Hi guys!

I have read similar threads in this forum with regards to this topic. But their questions are mainly focusing on the probability of replacing the AC source with a DC source. As for me, I am curious to know what will be the output voltage if I input a DC voltage of say 230V into the diode bridge. I did a simple simulation of this and I got an output of 230-1.4= 228.6V and the result seems about right (I attached the result). Practically this should be true as well right?

I am trying to feed a DC input voltage into the input of an AC laptop adaptor instead of using an AC source.

Usually a 230VAC source will be used an input for the AC adaptor, but if I use DC instead, should I be using 230VDC as an input to the AC adaptor or do I have to treat the 230VAC source as an rms value and use 325VDC (230*1.414) instead?

Any inputs are appreciated.

Thanks.

 

The reason I am asking this is that I am interested in feeding an DC source directly to the AC adaptor of the laptop instead of an AC source. As for the case of the AC adaptor, I will usually need an AC source of say 230VAC as input. But if I am interested in replacing the AC source with an DC source instead, what will be the DC input voltage required?

Not sure what you're asking.

Are you intending to feed DC into the input of an AC adapter?

 

With DC voltages, the full-wave rectifier bridge (assuming the bridge is rated for the input voltage), will only serve as reverse-polarity protection. Expect a voltage drop of around 1.4 volts DC or so because there will always be two forward-biased diodes.

The DC output of the bridge is connected to the (polarized) device you want to run... + to device positive, - to device negative. The input voltage should be at or very near the device operating voltage.

The input voltage is applied to the full-wave bridge ~ terminals (intending for AC voltage)... Input polarity to the full-wave bridge is irrelevant, the output polarity will always be correct.

 

Do I have to consider the peak DC voltage as compared to the rms value of the AC source or is it a 1:1 ratio?
1) So if the AC laptop adaptor requires an AC input of say 230VAC, the DC equivalent will be approx 325VDC?
2) Or can I just feed an DC source of 230VDC to replace the 230VAC directly?

When we look at DC Epeak = Eavg = Erms
The DC equivlent of 230 VAC RMS would be 230 VDC and the DC equivlent of 325.2 AC Peak would be 230 VDC and the DC equivlent of 650.4 VAC Peak to Peak would be 230 VDC.

The RMS value of a Sine Wave (we are talking about sine waves) is .707 * Epeak. Epeak being equal to 1/2 of E pk to pk.

Make sense?

Ron.

 

Not sure what you're asking.

Are you intending to feed DC into the input of an AC adapter?

Yes. Sorry if it's unclear.

 

Yes. Sorry if it's unclear.

Wouldn't it be simpler to just feed an appropriate DC voltage into whatever the adapter powers? Lower operating voltage, avoid inefficiency in the adapter, ...

 

When we look at DC Epeak = Eavg = Erms
The DC equivlent of 230 VAC RMS would be 230 VDC and the DC equivlent of 325.2 AC Peak would be 230 VDC and the DC equivlent of 650.4 VAC Peak to Peak would be 230 VDC.

The RMS value of a Sine Wave (we are talking about sine waves) is .707 * Epeak. Epeak being equal to 1/2 of E pk to pk.

Make sense?

Ron.

Shouldn't it be the reverse way? 230VAC will be 325VDC? Since 230VAC is in rms value and VDC = VRMS * 1.414

 

Wouldn't it be simpler to just feed an appropriate DC voltage into whatever the adapter powers? Lower operating voltage, avoid inefficiency in the adapter, ...

Unfortunately that's a constraint I have to work with.

 

With DC voltages, the full-wave rectifier bridge (assuming the bridge is rated for the input voltage), will only serve as reverse-polarity protection. Expect a voltage drop of around 1.4 volts DC or so because there will always be two forward-biased diodes.

The DC output of the bridge is connected to the (polarized) device you want to run... + to device positive, - to device negative. The input voltage should be at or very near the device operating voltage.

The input voltage is applied to the full-wave bridge ~ terminals (intending for AC voltage)... Input polarity to the full-wave bridge is irrelevant, the output polarity will always be correct.

Thanks for the input. But in this case, should I be feeding 230VDC or 325VDC? If I consider the initial AC source to be 230VAC.

 

Thanks for the input. But in this case, should I be feeding 230VDC or 325VDC? If I consider the initial AC source to be 230VAC.

Already answered by @Reloadron; 230VDC. And that's assuming the adapter is a transformerless switching regulator.

But, 230VDC and 230VAC are potentially lethal voltages. Do you know what you're contemplating?

 

Shouldn't it be the reverse way? 230VAC will be 325VDC? Since 230VAC is in rms value and VDC = VRMS * 1.414

That gets you Epeak, Vrms * 1.414 = Epeak.

Generally when not specified we assume Vrms. For example the US standard is 120 VAC or 240 VAC 60Hz and this is RMS while not called out. Back to what I mentioned with DC, Epeak = Eaverage = Erms. The RMS value of a sine wave is the DC equivlent.

Ron

 

Already answered by @Reloadron; 230VDC. And that's assuming the adapter is a transformerless switching regulator.

But, 230VDC and 230VAC are potentially lethal voltages. Do you know what you're contemplating?

Keep that in mind also.

Ron

 

How do you know the AC adapter can tolerate DC on its input?

 

More to the point WHY do you want to replace AC for DC???

 

With regard to tolerating DC on the AC inputs? I am assuming that you are also referring to the voltage regulator subcurcuir?

 

When we look at DC Epeak = Eavg = Erms
The DC equivlent of 230 VAC RMS would be 230 VDC and the DC equivlent of 325.2 AC Peak would be 230 VDC and the DC equivlent of 650.4 VAC Peak to Peak would be 230 VDC.
The RMS value of a Sine Wave (we are talking about sine waves) is .707 * Epeak. Epeak being equal to 1/2 of E pk to pk.

Yes, but -
That is all true for equivalent energy, but not necessarily for what the power supply controller wants to see as a viable input. If it is not of the universal input type, then it might require a minimum input peak voltage that is equivalent to an input sinewave of 240 Vac minus a design margin of maybe 20%, or 271 Vdc. More and more industrial power supplies are rated for both AC and DC inputs, but older power-factor corrected front ends still want to see a sine wave input and might shut down without one.

ak

 

Yes, but -
That is all true for equivalent energy, but not necessarily for what the power supply controller wants to see as a viable input. If it is not of the universal input type, then it might require a minimum input peak voltage that is equivalent to an input sinewave of 240 Vac minus a design margin of maybe 20%, or 271 Vdc. More and more industrial power supplies are rated for both AC and DC inputs, but older power-factor corrected front ends still want to see a sine wave input and might shut down without one.

ak

And that would be true and need considered, so it all depends on what you have.

Ron

 

More to the point WHY do you want to replace AC for DC???

Don't know about the TS, but military aircraft can have high voltage AC and/or DC power buses. A truly universal input supply handles 85-265 Vac at 45-440 Hz *and* 270 Vdc +/- 20% (working from memory).

ak

 

Feeding 320 VDC into the 230 VAC input of a computer power brick is not a good idea. I am fairly confident it will not work. Wear safety glasses and stand back when you flip the switch. I am curious to see the results... can you post a video on youtube?

 

Here is an example of a newer universal input supply with DC input capability:

https://www.artesyn.com/power/assets/imp_ds_28may2015_1432909865.pdf

ak