Variable LDO implementation - capacitor size

Question

I am designing two linear power supplies that will:

1. Drop 9V DC to 5V DC to power a Raspberry Pi around 2.5A
2. Drop 9V DC to 7.5V DC to power a DAC under 3A

I plan to use Texas Instruments LDO UCC283-ADJ. An adjustable LDO.

I have been reading the datasheet:

http://www.ti.com/lit/pdf/slus215

Which advises (see page 5) that a timing capacitor, C, is required:

CT: Short-circuit timing capacitor and shutdown input for the UCC283−ADJ version. Pulling CT below 0.25 V turns off the regulator and places it in a low quiescent-current mode. A timing capacitor, C, from CT to GND programs the duration of the pulsed short-circuit on-time. On-time, t_ON, is approximately given by:

t_ON = 750 k * C

Whilst the formula is given, I am at a loss as to what size capacitor I should use as C, for each of my implementations.

I was planning R1 to be 50kOhms and R2 to be 10kOhms.

Can anyone advise me on C?

Thanks.

Answer 1

For the bypass cap on vout they say this:

VOUT: Regulated output voltage. A bypass capacitor is not required at VOUT, but may be desired for good transient response. The bypass capacitor must not exceed a maximum value in order to insure the regulator can start.

They then say this:

A capacitive load on the regulator’s output appears as a short-circuit during start-up. If the capacitance is too large, the output voltage does not come into regulation during the initial tON period and the UCC383 enters pulsed mode operation. The peak current limit, tON period, and load characteristics determine the maximum value of output capacitor that can be charged. For a constant current load the maximum output capacitance is given as follows:

$$ C_{out(max)} = (I_{CL}-I_{LOAD}\frac{t_{on}}{V_{out}} $$

So choose how long you want the part to turn on with the first equation, then calculate the maximum capacitance and choose a value of capacitance lower than Coutmax.