• State
  • Locked Locked
  • Replies 5 replies
  • Subscribers 46 subscribers
  • Views 3120 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.

INA116 Buffer Amplifier

Arvo Palm-Leis
Prodigy 80 points
Other Parts Discussed in Thread: INA116

We are trying to build a 1:1 very high input impedance/low output impedance buffer amp for DC signals based on the INA116.  We built the circuit from the datasheet at http://focus.ti.com/lit/ds/symlink/ina116.pdf using a ±5V power supply.  When we check the input impedance we get ~4.5M, using a DMM in the resistance mode.  We're expecting a lot higher than that, hoping to be in the gigaohm range.

Should the INA116 input impedance appear higher than 4.5M?

  • 0
    Guru 21975 points

    Arvo,

    The input resistance of the INA116 is approximately 10^15 ohms. You cannot make a measurement of the input resistance using a conventional ohmmeter. We measure input resistance by measuring input bias current at various input voltages. The input resistance is the slope of the curve of input bias current vs. input voltage. Measuring input bias current in the femtoamp range is tricky business.

    In any event, measurements with a DMM may force current through protection diodes yielding totally bogus results. I don't recommend it.

    Regards, Bruce.

  • 0 in reply to Bruce Trump
    Prodigy 80 points

    OK, thanks, Bruce.  Femtoamp currents are tricky, I'm sure.

    What is the property of the amplifier input called, when a source is connected to the inputs, that is the ratio of the source's voltage to the amount of current the amplifier draws from that source?  Whatever that property is called is what we need to get up to >1G.

    Next time we'll apply a known voltage and measure the current through a shunt to see how much is going through the amplifier.

  • 0 in reply to Arvo Palm-Leis
    Guru 21975 points

    The behavior you describe is the input resistance and is somewhere in the area of 10^15 ohms, one-million times larger than your requirement. The load on your signal source really tends to be more like a tiny constant current rather than a resistance. See the curve below. In any event, the load on your input source is EXTREMELY small.

    Regard, Bruce.

  • 0 in reply to Bruce Trump
    Prodigy 80 points

    Thanks, Bruce.

    I put the INA116 on a solderless breadboard, used two 9V batteries for power, and did not add any pots, caps, or trim circuitry.  It followed voltage well, and when I put 8V in and used a 10M ohm shunt, there was <1 mV on the shunt, so our input resistance is >80G.  We'll be giving our previous circuit a thorough going over.

  • 0 in reply to Arvo Palm-Leis
    Prodigy 80 points

    Today I got another look at the circuit.  With the ±9 V above, the impedence is very high, more than I can measure.  On the circuit with the problem, we are using a ±5V power supply, and the impedance is around 4 M.  With no power the impedance is right at 2.5 M.  Am I underpowering the INA116?