Common mode filter assistance for the PM8805

Christopher1971 · ‎2025-04-22

Hi All,

To start, I'm a hobbyist and although I am OK with digital, analogue and the maths surrounding design is beyond me.

I am seeking assistance in specifying the values and component suggestions for the CM Filter as noted in figure 7, page 19 of the PM8805 data sheet. If someone has a ready made solution for the following, it would be very much appreciated.

My circuit will need two power rails which I will create using two buck convertors. A 12V rail as well as a 3.3V rail. The 3.3V will only draw a constant current of around 600mA, the 12V rail will spike on occasion to 2Amps. I have designed these before and I am happy with this area.

The 12V circuit will be driving a Texas Instruments TAS5670L Class D I2S Amplifier, while the 3.3V will be driving a Microcontroller.

Thanks in anticipation.

Peter BENSCH · ‎2025-04-22

I assume that your two buck will be behind the PM8805, right?

According to IEEE802.3bt, the buck or at least the first one must be selected for an input voltage of at least 57V, which can be handled by an L7987L (2A) or L7987 (3A) with a maximum input voltage of 61V, for example. Let's do the maths:

Assuming efficiencies of 80% (12V) and 76% (3.3V), your two rails draw a total power of (12V*2A)/~80% = ~30W and (3.3V*0.6A)/~76% = ~2.6W, i.e. a total of approx. 33W. The minimum voltage at the PD is 41.1V according to IEEE802.3bt, which results in a maximum current of 33W/41.1V = 803mA. For this current, you can use the common mode choke 744272102, for example.

Hope that helps?

Good luck!
/Peter

In order to give better visibility on the answered topics, please click on Accept as Solution on the reply which solved your issue or answered your question.

View solution in original post

Peter BENSCH · ‎2025-04-22

I assume that your two buck will be behind the PM8805, right?

According to IEEE802.3bt, the buck or at least the first one must be selected for an input voltage of at least 57V, which can be handled by an L7987L (2A) or L7987 (3A) with a maximum input voltage of 61V, for example. Let's do the maths:

Assuming efficiencies of 80% (12V) and 76% (3.3V), your two rails draw a total power of (12V*2A)/~80% = ~30W and (3.3V*0.6A)/~76% = ~2.6W, i.e. a total of approx. 33W. The minimum voltage at the PD is 41.1V according to IEEE802.3bt, which results in a maximum current of 33W/41.1V = 803mA. For this current, you can use the common mode choke 744272102, for example.

Hope that helps?

Good luck!
/Peter

In order to give better visibility on the answered topics, please click on Accept as Solution on the reply which solved your issue or answered your question.

Christopher1971 · ‎2025-04-22

Dear Peter,

Firstly, thank you for replying to my post.

Yes, they will be. I am planning on using the L7987 to drop the POE voltage to 12V for the amplifier, then use another L7987L connected to the 12V output, to provide the 3.3V to the digital circuits.

It has certainly helped thank you.

Regards,

Christopher