Because of the larger voltage and current stresses in an IBB, when compared to the buck, the power loss of the IBB will be greater. This means that the efficiency of the IBB will be less than that of a buck under similar conditions. As an example the LMR33630 achieves nearly 95% efficiency when converting +12 V to +5 V at a load current of 2 A. This gives a loss of about 0.53 W. This loss could be 3 to 4 times larger when the buck is used as an IBB converting to –5 V. This would reduce the efficiency to about 85%. This reduction in efficiency needs to be taken into account when calculating the device currents as above. Unfortunately, it is not easy to estimate the efficiency before the IBB is designed and tested. The best plan is to take a conservative approach to calculating the maximum operating currents when choosing a candidate buck.

The increased power dissipation also has consequences for die temperature. Every regulator has a maximum rated die temperature that must not be exceeded. Since the IBB has more dissipation than the equivalent buck, the extra heat will need to be removed or the die temperature may get too high. This means that the total θ_JA of the application will have to be lowered. With modern device packaging, most of the heat will flow out of the bottom of the device and into the PCB. Therefore, the best way to reduce the θ_JA is to increase the PCB copper area and choose a device with a die attached paddle (DAP) to help dissipate the heat. The application report AN-2020 Thermal design by insight, not hindsight provides guidance for achieving good thermal performance from your PCB layout.