(WIP) 48W MP8030 PoE PD
A 12V 48W Power over Ethernet PD Module
There are quite a few off the shelf Power over Ethernet PD modules available, but they usually have somthing anoying about them such as:
- some need an external full bridge rectifier (the whole point of using one of these modules is that everything is integrated into one bom line...) and this is also not as efficent as a active bridge rectifier.
- some require extra output capacitance
- the poe class id pins are not isolated from the ethernet interface, so you need to add external optoisolators for them
The specs I am targeting:
- 12V 48W output
- 41-57V input range
- PoE and AUX input, both with mosfet load switches, and not diodes for oring
- Active Bridge rectifier
- 93%+ efficiency at full load
Current Status: First 2 Boards built, and powered up. TODO: feedback loop compensation, and load / efficiency testing.
KiCAD files: https://gitlab.com/qfreiman/poe-bt12v50w-sip
Design Notes
Design is based around the MP8030 802.3bt PoE PD + DCDC controller ic
The controller supports flyback or forward operation, and I am doing an Active clamp forward design
Vin: 41-57V
Vout: 12V @ 4A
Duty Cycle
Forward Converter Duty Cycle: \[ D = \frac{V_{out} \cdot N}{V_{in}} \]
Where \(N\) is the turn ratio \(\frac{T_{pri}}{T_{sec}}\)
Duty @ 41V: \[ \frac{12 \cdot 1.71}{41} = 50% \]
Duty @ 57V: \[ \frac{12 \cdot 1.71}{57} = 36% \]
MOSFET Selection
Primary Side:
Max Voltage: \[ \frac{D \cdot V_{in}}{1 - D} + V_{in} \]
\[ \frac{0.50 \cdot 41}{1 - 0.50} + 41 = 82V \]
\[ \frac{0.36 \cdot 57}{1 - 0.36} + 57 = 89V \]
Secondary Side Low Side Rectifier:
Max Voltage: \[ \frac{D \cdot V_{in}}{N \cdot (1-D)} \]
\[ \frac{0.5 \cdot 41}{1.71 \cdot (1-0.5)} = 24V \]
\[ \frac{0.36 \cdot 57}{1.71 \cdot (1-0.36)} = 18.75V \]
Secondary Side High Side Rectifier:
Max Voltage: 57 / 1.8 = 33.3V
AUX Winding output Inductor
The inductor needs to be large enough to ensure it does not enter DCM at the minimum VCC load current (otherwise the voltage will just keep rising until OVP trips)
Total AUX Current: IC Draw + OVP Divider Draw + N-FET gate drive current + P-FET gate drive current:
\[ 1.1mA + \frac{12V}{120K} + (250k \cdot 14.8nC) + (250k \cdot 2.9nC) = 5.625mA \]
So the inductor must be large enough such that the peak to peak ripple current is less than \( 2 \cdot 5.625 mA \)
The transformer I am using has 0.86 windings for every winding on the secondary, so the output voltage wil be \( 0.86 \cdot 12V = 10.32V \)
\[ L = \frac{V_{out}}{f_{sw} \cdot \Delta I_L} \cdot (1 - \frac{V_{out} \cdot N}{V_{in}}) \]
\[ L = \frac{10.32V}{250k \cdot (2 \cdot 5.625mA)} \cdot (1 - \frac{10.32V \cdot 1.71}{57V}) = 2.53mH \]
Main Transformer
Using PA3856.003NL
Peak Primary side current: \[ L_{EXC} = \frac{V_{out} \cdot N}{2 \cdot L_p \cdot F_{sw}} \]
\[ L_{EXC} = \frac{12 \cdot 1.8}{2 \cdot 100uH \cdot 250k} = 0.432A \]
Output Inductor
Inductor ripple current should ideally be in the 30-40% range.
Iout is 2.5A, 30% of 2.5A is 0.75A, and 40% is 1A
\[ L = \frac{V_{out}}{f_{sw} \cdot \Delta I_L} \cdot (1 - \frac{V_{out} \cdot N}{V_{in}}) \]
At 1A P-P and 57V In:
\[ L = \frac{12}{250k \cdot 1} \cdot (1 - \frac{12 \cdot 1.8}{57}) = 30uH \]
At 0.75A P-P and 57V In:
\[ L = \frac{12}{250k \cdot 0.75} \cdot (1 - \frac{12 \cdot 1.8}{57}) = 39.8uH \]
At 1A P-P and 41V In:
\[ L = \frac{12}{250k \cdot 1} \cdot (1 - \frac{12 \cdot 1.8}{41}) = 22.7uH \]
At 1A P-P and 41V In:
\[ L = \frac{12}{250k \cdot 0.75} \cdot (1 - \frac{12 \cdot 1.8}{41}) = 30.3uH \]
So somthing in the 22-39uH range will be good
Saturation current rating needs to be 2.5A + half of worst case peak-to-peak ripple + some margin
Feedback Loop
https://www.ti.com/lit/an/slua535a/slua535a.pdf was used as a reference
Assuming that the 0V to 5V COMP pin range is linear with duty cycle, and that the opto output does not saturate minimum duty cycle would be at 0V on the COMP pin, which would require 5V across the 10K resistor in the MP8030, requiring: \( 5V / 10K = 500uA \) of current flowing through the optoisolator output.
And the maximum led current on secondary side after assuming the worst case CTR of 63% :
\[ 500uA / 63\% = 794uA \]
The ATL431 has a minimum output voltage of 2.5V, plus the worst case 1.2V forward voltage of the optoisolator means that the LED current limiting resistor needs to be at most:
\[ (12V - 2.5V - 1.2V) / 794uA = 10863 Ohms \]
10K is close enough
