Pressure Half Time (PHT)

• Related:
  • Aortic Regurgitation
• The time interval required for the initial pressure gradient to be reduced to its half
• Obtained using Continuous Wave Doppler

In the following example from the Mayo Board Review course, the initial $V_{\text{max}}=19$ mmHg. Half of that is 9.5 mmHg. Pressure Half Time (PHT) is the time it takes for the initial pressure gradient to be reduced by half, i.e. the time it takes for the pressure to drop from 19 mmHg → 9.5 mmHg in the provided example.

The above is a tedious method for calculating PHT that isn’t practical for the board exam. Instead, for the Echo board exam, you’ll calculate PHT using deceleration time with the following equation:

$$\text{PHT}=0.29\times \text{Deceleration Time (DT)}$$

This is based on the observation that PHT is always 29% of your deceleration time (DT). DT is defined as the time from the peak signal of your CWD with a line drawn down your doppler signal until you reach the baseline. In the example from the Mayo Board Review video, DT was 700 msec, ∴ the PHT was 210 msec.

Mitral Valve Area (MVA) from PHT

PHT to assess MVA doesn't work as well in patients with MAC

$$\text{Mitral Valve Area (MVA)}=\frac{220}{\text{PHT}}$$

📝 A long PHT → small valve area

Question: 48 year old female with PMH of rheumatic mitral stenosis. Echo shows a peak gradient of 19 mm Hg and mean gradient of 9 mm Hg (heart rate 85 BPM). Doppler tracing is shown below. What is the mitral valve area?

$$\begin{array}{rl}\text{PHT}=0.29\times \text{PHT}& =0.29\times 800\\ & =232\text{ ms}\\ \text{MVA}=\frac{220}{\text{PHT}}& =\frac{220}{232}\\ & =0.948{\text{ cm}}^2\end{array}$$

PHT in Aortic Regurgitation

See note

Pulmonary Regurgitation

• Short PHT with pulmonary regurgitation can indicate the following:
  • Severe Pulmonary Regurgitation
  • restrictive cardiomyopathy
  • constriction