Billable / Specific Code
ICD-10-CM S82.145 is the diagnosis code for Nondisplaced bicondylar fracture of left tibia. This code falls under the section "Injuries to the knee and lower leg" within Chapter 19 — Injury, poisoning and certain other consequences of external causes (S00-T88). It is a billable/specific ICD-10-CM code that can be used to indicate a diagnosis for reimbursement purposes. Medical coders and healthcare providers use this code to document and classify diagnoses in electronic health records, insurance claims, and clinical databases.
Always refer to the official ICD-10-CM Tabular List for complete coding guidelines. Ensure documentation supports the specificity of the code selected. When in doubt, consult a certified medical coder or the latest CMS guidelines.
Understanding where S82.145 sits in the ICD-10-CM classification helps ensure proper coding:
Yes, S82.145 is a billable/specific ICD-10-CM code that can be used to indicate a diagnosis for reimbursement purposes.
S82.145 is the ICD-10-CM diagnosis code for "Nondisplaced bicondylar fracture of left tibia". It is used by healthcare providers to classify and document this condition in medical records and insurance claims.
The parent code of S82.145 is S82.14 ("Bicondylar fracture of tibia"). S82.145 provides a more specific classification within this category.
S82.145 is located in Section S80-S89 — "Injuries to the knee and lower leg" within Chapter 19 of the ICD-10-CM Tabular List.
Use S82.145 when the patients documented diagnosis matches "Nondisplaced bicondylar fracture of left tibia" and the clinical documentation supports this level of specificity. Always verify with the latest ICD-10-CM guidelines and payer requirements.
The ICD-10-CM code for Nondisplaced bicondylar fracture of left tibia is S82.145.
Yes, S82.145 can be used as a primary diagnosis code since it is billable and specific.
S82.145 is in Chapter 19 of the ICD-10-CM Tabular List.
Yes, S82.145 is a valid ICD-10-CM code for the 2026 fiscal year, subject to official CMS updates.