Voltage Drop

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W6SJK

Senior Member
Me thinks you're all wrong. :) Remember this is an unbalanced load with current in BOTH wires so the voltage drop is 2x what it would be with a balanced load.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
091026-2237

cowboyjwc:

Some assumptions:
I am going to reduce your current to 15 A, but if necessary raise it to 20 A.
The lights are incandescent and are 120 V rating.
A very low current timer can be used, maybe 10 MA at 5 V DC.
The solenoid valve is moderately small.

The loop resistance for a 2800 ft run of #12 solid wire is 5.600*1.588 at 20 deg C, or approximately 10 ohms for easy multiplication. At 15 A this is a voltage drop of 150 V.

So we need to consider #10. This produces a drop of 5.6*15 = 84 V. Feed 240 V to the input. Put a phase shift regulator at the destination to provide the lamps with 120 V. This could include soft start for improved lamp life. Use a Phihong semi-regulated power source for the electronics of the timer. A PLA05A-120-R can supply approximately 12 V DC at 0.4 A from an AC input anywhere from 90 to 264 V AC. A little work may be needed on the supply to the solenoid valve. Will the NEC allow this?

.
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
If you have a 30 volt voltage drop at 120v with lets say 10 amps of load, lets say a 3 watt timer, and a 1250 watt motor, so your voltage has dropped to 90 volts, now you use a BB to do a 32 volt boost, so you now have 122 volts with both loads running, let the motor turn off, and your timer will see 152 volts. Not good.

The BB will boost what ever voltage that is fed into it, by the adder voltage, so if you are trying to over come a voltage drop, the input voltage to the BB will rise as the load gets less.
I agree with the first paragraph, but I would imagine one would boost the voltage at the source with a BB, not at the load.


In my opinion, the two idea to consider are the PV-and-battery and the two-transformer step-up/step-down methods.
 

cowboyjwc

Moderator
Staff member
Location
Simi Valley, CA
Call me crazy, but I was thinking a battery and solar charger.

Actually after I told them what they would need to do, they are thinking of going with this option.

The problem we seem to have with any solar lights, that are not in a well secured area, is that they seem to disappear fairly often.

Oh and then after all of this, they inform me that they want to put this just a little ways down from a traffic signal, WHERE THERE'S POWER! This is why I wish they would get me involved sooner in these projects.
 

hurk27

Senior Member
If you have a 30 volt voltage drop at 120v with lets say 10 amps of load, lets say a 3 watt timer, and a 1250 watt motor, so your voltage has dropped to 90 volts, now you use a BB to do a 32 volt boost, so you now have 122 volts with both loads running, let the motor turn off, and your timer will see 152 volts. Not good.

The BB will boost what ever voltage that is fed into it, by the adder voltage, so if you are trying to over come a voltage drop, the input voltage to the BB will rise as the load gets less.

I agree with the first paragraph, but I would imagine one would boost the voltage at the source with a BB, not at the load.


In my opinion, the two idea to consider are the PV-and-battery and the two-transformer step-up/step-down methods.

Oh sure don't agree with the second paragraph:cool:
But I believe it:D

with voltage drop, it doesn't matter which end you boost from, when the load turns off the voltage will rise to what ever is fed in at the start of the conductors.;)
 

hurk27

Senior Member
Anyone got schematics of the step-up xformer compared to the bb? I don't follow why it's the way it is..

Thx
Jason

Vary the voltage into any transformer and it will vary the output right?

With voltage drop the voltage at the end of the line will vary in proportion to the current being drawn ok

now install any type of transformer that brings a voltage drop back up to nominal voltage and as long as the load stays the same it will be ok, but have the load turn off like in my example, and the voltage will rise proportionally to the current. thus allowing a higher voltage then expected.
 

gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
091027-1915 EST

lakee911:

A typical transformer consists of a primary and one or more secondaries on a common magnetic core. Inherently the primary and secondary are isolated. Thus, no DC path (low resistance) between them. There could be a potential difference between the primary of any value that does not breakdown the insulation between the primary and secondary. You can deliberately provide a low resistance path, bond, one end of the primary to one end of the secondary if desired. Now no isolation. This is done internally on a single post utility transformer.

An autotransformer is a transformer with one coil and a tap somewhere on the coil. The wire size may be different on each side of the tap. If you classify the primary as one outer end and the tap, and the secondary as said outer end the other outer end, then you have a step-up transformer. Interchange primary and secondary and it is a step down. No DC isolation possible.

A commercial product called a Variac is a variable autotransformer.

If instead of the autotransformer we have a standard step-down transformer, for example 110 V to 10 V, a 110 V AC source, might be the secondary of a transformer, but just call it a voltage source, then if the voltage source is connected to the primary of the step-down transformer there will be 10 V output from the step-down transformer. If one side of the secondary is connected to one side of the source voltage, then the sum of the source and the secondary is 100 V or 120 V depending upon the phasing of the connection. Functionally looks very much like an autotransformer. When used this the transformer is called a buck-boost transformer.

.
 

hurk27

Senior Member
So it's only a big deal here because the load is fluctuating so much, right? Step it way up and it won't fluctuate much at all...gotcha

Not exactly

lets say you have a 20 amp load at 120 volts, you step it up to 480, now it's only a 5 amp load, at the other end you step it back down to 120 volts, your back to 20 amp at 120 volts but now you only have 5 amps on the long run wires so the voltage drop is less.

Power companys do this all the time. but at much higher voltages;)
 
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Not exactly

lets say you have a 20 amp load at 120 volts, you step it up to 480, now it's only a 5 amp load, at the other end you step it back down to 120 volts, your back to 20 amp at 120 volts but now you only have 5 amps on the long run wires so the voltage drop is less.

Power companys do this all the time. but at much higher voltages;)

You will have to use a transformer with taps to raise the voltage at the far end. Putting 456v (5% drop on 480) into a transformer with a 4:1 ratio will get you only 114v on the secondary. This doesn't account for any added loss of the transformers.

It's early let me rethink this...
 
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tepres

Member
Vary the voltage into any transformer and it will vary the output right?

With voltage drop the voltage at the end of the line will vary in proportion to the current being drawn ok

now install any type of transformer that brings a voltage drop back up to nominal voltage and as long as the load stays the same it will be ok, but have the load turn off like in my example, and the voltage will rise proportionally to the current. thus allowing a higher voltage then expected.

Building on the above...

Kind of like reading an open circuit. You will measure source voltage at an open. If you increase the voltage to 150V, and turn off your load, you will read 150V at the load (zero current - zero voltage drop).

Step up - Step down solves this problem...
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
with voltage drop, it doesn't matter which end you boost from, when the load turns off the voltage will rise to what ever is fed in at the start of the conductors.;)
I absolutely agree with that, but here's the difference: when you use a BB to boost, the power for the extra voltage is seen as extra line-side current, which adds to the voltage drop.

With the BB at the source end, the extra current does not have to be carried by the long section of the total circuit run. This way, the advantage of the higher voltage is maximized.
 
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