Hello,
I am designing an input filter for my circuit’s power rail. I plan to power the circuit using a laboratory power supply, and for additional protection and noise reduction I want to add a filter at the input. I know that I need a low‑pass filter, most likely a π‑filter or an LC filter, but I am unsure of the source impedance of the lab supply and the effective input impedance of the linear regulator that the filtered power will feed. Below is a picture of my linear regulator configuration.


Thank you for you're advice

 

What frequency of noise are you thinking about?

 

What makes you think you need filtering at the inputs?

What is the range of VPP? What type of supply is your "laboratory" supply? Switching or linear? What will you be powering from the regulators?

 

What frequency of noise are you thinking about?

I don't know I am pretty new to circuit design

 

What makes you think you need filtering at the inputs?

What is the range of VPP? What type of supply is your "laboratory" supply? Switching or linear? What will you be powering from the regulators?

I am powering two fast comparators (ADCMP580), a digital delay and clock generator. I am afraid if I don't filter the input power rail I will have problems. My LDO regulator input voltage range is from 15V to 6.5V and I will use a linear power supply

 

I am afraid if I don't filter the input power rail I will have problems. My LDO regulator input voltage range is from 15V to 6.5V and I will use a linear power supply

The dropout voltage is 1.2V at 1A. You shouldn't have any issues requiring significant input filtering when powering the LDO regulators from a linear power supply. You'll just need decoupling caps on the regulator outputs.

 

The dropout voltage is 1.2V at 1A. You shouldn't have any issues requiring significant input filtering when powering the LDO regulators from a linear power supply. You'll just need decoupling caps on the regulator outputs.

Ok, I will do that.

Thank you for the help!

 

there is one more thing to keep in mind - heat dissipation.
those regulators are good for up to 1A but they are in tiny package that cannot dissipate heat well.
suppose your Vpp is 9V and your 3V3 regulator needs to provide 0.8A to some load.
voltage drop across regulator is 9V-3.3V=5.7V
and since current is 0.8A, power dissipation is P=V*I=5.7V*0.8A=4.56W
and that is a lot of heat... the common SOT-223 package is only good for up to 1.2W and you probably don't want more than 0.7W.
unless your load is really low, i would suggest to use DPACK (TO-252) and a lot of copper area as a heatsink. even this one is only rated for about 2.2W.
the obvious solution would be to reduce supply voltage (such as 7V) and use LM1117 which is available as TO220 version should be good for up to 15W (with a heatsink of course)