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Why dont we use 2 phase DC?


Direct.Dude

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Yes, I am aware of that. However, that is only if the DC passes through him. AC can pass easily due to the capacitance of the human body but DC can't. 340 volts DC hurts much less than 120 volts AC. If you take a bug zapper and touch the inner net and the outer net, you will find out how much 170VDC hurts. And don't even think about doing that with 120VAC.

 

Please don't anyone try these experiments.

 

Old time service engineers were taught to keep one hand in their pocket when probing around electronic circuits in case they accidentally touched a live (at DC) terminal.

 

I can tell you that even the bite from the 90volt battery they used to use in valve portable radios hurts and has been known to kill.

 

Directdude, electricity is dangerous.

 

But properly treated it is also useful to essential so let us concentrate on your 3 wire distribution system.

You need to provide proper details to discuss, it is up to you, not me, to work it out for you.

 

I note that 3 wires is one more wire than 2, which represents a greater than 50% cost increase since it would also require stronger support structures.

I can't comment more without knowing what the proposed voltages at all stages are.

 

Please also note that most consumers receive their current AC supply over 2 wires.

In the US this is transformed to what is correctly known as split phase, although too many call it two phasse, which it is not.

 

There were once DC supplies in both the US and Europe, which were abandoned for very sound technical reasons.

 

There have been experiments in the US with superconducting primary distribution extra high voltage lines at DC.

 

But at the consumer end, the size, weight, cost and efficiency of machinery and some circuitry is least at DC and rises with increasing frequency at AC.

Why do you think most vehicles now have alternators rather than dynamos?

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Please don't anyone try these experiments.

 

Old time service engineers were taught to keep one hand in their pocket when probing around electronic circuits in case they accidentally touched a live (at DC) terminal.

 

I can tell you that even the bite from the 90volt battery they used to use in valve portable radios hurts and has been known to kill.

 

Directdude, electricity is dangerous.

 

But properly treated it is also useful to essential so let us concentrate on your 3 wire distribution system.

You need to provide proper details to discuss, it is up to you, not me, to work it out for you.

 

I note that 3 wires is one more wire than 2, which represents a greater than 50% cost increase since it would also require stronger support structures.

I can't comment more without knowing what the proposed voltages at all stages are.

 

Please also note that most consumers receive their current AC supply over 2 wires.

In the US this is transformed to what is correctly known as split phase, although too many call it two phasse, which it is not.

 

There were once DC supplies in both the US and Europe, which were abandoned for very sound technical reasons.

 

There have been experiments in the US with superconducting primary distribution extra high voltage lines at DC.

 

But at the consumer end, the size, weight, cost and efficiency of machinery and some circuitry is least at DC and rises with increasing frequency at AC.

Why do you think most vehicles now have alternators rather than dynamos?

Of course it is very dangerous to touch 340 VDC and don't try it, yes. But 120 VAC is still much more dangerous.

You ​can ​touch a bug zapper, though, as its current is very low.

Also, I have touched 170 VDC. It doesn't hurt very much.

Yes, electricity is very dangerous.

Since the electricity is coming half the time, thinner wires can be used which means that its not really a large increase.

I'm not sure about that alternator vs dynamo thing, but it is very stupid - instead of generating DC directly, the alternator generates AC and bridge-rectifies the outputs. I've been wondering why that's the case.​

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I'm not sure about that alternator vs dynamo thing, but it is very stupid

 

If you are not sure why do you think it is very stupid, or even a little bit stupid.?

 

We can discuss it if you wish, I am not trying to hide details.

 

 

Why are you avoiding my question about distribution details?

 

 

 

Since the electricity is coming half the time, thinner wires can be used which means that its not really a large increase.

 

Whatever the waveshape,AC or DC it is the RMS current that is important to determine the power.

 

Since we buy and use electricity in terms of power not current a given supply requires a specific RMS current and therefore the wires have to be the same.

Edited by studiot
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If you are not sure why do you think it is very stupid, or even a little bit stupid.?

 

We can discuss it if you wish, I am not trying to hide details.

 

 

Why are you avoiding my question about distribution details?

 

 

 

Whatever the waveshape,AC or DC it is the RMS current that is important to determine the power.

 

Since we buy and use electricity in terms of power not current a given supply requires a specific RMS current and therefore the wires have to be the same.

 

 

If you are not sure why do you think it is very stupid, or even a little bit stupid.?

 

We can discuss it if you wish, I am not trying to hide details.

 

 

Why are you avoiding my question about distribution details?

 

 

 

Whatever the waveshape,AC or DC it is the RMS current that is important to determine the power.

 

Since we buy and use electricity in terms of power not current a given supply requires a specific RMS current and therefore the wires have to be the same.

