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.
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.
You need to respect the maximum output current parameters.
Note that the number given is for a short-circuit. It doesn't mean that the op-amp will perform well at that current.