Capacitance between parallel conductors in a cable

[JohnW2]

Don't try to apply RF transmission line theory to internal 50Hz mains wiring.

To be fair, Eric was talking about 'extremely long distances', not internal 50Hz mains wiring. Admittedly, within the UK one would be hard-pressed to get a line more than about 0.15 of a wavelength (at 50Hz) in length, but even that may have to be taken into consideration .... and, of course, there are countries (and groups of countries which exchange power) much larger than the UK.

Kind Regards, John

 

[bernardgreen]

Many international power linkages use very high voltage DC for the long link.

The first HVDC Cross-Channel scheme was built by ASEA and went into service in 1961 [1] between converter stations at Lydd in England (next to Dungeness Nuclear Power Station) and Echinghen, near Boulogne-sur-Mer, in France. This scheme was equipped with Mercury arc valves, each having four anodes in parallel.[2]

The main reason is the near impossibility of synchronising the phases of the two countries. There is also the advantages of, among others, the avoidance of any problems with standing waves and capacitive effects in the very long cable.

 

[JohnW2]

Many international power linkages use very high voltage DC for the long link. ... The main reason is the near impossibility of synchronising the phases of the two countries. There is also the advantages of, among others, the avoidance of any problems with standing waves and capacitive effects in the very long cable.

That makes sense, particularly from the synchronisation (and potentially differing voltages) point of view. The link across the channel is obviously not, per se, particularly long (about 0.008 of a wavelength at 50Hz) compared with the rest of the networks, so that transmission line consdierations (standing waves etc.) are probably not so much of an issue.

I suppose that very long DC transmission lines would result in some significant 'inrush currents' due to the capacitance?

Kind Regards, John

 

[Porque223]

I suppose the modern meter is more like the old valve volt meter which did not draw any current. I have not seem a ohms per volt written on a meter in ages.

The concept of 'ohms per volt' (strictly 'ohms per volt range') is really only applicable to moving coil meters, where it is simply a way of expressing the full-scale-deflection current of the meter. Hence, a simple moving coil multimeter with a 100 μA FSD meter will have an input resistance of 10,000 ohms per volt (i.e. an input restistance of 1MΩ on a 100V range, 5MΩ on a 500V range etc.).

With a valve voltmeter or modern DVM equivalent, the input resistance/impedence is not only very high, but is also 'fixed' (i.e. the same, regardless of any issues of 'range') - so the concept of 'ohms per volt' does not really arise.

Kind Regards, John

That's the other thing.
Since the phantom voltage really acts like a current source the D'Arsonval readings change with the voltage scale you are on.

 

[bernardgreen]

Since the phantom voltage really acts like a current source the D'Arsonval readings change with the voltage scale you are on.

More like a voltage source with a very high internal impedance. Any current drawn has to flow through the internal impedance and thus creates an internal potential drop reducing the voltage.

A current source normally describes a source of constant current irrespective of the load.

 

[Porque223]

More like a voltage source with a very high internal impedance. Any current drawn has to flow through the internal impedance and thus creates an internal potential drop reducing the voltage.

A current source normally describes a source of constant current irrespective of the load.

It's more or less the same thing. An ideal current source can be a voltage source of value = infinite, in series with a resistor of infinite value.
IIRC, the compliance of electronic current sources sort of addresses the "resistor" value.

BTW, the highest internal impedance of a power source that was being sold as a voltage source that I could find was a 120v:9v transformer that had ~40 ohms.
I haven't done a comparable search for the lowest compliance value for a current source; it might be one of these two terminal JFET sources.

 

[bernardgreen]

An ideal current source can be a voltage source of value = infinite, in series with a resistor of infinite value.

Infinite resistance (which an open circuit is) means no current can flow so not a current source.

That said the result of infinity divided by infinity can be either infinite or 1 so the output would be either infinite current or one amp. ( assuming volts and ohms are used in the equation )

Luckily it is impossible to have an infinite voltage source ( in all practical circumstances )

 

[JohnW2]

An ideal current source can be a voltage source of value = infinite, in series with a resistor of infinite value.

Infinite resistance (which an open circuit is) means no current can flow so not a current source.

He was a bit too free/loose in his use of 'infinity' but, in terms of what I think he was trying to say, he is right. A very high voltage source with a very high internal resistance is one type of quasi-constant current source ... consider a 1000V source in series with a 1MΩ resistor. That will deliver a current within 1% of 1mA through a load over a wide range of load resistances(from zero to about 10.1k&#937

That said the result of infinity divided by infinity can be either infinite or 1 so the output would be either infinite current or one amp. ( assuming volts and ohms are used in the equation )

Just for the record, infinity divided by infinity, just like 0/0, can have any value (from zero to infinity) - but that is just esoteric math/maths, of no relevance to this discussion!

