How to boost output hold-up time in power supplies
Many supplies have outputs that don’t stay up long enough in the event of a power loss. Here are some ways to remedy the problem.
An ac-dc power supply’s input voltage can be interrupted during circumstances such as a brown out condition or a brief power failure. When this happens, the dc output will only remain within regulation for a short period of time. The time period 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.
In a typical ac-dc supply, the ac input voltage is filtered, rectified and boosted to provide a dc bus voltage of around 390 V. The dc-dc converter section of the supply provides primary-secondary isolation and reduces the 390-V bus down to the desired dc output voltage. Typically, there is a high-voltage bus electrolytic capacitor that reduces the ripple voltage on the 390-V bus and stores 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 typically varies from a half to a full cycle of the incoming ac 50/60-Hz voltage. This period is typically 8 to 20 msec at 100% load but is normally sufficient to keep the attached electronic equipment from having to restart or reboot.
However, some applications demand an extended hold-up time. The medical industry’s concern regarding hold-up time has risen 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 20 msec to five seconds. The longer outages are generally addressed by including batteries or otherwise ensuring that no harm will befall 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 200 msec.
How to increase hold-up time
There are various methods to extend power supply hold-up time, each with advantages and disadvantages. The amount of energy E stored in a capacitor C is E = ½ CV2. To increase that energy storage and hence the hold-up time, either the amount of capacitance or the voltage on the capacitor must rise. Because 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 150 W at 12 V, with a minimum output voltage requirement of 11.5 V as the output decays to zero; 200 msec is the desired hold-up time.
1. Using a higher-rated power supply and operating it at a reduced load. Consider the example of using the 150-W 12-V output TDK-Lambda RWS150B-12 power supply and operating it at 100% load. Based on test data, the supply would exhibit a hold-up of just over 30 msec.
As expected, the hold-up time depends on the amount of output load. The greater the load, the quicker the high-voltage bus capacitor’s stored energy depletes. To get a 200-msec hold-up, we could switch to another supply such as the TDK-Lambda 1,500-W- rated CUS1500M-12. This supply has a larger bus capacitance which would provide enough energy to hold-up a 150-W load for over 200 msec.
Thus swapping supplies solves the problem, but at the cost of using a much larger, more expensive power supply.
2. Adding capacitance across the power supply output terminals or load. At first glance, adding additional capacitance across the output seems like an easy solution. But consider the size of the necessary capacitor.
Suppose stored energy E in Joules is the product of the output power Pout in Watts and time t in seconds. Then E = Pout × t.
As called out earlier, E is half the product of the capacitance, C, and the square of the voltage, V2: E = 0.5×C×V2. Rearranging for C we get C = 2×E/V2.
Note: The voltage on capacitor C will drop as it discharges into its 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 here as Vend. In our scenarios we’ll use a voltage of 11.5 V for Vend.
The useful energy for hold-up will be the initial minus the remaining energy: E = (0.5CV2) – (0.5CVend2). Factoring the equation results in E = 0.5 C(V2 – Vend2) or C = 2×E / (V2 – Vend2). Substituting Pout × t for E we have C = 2 × Pout × t / (V2 – Vend2).
Now we can determine the size of the output capacitor needed on a 150-W supply for a 200 msec holdup time. Here, the output capacitor would have to be a massive 4,595,745 uF:
C = 2 × Pout × t/ (V2 – Vend2), C = 2 x 150 x 0.18 / (122 – 11.52) = 0.4595745. Note that as the power supply already has 20 msec of hold-up capability, t is reduced to 180 msec (0.18 sec).
Even a sufficiently large supercap of this value would consume a large amount of space. One other major concern is the over-current of the power supply during initial turn-on. An uncharged output capacitance 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 12-V 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 E = 0.5CV2, so adding capacitor C1a across the high 390-Vdc bus has a greater effect than adding capacitance across the 12-V output.
As the dc bus is not accessible on most power supplies, the adding of 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 20 msec, increasing it to 200 msec 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, this strategy would increase the size of the power supply by around 50% for the same product height.
A power-module-based approach could also employ a non-isolated ac-dc 360-Vdc-output converter and a high-voltage input isolated dc-dc 12-V-output converter. This approach again would require a custom board design with engineering and safety certification charges.
4. A 48-V or 60-V-output ac-dc power supply and a wide range input dc-dc converter. For an off-the-shelf standard product, a 48-V-output ac-dc power supply could be used with an isolated wide-range input 18-to-75-V dc-dc converter with a 12-V output. The dc-dc converter may require a heatsink or cold plate for cooling.
In addition to providing the hold-up inside the 48-V output ac-dc supply, capacitor C1 can be used for additional energy storage. We have taken advantage of the energy storage formula 0.5×C×V2 as V is now 48 V rather than 12 V. The efficiency of the dc-dc converter is considered to be 90%, typical for high-quality converters. The hold-up of the dc-dc supply is 20 msec, requiring an additional 180 msec (t): C = 2 ×P × t / (Eff x (V2 – Vend2)).
For our example, the required value of C1 would be 2 × 150 × 0.18 / (0.9 × (482 – 182)) = 30,300 uF.
If the 48-V ac-dc supply was replaced by one having a 60-V output, the additional capacitance could be reduced to 18,315 uF. A 72-V output power supply would further reduce that to 12,345 uF.
5. A 48-V output ac-dc 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) and has a wide input range of 9 to 53 V. A non-isolated dc-dc converter would not require heatsinking.
The efficiency of the non-isolated dc-dc converter is assumed to be 96%. The hold-up of the ac-dc supply is 20 msec, requiring an additional 180 msec (t): C = 2 × P × t / (Eff x (V2 – Vend2)). For our example, the required value of C1 would be 2 ×150 × 0.18 / (0.96 × (482 – 92)) = 25,304 uF.
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