"Dying Gasp" from a Circuit Standpoint


The Metro Ethernet Forum (MEF) specifies requirements for carrier-grade Ethernet services. One the features they define is the "dying gasp". A dying gasp is defined as follows:

Dying Gasp is a message (or signal) sent by the Customer Premises Equipment (CPE) to the service provider when a power outage occurs.

In most cases, the hardware does not know that the power is failing until the power is already starting to go away. The dying gasp protocol requires that three "I am dying" messages be sent before the CPE gear goes down for good, which means the CPE must stay operating for about 3 msec after the power has gone away. To accomplish this feat, we need to include some local energy storage with MEF-compliant Ethernet hardware. This post will examine how we can estimate the amount of local energy storage required.


We will begin by assuming the power supply architecture shown in Figure 1.

Figure 1: Circuit Model for "Dying Gasp" Capacitive Backup.

Figure 1: Circuit Model for "Dying Gasp" Capacitive Backup.

There are a few points that should be made about this power supply architecture:

  • The input circuit will only conduct current one way.
    We need to make sure that as the power drops away, the power we need for our Ethernet interface is not drawn out of the input. In Figure 1, I show a diode, but there are number of ways of isolating the backup capacitor from the dying power source. In most cases, FETs would be more efficient than diodes, but diodes are simpler to understand. However, all this is an implementation detail.
  • The load requirements are modeled as a current source.
    In most cases, we are trying to provide backup energy for a switching power supply, which draw relatively constant power over a range of voltages.
  • Energy is stored in capacitors.
    In most cases, capacitors are the cheapest and easiest way of storing backup energy. This assumes the required backup time is small (~msec). As times get longer, batteries look better.

Energy-Based Capacitance Requirement Determination

If first worked the problem from an energy standpoint. Recall from basic physics that energy stored on a capacitor is given by Equation 1.

Eq. 1 {{E}_{Capacitor}}=\frac{1}{2}\cdot {{C}_{Backup}}\cdot V_{Capacitor}^{2}

The energy needed to power our MEF-compliant device is given by Equation 2.

Eq. 2 {{E}_{Total}}=P\cdot {{T}_{DyingGasp}}=\frac{1}{2}\cdot {{C}_{Backup}}\cdot \left( V_{Initial}^{2}-V_{Final}^{2} \right)


  • P is the power draw of our MEF-compliant device.
  • TDyingGasp is the time required to transfer the three dying gasp messages.
  • VInitial is the initial voltage on the backup capacitor. This is the voltage on the backup capacitor at the moment power drops away.
  • VFinal is the final voltage on the backup capacitor and is the minimum voltage at which load input can function.
  • CBackup is the capacitance of the backup capacitor.

We can use Equation 2 to solve for CBackup (Equation 3).

Eq. 3 \therefore {{C}_{Backup}}=\frac{2\cdot P\cdot {{T}_{DyingGasp}}}{V_{Initial}^{2}-V_{Final}^{2}}

Current-Based Capacitance Requirement Determination

The backup capacitance requirement is given by Equation 3, but some electrical engineers do not like energy-based arguments. They prefer arguments based on circuit parameters like voltage and current, which can be measured directly. We can obtain the same result using circuit parameters as follows.

The current drawn by the load and the current drawn from the capacitor are given by Equation 4.

Eq. 4 {{I}_{Load}}=\frac{{{P}_{Load}}}{{{V}_{Load}}}={{C}_{Backup}}\cdot \frac{d{{V}_{Load}}}{dt}

Equation 5 shows Equation 4 broken up into its differential form.

Eq. 5 {{V}_{Load}}\cdot d{{V}_{Load}}=\frac{{{P}_{Load}}}{{{C}_{Backup}}}\cdot dt

We know can integrate both sides of Equation 5, as is illustrated in Equation 6.

Eq. 6 \int\limits_{{{V}_{Initial}}}^{{{V}_{Final}}}{{{V}_{Load}}\cdot d{{V}_{Load}}}=\int\limits_{0}^{T}{\frac{{{P}_{Load}}}{{{C}_{Backup}}}\cdot dt}
\left. \frac{V_{Load}^{2}}{2} \right|_{{{V}_{Initial}}}^{{{V}_{Final}}}=\frac{V_{Initial}^{2}-V_{Final}^{2}}{2}=\frac{{{P}_{Load}}}{{{C}_{Backup}}}\cdot T
\therefore {{C}_{Backup}}=\frac{2\cdot P\cdot {{T}_{DyingGasp}}}{V_{Initial}^{2}-V_{Final}^{2}}

As we would expect, both methods yield the same result.

Empirical Example

I threw the calculations into Mathcad and put the screenshot into Figure 1.

Figure 1: Calculation Used for Actual Product.

Figure 1: Calculation Used for Actual Product.


One of the more difficult parts of product development is determining requirements, a process I call "The Treasure Hunt." The analysis shown here describes a very common type of analysis that is done to determine capacitor requirements for backup systems.

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6 Responses to "Dying Gasp" from a Circuit Standpoint

  1. Gracias por esto, la verdad que es bueno conseguir cosas asi, ahora mismo iniciaré un trabajo del cual este tema me va como anillo al dedo.

  2. Pingback: Calculating a Parameter’s Worst-Case Range | Math Encounters Blog

  3. Vishal says:

    I have designed a Dying Gasp circuit using same circuit and formula.
    My requirement is 25W for 10msec. And input voltage range is 48V to 36V.
    For that Cbckp= 5mF.

    But My voltage is dropping to 24V. After that, it is maintaining the voltage to 24 V. Can someone tell me why is this happening and what can be done to rectify this?

    • mathscinotes says:

      Hi Vishal,

      I do have a few questions for you first:

      • Did you verify that the capacitor bank is being charged to the expected voltage? 48 V?
      • Is the system working? You mention that the capacitor bank sits at 24 V, but do not say whether the system runs at all.

      If the system is not working at all, it may be because your capacitor bank has a high output resistance. The output resistance will limit the current available and prevent the system from working. What kind of capacitor are you using? The design formula assumes an output resistance that is insignificant, which may not be true. I sometimes have to parallel capacitors to reduce the effective output resistance. You may need to simulate your system in LTSpice or equivalent to determine the effect of the series resistance.

      I have seen capacitor-driven systems drop to an odd voltage that is out-of-range and sits there before. I am going to speculate that your capacitor energy source is driving a small switching power supply – correct? While the load input is rated to run from 36V to 48V, it likely operates significantly outside this range. If so, 24 V could be the input voltage at which the switching supply no longer presents a significant current load. The capacitor bank will sit there because the load is no longer drawing an input current.

      I took a quick look at what the required capacitance should be. I expected that 0.5 mF should handle it (see calculation below). A 5 mF capacitor has plenty of capacitance, but may have too much output resistance.


  4. Vishal says:

    Hello Sir,

    Thank you for your response, There was a mistake in my circuit, I choose a wrong diode. I have corrected it. Now my simulation results are matching the requirements.


    • mathscinotes says:

      Excellent! Thanks for telling me how it turned out. Good luck on your future circuit design work.


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