Let's say I have an INA134 (http://www.ti.com/product/ina134) from Texas Instruments in my design. I want to take the output of that IC and feed it into a resistor network which can have a flexible input resistance (from 100 ohm to 100k ohm).

What is the reason that the output impedance of the IC is not specified in the datasheet? It's based on a simple opamp circuit after all.

Does it make sense to follow the INA134 by a OPA134 to allow for more drive capability?


1 Answer 1


The output drive capability is defined in a few places in the datasheet, but not guaranteed except at 2K load. The below graph is based on +/-18V supplies.


As you can see you can get 20mA easily out of it, with 15V swing. With a 100R load, the output capability would be a couple of volts or so safely.

I don't see why you'd want to follow this part with another such as the OPA134 that has similar or less current capability.


Keep in mind that the closed-loop output impedance on both amplifiers will appear very low, up until the point where you hit the current limit (or when thermal protection kicks in).

  • \$\begingroup\$ I know that the OPA134 has a pretty small output impedance so I thought I'd use that as a voltage follower. My main question here is: How do I deduce the output impedance from the datasheet of the INA134? \$\endgroup\$
    – apriori
    Sep 10, 2014 at 21:37
  • \$\begingroup\$ You can't really. The typical error with a 2K load is 0.02%, so the closed-loop impedance should be < 0.4 ohm, but that error includes resistor tolerance. Close enough for audio work, in most cases. \$\endgroup\$ Sep 10, 2014 at 23:38

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