SN65HVD3080E: (Related question) Calculation review for SN65HVD3080E

Hi Muk,

Should they use 0.13UL that is higher one?

SN65HVD3080E features a 1/8 Unit-Load allowing 256 Nodes per Bus

SN65HVD3080E don't need the Fail safe circuit like as application note

This device uses internal failsafe biasing to keep the receiver in a known state. This does not provide failsafe biasing to rest of the bus and does not provide as much noise immunity as an external failsafe network. Depending on your application, SN65HVD3080E can operate with or without external failsafe biasing.

because input hysteresis is -0.2V to -0.08V

Hysteresis voltage Vhys is 30mV. This is a characteristic of the receiver and not of the internal failsafe circuitry.

And, Can it drive 32UL/0.13UL = 246 node for BUS?

SN65HVD3080E features a 1/8 Unit-Load allowing 256 Nodes per Bus

Regarding Attenuation

If they use TP cable of Figure 3. Total 4pcs (master, slave) /1Bus.

Could you please share a diagram of the node layout? What bus configuration is being used? Where is termination to be located? Where is the external failsafe biasing network intended to be placed?

(Attenuation allowed in transmission line) = -6 - (-1)x4 = -2dB

This does not seem correct for a four-node system. The total attenuation allowance between any transmitter and receiver should be no more than -6dB. In a point-to-point system (2-nodes) with a failsafe biasing network implemented at both nodes with -1dB attenuation, the remaining attenuation allowance for cabling is -4dB. 

(-6dB) - 2*(-1dB) = (-4dB)

This calculation changes when more nodes are added to the bus. Depending on layout and biasing network implementation, more attenuation allowance can be budgeted for cabling. For example, in a three-node system, the middle node will not see attenuation caused by an end node that it is not communicating with. Please share more information on your application for a more relevant explanation.

From figure3 [link], If they want to use 100m cable, they can use until 650kbps, 100m.

Using the above example with a cable attenuation allowance of -4dB and 100m worth of cable,Figure 3 from the App Note shows a maximum signaling rate between 2MHz and 3MHz.

-4dB * 30m / 100m = -1.2dB -> ~2.5MHz

Let me know if this makes sense.

Regards,
Eric

Hi Muk,

1)      I believe the two equations in question are equal. You may choose whichever is easier to calculate given your values.

2)      Signal attenuation is caused by voltage drop across the transfer medium. In this case, the signal must travel through the resistance (and impedance) of the cable as well as the series resistance of R3. The attenuation will be related to the resistance to AC ground on the receiver-side. This includes the receiver load (Rin) and fail-safe resistors (R1). Because the differential signal does not travel through termination (R2) these values are not taken into account for this calculation. Circuits b) and c) in the application report show an attenuation value of 0dB because there is no series resistance between the driver and receiver (cable losses are not taken into account).

The values on the table for circuit a) may be misleading because I believe they are rounded up to the nearest 0.5dB. The simplified calculation for a DC case estimates attenuation using the voltage divider form.

Using this equation, the actual attenuation values for the circuits are approximately -0.449 and -0.987 respectively. Using the same equation, we can add the cable resistance and any other series resistance to R3 to approximate total system attenuation. Note this only includes the series resistance between the driver and receiver, therefore for devices with separate ports for driver and receiver, R3 can be included only on the receiver ports.

Assuming two series resistances R3 and the following values, our calculations would yield:

With a minimum driver output differential of 1.5v, our resulting input differential with our calculated attenuation would be

This value is much greater than the required 200mV differential need at the receiver and indicates an acceptable attenuation amount for the described system.

3)      Attenuation may be calculated from driver to receiver without including peripheral series resistances. This means in a system with 4 nodes, attenuation can be estimated only between the two farthest away nodes (assuming all fail-safe series resistances are the same) as these nodes will have the most cable losses between them. Adding additional nodes to the center of the bus does not increase signal attenuation (not taking signal reflections and additional loading into account).

When providing fail-safe biasing to multiple nodes, it is important to ensure that the bus is not overly-biased and will still allow all drivers to pull a valid differential on the bus (primarily of opposite polarity than the fail-safe condition). This is reflected in the unit load calculation. The values currently selected would add 22.4 unit loads to the bus with the depicted fail-safe circuitry. Decreasing the load on the bus will likely provide better performance. An ideal bus would be able to operate with 32 unit loads, but many real systems would struggle under such conditions. You may consider using different resistor values to decrease the load presented by these circuits.

Let me know if this information makes sense.

Regards,
Eric

Hi Muk,

The values in the application report are only used to show the attenuation added by the fail-safe circuitry. Because these values are given in dB, they can be added to the attenuation value calculated for your system based on cable losses. Note that the series fail-safe resistance (R3) should only be added at the receiver (A and B pins). If these resistors are included between the bus and the driver, the signal will be highly attenuated. 

The resulting schematic for an equivalent circuit would look as follows:

Note that no Vcc supply is present - AC signal ground is used instead. 

The simulations I ran with the above model resulted in a total system attenuation of near -1.7dB. When the failsafe network is removed (R3, R8 shorted; R4, R5 open) the system attenuation showed near -1.2dB. Simulating only the fail-safe circuitry (R1, R9 shorted; R2, R12 open) yields attenuation of -0.45dB. Note these values sum to the total system attenuation. The equation used in the my previous reply will show the same attenuation due to the fail-safe network.

If customer use 176, also In case of having some nodes, the customer believe R3 is needed for all nodes for the design of fail safe including short protection. Is it correct?
All nodes that require bus-short protection will require R3. Again, this series resistance should not be included on the driver-side of the network - only before the receiver. 

Let me know if this makes sense.

Regards,
Eric