Is there a minimum current drain to use LM25011 of TI?
This particular part is recommended to have a minimum of about 1mA output current because of considerations regarding the boostrap capacitor charge. You can select the feedback resistor values to guarantee that current. See 7.3.5. Efficiency will be miserable, of course, at low currents. 


  The LM25011 integrates an N-Channel buck switch and associated floating high-voltage gate driver. The gate
  driver circuit works in conjunction with an external bootstrap capacitor (CBST) and an internal high-voltage diode.
  A 0.1-µF capacitor connected between BST and SW provides the supply voltage for the driver during the ontime. During each off-time, the SW pin is at approximately –1 V, and CBST is recharged from the internal 5-V
  regulator for the next on-time. The minimum off-time ensures a sufficient time each cycle to recharge the
  bootstrap capacitor.
  In applications with relatively high output voltage and low minimum load current, the internal pre-charge device of
  the LM25011 may not pull the SW pin sufficiently low during the off-time to maintain the voltage on the bootstrap
  capacitor. If the bootstrap capacitor (CBST) discharges during the long off-times, and the regulator will cycle on
  and off at a low frequency. Decreasing the values of the feedback resistors RFB1 and RFB2 to provide a minimum
  load of typically 1mA at nominal VOUT will increase the minimum switching frequency and maintain sufficient
  bootstrap capacitor voltage.


If you want high efficiency at low load currents as well as high load currents (for example, for battery operation with intermittent high current draw) you could consider something like an LT8608 which will operate with better than 75% efficiency above 1mA or so, and 60% at 100uA. This is done by introducing burst mode operation at low currents, as well as designing the part to have a very low quiescent current.  

Of course that wonderfulness comes at a price, about double the cost for the chip in 1K quantity.