Meter shunts are a useful means of extending the current range of an ammeter. As current divides between two resistors in parallel it is possible to increase the range of a DC microammeter or milliammeter by paralleling an additional resistance with the inherent DC resistance of the meter itself. This is called a meter shunt.

# Meter Shunts

### What is a meter shunt?

A meter shunt is a useful means of extending the current range of an ammeter. As current divides between two resistors in parallel it is possible to increase the range of a DC microammeter or milliammeter by paralleling an additional resistance with the inherent DC resistance of the meter itself. This is called a meter shunt.

### Properties of Meters

It would prove helpful in considering meter shunts to consider the topic of meters first if you haven't already done so.

### Meter Shunt calculations

Typical ammeter ranges available to hams and electronic hobbyists are usually the 100 micro-amperes and 0 - 1 milli-ampere types. Obviously there are other ranges, especially when we consider surplus sales of meters.

Let's assume we have a typical 0 - 1 milliampere millammeter and want to extend it's useful range to record current up to 5 amps for a power supply project.

In figure 1 is a typical panel milliammeter of the MU45 type (45mm).

Figure 1 - a 0 to 1 mA ammeter requiring a meter shunt

So how do we make a meter shunt to go from 0 - 1 mA all the way out to 5A? Consider the schematic in figure 2 which gives you some idea on what we're about.

Figure 2 - schematic of a 0 to 1 mA ammeter with a meter shunt

The formula to determine the shunt resistance is:

R shunt = R meter / (n - 1)

Where R shunt is our meter shunt resistance, R meter is the inherent meter resistance and "n" is the multiplier. In our example the multiplier "n" is 5A divided by 1 mA or 5 / 0.001 which of course equals 5000. So (n - 1) must be 5,000 - 1 = 4,999.

For this exercise we will assume our internal meter resistance (R meter) was accurately measured  as 58 ohms.

Now substituting all those values into our formula we get a meter shunt resistance of:

R shunt = 58 / (5,000 - 1) = 0.0116 ohms

This meter shunt resistance must be capable of safely carrying our anticipated maximum current, in this case 5 amperes.

A fairly useful guide for current carrying capacity for ordinary copper wire of the type generally available to us is 250 circular mils per ampere. Mils in this context are thousandths of an inch and circular mils are simply the diameter of the wire in thousandths of an inch squared. So 5 amperes would require something in the region of 1,250 circ mils, the square root of which is about 35 mils diameter or about 0.9 mm.

Allowing a factor of safety we could look at AWG #19 which is described in wire tables as being 0.912 mm dia, 35.9 mils dia, 1288 circ. mils, and has a resistance of 8.21 ohms per 1,000' (305 metres) at 25 degrees centigrade.

It follows if the resistance is 8.21 ohms / 1000' then to obtain the required length of #19 copper wire we use the formula:

Length (in feet) = R shunt / [R wire / 1000]

Or length = 0.0116 / [8.21 / 1000] = 1.4129' ( always convert to metric by multiplying by 305 ) In this case in metric, we get 431 mm of #19 gauge copper wire. This length of wire should be wound on a suitably convenient form to withstand the heat generated and preferably spaced one turn apart (especially if you didn't use enamelled wire).

Be aware those figures are only likely to be valid at 25 C or 77 F.

Professionals in the instrument field would of course use other materials. If you have access to an accurate milli-ohmeter consider using wire salvaged from electric heating appliances such as toasters, jugs or radiators. Bear in mind such wire has different temperature properties i.e., as it heats up the wire resistance can vary quite significantly.

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