|Jul25-12, 10:52 PM||#1|
Eliminating Voltage Drop
I am new here so I figured I would post a problem I am having and see what comes out. I am designing a circuit that will force a voltage across a proprietary load that I cannot divulge on, but for this example we will assume it's resistors. So far everything works great but I am still running into an issue, see circuit diagram below:
Due to the high current draw (10A) and the distance between loads (1') I see around 100mV voltage drop at each load so I am reading 700mV at Point A. RL has exactly 1v so keep that in mind because I had to leave the drive circuit out of this post to protect our proprietary work.
What makes the problem worse:
The use of buffers between loads is absolutely out of the question but we are entertaining alternatives... Any Ideas?
|Jul26-12, 01:03 AM||#2|
Use seperate identical wires for each load (and return) that are long enough to reach your most distant load and increasing the drive voltage to 1.4V (or whatever it needs to be to compensate for the drop in the wire)?
|Jul26-12, 01:09 AM||#3|
You are correct, one minor detail however;
If I increase the load drive to 1.4v so that RL5 has a drop of 1v then RL will then have a voltage drop of 1.4v...
The drive sense provides feedback which monitors that specific point. RL is actually about 10' from the drive circuit so the sense line is critical where it's length compensates for the impedance of the drive line, if that makes sense.
Good idea though, thanks for the reply.
|Jul26-12, 06:39 AM||#4|
Eliminating Voltage Drop
Go ahead along gnurf's line but move your sensing point to a position so that there is the same cable length from the sensing point to the positive of each RL. Or beef up your cabling to reduce the resistance to an acceptable level.
|Jul26-12, 09:19 AM||#5|
Back to basics here:
If the goal is to provide equal voltage to each load,
they have to be connected in parallel.
Right now they are NOT in parallel because the wire connecting them has resistance that is NOT insignificant compared to the 0.5 ohm loads, so your interconnecting wires should be drawn as resistors not wires.
You have not a parallel connection but a series-parallel network .
what's the milli ohms per foot of your wire?
Lessee - it's a tedious arithmetic problem to solve your network
but if i do a back-of-napkin approximation, to get in the ballpark:
Let ohms/ft = R
8 amps in first segment
and 6 amps in second segment
and 4 amps in third
and each segment is 2 feet long (one foot coming and one foot returning)
EDIT dumb arithmetic mistake
and you're dropping 0.3 volts
so 0.3 volts = 2RX8+ 2RX6 + 2RX4
R = .3/36 = 8.3 milliohms per foot?
That's like #19 or 20 gage wire!
Is first section melting its insulation?
Try some #10, it's 1 milliohm per foot.
Solid copper wire on standoffs makes a nice busbar that's easy to solder to.
Go to hardware store and buy some solid #10
clamp one end in your workbench vise, wrap other around a tool handle and pull until it just yields. That stretching makes it a tiny bit smaller so the insulation slides right off . You can cut insulation into shorter lengths and use it to cover your busbar between solder connections; we call that "Spaghetti". Just dont forget to put spaghetti back on before soldering.
or else switch to a star connection with equal length feeds, as gnurf and jobraq suggested
|Jul28-12, 03:03 PM||#6|
Hats off to you good sir,
I guess I was having a hard time under standing what you guys were talking about but that was a great break down.
After running multiple tests using your guys suggestions I was above to average the voltage at each load to 1.051v which is one heck of a tolerance. Thanks guys!
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