Other Parts Discussed in Thread: TPS92518
Tool/software:
Hi, page 15 of datasheet for TPS92518HVPWPR notes that the IC has a tleb (min on time) that must be considered for every application to avoid a inductor saturation condition. I have a question as to what is meant by the parameter t_off_min. From my understanding it logically follows that equation 10 can be rewritten as :
V_led_min / V_in_max = (t_leb) / (t_leb + t_off_min).
If this is the case, and we solve for t_off_min, we obtain:
t_off_min = (V_in_max - V_led_min) * t_leb / V_led_min
Considering the example where L = 100u, V_in_max = 65V, V_led_min = 6V, t_leb = 250ns (to calculate with extra margin), we then find that t_off_min = 2.4583 us. However, the example in the datasheet instead suggests that 4.17us is your t_off_min. From my interpretation, while 4.17us is a valid off time (because it is greater than the minimum I calculated which is 2.4583us), one cannot say that it is the minimum off time. Yes, the 4.17us will achieve an inductor ripple of 250mA peak to peak, but there is no explanation on the datasheet that I can find that suggests a lower ripple cannot be safely achieved. For example, using 3us as the t_off_min as to give a margin to my calculated value will result in a ripple of 180mA peak to peak. With such a calculation, the minimum TOFF_DAC val ends up being 39, which from my understanding could be increased for extra safety but does not need to be because of the margin already built into choosing 3us as opposed to 2.4583us as my min off time. Where am I going wrong with my logic, and why is the datasheet telling me that the DAC needs to be set at 55 or greater? I either misunderstand or the datasheet has an error.
