0
\$\begingroup\$

I am studying about power supplies and I need some help with the following. I have a power supply with a 12 V DC output to power the lower power controller electronics. This power supply has two inputs as seen in the figure below. What components do I need to put in the box to make sure that the AC source is the main primary input and the DC source is only used when the AC source is not available?

Will a diode-based ORing circuit help for such high power levels? Or are there alternate ways that are better than the diode-based ORing circuit?

AC and DC inputs to Power Supply

Updated: Thank you everyone for your valuable comments. I tried something on MATLAB (see following pictures) and I would like to know if this is practical and can actually be implemented. I used a breaker which will look at the ac voltage and determine if the 400 V DC should be used or not. If the AC voltage is close to zero, the breaker will close and 400 V will be injected into the system from the DC source.

enter image description here

Output Voltage waveform when ac voltage is turned off and the 400 V DC input voltage supplies power to the 12 V buck converter.

enter image description here

\$\endgroup\$
4
  • \$\begingroup\$ Diode OR-ing won't work for AC. What is the input power/current? How fast do you need the changeover to occur i.e. what is the holdup time of the power supply? Or do you not care if there is a short loss of power? \$\endgroup\$
    – vir
    Commented Dec 2, 2023 at 7:57
  • \$\begingroup\$ "Will a diode-based ORing circuit help for such high power levels?" What is the power level here? If the power levels are high, say, over a few hundred watts, then you may need to consider the distortion (and power factor) of the current in the AC line, in which case a simple diode bridge, as shown here in your post, will not suffice. \$\endgroup\$ Commented Dec 2, 2023 at 8:09
  • \$\begingroup\$ @vir, the input power is a few hundred watts maybe up to a kW as well. I don't care if there is some loss. \$\endgroup\$ Commented Dec 2, 2023 at 8:30
  • \$\begingroup\$ Do you need galvanic isolation between the 480V AC line and the 400VDC supply? Does the 12V power supply provide galvanic isolation between input and output? \$\endgroup\$ Commented Dec 2, 2023 at 9:11

2 Answers 2

1
\$\begingroup\$

I suggest buying an off-the-shelf PSU to generate an isolated 420VDC from the AC supply (assuming the 12V PSU can handle 420V at its input). I suggest this because the power level is up to 1kW, which is well in the range where sinusoidal line current is required on the AC supply, which is non-trivial to design and build as a one-off, and more cost-effective to just buy an existing commercial product.

Connect the 420VDC negative to the 400VDC negative, and connect the 400VDC positive to the 420VDC positive via a diode, refer schematic below:

enter image description here

Alternatively, get a second +12V PSU, so that each main supply feeds just one +12V PSU, and use diodes to OR the outputs of these +12V PSUs to the load. Set the output voltage of one PSU a little higher than the other to select which power source is used by default.

Edited: Update After OP Added Change-Over Contactor to Original Question
The solution suggested by the OP adds a mechanical device (called a "circuit breaker", but I suspect the correct term is "contactor") that is controlled by a comparator. Both of these devices introduce some risk of failure from various mechanisms, for example:

  1. The contactor has "contact bounce", which can last up to 20 milliseconds, see refs below.
  2. The voltage comparator requires its own control power supply, and means of measuring the voltage it is monitoring.
  3. There will be a time delay from when the AC voltage fails, to when input to the 12V PSU is switched over to the DC voltage source. Does the +12V output have sufficient "hold-up time" to "ride-through" this power loss?
  4. Switching from the DC source to the AC source requires a contactor that can safely break the DC current, before making contact with the supply from the AC source.

I would suggest the simple solution as presented in my answer here, introduces far less risk of failure, with far less complexity.

References for Contact Bounce:
Physics behind switch bounce when opening?

http://www.ganssle.com/debouncing.htm

\$\endgroup\$
0
\$\begingroup\$

to make sure that the AC source is the main primary input and the DC source is only used when the AC source is not available?

The word "only" makes me consider using a relay:

schematic

simulate this circuit – Schematic created using CircuitLab

The small converter takes its power from the AC and drives the relay. The relay's NC (normally closed) contacts connect to the DC input. If the relay is not energised (i.e. AC is not present) then the main converter's input will come from DC. If AC is present, the relay's contacts switch and direct the main converter's inputs to the rectified AC.

NOTE: The small converter can be AC-to-DC, in this case, it should take its inputs from the AC directly instead of rectified AC.

\$\endgroup\$

Not the answer you're looking for? Browse other questions tagged or ask your own question.