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LM317 current issues

Madhuri Akkenepalli
Expert 1215 points
Other Parts Discussed in Thread: LM317, LM317L-N

The following are questions on the LM317 from my customer. Please help answer these questions.
We are experiencing problems with using the TI/National power supply LM317MDCY in our Powersafe board. The board (see attached schematics) has a block of 12 470-uF 10-V low-ESR tantalum capacitors connected in parallel; their total capacitance is 5600 uF. When the board power (12V) turns on, these capacitors immediately start charging  from the LM317MDCY power supply with the output voltage of 5V (through some diode that prevents discharging when the board turns off). There is no current limiter in the circuit, so the current through the capacitors is limited only by the current protection circuit of LM317MDCY that, as specified in the data sheet, limits the current at 1A level.
 
When we put the current probe we have (unfortunately, with only 100-kHz bandwidth) around the wire that connects the diode to the capacitors, we observe a sharp peak of the current (see the picture above) in the very beginning of the charging process. Because of the narrow bandwidth of the current probe, we cannot really say what the amplitude is; we still see the initial current of 1.5 A that is much higher than the power supply current limit. It looks like the current limit circuit of the LM317MDCY does not immediately turn on. In some very rare cases, we observe the capacitors blowing off, probably because of this peak.
 
Can you answer the questions below, please?
 

1.       Is it safe for the power supply to work the way I have described above (in almost the short-circuit mode in the very beginning of the charging process)?

2.       Are we right to suppose that the initial current peak appears because of some delay in the LM317MDCY current protection circuit? Is there any way to eliminate this delay?

3.       How high can be the real value of the initial current (measured by the wide-bandwidth current probe we do not have)?

4.       Is there any simple way to introduce the “soft-start” in the LM317MDCY power supply (or in any other TI LDO power supply with similar characteristics)?

5.       As a temporary solution, we tried to limit the charge current by a small-value resistor in series with the diode. This totally eliminates the initial current peak but substantially increases the charging time. Is this a good solution of the problem, or do you have a better one?

31473 - Kinder care

  • 0
    Expert 1215 points

    Please find the schematic and the screen capture attached here.

    Current trough D7 of the Powersafe board.pdfLM317 charger.pdf

  • 0 in reply to Madhuri Akkenepalli
    TI__Mastermind 29655 points
    Hi Madhuri,

    This post was originally placed in the High Speed Interface Forum. Relocating to the Linear Regulators Forum for better support.

    Michael
  • 0 in reply to Michael Lu (Santa Clara)
    Expert 1215 points

    Thanks Michael


    Please advise on your feedback on the questions.

    thanks for your support.

    regards

    Madhuri

  • 0
    TI__Intellectual 2010 points

    Madhuri,

    The over current circuitry normally kicks in once the OC condition occurs. There is most likely a delay time for the over current circuit to initiate. If a customer doesn't want to use a sense resistor, there is way to slow turn on the output. This will avoid any inrush current. A PNP transistor combined with a capacitor at the output will help the circuitry to slowly start supplying voltage.The circuit shown below is from 317N datasheet. It's very effective for slow turn on. 

     

  • 0 in reply to Kaisar Ali
    TI__Mastermind 22610 points
    No matter what, reducing the charge current will extend the charge time. Simple math.
    For Figure 20 at power-on (t=0), the Vout needs to get to up about 2V before the soft-start circuit kicks in. Rise time when Vout > 2V will be slowed. Since the output capacitance appears as dead-short at power-on there will will still be in-rush, but duration will be limited/reduced.
    See E2E Thread ID: 414351 "LM317L-N Slow Turn-on Time and Equations"
  • 0
    TI__Mastermind 22610 points

    In Q5 you said "... a small-value resistor in series with the diode ...", how ~small~ is ~a small value~ ?

    Yes, any resistor at the diode D7, after the regulator output, would reduce the charge time to RC time constants, and that could become a long time.

    What is the maxium allowed charge time?

    I'd suggest a combination of the Battery-Charger Circuit with Rs set to 1.2 to 2.4 ohms (with Rtop= 243 ohms and Rbottom=768 ohms for 5.25V), PLUS a series R at the LM317 input of about 4 ohms to 5 ohms.

    Are the R values shown in the 'LM317charger.pdf'  (R32=240, and  R31=75) correct? Schematic says "5.25V", but I calculate Vout=1.64V with those values.

  • 0 in reply to Donald Jones
    Expert 1215 points

    Hi Donald, Please check the following response to your questions. Please let me know your feedback.

    Above all, the customer needs the recommended values  of the resistors and capacitors for the output voltage of 5.25V and the soft-start time of about 5-10 ms

    In Q5 you said "... a small-value resistor in series with the diode ...", how ~small~ is ~a small value~ ?

    The “small value” is 2.43 Ohm.

    Yes, any resistor at the diode D7, after the regulator output, would reduce the charge time to RC time constants, and that could become a long time.

    What is the maximum allowed charge time?

    There is no pre-set limit to charging time; with 2.43 Ohm resistor, it increases from 26 ms to about 100 ms that is also quite acceptable.

    I'd suggest a combination of the Battery-Charger Circuit with Rs set to 1.2 to 2.4 ohms (with Rtop= 243 ohms and Rbottom=768 ohms for 5.25V), PLUS a series R at the LM317 input of about 4 ohms to 5 ohms.

    Are the R values shown in the 'LM317charger.pdf'  (R32=240, and  R31=75) correct? Schematic says "5.25V", but I calculate Vout=1.64V with those values.

    The values are correct, but we forgot about some voltage drop on the diode; therefore, actual voltage is just below 5V.

  • 0 in reply to Madhuri Akkenepalli
    Expert 1215 points
    Greetings E2E team. Please advise if you can help asap with the component suggestions to the customer requirements in the post.
    Appreciate your help.
    regards,
    Madhuri
  • 0 in reply to Madhuri Akkenepalli
    TI__Mastermind 22610 points

    To charge Cout=5600uF from 0V to 5V in 100ms requires an average current of 280mA.

    You can insert an additional LM317, configured as a Current Limiter, at the circuit input. R1 would be equal to Vref / 280mA, or about 4.5 ohms.

    If the 2.43 Ohm series resistor gives satisfactory behavior, then that would be a solution. RC time constant is 13.6ms, and 7 time constants (i.e. 99.9%) is 95ms. Peak current at t=0 would be 5V/2.43 ohms = 2A

    You could increase the sereis R to 5 ohms which would cut the peak current to 1A, but would extend the time to charge to 99.9% out to 196ms.