Kiund Regards, John

 

[Porque223]

And for a 20 mA red indicating LED supplied from a 5 +/- 0.2v source and having its own tolerance on the voltage across the LED, 150 ohms serves as a close-enough current source.

I suppose 0/0 is as undefined and indeterminate as infinity/infinity.
With spreadsheets you can determine the limit of a function without knowing the math and these things have made me lazy in more ways than one. Once you program in the formula you can tweak values all over the place.
However, John Allen Paulos has a simple looking formula that gives very weird results. There were forbidden regions in the input/output values.

 

[JohnW2]

And for a 20 mA red indicating LED supplied from a 5 +/- 0.2v source and having its own tolerance on the voltage across the LED, 150 ohms serves as a close-enough current source.

I really don't understand what you're talking about there. Like any other forward-biased diode, the relationship between voltage across an LED and the current is very flat (i.e. voltage is roughly constant, regardless opf current). I don't understand in what sense you think of it as a 'current source'

I suppose 0/0 is as undefined and indeterminate as infinity/infinity.

You don't really need to suppose. Since the reciprocal of infinity is zero, 0/0 and ∞/∞ they are the same thing

With spreadsheets you can determine the limit of a function without knowing the math and these things have made me lazy in more ways than one.

Indeed you can, but "numerical methods" were around long before spreadsheets, or even electronic computers - but, in the absence of such aids, they certainly are not for the lazy!

Kind Regards, John

 

[Porque223]

And for a 20 mA red indicating LED supplied from a 5 +/- 0.2v source and having its own tolerance on the voltage across the LED, 150 ohms serves as a close-enough current source.

I really don't understand what you're talking about there. Like any other forward-biased diode, the relationship between voltage across an LED and the current is very flat (i.e. voltage is roughly constant, regardless opf current). I don't understand in what sense you think of it as a 'current source'

Yes, dV/dI can be close to zero but what is the value of V so you can determine the resistor value? LED makers don't seem to control this V very well.

LED brightness vs LED current is predictable but much less so for LED brightness vs LED voltage so you want to drive an LED with a current source, and this 150 ohm resistor [R]/5 vdc supply [Vs] seemed to work well enough in many cases, based on brightness.
For a small LED, IIRC assuming 1.8v +/- 0.3 vdc drop [VLED] @ 20 mA [ILED] may work, but many datasheets do not post the lower limit for LED voltage @ some current.

For the tolerance variation for approximating the range of resulting LED current due to the variation of both voltages and the resistor value there is the Root Sum Square (RSS) method.
There are also probably Web circuit calculators that use a Monte Carlo method to do this tolerance analysis.

ILED = (Vs-VLED)/R

Of course, with onesy/twosy you can tweak each LED circuit.

 

[bernardgreen]

and this 150 ohm resistor [R]/5 vdc supply [Vs] seemed to work well enough in many cases, based on brightness.

More accurately the current is ( 5 - Vf ) / 150 where Vf is the forward voltage of the LED.

 

[ban-all-sheds]

I suppose that very long DC transmission lines would result in some significant 'inrush currents' due to the capacitance?

2,375 km

 

[JohnW2]

Yes, dV/dI can be close to zero but what is the value of V so you can determine the resistor value? LED makers don't seem to control this V very well.
LED brightness vs LED current is predictable but much less so for LED brightness vs LED voltage so you want to drive an LED with a current source, and this 150 ohm resistor [R]/5 vdc supply [Vs] seemed to work well enough in many cases, based on brightness.
For a small LED, IIRC assuming 1.8v +/- 0.3 vdc drop [VLED] @ 20 mA [ILED] may work, but many datasheets do not post the lower limit for LED voltage @ some current.
For the tolerance variation for approximating the range of resulting LED current due to the variation of both voltages and the resistor value there is the Root Sum Square (RSS) method.
There are also probably Web circuit calculators that use a Monte Carlo method to do this tolerance analysis.
ILED = (Vs-VLED)/R
Of course, with onesy/twosy you can tweak each LED circuit.

I think that your life would be much simpler if you moved your mind out of the academic classroom environment into the real/practical world

As Bernard has implied, in practical real world terms, all you usually need to do is subtract the nominal Vf of the LED (at the desired operating current) from the nominal supply voltage and divide by the operating current. That will nearly always be adequate. If, exceptionally, you had a situation in which LED brightness was very critical, you would then, as you say, simply 'tweak' the resistance value on an individual basis.

Kind Regards, John

 

[JohnW2]

I suppose that very long DC transmission lines would result in some significant 'inrush currents' due to the capacitance?

2,375 km

Thanks - but am I missing something? I can find no reference to inrush currents in that article.

Kind Regards, John