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LM258: LM258 Output phase abnormal analysis

Part Number: LM258

Dear TI expert,

Pls check the output voltage(CH1) of a comparator LM258(Pin1), the voltage pulse width is small and small with time.

Background information: I use LM258 as a comparator with signle voltage power supply, +12.6V (V-=0, V+=12.6V).  IN- connected to 8V source.

IN+ connected to a signal varying from -4V to +5.6V(Not CH2 below but similar as it, the negative side is only -4V ),  there is a 10MΩ resistor between IN+ and input signal.
The IN+ input is around -0.45V and clamped at negative half-cycle.

Analysis: In the datasheet,   either input more than 0.3 V below V– then input current should be limited to 1 mA and the output phase is undefined.

I don't know if the negative input voltage casues this output abnormal or other reason. Pls help me to check.

If like this, why the abnormal happened in the positive half-cycle? I assume the output voltage should be Zero in the positive half-cycle coz 5.6V< 8V.

BTW: I already used IN4148 in the front of 10M resistor, so the negative voltage is not happened again. But I am not sure the abnormal is due to the -0.45V voltage.

Thanks!

  • Hi Hong,

    the common mode input voltage range is what counts. See section 7.3 of datasheet. If this is violated, the OPAmp will not work properly, which exactly is what you observe. So avoid any negative input voltage.

    The current limting thingy -if the input voltage is going negative- is only a measure to protect the OPAmp against destruction. But this is no measure to make the OPAmp work properly. Again, avoid any negative going input voltage Relaxed

    Kai

  • Hi Kai, 

    Thanks for your reply. Could you pls check if the output voltage phase is normal or not when 0V  > IN+ > - 0.3V? IN- is around +8V.

    Attached is the schematic.

  • Hi Hong,

    again, a negative input voltage is not allowed:

    "(V-)" here means the negative supply rail which is 0V in your case.

    Do you want to build a zero crossing detector? Here's a useful scheme:

    Kai

  • Hi Kai,

    Thanks for your suggestion. I also find this in DS, but there is no information about 0> IN+ > -0.3V when IN- in the normal CM range and all input current limited to1mA.

    I need the definition about this violation, and you although defined the more serious working condition when IN+ lower than -0.3V.

    Some reason: the leakage current (Ir) of D2 is much higher in high ambient temperature, Vr = 12.3V.

    I observed the input voltage(in front of 10M) could be -0.18V when Tc=90℃ of D2.

    I assume there is no abnormal status when 0>IN+ >-0.3V and IN- in the normal CM voltage range, pls double check.

  • Hong,

    What is the purpose of this circuit? Please explain with sentences or math equations.

    X = input, Y = output,  If  X>8.4V then Y = 11V else Y = 0V ; domain (X) is [-14V to +4V]

    I think now is the time to completely start over.

  • HI Ron,

    We can discuss here together with you, thanks for your support.

    This is an signal detecting circuit for EV charging pile, the output +6V/-12V or +9V/-12V 1kHz PWM signal is a standard voltage.

    Input D2 is used to isolate the minus voltage.

    This comparator circuit is designed to detect the peak volage of the waveform.

    X = input, Y = output,  If  X>8.4V then Y = 11V else Y = 0V ; domain (X) is [-0.4V to +12V], assume Vcc=12.6V.

    D2 almost delete all the minus voltage, only the leakage current which will lead -0.4V input voltage.

  • Hong,

    This outputs low when input is [1kHz -12V, +6V] and outputs high when input is [1kHz -12V, +9V] 

    Is the input clock continuous? 

  • Yes, continuous. You can only test one case here, +9V/-12V, then input is around +9V/-0.4V behind D2. When EV charging gun is not inserted, input is +12V constant voltage. We have a group of comparator circuits behind D2, above is only one comparing point with 8.4V. No need to test all cases.

  • Hong,

    I see three cases so the minimum solution is two comparators. Wouldn't be nice if these two comparators had a DC output instead of a pulsing output?

    Your implementation is still too negative. You can add pull up resistor to keep the lowest level zero or slightly positive. 

  • Hi Ron,

    Thanks for your sugesstion. Currently, it is not good to change the circuit coz it is under mass production.

    My concern is only the minus voltage on IN+,  0>IN+>-0.3V and IN- =8.4V and all current limited to 1mA, the output is Zero. Could you confirm this?

    BTW: I will optimize the circuit in next version with your suggestion.

    Have a good weekend.

  • Hong, with less than 1mA the op amp will not be damaged, but the output could be incorrect.

    Phase reversal requires some amount of current (well under 1mA), So your R3 helps reduce input current and normal IIB add some voltage.

    If it is already in mass production and not broken, then don't change it now. If it is broken, then make a small change as to not break something else.

    Is the mass production going well?

  • I can't confirm -0.3V will also be fine across temperature . I can say that high temperature makes the phase reversal happen at lower voltage. However D1  voltage should also drop with high temperature. D2 leakage will increase with high temperature.