The power supply unit is set to 24.0VDC, measured at its output. The unit that uses the power draws 100mA. With that unit connected the supply is only 19V. Connecting the two is a series of components. Normally the voltage drop is negligible. But now it's 5V. What value of resistance along the line would cause that voltage drop?
Bill
E=IR so 50 ohms
V = IR so R = V/I = 50 ohms
So how long is a length of wire? (:-{)
Why the secrecy?
You seem to be assuming that it is the 'series of components' that has changed, and that the PSU is fine. This may be true, but an alternative is that the power supply is failing, and the 100A draw is now causing a larger drop from the PSU than before.
A 'series of components' a bit vague though ... it may be unwise to assume that there is just a resistive element to the situation.
jkn <jkn snipped-for-privacy@nicorp.f.co.uk> wrote
That can't happen if the output of the PSU is still 24V
The PSU output remains at 24V as long as the load withing it's capabilities. Which is is, by a factor of fifty.
It's 100mA not 100A.
We have 24V from the psu and 19V at the load, simultaneously. The load is 100mA. There has to be some resistance along the line. We have access issues all along the line. But knowing the total line resistance would help. So what is it?
Bill
Didn’t the other John answer that?
Tim
What secrecy? Bill
It'd be normal to say there was "a wire" between the PSU and the load, not "some components", what are they? if it's just wire is it hundreds of metres?
the "series of components"
Wit doesn't work on newsgroups! Sorry, Bill ;-}
"How long is a bit of string?", a standard English expression meaning "without more information, no-one can give you an answer" - reworked.
Andy got it!
PA
It's a very old TV distribution system that's on life support. It uses 24V DC line power. (On large systems it is not possible to supply every dwelling with adequate signal strength from one amplifier. It is necessary for there to be ‘repeater’ amplifiers at various locations to boost the signal. To avoid the need for a multiplicity of mains electric supplies such systems have electrical power carried on the same cables as the signal. This power supplies the repeater amplifiers. This practice is called ‘line powering’.) The components between the PSU and one of the repeaters are several two-way splitters and a large number of tap-off units. All these are designed for through powering. It is bad practice to feed repeaters from tap-off lines, but that's what we're faced with. It would be a lot of hassle to disconnect every branch and then check resistance at the head-end with a short-circuit at the relevant repeater position.
Bill
your original sentence says "With that unit connected the supply is only 19V" (or, later, 5V). It is not clear, to me at least, where you are measuring that. From what you say, is this at the junction of the 'series of components' and 'the unit'?
Thanks for correcting my typo.
What's the characteristic impedance of the transmission line? 50 ohms?
75? Do the repeaters have terminating resistors on their inputs? Would it be feasible for a fault to develop whereby one of these resistors somehow ended up in series with the DC feed? Or have I misunderstood the problem?
A TDR or a VNA with that capability would probably let you localise the problem more quickly. Not sure if the cheap Chinese "skeleton" jobbies are up to it.
The power supply unit is set to 24.0VDC, measured at its output. The unit that uses the power draws 100mA. With that unit connected the supply is only 19V. Connecting the two is a series of components. Normally the voltage drop is negligible. But now it's 5V. What value of resistance along the line would cause that voltage drop?
So it is a network of amplifiers all drawing their power from one power supply? There was a time when the power supply could maintain 24vDC across the network; now it can only maintain 19vDC across the same network?
You say that "The unit that uses the power draws 100mA". So is this "unit" just the first amplifier or the whole network?
Where do you see 19v? At the power supply? At the power in to the "master" amplifier? At one of the "repeaters"? Or does the entire power supply and network show 19vDC?
"The power supply unit is set to 24.0VDC, measured at its output." So the power supply unit is a discrete item and can be adjusted that precisely?
Still very much in "How long is a piece of string?" territory, Bill.
PA
Just a suggestion. Can you look at the 19vDC with an oscilloscope? (to see if it shows significant ripple on the "DC")
Yes, 75.
Do the repeaters have terminating resistors on their inputs?
They aren't fed directly from the line power. The latter feeds stabilised voltage converters.
The tests are done with the load disconnected.
Bill
Voltage drop 5 Volts divided by current 0.1 amps is 50 Ohms.
Thomas Prufer
In article snipped-for-privacy@4ax.com, Thomas Prufer snipped-for-privacy@mnet-online.de.invalid> scribeth thus
Of course make sure that the DC resistance of what 50 ohms, isn't confused with the RF Impedance of 75 Ohms!, just saying;)...
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