformer has max power capability, the output inductor (when used for DC function) has max Isat...these are the main power stage elements protected by the SN6507 OCP setting.

    Steve