VS-min = VF-max + VDO-max + VO-max

(5)

Q

VIN

VF

RL

RDS

VDS

VDO

VP

VI

VO

VS

Figure 8-5. Establishing the Required Minimum Turns Ratio Through Nmin = 1.031 × VS-min / VP-min

Then calculating the available minimum primary voltage, VP-min, involves subtracting the maximum possible

drain-source voltage of the SN6501, VDS-max, from the minimum converter input voltage VIN-min:

VP-min = VIN-min – VDS-max

(6)

VDS-max however, is the product of the maximum RDS(on) and ID values for a given supply specified in the

SN6501 data sheet:

VDS-max = RDS-max × IDmax

(7)

Then inserting Equation 7 into Equation 6 yields:

VP-min = VIN-min - RDS-max x IDmax

(8)

and inserting Equation 8 and Equation 5 into Equation 9 provides the minimum turns ration with:

F-max

DO-max

O-max

min

IN-min

DS-max

D-max

V

+ V

+ V

n

= 1.031

V

R

I

´

-

´

(9)

Example:

For a 3.3 VIN to 5 VOUT converter using the rectifier diode MBR0520L and the 5 V LDO TPS76350, the data

sheet values taken for a load current of 100 mA and a maximum temperature of 85°C are VF-max = 0.2 V,

VDO-max = 0.2 V, and VO-max = 5.175 V.

Then assuming that the converter input voltage is taken from a 3.3 V controller supply with a maximum ±2%

accuracy makes VIN-min = 3.234 V. Finally the maximum values for drain-source resistance and drain current at

3.3 V are taken from the SN6501 data sheet with RDS-max = 3 Ω and ID-max = 150 mA.

Inserting the values above into Equation 9 yields a minimum turns ratio of:

min

0.2V + 0.2V + 5.175 V

n

= 1.031

= 2

3.234 V

3 Ω

150 mA

´

-

´

(10)

Most commercially available transformers for 3-to-5 V push-pull converters offer turns ratios between 2.0 and 2.3

with a common tolerance of ±3%.

SN6501

SLLSEA0I – FEBRUARY 2012 – REVISED JANUARY 2021

www.ti.com

20

Submit Document Feedback

Copyright © 2021 Texas Instruments Incorporated

Product Folder Links: SN6501