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:


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, tON, is approximately given by:

tON = 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?


  • \$\begingroup\$ Look at Eq (1) in the datasheet. The t_on you need depends how much capacitance is loading the regulator's output. \$\endgroup\$
    – The Photon
    Commented Sep 16, 2018 at 0:07
  • \$\begingroup\$ I just want to strongly recommend that you use a switching regulator. You're using 49.5W input for 35W output, and the efficiency of everything in proximity to your linear regulators will be reduced because they are dissipating up to 15W as heat, making things worse. If you desire the accuracy of an LDO linear regulator(For the Pi at least, a decent switching regulator should be more than accurate enough), consider using a switching regulator to drop most of the voltage and use the LDO just to chop off the last little bit. Switch from 9 to 5.2V for example, then LDO from 5.2 to 5. \$\endgroup\$
    – K H
    Commented Sep 16, 2018 at 4:37
  • \$\begingroup\$ @KH it is also possible to make a Buck regulator using an LDO , FET, L , hysteresis comparator \$\endgroup\$ Commented Sep 16, 2018 at 4:54
  • \$\begingroup\$ @TonyEErocketscientist thank you. I'm just learning about switching regulator design right now, so I'm hardly the most informed. Just hate to see someone burning off 15W when they could reclaim a lot of it with a $2 part. \$\endgroup\$
    – K H
    Commented Sep 16, 2018 at 5:05
  • \$\begingroup\$ All helpful and will consider this, but until I figure out how to setup the LDO I cannot really progress. \$\endgroup\$
    – user32407
    Commented Sep 16, 2018 at 9:15

1 Answer 1


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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.

  • \$\begingroup\$ Thanks. I think that C on the CT pin which is where I am struggling and it is not the same as the Cout capacitor. For Cout using eq 1 ILoad I think is 2.5a L I assume is ILoad + 3a so 5.5a based on Pg5 "During an overload, the regulator actively limits the maximum current to the peak current limit value. The peak current limit is nominally 3 A greater than the overcurrent threshold." VOut will be 5v. Ton unfortunately I have no idea, I know it is 750,000* C - but still no closer to determining C on the CT pin. Any help appreciated. \$\endgroup\$
    – user32407
    Commented Sep 16, 2018 at 9:22

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