Extended HoldUp for 1500 watt Power Supply

Once I was stated an interesting problem:
1500wat ACDC power supply I was supposed to design had to have hold up energy of 600 joules.
Nominal Power output of the power supply was 1500 watt. After losing external AC power the unit should provide 1200 watt during 0.5 second, or 600 joules.
The unit consisted of 2 main parts: PFC with output voltage of 420vdc and phase shifted full bridge. The bridge provided isolation and several output voltage sources of different voltage and current.
The bridge worked efficiently by Vin from 360 vdc to 420 vdc.
So how to solve this problem?

1

Conventional solution

Usually required hold up time is equal to one period of AC voltage, or 20millisecond and is provided by placing an electrolytic capacitor on the output of PFC. In our case we needed 25 times more: 500millisecond.
Using maximum space available I could place 12 pcs of 680 uF 450v capacitors, with total capacitance of 680 x12 = 8160 uF. Maximum Energy by charged voltage 420v would be E = 0.5*V2*C = 0.5*4202*0.00816 = 720 Joules. It looks certainly enough, more than 600 joules required.
There is one problem though: The PFC voltage supplies the full bridge converter and it cannot be lower than 360v. If we leave the PFC capacitor discharge naturally, its voltage will decrease exponentially:

By constant power:
Discharge time tdsch = 0.5*C*(V12- V22)/ P
Let’s see in how much time the capacitor voltage will drop to 360v – minimum voltage required by bridge converter.
t = 0.5*0.00816*(4202-3602)/1200 = 159 millisecond only.
Energy supplied to the bridge converter during discharge time to 360v will be only 191 joules or 26.5% from total available energy of 720 Joules.
So how we utilize the most of energy accumulated in the PFC capacitor bank?

2

Switched PFC capacitors bank

There was a working solution used in one of my company previous design:
Switched PFC capacitors bank
When power is lost, PFC capacitors bank is switched from output of the PFC circuit to its input. At the same time PFC function of the PFC converter is switched off and it works as a boost converter. Control loop parameters also have to be adjusted. It requires precise timing. Therefore, a microcontroller was used to control the process. Several high side switches and additional current and voltage sensors were needed. Also, this solution did not allow the whole capacitor bank to get switched to the input side. Some capacitance is needed to remove voltage ripple on the output of the boost converter. Pretty complex and not always stable solution.
One of the variants of such solution see below:

Therefore, in search for something simpler and more reliable I encountered another idea:

3

Additional holdup boost converter.

It is placed between PFC converter and DCDC boost. It sleeps during normal conditions when AC power is present. When AC power is removed it wakes up and utilizes almost all PFC capacitor bank energy supplying the DCDC bridge with constant voltage.

This solution work very good. Unfortunately boost ratio of a boost converter is limited to maximum 10. In reality it rarely exceeds 5 or 6, limited by losses in inductor and mosfet. In order to support high boost ratio boost inductance has to be made small. It causes very high peak currents in mosfet when input voltage is low.
Let’s calculate minimum voltage required to provide constant power of 1400watt during 500milliseconds:

As you can see from the chart the holdup boost has to keep working until 40-50vdc.
Therefore, I decided to modify the holdup boost to Tapped Inductor Boost Converter.

4

Tapped Inductor Boost Converter.

Regular boost converter requires a very high duty ratio to achieve high voltage gain, causing high reverse-recovery current and high device stress. The tapped inductor boost converter achieves high output voltage/ boost ratio with a low duty cycle

The inductor consists of 2 part: L1 (primary side) and L2(secondary side). If n ratio = L2/L1,
Then during on time period Vl1= Vin, Vl2 = n*VL1
During off time period total voltage on inductor is Vl1+n*Vl1=Vl1(1+n)
Transfer ratio becomes Vo/Vi = 1+n*D/(1-D)
If n=0, the converter becomes a simple buck converter with ratio Vo/Vi = 1/(1-D)
Let’s calculate required boost duty cycle:

Vin: 40 volts
Vout: 385 volts

For simple boost converter Vo/ Vin =1/(1-D); D =1 – Vin/Vout
In our case D = 1 – 40/385 = approx. 0.9
We can see from the chart below how tapped inductor ratio influences required duty cycle:

Nothing comes free as the average input current in tapped inductor boost becomes: Iin = Il1(n*D +1)
With the required output power of 1400 watt, the current through mosfet switch can become excessively big, which will require use of expensive mosfet. Therefore, a compromise, as always in power electronics design, is needed.
I have chosen for ratio n = 1. In this case, the required duty cycle is 0.81, which is acceptable.
On the next page you can see the real schematic of the tapped boost.
As you can see, there are 2 mosfets working in parallel, as the peak current is very high, in excess of 50 ampere.
Two protections are implemented here:
Overcurrent protection: as current through mosfet reaches certain level, the converter is switched off.
Undervoltage protection: as input voltage from PFC capacitor bank falls below 40 volt the converter is switched off. These protections are in fact similar, as the highest current occurs by lowest voltage.