 

 

 

If you are not sure why do you think it is very stupid, or even a little bit stupid.?

 

We can discuss it if you wish, I am not trying to hide details.

 

 

Why are you avoiding my question about distribution details?

 

 

 

Whatever the waveshape,AC or DC it is the RMS current that is important to determine the power.

 

Since we buy and use electricity in terms of power not current a given supply requires a specific RMS current and therefore the wires have to be the same.

 

"instead of generating DC directly, the alternator generates AC and bridge-rectifies the outputs. I've been wondering why that's the case."

 

Yeah, I know about that. However, the RMS/wire will be half the actual voltage if I'm right. Because its basically like a 1/2 PWM which means that the only cost that is more is for insulation.

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Yeah, I know about that. However, the RMS/wire will be half the actual voltage if I'm right. Because its basically like a 1/2 PWM which means that the only cost that is more is for insulation.

 

 

I don't quite get what you actually 'know about' after repeating my post 3 times,

 

But I respectfully suggest your second sentence indicates you need to know more about electrical power theory and practice before making judgements.

Edited by studiot
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I mean about RMS. (and surprisingly I found out about it just one day before you posted that comment)

Can you please tell me where I am wrong? Thank you. By the way if you want to know why I posted that statement, RMS is just some way of finding a mean for a sine as far as I know. However in a square wave equivalent if it is on 1/2 of the time, its actual power is 1/2 its v*a but connecting them both will power of v*a.

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I mean about RMS. (and surprisingly I found out about it just one day before you posted that comment)

Can you please tell me where I am wrong? Thank you. By the way if you want to know why I posted that statement, RMS is just some way of finding a mean for a sine as far as I know. However in a square wave equivalent if it is on 1/2 of the time, its actual power is 1/2 its v*a but connecting them both will power of v*a.

 

Connecting them both to what?

 

That is why I keep asking for a full circuit digram.

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Now that we have the beginning's of a circuit, I can comment on Stranges post#17 and your over-casual dismissal of it.

 

In a perfect world Strange's comment would be correct.

You have gained nothing in power efficienty terms over a straightforward DC supply at the same voltage.

 

However in the real world your configuration has reduced power efficiency significantly since all real world supplies have an internal impedence.

The internal impedence of the 'off' supply would appear as a load in parallel to the load resistor for the 'on' supply.

In other words the on supply would try to drive current back through the off supply.

 

In addition to ineficiency, this is likely to lead to the early demise of one or both supplies.

 

I forget which but it is rule 0 or rule zero that you do not connect two outputs together without very special measures.

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29nauc6.png

At the top, the power through the resistor is (v*a)/2

At the bottom, the power through the resistor is v*a

 

Are the two "-" wires the ones with your square waves on? Doesn't that need a couple of diodes to stop current flowing back down the other "-" wire when it is at 0V?

 

It's the same for wires - they are split in two but the two also have to be half in size.

 

You seem to think that the advantge is that you can put your square waves through a transformer.

 

Have you calculated the effect (of having a broadband signal) on the efficiency of the transformer?

Have you looked at the amount of noise (RF interference) that this would generate?

Have you worked out what sort of circuitry would be needed to regenerate a square wave from the output of the transformer? (Which will, very obviously, not be a square wave any more)

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@studiot it can be transformed

 

 

Are the two "-" wires the ones with your square waves on? Doesn't that need a couple of diodes to stop current flowing back down the other "-" wire when it is at 0V?

 

 

You seem to think that the advantge is that you can put your square waves through a transformer.

 

Have you calculated the effect (of having a broadband signal) on the efficiency of the transformer?

Have you looked at the amount of noise (RF interference) that this would generate?

Have you worked out what sort of circuitry would be needed to regenerate a square wave from the output of the transformer? (Which will, very obviously, not be a square wave any more)

Yes, look at my posts above. Either a diode or a different instance (using a transformer).

I look at these things, but if the impendance of the transformer is low, this will be too small to be noticeable - plus it'll still be able to be transformed.

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I look at these things, but if the impendance of the transformer is low, this will be too small to be noticeable - plus it'll still be able to be transformed.

 

Can you show your calculations? (It may be over my head, but interesting anyway.)

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Can you show your calculations? (It may be over my head, but interesting anyway.)

 

Unfortunately I don't know much formulas for calculating. However, I know how inductance works.

So if the input is a pure square, the output will be the rated voltage first, then drop a bit, how much depends on the frequency, and when the phase changes it will spike up and then down to 0.

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Unfortunately I don't know much formulas for calculating. However, I know how inductance works.

So if the input is a pure square, the output will be the rated voltage first, then drop a bit, how much depends on the frequency, and when the phase changes it will spike up and then down to 0.

 

So you agree you won't get a square wave out. How do you plan to regenerate a square wave after this?

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