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OPA561 gets hot when e/s pin is pulled up thru 402k to +12v

Other Parts Discussed in Thread: OPA561

Hi Mark,

Upon review of your reply including the specifications for the maximum voltage to the E/S pin

I can see your point. So I then added a 100k ohm resistor from pin 8 (E/S) to ground which forms

a voltage divider (402k to +V (+12V), 100k to -V (GND)) which should allow for +2.39V at pin 8.

Instead i am reading 2.89V at this point in reference to -V (GND) (pin 5) and the resistors are +/- 1%. Besides that the OPA561

is still getting quite hot. It seems to me that since the threshold above -V for E/S to Enable the

device is > 2V it should safely enable it but without this devestating side affect.

I am thinking that just maybe I could have damaged the device by having +12v at this pin in the first place?

Do you think that is why it could be getting hot? Then why is it that when just pulling it down to -V (GND)

I do not get this heating affect and the correct output voltage is present?

Another question. Why do you suppose a 402k (specific very high value) is reccomended as a pullup rather

than lets say 10k or 100k which is more common? Is there anything wrong with my resistor voltage divider approach?

Thank you

Les

 

 

 

 

 

  • Les,

    At this point I am unsure anymore what your power supply voltages are; if you use (V+)=+12V and (V-)=0V, you should NOT connect E/S pin to GND thru 100k resistor because Enable pin also draw current (~20uA) and that is the reason why your voltage divider calculation gives you erronious value  - 100k connection to GND only works if (v-) is at -5V.  402k resistor between (V+) and E/S pin, as shown in the circuit diagram below, permanently enable OPA561 regardless of supplies used as long as the total power supply voltage, (V+) - (V-), is between 7V to 15V. 

      

    If this does not work, it is most likely that the part has been damaged.  Connecting E/S directly to 12V would damage the part instantly but connecting it thru 402k resistor assures enough voltage drop across it so the E/S pin is within the voltage range of 2V to 5V on 0V to 12V power supplies.

  • Hi Marek,

    I am definitely connecting +V to +12v and -V to GND (0v) so there should be no confusion on this.

    I was originally connecting +12v thru a 402k resistor to the E/S pin.

    In your initial email you advised me that this superceded the +5v maximum rating of the E/S pin

    and you sent me a copy of that specification.

    So then I used a voltage divider instead to affect a 2.39v at the E/S pin thru a voltage divider of 402k and 100k

    from +V to -V with the center node connected to the E/S pin. Of course as I mentioned I read a higher voltage than expected which you attributed to the internal current drain of about 20ua. If you multiply 20ua x 100k = 2v. Hmmm.

    Does not sound right. Putting that peripheral aside I don't see a problem.

    However, you surprised me when you said

    "If this does not work, it is most likely that the part has been damaged.  Connecting E/S directly to 12V would damage the part instantly but connecting it thru 402k resistor assures enough voltage drop across it so the E/S pin is within the voltage range of 2V to 5V on 0V to 12V power supplies."

     

    I never connected the E/S pin directly to +12v. Only thru a 402k resistor. I have since measured the voltage at

    the E/S pin with this connection (no 100k pulldown) and it read 5.3V. However this is indeed over the 5v maximum spec. So, is this bad or good? The part still gets hot.

    Upon further study I noticed that although the Basic connection shows the 402k pullup resistor to V+.

    In single supply operation which I am using thedatasheet range is limited to +7v to +15V. However the Output Enable/Status specification shows  a maximum VE/S High of (V-) +5V.  This spec seems to imply that

    +V cannot be connected to the E/S pin in single supply operation since the minimum single supply voltage is listed as +7V  (> 5v)!

    The data sheet is confusing. Can I ssume then that according to your latest post that if I use a 402k

    ohm series resistor pullup that I need not worry about this since it limits the voltage to 5v (even though I am reading 5.3v  which is over the spec???? Even so the thermal problem is not solved.

    Another question: The Flag pin is mentioned in the datasheet but no pin number is assigned anywhere.

    I assume that it is the thermal pad which must be connected to -V which is the case. But why do they then allude to it as a "FLAG" pin without any pin number? Why would you call a thermal pad "FLAG" anyway?

    Sorry for any  confusion.

    Les

     

    Hi Les,

    Before I answer your questions, please do not initiate a new forum thread each time you write a response-you need to go back to the original post by clicking on the link at the bottom of the email (here) and  reply from within the forum post.

    Here are my answers:

    I was originally connecting +12v thru a 402k resistor to the E/S pin.

    As long as (V-) was at 0V, this alone should never damage the part

    In your initial email you advised me that this superceded the +5v maximum rating of the E/S pin

    and you sent me a copy of that specification.

    So then I used a voltage divider instead to affect a 2.39v at the E/S pin thru a voltage divider of 402k and 100k

    from +V to -V with the center node connected to the E/S pin. Of course as I mentioned I read a higher voltage than expected which you attributed to the internal current drain of about 20ua. If you multiply 20ua x 100k = 2v. Hmmm.

    Does not sound right. Putting that peripheral aside I don't see a problem.

    The E/S pin sinks about 20uA current (IE/S HIGH) on Vs=15V in an enable mode and 0.1uA (IE/S LOW) in a shutdown (see PDS table below) - IE/S High current should be somewhat lower with lower supply voltage.

    However, you surprised me when you said

    "If this does not work, it is most likely that the part has been damaged.  Connecting E/S directly to 12V would damage the part instantly but connecting it thru 402k resistor assures enough voltage drop across it so the E/S pin is within the voltage range of 2V to 5V on 0V to 12V power supplies."

     I never connected the E/S pin directly to +12v. Only thru a 402k resistor. I have since measured the voltage at the E/S pin with this connection (no 100k pulldown) and it read 5.3V. However this is indeed over the 5v maximum spec. So, is this bad or good? The part still gets hot.

    5.3V on E/S pin would NOT damage the part so the part must have been previously damaged.  Did you connect Ilim (pin 4) - leaving it open could damage the part (see below).

     

    Btw, you may calculate here the IE/S HIGH current: (12V-5.3V)/402k=16.7uA (as expected it is lower than 20uA for Vs=15V)

    Upon further study I noticed that although the Basic connection shows the 402k pullup resistor to V+.

    In single supply operation which I am using thedatasheet range is limited to +7v to +15V. However the Output Enable/Status specification shows  a maximum VE/S High of (V-) +5V.  This spec seems to imply that +V cannot be connected to the E/S pin in single supply operation since the minimum single supply voltage is listed as +7V  (> 5v)!

    As I said before, you may never directly connect E/S pin to (V+) and connecting it thru a resistor results in a voltage drop across it so E/S is never at (V+) potential.  Thus, for example, for Vs=7V IE/S HIGH=10uA and thus connecting E/S thru 402k resistor would result in E/S pin at perhaps: 7V-10uA*402K=~3V (within specified 2V to 5V).

    The data sheet is confusing. Can I ssume then that according to your latest post that if I use a 402kohm series resistor pullup that I need not worry about this since it limits the voltage to 5v (even though I am reading 5.3v  which is over the spec???? Even so the thermal problem is not solved.

    Yes, this should be fine even if E/S is a little above 5V.  Thermal problem is indicative of a damaged part; use a new part and connect Ilim to (V-) for the max output current of 1.2A - do NOT float Ilim (pin 4).

    Another question: The Flag pin is mentioned in the datasheet but no pin number is assigned anywhere.I assume that it is the thermal pad which must be connected to -V which is the case. But why do they then allude to it as a "FLAG" pin without any pin number? Why would you call a thermal pad "FLAG" anyway?

    Flag is the thermal pad and thus need to be connected to (V-). 

    E/S pin 8 provides enable/disable and thermal shutdown function - see below.