The absolute easiest way is to purchase a prebuilt LCD Ammeter module. You could also use a shunt across one wire to the load, measure the voltage drop and compute the current draw.
Simplest way (to me) would be to pass your current trough a small resistor in series with whatever load. This will cause a small voltage drop over the resistor wich is relative to the current (more current, bigger voltage drop). Then feed this voltage into a A/D converter with direct lcd-drive and you're done.
If you're talking 'bout a HD44780 LCD then i think there's no other way then using a pic or some other form of control
You can get all kinds of LCD meters, many either come with (or have the option of) panel mounting. Most are usually based on something like the 7107 digital meter chip, you may need to use an external current shunt to use one as an ammeter - but again, these are usually available as an option with the modules.
... pass your current trough a small resistor in series with whatever load. This will cause a small voltage drop over the resistor wich is relative to the current (more current, bigger voltage drop). Then feed this voltage into a A/D converter with direct lcd-drive and you're done.
With this approach, you have to be careful if you intend to use the same circuit you are monitoring to also power the measurement circuit (and the supply you are monitoring is >5V). The easy way to do it is to put the power resistor in series with the ground path (so that the ADC has the same ground reference). The problem is that you have unwanted impedance in the return path. If you put the resistor on the high side, you need to supply a conditioning circuit to ground reference and limit your voltage before you input it to the ADC.
If you are using an independent supply .... no problems.
If you really want a good controllable power supply you should look for circuitry containing a voltage-regulator and a current-limiter. They will allow you to control voltage and current independent of each other.
So you can set a certain voltage & current and choose wich one of the 2 is in charge.
Adding a pot on the output will change voltage & current at the same time. Making it difficult to set a value. Also too much will depend on the load