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# How can I increase a power supply’s hold-up time?

April 15, 2020

### What is hold-up time?

When an AC-DC power supply’s input voltage is interrupted, during a brown out condition or a very brief power failure, the DC output will only remain within regulation for a short period of time. This is specified on the power supply datasheet as the hold-up time. During this hold-up time, the power supply relies on energy stored in its capacitors to maintain operation.

Figure 1 shows a simple block diagram of a power supply.

The AC input voltage is filtered, rectified and boosted to provide a DC bus voltage of around 390V. The DC-DC converter section of the power supply provides primary-secondary isolation and reduces the 390V bus down to the desired DC output voltage. The function of the high voltage bus electrolytic capacitor C1 is to reduce the ripple voltage on the 390V bus and store energy to keep the DC-DC converter operational during brief interruptions to the AC input.

Depending on the application, the length of the hold-up time varies from a half or full cycle of the incoming AC 50/60Hz voltage. This is typically 8 to 20ms at 100% load, but is normally sufficient to avoid electronic equipment from having to restart or reboot.

### When is extended hold-up time required?

The medical industry’s concern regarding hold-up time has increased since the release of the EN 60601-1-2; 2015 (Ed4) immunity standard. Primarily created to address the growing number of products used in home healthcare, this standard specifies multiple AC voltage dips ranging from 20ms to 5 seconds. The longer outages are addressed by batteries or ensuring that no harm will occur to the patient or operator if the power supply output voltage drops out of the regulation band.

Airborne equipment is covered by the DO-160 standard. Section 16 refers to power input, simulating conditions of aircraft power from before engine start (using auxiliary ground based power) to after landing, including emergencies. The requirement is for a hold-up time of at least 200ms.

### What techniques are available to increase hold-up time?

There are various methods to extend power supply hold-up time, each with advantages and disadvantages. As the amount of energy stored in a capacitor C is calculated as: ½ x C x V^{2}, to increase that energy storage and hence the hold-up time, either the amount of capacitance or the voltage on the capacitor has to increase. Since V is squared, an increase in the value of V will have a greater impact.

The following examples are based on a hypothetical system requiring a load of 150W at 12V, with a minimum output voltage requirement of 11.5V as the output decays to zero; 200ms is the desired hold-up time.

### 1. Using a higher rated power supply and operating it at a reduced load

If we were to use the 150W 12V output TDK-Lambda **RWS150B-12** power supply and operate it at 100% load, we would have a hold-up of just over 30ms based on the evaluation test data. See Figure 2.

As expected, the hold-up time is dependent on the output load drawn. The greater the load, the quicker C1’s stored energy is depleted. To obtain 200ms hold-up, we could use the TDK-Lambda’s 1500W rated **CUS1500M-12** power supply. See Figure 3.

The larger value of “C1” in the CUS1500M-12 would provide enough energy to hold-up a 150W load for over 200ms. This would be a simple off the shelf solution, but a much larger, more expensive power supply would be required.

### 2. Adding capacitance across the power supply output terminals or load

At first glance adding additional capacitance (C3) across the output seems like an easy solution. Based on the equation 2 in the appendix, however, C3 would have to be a massive 4,595,745uF.

C = 2 x Pout x t / (V^{2} – Vend^{2})

C = 2 x 150 x 0.18 / (12^{2} - 11.5^{2})

Note as the power supply already has 20ms of hold-up capability, t is reduced to 180ms (0.18s).

Even with a supercap, this would consume a large amount of space. One other major concern is the over-current characteristics of the power supply during initial turn on. An uncharged C3 would appear to the power supply control circuit as a dead short across the output. The power supply would most likely fail to establish a 12V output when initially turned on.

### 3. A customized power supply with a larger high voltage bus capacitor (C1)

As previously mentioned, the energy stored in a capacitor is equal to ½ x C x V^{2}, so adding capacitor C1a across the high 390Vdc bus has a greater effect than adding capacitance across the 12V output. See Figure 5.

As the DC bus is not accessible on most power supplies, adding capacitors would involve creating a custom or modified standard design. This would involve engineering charges and safety re-certification fees, plus time to make the modification.

If the power supply to be modified had a hold-up time of 20ms, increasing it to 200ms would require adding the equivalent of nine more C1 capacitors. As C1 typically occupies 5 to 6% of the internal space of a power supply, it would increase the size of the power supply by around 50% for the same product height.

