I am looking at the best method to control an LDO enable pin using a negative voltage current limit circuit. The LDO is disabled when the enable pin is 0 to 0.4V. The LDO is enabled when the enable pin is 2 to 10V (max).

The current limit circuit that i am using is below. R3 models the load. Under normal condition the load current is 0.2mA or less. Under fault conditions the load current is 2mA.

enter image description here

The basic idea is that the LDO is disabled (LDO enable pin 0 to 0.4V) when the load current reaches 2mA. Under 2mA the LDO is enabled (LDO enable pin 2 to 10V (max)).

Apart from the -32V input there is also 3.3V.

Look forward to your replies.

Regards Joe

  • \$\begingroup\$ Which LDO, and what is the maximum supply voltage you will give it? \$\endgroup\$ – Bruce Abbott Oct 14 '16 at 18:00
  • 1
    \$\begingroup\$ @JMiller This is a Q&A site, not a discussion site. It may be helpful to the readers to post a question for them to answer, its a little unclear what your question post is about. electronics.stackexchange.com/help/how-to-ask \$\endgroup\$ – Voltage Spike Oct 14 '16 at 19:27
  • \$\begingroup\$ The best design is one that meets or exceeds all the specs. What are your specs? ** Lowest cost, lowest dropout? smallest area, fewest parts? easiest to undestand, uses parts in your local shop? cheapest in 1k volume? OR do you have real numbers for a spec for dropout, current, voltage range \$\endgroup\$ – Tony Stewart EE75 Oct 15 '16 at 0:53

enter image description here Here is one way.

To get an idea how this works, assume the betas are very large. When the current limit (of around 2mA) is reached, Q1 is held off from passing more current. Your current limit circuit then functions with R1 as the equivalent resistance.

With R1 being 100K, with more than 0.006mA above the current limit, >0.6V develops across R1 to turn on Q10. Q11 and D1 prevents LDO_EN from dropping below 0V. Q11 insulate Q10 from seeing V_PLUS, allowing Q10 to work under 32V.

The current limit 2mA is low. And the transistor betas are unfortunately not infinite, so you have to juggle the reistances and currents to get limits acceptable to you.


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