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I am trying to build a power supply circuit which accepts two phases (220 VAC 50 Hz) and outputs 5V 1A. It must keep working if there is at least one phase present. Both phases can be coming from separate sources so I have taken a worse case phase difference of 180 degrees. Essentially the phase difference can be 0-180 degrees.

The two phases can be any two out of the following:

phases

Phase-1,2,3 are three phases of a 3 phase 4 wire system. Inverter is common household inverter. The household appliances are usually wired like as shown above - Common neutral for all loads. Inverter is usually powered using one of the phases (phase-1,2,3). When that phase goes off, the inverter kicks in. When that phase is present, the power to all loads connected on inverter phase essentially is provided by that phase.

I could think of two ways to tackle this:

1) Use a separate bridge rectifier for both phases as shown below:

bridge

However the output voltage on bulk cap varies a lot depending upon whether both phases are present or just a single one is present. Please check the waveforms below:

Both phases present

Fig (above) : Both phases present (180 degrees phase difference, 350 V amplitude)

one phase present

Fig (above) : Single phase present (180 degrees phase difference, 350 V amplitude)

I see two problems with this solution:

(a) 700 V bulk capacitor voltage will be difficult to handle with regular offline switching ICs.

(b) A voltage difference of 350 V whether both phases are present or not will add further difficulty. Since I'd also want it to work in universal voltage range 90-270 VAC, I guess it will add further complexity to the circuit and part selection. Cost is another concern.

2) Use a half wave rectifier for both phases.

Half wave

Fig (above) : Half wave rectifier for both phases

(Resistor represents the complete SMPS circuit)

Waveforms are as shown below:

both present

Fig (above) : Both phases present (180 degrees phase difference, 350 V amplitude)

one present

Fig (above) : Single phase present (180 degrees phase difference, 350 V amplitude)

This looks like an easier thing to handle. Voltage levels are same in both cases. The only difference is - one has higher ripple than the other.

As such I am leaning towards going the half wave rectifier way. Am I doing this right? Am I missing something here?

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  • \$\begingroup\$ That ripple is negligible compared to the wide-range input voltage variation. \$\endgroup\$
    – winny
    Commented Jan 2, 2018 at 11:18
  • \$\begingroup\$ Are you sure you can safely connect those two neutrals together in your situation? \$\endgroup\$
    – Finbarr
    Commented Jan 2, 2018 at 11:31
  • \$\begingroup\$ @Finbarr - Yes, for all other appliances, same neutral is used for whatever phase is being used for powering up the device. So, I think I can use the same neutral in this case as well. \$\endgroup\$ Commented Jan 2, 2018 at 11:57
  • \$\begingroup\$ Do these two AC source floating with respect to each other, or do they have some common connection? This is very important in deciding the input architecture. \$\endgroup\$ Commented Jan 2, 2018 at 12:05
  • \$\begingroup\$ @olin - added those details in the question. I hope I am clear. All phases are referenced to a common neutral. There can be some cases where an appliance might need to powered using phase and phase but my power supply won't be doing that. \$\endgroup\$ Commented Jan 2, 2018 at 13:18

1 Answer 1

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It seems your question comes down to how to get rectified DC from two AC voltages with a common neutral. Here is the general case way:

The resulting DC is not referenced to the incoming neutral or any phase. Both DC+ and DC- should be considered live and dangerous. However, if this is the front end of a power supply, then the result will be chopped to drive a transformer anyway. That provides the isolation, and allows you to connect any one point of the output to ground.

The rectified DC voltage is the instantaneous maximum difference between any of the three input lines. With a single phase connected, or both AC1 and AC1 in phase, that will be the AC peak voltage. With AC1 and AC2 out of phase, that will be the peak to peak voltage.

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  • \$\begingroup\$ Hey Olin, the first method was about the idea that you shared. I have added the circuit image in my question. The problem is that if only one phase is present, I will get around 350 volts on bulk cap. When both are present, I will be getting something around 600 V on the cap. And my circuit will have to handle a huge voltage range. As such design process and part selection will get very difficult. That's why I came up with half wave rectifier method where the cap is referenced to Neutral. Is half wave method safe or am I missing some potential danger here? \$\endgroup\$ Commented Jan 3, 2018 at 8:02
  • \$\begingroup\$ @Whis: Half wave is safe. The downsides are more ripple and more uneven about drawing power across a whole cycle. If you don't care about power factor, and you can use a big enough cap to get ripple to where you can tolerate it, then half wave rectifying will work, and gets around the large voltage range due to phase shift. \$\endgroup\$ Commented Jan 3, 2018 at 11:39
  • \$\begingroup\$ I think I am at a stage where I need to choose the lesser of two evils. I did a quick search on digikey and most of cheap SMPS ICs have a mosfet which can tolerate 600V - 700V. There is none having >1000V. From a design perspective, all of these are supposed to be fed 90-260 VAC and nothing higher. I don't think I will be able to handle the phase to phase using these parts. As such solving the ripple problem appears to be the only way out. Using a relatively larger cap is possible. Can you please elaborate more on the power factor part? How will half wave give me a poor PF as compared to BR? \$\endgroup\$ Commented Jan 3, 2018 at 13:56
  • \$\begingroup\$ @Whis: Half wave draws all the current in one spike per cycle. Full wave draws it in two spikes per cycle, which is only marginally better. However, full wave allows for active power factor management more easily. \$\endgroup\$ Commented Jan 3, 2018 at 14:15

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