A power module based solution could also be created using a non-isolated AC-DC 360Vdc output converter (TDK-Lambda’s **PF-A series**) and a high voltage input isolated DC-DC 12V output converter (TDK-Lambda **PH-A280 series**). This again would require a custom board design with engineering and safety certification charges.

### 4. A 48V or 60V output AC-DC power supply and a wide range input DC-DC Converter

For an off-the-shelf standard product solution, a 48V output AC-DC power supply could be used with an isolated wide-range input 18 to 75V DC-DC converter with a 12V output. See Figure 6. The DC-DC converter may require a heatsink or cold plate for cooling.

In addition to the hold-up inside the 48V output AC-DC power supply, capacitor C1 can be used for additional energy storage. We have taken advantage of the energy storage formula ½ x C x V^{2} as V is now 48V rather than 12V in scenario 1. The efficiency of the DC-DC converter is considered as 90%. The hold-up of the AC-DC power supply is 20ms, requiring an additional 180ms (t).

C = 2 x P x t / (Eff x (V^{2} – Vend^{2}))

For our example, the required value of C1 would be 2 x 150 x 0.18 / (0.9 x (48^{2} – 18^{2})) which calculates as 30,300uF.

If the 48V AC-DC power supply was replaced by one that had a 60V output, the additional capacitance could be reduced to 18,315uF. A 72V output power supply would reduce that further to 12,345uF.

### 5. A 48V output AC-DC power supply and a wide range input non-isolated DC-DC Converter

In this case, the chosen DC-DC converter is non-isolated (for higher efficiency, smaller size and lower cost) having a wide input range of 9 to 53V. See Figure 7. A non-isolated DC-DC converter would not require heatsinking.

The efficiency of the non-isolated DC-DC (TDK-Lambda's **i6A series**) converter is considered as 96%. The hold-up of the AC-DC power supply as 20ms, requiring an additional 180ms (t).

C = 2 x P x t / (Eff x (V^{2} – Vend^{2}))

For our example, the required value of C1 would be 2 x 150 x 0.18 / (0.96 x (48^{2} – 9^{2})) which calculates as 25,304uF.

### Summary

Method | Additional capacitance | Pros | Cons |
---|---|---|---|

Larger power supply | None | Standard part | Cost and size of bigger supply |

Larger output cap | 4,595,745uF | Standard part | Size and potential start-up issues |

Larger HV bus cap | 9 x C1 (~9,000uF) | Custom solution | Eng. cost and development time |

48V AC-DC + isolated DC-DC | 30,300uF | Standard parts | Space for DC-DC and heatsinking |

60V AC-DC + isolated DC-DC | 18,315uF | Uncommon 60V AC-DC Standard DC-DC | Space for DC-DC and heatsinking |

72V AC-DC + isolated DC-DC | 12,345uF | Uncommon 72V AC-DC Standard DC-DC | Space for DC-DC and heatsinking |

48V AC-DC + non-isolated DC-DC | 25,345uF | Standard parts | Space for DC-DC |

### Appendix - Calculating the required capacitance

Below are the equations for calculating the required hold-up capacitance. The units of measure are Energy – Joules, Power – Watts, time – seconds, Voltage – Volts, Capacitance – Farads, Efficiency – Percentage.

Energy E is the product of the output power Pout and time t.

__Equation 1:__

E = Pout x t

Stored energy (E) is half the product of the Capacitance (C) and the voltage (V) squared.

E = 0.5 x C x V^{2}

Rearranging for C we get:

C = 2 x E/V^{2}

Note: The voltage on capacitor C will drop as it discharges into the load. At some point the voltage will be too low for the load to function leaving unused energy in the capacitor. That voltage point will be referred to as “Vend”. In the scenarios 2, 4, 5, 6 a voltage of 11.5V is used.

Our “useful” energy for hold-up will be the initial minus the remaining energy.

E = (0.5 x C x V^{2}) – (0.5 x C x Vend^{2})

Factoring the equation we get:

E = 0.5 x C x (V^{2} – Vend^{2}) or C = 2 x E / (V^{2} – Vend^{2})

__Equation 2:__

Substituting for E from equation 1 we have:

C = 2 x Pout x t / (V^{2} – Vend^{2})

In cases where a separate DC-DC converter is used, that converter will not be 100% efficient (Eff) and we need to adjust the value of our AC-DC output power Pout to compensate:

P = Pout / Eff

__Equation 3:__

C = 2 x P x t / (Eff x (V^{2} – Vend^{2}))