1. RJ45 port, connection attributes:
RJ45 connection RJ45.1, RJ45.2, RJ45.3, RJ45.6 do the signal line connection, the connection is as follows:
RJ45.1----RX+
RJ45.2----RX-
RJ45.3-- --TX+
RJ45.6----TX-
This connection method is 802.3 standard wiring. Many chips now support the MDI/MDIX function, that is, the parallel cross-adaptive function. In this case, the following wiring method of the chip is also possible.
RJ45.1----TX+
RJ45.2----TX-
RJ45.3----RX+
RJ45.6----RX- the
remaining RJ45.4, RJ45.5, RJ45 7. The general connection method of RJ45.8 is to put a 1000PF high voltage capacitor through a 75 ohm resistor. It is called BOB-Smith circuit, which acts as electromagnetic compatibility.
2. Connection of network transformer and RJ45 The
structure of network transformer is as follows:
3. Signal part:
RX+ requires the end of the same name of the RX part of the network transformer, generally refers to the end of the transformer with a black dot, as shown in the figure, 6-pin
RX- requires the other end of the same group of coils in the RX part of RX+ The 8-pin TX+ as shown in the figure requires the same-named end of the TX+ part of the network transformer, which generally refers to the end of the transformer with black dots. The 6-pin
TX- as shown in the figure requires the other end of the same group of coils connected to the TX+ part of the TX+. Pin 8 as shown in the figure.
For RX+, RX- differential signals require that each signal line pull a 49.9 or 51 ohm resistor to ground through a 0.1uF capacitor. The specific connection is as follows:
The differential pair needs to be connected to 49.9 ohms or 51 ohms: because the resistance of a single line on the PCB is generally 50 ohms, the resistance between the two differential lines is 100 ohms, in order to make the signal output the strongest, it is required to output on the differential line Add a 100 ohm resistor between the two wires at the end, 49.9+49.49=99.8 ohm, 51+51=102 ohm and 100 ohm resistance are similar, which is why these two resistors are added. The purpose of adding 0.1uF capacitor is to remove the noise signal of the signal signal.
The waveform of the differential signal is as shown in the figure below. If the following standard waveform can be measured with an oscilloscope, there should be no problem with the data connection.
Note: Some chips now use current drive mode. If it is current drive mode, then the terminal does not need to add a 49.9 ohm resistor, because the chip is already integrated.
TX+ waveform:
TX- Waveform:
If the design engineer uses a chip with MDI/MDIX function automatic recognition, then RX+ appears with the waveform of TX+, and RX- appears with the same waveform as TX-.
The tap voltage of the network transformer should be determined according to the chip. Generally, the voltages appearing are:
2.5V, 1.8V, 1.2V, 3.3V
The selection of the voltage connection of the two taps of VDD_RX & VDD_TX can be divided into two situations:
1. If the chip has automatic identification function of MDI/MDIX function, VDD_RX=VDD_TX
2. If the chip does not have the MDI/MDIX function, the VDD_TX, which has the automatic identification function, should be connected to the power terminal, and VDD_RX should be directly connected to the signal ground.
Note: Many chips now require a 1:1 transformer turns ratio. If it is a 1:1 transformer, RX and TX can be interchanged. If some chips require special transformers with turns, which RX and TX are definitely not interchangeable.
Problems encountered during debugging:
1. Can't connect
If the designer has an oscilloscope, let him test the TX+ and TX- signals. If there is the LINK signal described above, it means that the chip is working normally, and the problem is not in the chip design.
If there is no oscilloscope, first let the designer test whether it can be connected at 10M speed, the specific method is as follows:
Open the network connection
Right-click on the properties and the above picture will appear, click on the configuration and the following interface will appear
Click Advanced
After clicking the link speed and duplex column, change the status value on the right, select 10M/full duplex and click OK. See if it can be connected in this state, if possible, it should be a welding problem of the chip or other components, or a problem with the PCB wiring.
Another way to confirm the link is to connect TX+ to RX+ through a network cable, and connect TX- to RX- to do a loopback test. If it can be connected normally, it means that there is no problem with the design.
2. Can connect and communicate under 100M state; can not link and communicate under 10M state
This problem is mainly to check the welding problem. Generally, one of the differential lines is not welded well.
3. The status of the link is unstable, there will be one, and there will be no
Please check whether the chip's power supply is stable, ripple, whether there are power traces or the device's rated current is too small. Is the reset time sufficient?
The Ethernet differential signal is not good-
please confirm a few points:
1. The 49.9 resistor on the secondary side of the network transformer is pulled up to 3.3V, and some are designed to pass a capacitor and then ground.
2. Please confirm the accuracy of the resistance of REXT (PIN37) and whether it is placed close to the pin on the PCB.
3. Ripple of the power supply.
4. Whether the accuracy of the clock meets the requirements.
5. Add a few capacitors to the PLL power supply to see if there is any improvement.
6. After C1 and C2 are soldered close to the PIN pin, confirm whether there is any improvement.
7. Check whether the reset time is sufficient.
8. Whether the unused input pin is pulled up.