I am trying to simulate frequency response of TLV07 in non-inverting configuration for gain 10 as shown below: enter image description here

The frequency response shows peaking as shown below: enter image description here

It is worse for smaller values of resistors like 1 and 10: enter image description here

The response is considerably better for 1k and 10k: enter image description here

What is the reason for this? In my understanding, peaking is generally seen at higher values of resistances because of influence of parasitic capacitors but here I am observing the opposite behaviour.

  • 2
    \$\begingroup\$ Why are you performing a simulation on a circuit that cannot work properly in real life with the low resistance values you have chosen? Do you expect a simulation model to exactly mimic a real IC in all abnormal configurations including abuse? All bets are off when you do this. \$\endgroup\$
    – Andy aka
    Jul 4, 2021 at 10:03
  • \$\begingroup\$ I just wanted to understand the effect of resistor values on frequency response out of curiosity. Obviously in real life I would use higher values of resistances. \$\endgroup\$ Jul 4, 2021 at 10:24
  • 1
    \$\begingroup\$ Then you are using the wrong tool. \$\endgroup\$
    – Andy aka
    Jul 4, 2021 at 10:35
  • \$\begingroup\$ What's your simulator? Where's the opamp component model taken from? If AC analysis really can take into the account slowness effects caused by saturation inside components it would be useful and I will try the same simulator. Saturation makes transistors slow and the feedback will have phase lag in higher frequencies, it does not reduce gain as much as in lower frequencies. \$\endgroup\$
    – user136077
    Jul 4, 2021 at 10:51
  • 1
    \$\begingroup\$ Models are only "reasonably" accurate when they are used in a way that a valid real circuit can be used. You are not doing this. \$\endgroup\$
    – Andy aka
    Jul 4, 2021 at 11:41

2 Answers 2


It is due to the opamp's open-loop output impedance. As you can see in Figure 21 of the datasheet, the open-loop output impedance is inductive from about 100Hz to 10kHz. I modeled it using this circuit and got the same results.

* sham opamp
* GBW is 1MHz
* open-loop gain at DC is 240dB
* open-loop output resistance is 900Ω // 2mH
G1 0 1 vinp vinn 1
C1 1 0 159n
R1 1 0 1T
E1 vout1 0 1 0 1
R2 vout1 vout 900
L1 vout1 vout 2m

V1 vinp 0 DC 0 AC 1

R3 vinn vout 10
R4 vinn 0 1

An inductive output impedance and the load resistance form another pole in the loop gain. We usually want the loop gain to have only one pole before its magnitude reaches 1. Here the pole brought by the opamp output impedance is at 800Hz, which means the loop gain will almost have a phase shift of 180° when its magnitude reaches 1, which means ringing is likely.

As of why the open loop output impedance is inductive at that frequency. This is a rail-to-rail output opamp. Its final stage is common-source. It has a large output impedance. But when people use opamps they usually want a small output impedance. Probably the guy who designed it added feedback in the output stage to lower the open-loop output impedance. However, the feedback works better at low frequencies, and thus the open-loop output impedance looks inductive. Question is where to put the inductive part. For stability with a capacitive load, the open-loop output impedance should look resistive around the GBW. That explains why the output impedance is what it is from 100Hz up. For lower frequencies, there is probably abundant loop gain, so there is no need to lower the open-loop output impedance.

We can see that the simulation is correct, and the people at TI have done a good job designing the SPICE model and the component.

  • 1
    \$\begingroup\$ Yes - and the additional (unwanted) pole has a frequency of wo=R/L. As we can see, for low R values (R=R1+R2) the pole frequency can be rather small and will contribute to the loop gain phase in a region where the loop gain is still above unity. \$\endgroup\$
    – LvW
    Jul 5, 2021 at 7:13

You need to respect the maximum output current parameters.

enter image description here

From datasheet.

Note that the number given is for a short-circuit. It doesn't mean that the op-amp will perform well at that current.

  • \$\begingroup\$ Oh this makes sense. But I am not able to understand how current may directly relate to frequency response. \$\endgroup\$ Jul 4, 2021 at 8:53
  • \$\begingroup\$ I don't know either but have a look at the output waveforms and see if they're distorted. It will depend on how good the simulation model is, I suppose. \$\endgroup\$
    – Transistor
    Jul 4, 2021 at 8:56
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    \$\begingroup\$ The effect on frequency response is completely dependant on how the SPICE model for the op-amp in question is made. As it is not an 'ideal' opamp several effects can cause this. \$\endgroup\$ Jul 4, 2021 at 11:16
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    \$\begingroup\$ The DC output current in this case is the opamp's offset voltage divided by the resistor between the + and - inputs. It is less than 1mA. SPICE AC analysis does not affect the DC operating point. Thus this simulation is within the recommended operating conditions. Current limiting does not come into play here. \$\endgroup\$
    – 7efkvNEq
    Jul 5, 2021 at 5:35

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