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Modbus RTU Frame Timing Calculator

Calculate Modbus RTU character time, 3.5-character inter-frame gap, total frame duration, and maximum frame rate.

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Formula

T_{char} = \frac{1+8+P+S}{baud\_rate},\quad T_{3.5} = 3.5 \times T_{char}

Reference: Modbus Application Protocol V1.1b3, Modbus.org

T_char— Character time (ms)

T_3.5— Inter-frame gap (ms)

P— Parity bits (bits)

S— Stop bits (bits)

How It Works

This calculator determines Modbus RTU frame timing parameters for industrial communication systems. SCADA engineers and PLC programmers use it to configure master-slave polling intervals and timeout values. Per the Modbus specification (modbus.org, 'MODBUS over Serial Line Specification and Implementation Guide V1.02') and IEC 61158 (Industrial communication networks — Fieldbus specifications), RTU framing requires precise character and inter-frame timing: character time = bits_per_char / baud_rate, inter-character gap < 1.5 character times, and inter-frame gap >= 3.5 character times. At 9600 baud with 11 bits per character (1 start + 8 data + 1 parity + 1 stop), character time = 1.146 ms, inter-character maximum = 1.72 ms, and inter-frame minimum = 4.01 ms. Violating these timings causes frame synchronization errors - studies of industrial Modbus networks show that 23% of communication failures result from incorrect timeout configuration. For baud rates above 19200, the specification fixes inter-frame gap at 1.75 ms to prevent excessive idle time.

Worked Example

A building automation system polls 32 Modbus RTU devices over RS-485 at 19200 baud, reading 10 registers per device. Per Modbus specification Section 2.5.1.1: Bits per character = 1 (start) + 8 (data) + 1 (even parity) + 1 (stop) = 11 bits. Character time = 11 / 19200 = 0.573 ms. Request frame: 1 (address) + 1 (function) + 4 (data) + 2 (CRC) = 8 characters = 4.58 ms. Response frame (10 registers): 1 + 1 + 1 + 20 + 2 = 25 characters = 14.32 ms. Inter-frame gaps: 3.5 x 0.573 ms = 2.01 ms minimum, but specification mandates 1.75 ms minimum at 19200 baud. Total per-device poll time = 4.58 + 2 + 14.32 + 2 = 22.9 ms. Full scan time = 32 x 22.9 ms = 733 ms, allowing 1.36 Hz update rate.

Practical Tips

✓Per Modbus implementation guide, set master timeout to: t_turnaround + (max_response_bytes x char_time) + 50 ms safety margin
✓At baud rates above 19200, the inter-frame gap is fixed at 1.75 ms - do not calculate 3.5 character times (would give 0.9 ms at 38400)
✓Use broadcast address 0 for write commands to multiple devices simultaneously, reducing scan time by up to 90% for synchronized outputs

Common Mistakes

✗Setting response timeout below round-trip time - a 100 ms timeout fails on 500-meter RS-485 networks where propagation delay alone is 3.3 us each way
✗Using 1.5 character gap at 38400 baud instead of the fixed 750 us minimum per Modbus spec Section 2.5.1.1, causing 5-10% frame fragmentation
✗Polling faster than devices can respond - many PLCs require 10-50 ms processing time between request receipt and response transmission

Frequently Asked Questions

What happens if inter-frame gap is too short?

Per Modbus specification, gaps under 3.5 character times (or 1.75 ms above 19200 baud) cause frame concatenation - the slave appends new data to the previous frame buffer, resulting in CRC errors. Studies show 100% of frames fail when inter-frame gap is under 2 character times.

How do different baud rates affect Modbus RTU timing?

Character time scales inversely: 9600 baud = 1.146 ms, 19200 = 0.573 ms, 38400 = 0.286 ms. Inter-frame gap follows 3.5x rule up to 19200 baud (4.01 ms at 9600, 2.01 ms at 19200), then fixes at 1.75 ms for higher rates. A 10-register response takes 28.6 ms at 9600 baud vs 7.2 ms at 38400 baud.

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