Table of Content

Toggle## What is a Hold-Up Capacitor & When Do We Need It?

In real-time critical applications such as avionics and life-support systems, it is very important to have an uninterruptable power supply supplying adequate power to the system.

Any sudden drop in power can jeopardize the system’s working, leading to fatalities in the worst-case scenario. Here is a typical switched-mode power supply block diagram:

A capacitor is placed right after the rectification block to cater for this sudden drop (usually from a main line power outage). In case of a power drop, the capacitor can kick in and maintain power levels for the load until the power comes back.

### What is Hold-Up Time?

Hold-Up is when the power supply keeps supplying power to load before it drops below the specified voltage level. This voltage level and the corresponding hold-up time differ for different power supplies.

Generally, linear supplies have very short hold-up times and go to zero almost instantaneously. However, the switched-mode supplies can hold up until the power comes back on or the UPS gets connected.

This is because switched-mode power supplies have hold-up capacitors after the rectification stage that keep the power within given levels.

### How to Calculate the Value of Hold-Up Capacitor?

It has been established earlier that switched-mode power supplies use hold-up capacitors to hold the power level above a specific level in case of power cuts.

Now to calculate the value of such a capacitor, the supply’s hold-up time is set first. Secondly, the output voltage and current values are to be known to calculate power. This power value is to be used later for capacitor value calculation.

Using the power relation, we can calculate the output power that the circuit provides to load:

**P[W] = V . I [V,A]**

We also calculate the energy stored in a given capacitor:

**E[J] = ½ . C . ( V ^{2}_{U } – V^{2}_{L} )**

Here ** V_{U}** is the voltage at the beginning of discharge time, i.e. the nominal output voltage, and

**is the voltage at the end of cap discharge time, i.e. the safe limit for hold-up time.**

**V**_{L}We know that power is energy per unit of time, so we can use this relation and write the energy relation in terms of power and time. After that, the equation can be rearranged to find the capacitance value:

**C[F] = ( 2 . P[W] . t[s] ) / ( V ^{2}_{U } – V^{2}_{L} )**

Let’s do an example and use this relation to find the hold-up capacitor for a system:

**Example:** The hold-up time of a switched-mode power supply is 10ms. The output voltage is set to 5V, and the current is 1A. The safe voltage limit for the hold-up is 4.8V. Find the value of the hold-up capacitor.

We can first find the power supplied by the capacitor:

**P[W] = V . I [V,A]**

**P = 5W**

The relation for energy is given here:

**E[J] = ½ . C . ( V ^{2}_{U } – V^{2}_{L} )**

Here ** V_{U} **is the voltage at the beginning of discharge time, i.e. 5V, and

**is the voltage at the end of cap discharge time, i.e. 4.8V.**

**V**_{L}Now, we can rearrange this to find the capacitance value:

**C[F] = ( 2 . P[W] . t[s] ) / ( V ^{2}_{U } – V^{2}_{L} )**

Here P[W] = E[J]/t[s] relation is used to replace energy.

The final value for the capacitor comes out to be:

**C[F] = 0.0510204081633 F**

OR

**C[μF] = 51020.40 μF**

There you go!

Now we can understand the usage of a hold-up capacitor and incorporate it into our design decisions for switched-mode power supplies. Moreover, we can calculate the value of such a capacitor according to our specific application.

#### References

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