7805 switched mode replacement

[Pha5e]

I am currently looking at using a 7805 switched-mode replacement module to use instead of a traditional 7805 linear regulator.

To meet EMC, a PI filter is recommended on the input.

How do I calculate the appropriate component values to filter out the conducted switching noise?

 

[aliarifat794]

Here is a PI filter calculator that may help you.

Pi Filter calculator LPF | LC Low Pass Filter

This Pi filter calculator is used as LC low pass filter for impedance matching.It takes cutoff frequency (fc) and Z0 as input and calculates L (inductance) and C (capacitance) values.

www.rfwireless-world.com

 

[FvM]

You have 3-pin drop-in replacement like https://recom-power.com/pdf/Innoline/R-78B-1.5.pdf or TI PT78ST105. Due to room restrictions, they have no internal emc filters.

 

[D.A.(Tony)Stewart]

With ~1uH feed to input or equivalent in TP or UTP cable you can expect low emanation if you have a spec for rated back current.

For a system level EMC the use of DCDC converters depends a lot on too many system factors to guarantee. But attention to each can attenuate risk of egress from this low power regulator. Factors may include;

• The switched current frequency. e.g. 500 kHz
• The supplier quality e.g. OKI Murata have over 20 yrs experience on this design. https://www.murata.com/products/productdata/8807037992990/oki-78sr.pdf
• System wire shielding of DC+ GND,
• The source RLC characteristics of the path may have complex resonance gain between source and input C, ESR. It may be reduced with added R or L and synthesized to compare with test results
• The dynamic nature, if any of the load may impact quasi-peak results slightly.

1. Twisted pair (UTP) or STP wire is what I recommend to reduce EMI or proper layout on PCB.
2. Use a DCDC converter with a rated ripple input current.

OKI-Murata spec is ; https://www.murata.com/products/productdata/8807037992990/oki-78sr.pdf
Reflected (back) ripple current ② (Cin = 2 X 100uF, CBus = 1000uF, LBus = 1uH)
50 mA, pk-pk

If f = 0.5 MHz and V=LdI/dt then Vmin is 1uH * 0.05A / 1us = 0.05 V.