I am trying to design a communication system that will send data signals over a 400V DC voltage line by using line transceivers. I attached the LTspice schematics below. The transformer is for providing isolation between 400VDC and the transceivers.
The problem is the size of the inductor shown inside the red circle. If I decrease the inductance value, the square wave coming from the A-B input loses its shape and peak voltages occur. This causes false high voltages at the RO output of the receiver side. However, since the DC current is 3.5A at the power line. The required inductor size increase enormously with the inductance value in order to prevent saturation. In the attached plot the simulation results are shown while using 200 uH inductor. You can see the problem. So, how can I calculate the required inductance and is there any other configuration for this system that works with a smaller inductor?
Hi Gökhan Bayraktar ,
perhaps not the most elegant solution but it should work in any case:
Use bifilar inductors with suitable inductance. If the load is constant then send about 3 A through the second winding in order to avoid saturation. This current can be provided by a very low voltage source.
If the load isn't constant then you would need a control circuit at each inductor which adjusts the current through the second winding.
The circuit shown has the task of transmitting an 80 kHz signal. Why is the red curve the envelope curve of the signal even though no rectifier is installed? The cause is puzzling; it could be massive overdrive. It is also incomprehensible that the green curve is asymmetrical and apparently contains a strong DC voltage component, although the AC voltage is transmitted via transformers and capacitors.
I would not transmit a square wave signal, but rather transform it into a sinusoidal signal (f = 80 kHz) before transmission. Then the received signal would also be sinusoidal, can be selectively amplified and converted back into a square wave signal. The annoying increase in the high frequencies (white curve) would not exist.
Details: C2 and C4 are much too big, Z (80 kHz) = 2 ohms. 1 nF to 10 nF would be sufficient. It would be optimal if C2 / L3 and C4 / L5 were resonance circuits for 80 kHz. According to the circuit, now the resonance is 5035 Hz, so it has no selective effect. First correct this incorrect dimensioning.
Herbert Weidner Thanks for that correction about C2 and C4 capacitors. I did not show the "DK" input because of decreasing the picture size. Let me try to explain more.
The modulated input is given to "DI" input of the transmitter side LTC2862 ( left side is the transmitter) as shown in the first picture.
Then this LTC2862 has an output through A-B terminal as shown in the second picture. This is AC voltage since the features of A and B terminals.
This AB voltage travels through the transformers and 400V DC line then comes to the A-B terminals of LTC2862 at the right side (receiver side).
Similarly, RO terminals of the right side LTC has a digital output according to the voltage across its A-B terminals. It is like when A-B voltage is greater than 200mV RO is high and vice versa.
According to this, RO output is shown at the third picture.
Finally, the low pass filter is removing the 80kHz carrier signal.
That C2 and C4 capacitor are for decoupling the DC 400V and AB voltage becomes AC. The transformer is for isolation.
My problem is the component values. Especially, the inductor at the DC line. When it is too small the AB voltage loses its square wave shape and becomes sharper. This causes false high at the RO terminal of the right side LTC2862.
When that inductor has bigger inductance it is hard and expensive to design that kind of inductor.
I recommend leaving out both coils L1 and L4 (in your first picture). The reason: You have a low-resistance DC line between V1 and R2, the impedance of which is additionally reduced by C1 and C3. Nevertheless you try to feed in the signal capacitively via C2 and C4. To do this, you try to increase the impedance of the DC line through L1 and L4. This plan of transmitting the signal as AC *voltage* is not a good idea, it is what is causing your problems.
My suggestion: transmit the signal as alternating *current*, as an overlay of the direct current (3 A). The LTC2862 on the left delivers a signal current of approximately 2 mA. L2 / L3 is a current transformer with (500 turns: 5 turns). Therefore, a signal current of 200 mA flows on the DC line. This is only possible if neither L1 nor L4 are installed and the line resistance is very low!
The current transformer L5 / L6 is located on the receiver side and transforms the signal current down to 2 mA, whereby the signal voltage increases by a factor of 100. Ideally, the output voltage of the left LTC2862 is the same as the input voltage of the right LTC2862.
Both transformers L2 / L3 and L5 / L6 isolate the consumer current from the signal current.
Important: The DC circuit breaker must be on the left of C1 or on the right of C3 so that the signal can also be transmitted when R2 is switched off.
Herbert Weidner Thanks for the answers. I have seen your drawing just now I don't why I did not see it at the first place. I will made the change at your drawing about the transformer as you suggest at your last answer .
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