TIDA-00349: Power loss when in no load

Hi Zoe,
the power loss can be derived as PIN-POUT from the efficiency, VOUT and VIN values.
Regarding VOUT: The design doesn't use a closed loop control . As a result VOUT depends on VIN, the turns ratio of the transformer and the forward voltage drop of the secondary side diodes - see equation 19 of the TIDA-00349 design guide https://www.ti.com/lit/pdf/tidu813 . The forward voltage drop depends on the IOUT as well as on temperature - they are the main contributor to variation of VOUT at medium to high output currents.
At low and especially at very low IOUT there will be additional contribution of parasitic switching spikes which will be peak rectified by the diodes and leading to an increase of VOUT which isn't covered by the mentioned equation 19. This effect is however documented in figure 27 of  the TIDA-010018 design guide https://www.ti.com/lit/pdf/tiduen0 . This reference design uses more or less the same circuit for the isolated DC/DC as implemented on TIDA-00349. 

Specific values for TIDA-00349:
VIN = 4.5V, IOUT=100uA, resulting VOUT=5.16V, IIN=256uA, Efficiency 44.8%, PIN=1.15mW, POUT=516uW, PLOSS=634uW
VIN = 4.5V, IOUT=10uA, resulting VOUT=5.61V, IIN=144uA, Efficiency 8.7%, PIN=648uW, POUT=56.1uW, PLOSS=592uW
VIN = 4.5V, IOUT=1uA, resulting VOUT=6.86V, IIN=132uA, Efficiency 1.2%, PIN=594uW, POUT=6.86uW, PLOSS=587uW

There was no specific measurement for no load condition, but the 1uA or 10uA IOUT-condition could be used as a reasonable approximation to that.
Dummy load (resistor) or  VOUT-post-regulation (i.e by low Iq-LDO) can be used to get a  VOUT which isn't that much load dependent.  

Results for VOUT = 10V haven't be evaluated with the existing designs. I would expect that the switching losses of the secondary diodes will be slightly larger . No-load loss shouldn't change that much with 10VOUT compared to the 5V case.

Best regards

Juergen