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Negative current input to the LOG112

Jesus Vasquez
Prodigy 80 points
Other Parts Discussed in Thread: LOG112, ADS8505

Hello everybody. I recently build a system for reading a current analog signal using the LOG112 logarithmic amplifier and the 16-bit ADC ADS8505. I directly connected the current input to the first input of the LOG112 (I1) (a put a reference current on I2 as shown on FIGURE 3 of its datasheet) and the output of the LOG112 (using the internal o-amp A3 for level scaling) to the input of the resistive network of the ADS8505. The problem was that the LOG112 broke. My feeling is that when a power on the system, an initial negative current was generated and flowed through the input I1 of the LOG112. Is possible that a negative current directly applied to the input of the LOG112 breaks it?

Application Rreference: The input current source is a copper plate on a vacuum chamber that collects positive ionized particles. I’m using this system to read this current signal. Additionally, around this copper plate, there is a second metal plate, isolated from the first one and a distance more or less of 1 cm, which is held to -300V. I think that when I power on -300V power supply, some electrons passed from the second to the first plate generating a negative current. I verify the isolation of the two plates and it is ok, so I don’t think that the -300V were applied to the LOG112 input.

 

Thank you very much for your help in advance.

  • 0
    TI__Genius 12325 points

    Jesus,

    The datasheet for the LOG112 addresses negative input currents beginning on page 8 with potential application circuits in figures 6, 7 and 8.

    I suspect that the issue may be coming from the fact that your charged plate and your measurement plate form the two plates of a capacitor. Recall that the current through a capacitor is i = C (dV/dt), and therefore if you rapidly charge one plate of this capacitor to -300V, a large displacement current will be pulled through the other plate possibly damaging the LOG112 (which has an absolute maximum input current rating of +/-10mA). I recommend that you measure the capacitance between the two plates using a LCR meter or other impedance analyzer and use this value to determine a safe rate (dV/dT) at which you can charge the other plate to -300V without causing the maximum input current ratings of the LOG112 to be exceeded.

    For example, if the capacitance between the two plates is 100pF and we consider 1mA to be a "safe" input current then:

    1mA = 100pF (dV/dT)

    dV/dT should be less than 10V per microsecond. I would imagine that simply switching on a -300V supply could exceed this ramp rate or even worse if the power supply is switched on disconnected from the plate and then connected once it has reached its set voltage. If you have the ability to very slowly increase the voltage on the charged plate that may be the safest option.

    Physics is wild!

  • 0 in reply to John Caldwell
    Prodigy 80 points

    Hello John,

    thank you very much for your answer. I have made some measurement and it turns out that the capacitance is something like 0.2pF, a very small value, which will generate small current even rapidly charging one of the plate of the capacitor (a 300V change on the charge on 60ns will generate a 1mA current). Even though, I made sure to slowly increase the voltage until the final level. So, the problem is still a “mystery” for me, but I’m still looking into it.

    I would like to ask you however if you know any way of protecting the input of the LOG112 of over currents (or over voltages). I know some limit circuit based on transistor to prevent the current on a circuit to overpass a limit (using a couple of BJTs), but this limit is usually on the range of some cents of milliamps. I don’t know any circuit for limiting currents of some nano or microamps, and ever harder without introducing currents capable of interfering with measurements of some pA. So, do you have any recommendation for protecting the input of the LOG112 for possible currents over 10mA?

     

    Thanks once again for your help!

    Jesus