I'm having some problem comprehending the internal functions of op-amps and I'd appreciate any insight in to why my circuit behaves the following way:

I'm working with TI's OPA548 op-amp (PDF datasheet link) in a single-supply non-inverting voltage-follower configuration as shown here:


simulate this circuit – Schematic created using CircuitLab

The datasheet claims "In single-supply operation, the input common-mode range extends below ground." and the table in section 7.5 shows input voltage common mode range to be at least (V-)-0.1V.

For Vin > 0.45V the circuit works fine and the output tracks the input as expected. However, the Vout node never goes below 0.45V, even though I bring Vin to ground. It does not seem to matter whether I have R1 installed or not.

I do get that the op-amp datasheet does not claim rail-to-rail output but figure 13 on page 10 seems to indicate that for miniscule current the negative output should swing all the way to the rail.

My thought is that since the other side of R1 is grounded I'm not really asking the op-amp to pull anything down to the ground rail, I'm just asking it to stop supplying current through the top-portion of the output stage. Since the input range extends below ground I feel the op-amp should be able to "see" that its inverting input is at a higher voltage than its non-inverting input (when it's at ground) and cut off the top transistor.

What is it that I'm missing here? Is there any good way for me to remedy this?

The circuit above I've built on a breadboard but I have the exact same problem with TI's OPA548EVM evaluation module.

Edit: I even tried swapping R1 for a 1Ω resistor, but the output was still 0.27V above the op-amp's negative power suppply rail (ground). This presumably means that the top output transistor is supplying 270mA even though I'm asking the op-amp to lower the output voltage.

I also tried raising V1 to 60V to match the datasheet's Vs = +/-30V condition (should look identical to the op-amp). The result is the same (the voltage difference actually increased to 0.47V).

  • \$\begingroup\$ The op amp output stage transistors will have some saturation voltage and this is what you see. e2e.ti.com/blogs_/archives/b/thesignal/archive/2012/05/08/… \$\endgroup\$
    – G36
    Jun 11, 2017 at 14:12
  • \$\begingroup\$ Figure 13 on page 10 is explicitly stated to be valid for power supplies of +/- 30 volts. \$\endgroup\$ Jun 11, 2017 at 14:23
  • \$\begingroup\$ A 3A output stage is going to have several mA of quiescent bias that the low side driver has to sink to pull the output low. Try a much lower R value say 100 ohms, and you can measure the output short-circuit current with a multimeter for small vin \$\endgroup\$
    – sstobbe
    Jun 11, 2017 at 14:38
  • 1
    \$\begingroup\$ @sstobbe - Please read the figure referenced in the OP. What you say seems reasonable, but it is wrong. \$\endgroup\$ Jun 11, 2017 at 19:02
  • \$\begingroup\$ @WhatRoughBeast - Do you mean that if I were to supply +/-30V and then connect the load resistor between op-amp output and -30V that the output would be able to swing down to -30V? That would be identical (to the op-amp) as setting V1=60V in my circuit, which I just tested with the same result (output node never got closer than 0.4V above the op-amp's negative supply, despite not having to supply any current). \$\endgroup\$
    – GummiV
    Jun 11, 2017 at 21:06


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