# Overcoming a wire's internal resistance?

Let's say I have a very sensitive device that needs EXACT voltage. Let's say that the voltage it requires is 5 volts. Now let's say I'm running the aforementioned voltage over a wire with an internal resistance of 20 ohms.

How many volts do I need to put into the wire to ensure I get 5 volts out?

Based on my rudimentary understanding of Ohm's law, I would say 5 volts / 20 ohms = 0.25, so I need 5.25 to overcome the wire's resistance. No idea if that's right or not though.

Edit: The current drawn will be anywhere from 1 to 2 amps.

• And remember what goes out must come back - don't forget the ground return wire resistance as well because it won't be zero. Jul 28, 2013 at 18:21
• 5 volts x 20 ohms = 0.25 coming from the Ohm's law, which is V=IR. How many things are wrong with this? Jul 28, 2013 at 20:30
• @NickAlexeev In my own defence, that was supposed to be division... I'm not quite that dense. ;) Jul 29, 2013 at 1:16
• @KevinSoviero you can still see the error in your logic if you look at the units. $5V/20\Omega = 0.25A$ (because $\Omega=V/A$) Jul 29, 2013 at 1:18
• @KevinSoviero So far so good, we've established where the number 0.25 came from. 0.25... 0.25 what? 0.25 amperes. Then your imagination seems to have taken over. (1) 0.25A ain't the current through the wire. You know that the current will be 1A to 2A. (2) You do 5V + 0.25A = 5.25V. Vastly different measurement units can't be added frivolously like that. Jul 29, 2013 at 2:06

## Regulation is required with a dynamic load

If you do not know the current (load impedance) a priori (before-hand) and if this current is not constant from the moment your device begins operation to the moment it ceases, you will need to employ active regulation to achieve this.

That is, you will need a circuit element that will monitor the load voltage and adjust the source voltage in response to maintain the desired 5V load voltage.

## Here's why...

The problem is that Ohm's law applies to static conditions.

simulate this circuit – Schematic created using CircuitLab

You do not know the load impedance and you also do not know the source voltage. The load current is specified as a range (1A to 2A). If the current could be determined, and it is constant, you could solve it this way:

$$V_{wire} = I_{device} * R_{wire}$$

We want to know how much voltage to apply at V_source to overcome V_wire...

$$V_{source} = V_{device} + V_{wire}$$

## So...

• If current is 1A, the source voltage must be 25V.
• If the current is 2A, the source voltage must be 45V.

That's a big range and driving through a 20 Ohm resistance (which is a lot for a wire) will result in large changes (20x !!!) to the requisite source voltage for small changes in the source current.

## Your best options here are to

1. use a heavier gauge wire (wider diameter, or run more wires in parallel) which will lower the wire's resistance
2. or a point-of-load voltage regulator which will stabilize the voltage independent of current

Good luck with your project. Cheers.

Rather than make your device less sensitive (maybe not possible) or remotely regulating the power (placing a regulator closer to the sensitive load - again maybe not possible) the "old-timey" way of doing this is through using a kelvin connection. You bring a lead back to the power supply that samples the load at it's pin (i.e. after the wire resistance). What is important is that this kelvin connection should NOT support any current so that is doesn't suffer any IR related voltage drop. You now have a measurement basis from which you can drive your supply into the load (say at your referenced 5.25 V) and dynamically overcome the main power supply wire resistance.

The issues are that you require another wire and that wire may pickup stray fields/signals and generate noise.

However, this is a very common approach. Look at any power supply that has a "sense lead" input.

Here is a snip from an Agilent website:

The current drawn will be anywhere from 1 to 2 amps.

You cannot put more volts into one end of the wire, when the current varies, and expect to get a constant voltage out the other end. Using ohm's law for 1 amp and 20 ohms says that you will have a 20 volt drop across the wire-- and you'd have to put in 25 volts to get 5 volts out. But for 2 amps you will have a 40 volt drop and you need to put 45 volts into the wire.

So let's say that you put 45 volts into the wire, but only draw 1 amps. There will be a 20 volt drop in the wire and you will get 25 volts out the other end.

The solution is to redesign the end with the "very sensitive device" to be not so sensitive.

## Another powerful idea

There is another (really a bit extravagant way) to compensate for the line resistance - by an S-shaped true negative resistor with resistance -Rl connected in series. The two resistances are summed and the result is zero line resistance (Rl - Rl = 0).

## Conceptual implementation

I have illustrated this powerful idea by the picture below taken from my Circuit-fantasia story:

The source internal resistance is compensated in this picture but the line resistance is basically the same. Note that the true negative resistor is actually a variable voltage source producing a voltage V = I.R that compensates the voltage drop V = I.R across the "positive"resistor.

## Op-amp implementation

I have shown a possible op-amp implementation of this idea - the so-called voltage-inversion negative impedance converter (VNIC), in my Wikibooks story about S-shaped negative resistance:

The advantage of this solution is that it does not require additional sense wires. Its disadvantage is that the disturbing effects (temperature, etc.) are not taken into account. And, of course, the op-amp output current is limited; so, in this low-resistance application, a booster will be needed.

## Amazement

I have always wondered how such a brilliant idea is so difficult to understand although it is so simple:

There is a line resistance (Rl) >>> a voltage drop (I.Rl) appears across the line resistance >>> it is subtracted from the input voltage (Vin) >>> but we insert another voltage source with the same voltage I.Rl in series so this voltage is added to the input voltage >>> as a result, the voltage across the load (VL = Vin - I.Rl + I.Rl = Vin) and the current through the load (IL = Vin/RL) do not change. All the inverting op-amp circuits are based on this idea (aka "virtual ground").

## "Ideal" ammeter

By the help of this extremely simple and intuitive idea, a month ago, my students made an "ideal" ammeter in the laboratory using only a potentiometer and another but negative voltage source: