Connecting AC and DC voltages together

[PhaseShift]

What type of value would you get if you connected a 120V AC line to a 120V DC line? What reading would you get if you measured the voltage between the two? If you connected them together would there be a short circuit or would the DC voltage simply offset the AC voltage?

 

[gar]

101112-0854 EST

PhaseShift:

The following is not a complete answer.

If you use a meter with an input coupling capacitor, (Examples: Simpson 260 in the OUTPUT mode, or Fluke 27 or 87 in AC position), the AC and DC voltage sources are isolated and added together, then you will read the AC voltage.

If you use any of the above meters in the DC position, then the reading is the DC voltage.

If you have an RMS reading meter that works for DC thru an adequate upper frequency, then the reading will be the RMS value of the composite waveform.

If your meter is a peak reading type, then it's the peak that is measured. Note: + and - peaks could be different and you would need to switch polarity to get the two different values.

Undesired current will flow if you parallel a DC voltage source with an AC voltage source.

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[jghrist]

The dc source (battery?) will offer little resistance to reverse current flow and will essentially short-circuit the ac source.

 

[mull982]

101112-0854 EST



If you have an RMS reading meter that works for DC thru an adequate upper frequency, then the reading will be the RMS value of the composite waveform.


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Would the composite waveform simply be the RMS value of the AC waveform offest or shifted upwards by the amount of the DC voltage?

 

[mull982]

The dc source (battery?) will offer little resistance to reverse current flow and will essentially short-circuit the ac source.

O.k. so the battery will be seen as a low resistance to the AC source? What would be the return path in this circuit? Would it be through the - terminal of the battery which is grounded to the same ground as the AC source? What if the - terminal on the battery wasn't grounded? Would there still be a return path and therefore current flow?

Does the reverse current flow on the battery present a much higher impedane to another DC source since we are able to parallel batteries? In other words does it present a higher impedance to another DC source as it does and AC source?

 

[nollij]

If you shorted the two lines together...
The 120VAC rms source is actually 169V 0-peak. So I would suspect that for the time when the AC source voltage is higher than the DC, it would try to send reverse power flow into the source and vice versa. As for the loads, granted the sources can withstand this reverse flow, they would see a mostly DC voltage with a little "hill" every 17ms.

 

[K8MHZ]

The dc source (battery?) will offer little resistance to reverse current flow and will essentially short-circuit the ac source.

That was my thought as well, every half cycle. The reverse half of the cycle would present a huge current spike while the AC was being kept from going too far across the zero line.

Think of it as connecting two batteries together, pos to pos, (no current flow) and then reversing the leads rapidly to get 'AC'. Obviously there will be big smoke as soon as the two batteries are connected together backwards. Using 120 volts, that would mean a direct 240 volt short circuit.

 

[gar]

101112-1209 EST

mull982:

First you integrate from 0 to 2*Pi the square of (Vdc + Vacp*sin t). Note: the term 2*Vdc*Vacp*sin t averages to 0 and thus is dropped.

Then average this over 2*Pi. The result is ( Vdc^2*2*Pi + ( Vacp^2*(2*Pi/2) ) / 2*Pi . This reduces to Vdc^2 + Vacp^2/2 . Vacp is the peak amplitude of the AC sine wave. Vacp^2/2 = Vacrms^2 .

Next take the square root of the average. The result in other words is the square root of the sum of the squares of the separate RMS voltages.

The RMS value of 120 Vdc + 120 Vac sine wave is 170 V RMS if I haven't made any mistakes.

You may find it interesting to plot ( 1 + 1.414*sin x )^2 from 0 to 2*Pi.

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[LarryFine]

Would the composite waveform simply be the RMS value of the AC waveform offest or shifted upwards by the amount of the DC voltage?

If paralleled, you'd get what effectively would be a short circuit.

If connected in series, you'd get the AC wave with the DC offset.

 

[gar]

101112-1409 EST

Larry:

I believe mull982 wants to know what an unspecified meter would read on a waveform consisting of the sum of a DC voltage and a sine wave.

Where the DC voltage is what would be read on a normal DC meter or DC reading RMS meter. And the AC sine wave, if measured separately, would have some know measured RMS value.

Knowing these two values, then what would an AC-DC reading RMS meter display?

What the reading won't be is the DC voltage plus the separately read AC RMS voltage.

On a Simpson 260 meter in the AC position, but not thru the OUTPUT input, what will be read is the DC voltage modified by the meter scaling that converts full wave average to RMS. The reason is the average AC component is 0, and thus only the DC value is measured. Applies so long as the composite never goes negative.

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[LarryFine]

There could not be a single waveform if both sources have zero impedance; there would be an overcurrent every opposing cycle half.

If we assigned each source an equal impedance, one high enough to prevent overcurrent, then I'd say the waveform would be 60vac with a 60vdc offset.

In other words, a wave that varies from zero to +120v.

 

[LarryFine]

In other words, a wave that varies from zero to +120v.

Correction: from 30v to 90v. :roll: