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Absorption Models · Tutorial 63 of 797

Zero-Order Absorption as a Constant-Rate Input

Learn to express finite-duration zero-order absorption as a constant input rate and calculate the depot amount remaining at any time.

Intermediate 9 min read

What You'll Learn

  • Define a constant-rate absorption input that stops after a specified duration.
  • Relate dose, bioavailability, input rate, and absorption duration.
  • Calculate remaining depot amount before, at, and after the end of absorption.
  • Distinguish physical depot depletion from bioavailable systemic input.
  • Identify assumptions and practical signs that a zero-order input may be inadequate.
  • Compare finite zero-order input with first-order absorption and delayed input.

Constant-Rate Input Over a Finite Duration

Some formulations or delivery processes are represented as supplying drug to the body at a constant rate for a limited time. This is called zero-order absorption: while the process is active, the amount leaving the absorption depot per unit time is constant, rather than being proportional to the amount still in the depot. The input ends when the specified duration is reached. This tutorial defines that input and tracks the depot amount; it does not derive the resulting concentration profile.

Let \(D\) be the dose placed in the depot at the start of absorption, in units such as mg. Let \(T\) be the absorption duration, in hours, and let \(\tau\) be the absorption start time. The depot is assumed to contain \(D\) immediately before time \(\tau\). If there is no lag, \(\tau=0\). Let \(F\) be the fraction of the depot dose that reaches the systemic circulation, with \(0\leq F\leq1\). Separating depot depletion from systemic availability keeps the physical amount remaining in the depot distinct from the amount that ultimately enters systemic circulation.

Definition: For a dose \(D\) absorbed at a constant rate over duration \(T>0\), the depot-depletion rate is \(R_{\mathrm{dep}}=D/T\). The remaining physical depot amount at time \(t\) is \(A_{\mathrm{dep}}(t)=D\) for \(t\leq\tau\), \(A_{\mathrm{dep}}(t)=D[1-(t-\tau)/T]\) for \(\tau<t<\tau+T\), and \(A_{\mathrm{dep}}(t)=0\) for \(t\geq\tau+T\). If a fraction \(F\) reaches systemic circulation, the constant systemic input rate during absorption is \(R_{\mathrm{sys}}=FD/T\), and the total systemic input is \(FD\).

The rate units follow directly from the dose and duration: mg divided by hours gives mg/h. The depot amount is linear in elapsed absorption time, not exponential. Define elapsed absorption time as \(u=t-\tau\) after absorption begins. The piecewise form matters: before the start the depot has not begun to decline, and after the end it cannot become negative.

Why the Dose-to-Rate Relationship Holds

During the active interval, a constant rate \(R_{\mathrm{dep}}\) removes \(R_{\mathrm{dep}}\Delta t\) amount over any interval of length \(\Delta t\). If the entire depot dose \(D\) is depleted over duration \(T\), conservation of amount requires that the total removed amount equal \(D\).

$$ D=R_{\mathrm{dep}}T \qquad\Longrightarrow\qquad R_{\mathrm{dep}}=\frac{D}{T}. $$

For \(0<u<T\), the amount removed by elapsed time \(u\) is \(R_{\mathrm{dep}}u\). Subtracting that amount from the initial depot amount gives \(A_{\mathrm{dep}}=D-R_{\mathrm{dep}}u\). Substitution of \(R_{\mathrm{dep}}=D/T\) yields the linear depletion equation in the definition. At \(u=T\), it gives \(A_{\mathrm{dep}}=0\), as required. If only a fraction \(F\) of the depot outflow reaches systemic circulation, multiplying the depot-depletion rate by \(F\) gives the systemic input rate; it does not change the physical depot-depletion rate under this formulation.

$$ A_{\mathrm{dep}}(t)=D-\frac{D}{T}u =D\left(1-\frac{u}{T}\right),\quad 0\leq u\leq T, \qquad R_{\mathrm{sys}}=F\frac{D}{T}. $$

This formulation assumes the rate is constant from the start through the end of absorption, the full depot amount is depleted by time \(\tau+T\), and there is no interruption or additional input to that depot. It also treats \(F\) as a fixed fraction of the outflow. Those are modeling assumptions, not conclusions guaranteed by the dose and duration alone.

