• State
  • Locked Locked
  • Replies 3 replies
  • Subscribers 62 subscribers
  • Views 344 views
  • Users 0 members are here
Support feedback
Options
Options
Similar topics

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.

LM9036 with too high start up current

Martin74
Prodigy 60 points
Other Parts Discussed in Thread: LM9036, LM2936, LM334

Hi,

I want to use the LM9036 with a series resistor (variable between 1k and 10k) in the supply line.

For example:

24V --- [3,3k] ---- Vin[LM9036]          I have Cin=100nF  and  Cout=10µF.  No additional load. Later there will be a load of 100 to 150µA

If I turn on the 24V voltage, a current of 6mA flows into the regulator. The voltage at Vin breaks down to 2,8V. This situation stays  until I bypass the resistor a very short time. Then the supply current to Vin reduces to 15µA.

 

The same with series resistor 10k:  A Current of 2,2mA flows when I turn on the 24V, until I bypass the resistor.

Which regulators with Vinmax=40V  and a quiescent current of  max 100µA can start without problems with a series resistor of 10k (with Load max 150µA) ?

It doesn´t need to be a low drop regulator, any other is fine, too.

In the later design the regulator will be switched on and off about once a second.

I have not testet this with the LM2936. Makes it sense to buy/sample this, or will it have the same behavior?

Are there any other regulator that can solve this problem?

Thank you

Martin

 

 

 

 

  • 0
    TI__Mastermind 22610 points

    Sounds like you are trapped by the quiescent-current caret. When LM9036 PNP pass element is in saturation mode, as it would be at power-up, there is higher than normal current used to drive the base of the pass element.

    I would guess that LM9036 would need about 200mA for less than 1ms at start-up to get the 10uF output capacitor charged to regulated voltage.

    That would put a series resistor max value closer to 90 Ohms. ((24V - 5.5V) / 200mA) = 92.5 Ohms.

    Why would you want, or need, such a high value (k ohms) series resistor on the input voltage?

    As for LM9036 vs LM2936. They are the same design, with 99% of the die being the same, The only difference is the LM9036 is the ~value~ version where the bandgap and output voltage are not trimmed at test. Plus some other minor spec variations.

  • 0 in reply to Donald Jones
    Prodigy 60 points

    Hi,

    I want to design a open collector output for an industrial device with a check-back indication. The customer puts a voltage from 12 to 40 V to an external pull-up resistor from 1k up to 10k at the collector.

    The output need to be galvanic isolated from the rest of the circuit. I want to have  a check-back indication, if the output works. If the open collector is aktive, the voltage is lower than 1V and the feedback optocoupler is not activated. If there is any voltage higher than appr. 3V then the feedback optocoupler is powered by a constant voltage from the regulator and the rest of the circuit gets the feedback, that the output works.

    The current consumption of the additional feedback need to be smaller than approx  200µA, to keep the voltage drop on the pull up resistor lower than 1 to 2V

    Meanwhile, I realised that the soulution with the linear regulator is not the best solution. I think you will agree, that it would be much easier and cheaper to use a simple adjustable current source for example the LM334 to power the feedback optocoupler.

    I will try this next.

     

  • 0 in reply to Martin74
    TI__Mastermind 22610 points

    Should be easier to work with.