5G NR TBS vs CBG: Understanding the Differences

This page compares 5G NR TBS (Transport Block Size) vs CBG (Code Block Group) and mentions the difference between TBS (Transport Block Size) vs CBG (Code Block Group) used in the 5G NR system.

Introduction

• TB (Transport Block) is large in size, and hence 5G NR introduces small size units called code blocks to exploit a better mechanism for reliable transmission and reception. Code blocks are grouped into code block groups (CBGs).
• CBG-based transmission with single or multi-bit HARQ-ACK feedback is supported. HARQ-ACK multiplexing is supported for CBG-based retransmission.

TBS | Transport Block Size

• Similar to the LTE system, in 5G NR, TBs are also split into smaller size multiple CBs (Code Blocks).
• Moreover, a new concept of CBGs has been introduced in 5G NR, in which multiple CBs are combined to form a CBG (Code Block Group).
• 5G NR provides ACK/NACK for the delivery of each of the data units, which are CBGs after segmentation from TBS.
• CBGs are indicated using DCI formats viz. DCI format_0_1 and DCI format_1_1.

UE procedure for grouping Code Blocks to Code Block Groups (CBGs)

• If a UE is configured to receive code block group (CBG) based transmissions by receiving the higher layer parameter “codeBlockGroupTransmission” for PDSCH.
• 5G NR UE shall determine the number of CBGs for a transport block reception as follows:

$M = min(N, C)$

where,

• N = the maximum number of CBGs per transport block as configured by maxCodeBlockGroupsPerTransportBlock for PDSCH
• C = the number of code blocks in the transport block according to the procedure defined in Subclause 7.2.3 of 3GPP TS 38.212 document.

Define:

$M_1 = mod (C, M)$ $K_1 = \lceil C/M \rceil$ $K_2 = \lfloor C/M \rfloor$

• If $M_1 > 0$, CBG $m$, $m = 0, 1, ..., M_1 - 1$ consists of code blocks with indices $m \cdot K_1 + k$, $k = 0, 1, ..., K_1 - 1$.
• CBG $m$, $m = M_1, M_1 + 1, ..., M - 1$ consists of code blocks with indices $M_1 \cdot K_1 + (m - M_1) \cdot K_2 + k$, $k = 0, 1, ..., K_2 - 1$.