Worked Example: Depot Amount and Systemic Input Rate

Suppose a hypothetical 600 mg dose begins absorption at time zero and is absorbed uniformly over 4 hours. Assume \(F=0.80\). Find the depot-depletion rate, systemic input rate, and physical amount remaining after 1.5 hours.

The depot-depletion rate is the dose divided by duration:

$$ R_{\mathrm{dep}}=\frac{600\ \mathrm{mg}}{4\ \mathrm{h}}=150\ \mathrm{mg/h}. $$

The systemic input rate is the fraction \(F\) times the depot-depletion rate:

$$ R_{\mathrm{sys}}=0.80(150\ \mathrm{mg/h})=120\ \mathrm{mg/h}. $$

After 1.5 hours, the physical amount remaining is

$$ A_{\mathrm{dep}}(1.5)=600\ \mathrm{mg}-(150\ \mathrm{mg/h})(1.5\ \mathrm{h}) =375\ \mathrm{mg}. $$

Thus 375 mg remains in the depot, while systemic input has accumulated at 120 mg/h for 1.5 hours, or 180 mg in total. The latter is \(0.80(600-375)=180\) mg. It is not correct to report the remaining physical depot as 300 mg by multiplying 375 mg by \(F\); \(F\) scales systemic availability, not the physical amount remaining under the stated model.

Using the Piecewise Definition at Boundaries

A finite-duration input is active only during a specified interval. If its start time is \(\tau\), the active interval is from \(\tau\) to \(\tau+T\). The depot amount stays at \(D\) before the start, declines linearly during the interval, and is zero from the endpoint onward. This prevents two common arithmetic errors: subtracting absorption before it starts and extending a constant rate beyond the time needed to deplete the dose.

Worked Example: A Lagged Start and an Observation After Completion

A hypothetical 900 mg depot dose has \(T=6\) hours and begins absorption after a 2-hour lag. Find the depot-depletion rate and the depot amount at clock times 1 hour, 5 hours, and 9 hours.

The rate during absorption is

$$ R_{\mathrm{dep}}=\frac{900\ \mathrm{mg}}{6\ \mathrm{h}}=150\ \mathrm{mg/h}. $$

At 1 hour, absorption has not started because \(1<\tau=2\), so \(A_{\mathrm{dep}}(1)=900\) mg. At 5 hours, elapsed absorption time is \(u=5-2=3\) hours. Therefore,

$$ A_{\mathrm{dep}}(5)=900\ \mathrm{mg}-(150\ \mathrm{mg/h})(3\ \mathrm{h})=450\ \mathrm{mg}. $$

Absorption ends at clock time \(\tau+T=8\) hours. Since 9 hours is after that endpoint, \(A_{\mathrm{dep}}(9)=0\) mg. The rate is not continued to 9 hours: doing so would imply depletion of more than the original dose.

At exactly \(t=\tau+T\), the during-absorption expression also gives zero. The endpoint can therefore be included in either adjacent piece without changing the depot amount. For the input rate, it is convenient to say the rate applies during the active interval and is zero after completion; the single instant at the boundary does not change the total amount delivered.

Duration, Rate, and Dose Must Agree

Any two of dose, rate, and duration determine the third, provided they refer to the same amount and the same process. A proposed rate and duration imply a total amount \(R_{\mathrm{dep}}T\). If that amount differs from the specified depot dose, the proposed values do not describe complete absorption of that dose under a constant-rate model.

Worked Example: Checking a Proposed Rate and Duration

A hypothetical 720 mg depot dose is proposed to be absorbed at a constant rate of 120 mg/h. Check the implied duration. If \(F=0.75\), also calculate the systemic input rate and total systemic input.

