Question: If I were wanting to drive i.e. industrial relays which require 24v from a Microchip PIC what would I need to look at in terms of a general transistor circuit (and it's pitfalls when driving such a component).

Answer: 
I'm sorry, but I think Matt's answer is not a good one at all.

The MOSFET in his schematic is a *P-channel*, not an N-channel. The diode doesn't offer any protection for the FET; it may be destroyed together with the FET. Besides it's a 20 V diode, so even if it would protect against the induction voltage the 24 V supply may already kill it. The 7406 is superfluous, besides its maximum voltage is 30 V, not 40 V, and that 30 V is **Absolute Maximum Ratings**, not for continuous use. The circuit will also draw an unnecessary 5 mA with the relay on, and 10 mA no less with the relay off. Also the 100 Ω resistor doesn't "dampen oscillations".

What you need is a **logic level gate** MOSFET. You're using a PIC, which probably will have a supply voltage of minimum 3.3 V. Let us know if the voltage is lower. A logic gate FET will switch on with a 3.3 V gate voltage, so the PIC can drive it directly. No 7406 needed. 

A relay typically needs less than 500 mW, at 24 V that would be 20 mA, but this is an industrial relay, and will probably need more. Let's be generous and say it needs 1 A (that's 24 W!). If we can find a FET with an \$R\_{DS(ON)}\$ of less than 350 mΩ we'll be able to use an SMD; these are much cheaper than PTH parts. At the high 1 A it will dissipate 350 mW. What else? Power supply is 24 V, so let's take a maximum \$V\_{DS}\$ of minimum 40 V. One FET which fulfills these requirements is the [BUK98150](http://www.nxp.com/documents/data_sheet/BUK98150-55A.pdf):

Max. \$V\_{DS}\$ 55 V  

Max. \$I\_D\$ 5 A  

Max. \$R\_{DS(ON)}\$ < 200 mΩ @ 3.3 V  

Max. \$V\_{GS(th)}\$ 2 V

Looks good. The BUK98150 will sink 2 A at 2.6 V gate voltage.

![enter image description here](https://i.stack.imgur.com/DSVkd.png)

This graph shows an \$R\_{DS(ON)}\$ of 175 mΩ @ 3 V and 2 A, for 1 A it will be less. Then dissipated power will be 175 mW, which the SOT-223 package can handle easily. The 175 mV drop is negligible. 

![enter image description here](https://i.stack.imgur.com/ZjVyT.png)

This is the circuit. Contrary to Matt's it only consumes 0.1 mW. I've kept his 100 Ω resistor, which limits the short current spikes when switching; a microcontroller doesn't like capacitive loads much. The 100 kΩ ensures that the gate won't float if the PIC's I/O would be switched to input accidentally. 

As you can see the diode goes over the relay, not the FET. You can use a Schottky diode here. [This one](http://www.nxp.com/documents/data_sheet/PMEG4010EH_EJ_ET.pdf) has a maximum reverse voltage of 40 V.
A very simple and integrated solution may be the [NUD3160](http://www.onsemi.com/PowerSolutions/product.do?id=NUD3160). Basically the same footprint as a discrete MOSFET or transistor but with inbuilt protection for both the input and output.