RxDoctor Payments Data

Clinical Pharmacology: industry payments

Drug and device makers reported $1,072,964 in payments to providers registered in this specialty across Open Payments program years 20192025.

Providers
36
Received a payment
27 (75.0%)
Total reported
$1.1M
Average, those paid
$39,739

The average is computed only over providers who received something, because including the 9 who received nothing would produce a number that describes neither group. Totals are dominated by a small number of large royalty and consulting arrangements; the typical payment in this specialty is far smaller than the average implies.

What Medicare pays clinical pharmacology clinicians

Part B allowed per clinician, 2024
$10,548
Of which, services
$10,548
Clinicians billing Part B
2

Across 2 clinical pharmacology clinicians who billed Original Medicare Part B in 2024, Medicare allowed $21,096 in total — an average of $10,548 each, of which $10,548 was for services and the rest for drugs administered in the office.

This is not a salary. The rest is what Medicare allowed for services these clinicians billed — Original Medicare fee-for-service only. It excludes Medicare Advantage, which now covers more than half of Medicare enrollees, along with Medicaid and every commercial insurer, so for most clinicians it is a fraction of their practice. It is also gross revenue: staff, rent, supplies and malpractice premiums all come out of it before anyone is paid. Anyone quoting a figure like this as physician pay is misreading it.

By state

StateProvidersTotal reportedPer provider
Kansas1$493,190$493,190
California4$294,586$73,647
New Jersey2$196,779$98,389
Maryland2$46,141$23,071
Missouri2$12,963$6,481
Tennessee1$9,551$9,551
Washington1$7,982$7,982
Ohio1$5,674$5,674
Texas2$3,898$1,949
Delaware1$744$744
Louisiana2$427$214
Oklahoma3$247$82
Pennsylvania2$189$94
New Mexico1$181$181
Kentucky1$125$125
Nevada1$119$119
New York1$59$59
Minnesota1$36$36
Colorado1$32$32
Idaho1$22$22
Rhode Island1$21$21
Florida2$0$0
Arizona1$0$0
North Carolina1$0$0