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How critical is it for the A/B differential signal lines of SN65HVD1780DR to have matched trace lengths on PCB?

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  • 1
    \$\begingroup\$ Does it say anything in the datasheet? There should be a "Layout Specifications" section. \$\endgroup\$ – KingDuken Jun 5 at 15:20
  • \$\begingroup\$ What speed are you driving the RS485 bus at? \$\endgroup\$ – Voltage Spike Jun 5 at 15:45
  • \$\begingroup\$ Compute skew vs % jitter of bit @ C/√Er \$\endgroup\$ – Sunnyskyguy EE75 Jun 5 at 15:55
  • \$\begingroup\$ 9600 bps is the speed of data on the bus \$\endgroup\$ – be-ee Jun 5 at 15:56
  • \$\begingroup\$ not an issue.... consider <60 ps / cm \$\endgroup\$ – Sunnyskyguy EE75 Jun 5 at 15:57
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9600 bps equates to about 0.104 milliseconds per bit (1/9600).

The signal speed on a PCB microstrip (external trace) is calculated by:

enter image description here

where Vs is the velocity of the signal, c is the speed of light (11.8 inches per nanosecond), and Ereff is the effective dielectric constant (about 2.92 for FR-4).

The signal speed on a PCB stripline (internal trace) is calculated by:

enter image description here

where again, Vs is the velocity of the signal and c is the speed of light. Er is the dielectric constant of FR-4 which is usually around 4.

If you do the math out, the signal velocity on a microstrip is approximately 6.9 inches per nanosecond and the signal velocity on a stripline is approximately 5.9 inches per nanosecond. If your bit rate is such that each bit takes 0.104 milliseconds (104000 nanoseconds), then you would need a length difference of roughly 717,600 inches (11.3 miles) for a microstrip or 613,600 inches (9.7 miles) for a stripline to present enough delay in one of the differential lines for the receiver to miss an entire bit. You really don't need to worry about length matching on a PCB when dealing with only 9600 bps.

NOTE: Feel free to check my math, I rushed through this on my lunch break.

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