This has nothing to do with Ayrton Senna at all. Somebody set up you the bomb.

What is an Ayrton shunt?
Or a shunt circuit at all, for that matter?

An Ayrton, or universal shunt is used in ammeter circuits involving PMMCs to alter the sensitivity of the meter. Why use shunts? Since the meter will probably melt and hurt you if a large current goes through it, we use the shunt. A large part of the current is drawn across the shunt resistor, hopefully enough such that only the full-scale deflection current passes through the meter.

Before we consider the Ayrton shunt, though, let's have a quick look at a standard shunt circuit.

```

I
o---->----+-----+-----+-----+--------+
Ish V     |     |     |        |
\     \     \     \        |
/Ra   /Rb   /Rc   /Rd      V Im
\     \     \     \        |
/     /     /     /        |
|     |     |     |        |
|     O     O     |      -----
\               /      |     |
O             O       |Meter| Rm
o                    |     |
\                    -----
\  Switch             |
\                    |
\                   |
\                  |
|                 |
o------------------+-----------------+

Where Ra, Rb, Rc and Rd are all differing resistances;
Rm is the internal resistance of the meter;
I is the initial current supplied by the source;
Im is the full-scale deflection current through the meter;
and Ish is the current through the currently selected shunt resistor.```

Since the shunt resistance is in parallel with the meter, the voltage drop is therefore the same across the shunt resistor as it is across the meter. Thus:

```
Vsh = Vm

Using V = IR,

IshRsh = ImRm

ImRm
Rsh = -----
Ish

However, I = Ish + Im

ImRm
Rsh = -------
I - Im

```

This circuit seems to be working perfectly, and indeed it does, provided the switch is a make-before-break type. If it isn't, then the full current goes through the meter and we know what happens then. The Ayrton shunt circuit solves this problem by ensuring there is a shunt resistance at all times in the circuit. How? Like so:

```

/----+------------------+
/     |                  |
/      \        /|\       |
/       /         |        |
AO        \ Ra      |        V Im
/         |        |
I   Switch      |         |        |
+ve o--->---x----oO-----+         |      -----
B     |         |     |     |
\         |     |Meter| Rm
CO        /         |     |     |
\       \ Rb     Rsh     -----
\      /         |        |
\     |         |        |
\----+         |        |
|         |        |
\         |        |
/         |        |
\ Rc      |        |
/        \|/       |
Ish         |                  |
-ve o--------<----------+------------------+

Where Ra, Rb and Rc are resistances;
Rsh is the total shunt resistance;
A, B and C are the three switch positions;
I is the initial current;
Im is the FSD current through meter;
and Ish is the shunt current.
```
As can be seen, there is no possible way that the circuit can be left without a shunt resistance, saving you millions of pounds on replacement PMMC meters. However, nothing in life is free: as a result of these hijinxs, the overall meter resistance is increased.

Calculating resistance in an Ayrton shunt

To calculate the resistance valves, the best method is to start with the most sensitive range and work backwards to the least sensitive. The calculations are as follows:

```
Most sensitive range (position A):

Rsh = Ra + Rb + Rc

Vsh = Vm => IshRsh = ImRm

ImRm
Rsh = ------         (Remember: Ish = I - Im)
I - Im

Next range (position B):

Note that on this range, Rb + Rc are in parallel with Ra + Rm. So:

V(Rb + Rc) = V(Ra + Rm)

(I - Im)(Rb + Rc) = Im(Ra + Rm)

We know Rsh = Ra + Rb + Rc,
=>  Ra = Rsh - (Rb + Rc)

I(Rb + Rc) - Im(Rb + Rc) = Im(Rsh - (Rb + Rc) + Rm)

I(Rb + Rc) - Im(Rb + Rc) = ImRsh - Im(Rb + Rc) + ImRm

Cancelling out Im(Rb + Rc):
I(Rb + Rc) = ImRsh + ImRm

ImRsh + ImRm
Rb + Rc = ------------
I

Least sensitive range (position C):

On this range, Rc is in parallel with Ra + Rb + Rm. So:

VRc = V(Ra + Rb + Rm)

Rc(I - Im) = Im(Ra + Rb + Rm)

Again, Ra = Rsh - (Rb + Rc);
IRc - ImRc = Im(Rsh - (Rb + Rc) + Rb + Rm)

IRc - ImRc = ImRsh - ImRb - ImRc + ImRb + ImRm

Cancelling out:
IRc = ImRsh + ImRm

ImRsh + ImRm
Rc = ------------
I

```

"What? That's the same as last time!" I hear you scream. You're correct, but remember: Everything you change changes something else. In this case, since the resistance values have changed, the I is also changed, thus giving a different result.

Information from class notes/study, Instrumentation course, QUB Electrical and Electronic Engineering.