From \(D=R_{\mathrm{dep}}T\), the duration is

$$ T=\frac{D}{R_{\mathrm{dep}}} =\frac{720\ \mathrm{mg}}{120\ \mathrm{mg/h}}=6\ \mathrm{h}. $$

The systemic input rate and total systemic input are

$$ R_{\mathrm{sys}}=0.75(120\ \mathrm{mg/h})=90\ \mathrm{mg/h}, \qquad FD=0.75(720\ \mathrm{mg})=540\ \mathrm{mg}. $$

The depot is depleted by 6 hours, while the model assigns 540 mg of the dose to systemic availability over that period. For example, after 2 hours the physical depot amount is \(720-(120)(2)=480\) mg. The remaining physical amount is not obtained by subtracting systemic input from \(D\), because the stated model allows only the fraction \(F\) of depot outflow to reach systemic circulation.

Assumptions, Diagnostics, and Alternatives

A zero-order input is useful when the scientific model supports a nearly constant delivery rate over a finite interval. It is a compact idealization: the model does not explain why the rate is constant, nor does it accommodate a rate that gradually accelerates or slows unless the formulation is extended. Before using it, check that the dose, rate, and duration are compatible; that the start time is defined; and that the assumed depot amount cannot become negative over the times being modeled.

Potential warning signs include a rate-duration product inconsistent with the dose, evidence that input begins gradually, or systematic disagreement with observations near the presumed start or end of absorption. A flat input rate is an assumption about drug entry, not a claim that measured concentrations must rise linearly: disposition processes also affect concentration. The concentration consequences require a separate model.

The main alternative in the earlier material on first-order absorption is an input rate proportional to the amount remaining in the depot. Under first-order absorption, the depot declines exponentially and the input rate changes over time; under zero-order absorption, depot depletion is linear and the rate remains fixed until the finite endpoint. A lag-time model shifts when input begins but does not, by itself, specify whether the subsequent input is constant-rate or first-order. These features can be combined only by specifying both the start time and the post-start input mechanism.

FeatureFinite zero-order inputFirst-order absorption
Input rate while activeConstantProportional to amount remaining
Depot amount over timeDeclines linearly to zeroDeclines exponentially
Input endpointSpecified finite durationNo fixed depletion time in the usual model
Key parametersDose and duration (or rate)Initial depot amount and absorption rate constant

A practical pitfall is to equate a constant systemic input rate with a constant concentration increase. The former describes drug entering the systemic circulation per unit time; it does not account for drug distribution or elimination. Another is to confuse the dose \(D\) with the systemic amount \(FD\). State which amount is being used whenever rate units or remaining amounts are reported.

Key takeaway: For a dose \(D\) depleted uniformly over duration \(T\), the depot rate is \(D/T\), and the physical depot amount falls as \(D(1-u/T)\) during the active interval. Bioavailability \(F\) scales systemic input to \(FD/T\); it does not change the physical depot amount in this formulation.

Check Your Understanding

Answer each question using the finite-duration constant-rate model. Solutions are included below the questions.

  1. A 500 mg depot dose is absorbed over 5 hours. What is the depot-depletion rate, and how much remains after 3 hours?
  2. A 400 mg dose begins absorption at time 1 hour and has duration 4 hours. What remains at clock time 3 hours, and when does absorption end?
  3. If \(D=800\) mg, \(T=8\) hours, and \(F=0.60\), calculate the depot-depletion rate, systemic input rate, and total systemic input.
  4. Explain why a proposed dose of 600 mg, rate of 100 mg/h, and duration of 5 hours are inconsistent with complete constant-rate depletion of the dose.

Solutions. (1) \(R_{\mathrm{dep}}=500/5=100\) mg/h. After 3 hours, \(A_{\mathrm{dep}}=500-(100)(3)=200\) mg. (2) The active interval is from hour 1 to hour \(1+4=5\). At clock time 3, elapsed absorption time is 2 hours, so \(A_{\mathrm{dep}}=400-(400/4)(2)=200\) mg. (3) \(R_{\mathrm{dep}}=800/8=100\) mg/h, \(R_{\mathrm{sys}}=0.60(100)=60\) mg/h, and total systemic input is \(FD=0.60(800)=480\) mg. (4) The proposed rate and duration deliver \((100\ \mathrm{mg/h})(5\ \mathrm{h})=500\) mg, not 600 mg. For 600 mg at 100 mg/h, the duration must be 6 hours; for 600 mg over 5 hours, the rate must be 120 mg/h.