LM76005: Output capacitance required differs between Datasheet calcs and T.I. WeBench tool.

Hi,

Value of output capacitor to use differs between datasheet calculations and WeBench tool. (ambient of 80C max.)

Vin = 11 to 52 V

Vout = 5.667 to 5.881 V (using Rfb of 100k + 21k 0.1%). [Require Vout(min.) = 5.5 V and Vout(max.) = 6.0 V]

Iout(max.) = 4.163 A

f(sw) = 500 kHz

Inductor = 8.2µH (XAL6060-822MEB)

Output capacitor value using formulas in datasheet yields Cap(min.) of 445 µF

Workings as follows:

r = (Δi_L / I_OUT) = 0.833 A / 4.163 A = 0.200096

ΔV_OUT = 100 mV max. (to keep V_OUT within the limits of the downstream regulators)

D’ = 1 – duty cycle = 1 – 0.113 = 0.887 (for 52 V input) and 1 – 0.535 = 0.465 (for 11 V input)

f_SW = 500 kHz

I_OUT =4.163 A (max.)

D’ worst case is 0.887, so this will be used to work out the minimum capacitance required, as follows:

COUT              > (1 / (500 kHz x 0.200096 x 100 mV / 4.163 A)) x [((0.200096² / 12) x (1 + 0.887)) + (0.887 x (1 + 0.200096))]

                                    > (1 / 2403.267) x [((0.003336) x (1.887)) + (1.064485)]

                                    > 445.55 µF minimum

TI WeBench output capacitor value recommends 4x 47 µF, 16V, 20%, X7R, 2220 (TDK CGA9N3X7R1C476M230KB), so after derating comes out to an effective capacitance of 109.64 µF minimum!?

What output capacitor choice should I use here as something seems to be wrong by a factor of 4.

Regards,

Richard.

.WBDesign16_TINA-TI.zip