SLUSDD4 Data sheet

SLUSDD4B April 2019 – December 2020 UC1843B-SP

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8.2.2.5 Output Filter and Capacitor

The output capacitance value is picked such that there is enough capacitance for the required voltage ripple and output current load step. The UC1843B-SP design uses equations Equation 22 and Equation 24 to find a minimum capacitance.

Equation 22.

C_{o u t} > \frac{I_{o u t} \times D_{M A X}}{V_{R i p p l e} \times f_{o s c}}

Equation 23.

C_{o u t} > \frac{10 A \times 0.5}{50 m V \times 200 k H z} = 500 μ F

Equation 24.

C_{o u t} > \frac{Δ I_{s t e p}}{2 π \times Δ V_{o u t} \times f_{c o}}

Equation 25.

C_{o u t} > \frac{10 A}{2 π \times 0.7 V \times 2.2 k H z} = 1 m F

A value of around 1145 µF was chosen to keep output voltage ripple low. Note that the output voltage ripple in the design was further decreased by adding an output filter and by adding an inductor after a small portion of the output capacitance. Six ceramic capacitors were picked to be placed before the output filter and then the large tantalum capacitors with some small ceramics were added to be part of the output filter. The initial ceramics will help with the initial current ripple, but have a very large output voltage ripple. This voltage ripple will be attenuated by the inductor and capacitor combination placed between the ceramic capacitors and the output. The equations below allow for finding the amount of attenuation that will come from a specific output filter inductance. An inductance of 500 nH was chosen to attenuate the output voltage ripple and the attenuation was sufficient for the design.

Equation 26.

F_{r e s o n a n t} = \frac{1}{2 π \times}

LFilter×CoBulk

Equation 27.

F_{r e s o n a n t} = \frac{1}{2 π \times}

0.5 nH×1127 μF=6.7 kHz

Equation 28.

F_{Z e r o =} \frac{1}{2 π \times C_{o B u l k} \times E S R_{o B u l k}}

Equation 29.

F_{Z e r o =} \frac{1}{2 π \times 1127 μ F \times 0.009 Ω} = 15.69 k H z

Equation 30.

A t t e n u a t i o n_{f s w} = 40 \times \log_{10} (\frac{f_{o s c}}{f_{r e s o n a n t}}) - 20 \times \log_{10} (\frac{f_{o s c}}{f_{z e r o}})

Equation 31.

A t t e n u a t i o n_{f s w} = 40 \times \log_{10} (\frac{200 k H z}{6.7 k H z}) - 20 \times \log_{10} (\frac{200 k H z}{15.69 k H z}) = 36.88 d B

Sometimes the output filter can cause peaking at high frequencies, this can be damped by adding a resistor in parallel with the inductor. For the UC1843B-SP design, 0.5 Ω was used as a very conservative value. The resistance needed to damp the peaking can be calculated using the following equations:

Equation 32.

ω_{o} =

2(CoCerm+CoBulk)LFilter×CoCerm×CoBulk

Equation 33.

ω_{o} =

2(19 μF+1127 μF)500 nH×19 μF×1127 μF=463 kHz

Equation 34.

R_{F i l t e r} = \frac{R_{o} \times L_{F i l t e r} \times (C_{o C e r m} + C_{o B u l k}) - \frac{L_{F i l t e r}}{ω_{o}}}{}

Ro×(CoCerm+CoBulk)ωo-LFilter×CoCerm

Equation 35.

R_{F i l t e r} = 0.5 \times 500 n H \times (19 μ F + 1127 μ F) - \frac{500 n H}{463 k H z}

0.5×(19 μF+1127 μF)463 kHz-500 nH×19 μF=0.232 Ω