I need to power a board that works at 3.3V, and I have a regulated power of 12V.

I tried to use a single linear regulator like the TS2940 but the power required is so high that the heat produced was too high (melted the case).

So I decided to use a switching regulator (the LM2594) to go down to 5V and a LDO linear regulator to go down to 3.3V. I read everywhere that this is a common practice.

The switching regulator works fine but my problem is that the linear regulator does not work properly (2.8 Vmean, 1.2 Vpp). That is because by datashet the output capacitance of a switching regulatr should be around 100uF, and the input capacitance of a linear regulator should not be greater than 10uF.

What is the best way to adapt those two impedences??


  • 1
    \$\begingroup\$ There's nothing wrong with having a large cap in the input of a linear regulator. \$\endgroup\$
    – brhans
    Commented Jul 28, 2019 at 13:46
  • \$\begingroup\$ What device is your LDO? IS it really an LDO or just a linear regulator? \$\endgroup\$
    – user16324
    Commented Jul 28, 2019 at 13:48
  • \$\begingroup\$ the LDO is mcp170033. In its datasheet says that input impedance larger than 10uF will get the device unstable \$\endgroup\$
    – Dukenukem
    Commented Jul 28, 2019 at 14:13
  • \$\begingroup\$ This regulator? Which package? Can you post the reference of the caps you used, and how much current the load uses? What is the load? \$\endgroup\$
    – bobflux
    Commented Jul 28, 2019 at 14:50
  • 2
    \$\begingroup\$ To be honest, it's so rare that I'm not sure you're not misunderstanding something in the datasheet. Can you point out where exactly it says that 10 uF is the maximum input capacitance? \$\endgroup\$
    – The Photon
    Commented Jul 28, 2019 at 15:09

1 Answer 1


In the datasheet for the MCP1700 family, I found two paragraphs talking about the input capacitance requirement.

3.3 Unregulated Input Voltage Pin (VIN)

Connect VIN to the input unregulated source voltage. As with all low dropout linear regulators, low source impedance is necessary for the stable operation of the LDO. The amount of capacitance required to ensure low source impedance will depend on the proximity of the input source capacitors or battery type. For most applications, 1 μF of capacitance will ensure stable operation of the LDO circuit. For applications that have load currents below 100 mA, the input capacitance requirement can be lowered. The type of capacitor used can be ceramic, tantalum or aluminum electrolytic. The low ESR characteristics of the ceramic will yield better noise and PSRR performance at high frequency.


5.1 Input

The input of the MCP1700 is connected to the source of the P-Channel PMOS pass transistor. As with all LDO circuits, a relatively low source impedance (10 Ω) is needed to prevent the input impedance from causing the LDO to become unstable. The size and type of the required capacitor depend heavily on the input source type (battery, power supply) and the output current range of the application. For most applications (up to 100 mA), a 1 μF ceramic capacitor will be sufficient to ensure circuit stability. Larger values can be used to improve circuit AC performance.

Neither one says that there is a maximum input capacitance. In fact, section 5.1 says that higher capacitance values will improve AC performance.

I don't believe there is any need to reduce the input capacitance seen by your regulator.

You will probably want to have 1 uF ceramic (low ESR) capacitor immediately at the input pin of the linear regulator, in addition to the higher-value (but higher ESR and ESL) capacitors used at the output of your switching regulator.

If you need a really clean output voltage, you might want to pay attention to what harmonics of the switching frequency your switching regulator produces, and be sure to filter those out if they are above the frequencies where your linear regulator has good line regulation.


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