Electrical – Are “Apk” and “Amp peak-to-peak” the same thing

The real difference is in the power handling. The actual power (and therefore current) provided by the transformer is determined by the load on the transformer.

The 1000 mA transformer can handle a load that consumes more power and also more current.

For the sake of simplicity, I'm going to assume a resistive load with no reactance, and therefore a power factor of 1. You will probably likely want to look up those terms if you're interested in learning more about transformers and AC power in general*.

For example,

If you have a 5W load (device you want to power) connected, it will draw 5W / 12 V = 417 mA (approximately). If you connected this load to either tranformer, it would draw 417 mA.

If you have, say a 10 W load, it would draw (or attempt to draw) 10 W / 12 V = 833 mA (approximately) - it would draw this if connected to the 1000 mA transformer. If you tried to connect this load to the 600 mA transformer then one of the following would occur:

1. The load would not function (because the voltage droops or sags too low to power the device)

2. The load would draw more current than the transformer can handle and possibly damage the transformer (depending on the transformer and load, could be immediate, or could slowly heat up and deteriorate over time)

3. The transformer fuses would "Blow" if installed and properly size for the transformer (to prevent the damage in #2)

This is a simplified explanation. Most transformers can actually handle a little more than what their rating states - so they won't always get damaged, but you should always stay within the manufacturer's specifications to ensure maximum life of the equipment.

*If the loads are reactive and have a pf other than 1, you cannot simply divide power by voltage to get current.

The maximum current in a permanent magnet motor is governed by two things, which are in order of importance ...

a) demagnetisation of the permanent magnets

b) heating

Demagnetisation will kill your motor in an instant, if the current goes over this limit at all, it's game over. Fortunately, it takes a rather large current to do it. Unless the manufacturer specifies it directly, it can be rather difficult / impossible to infer from the other specifications. You can assume it's above the peak rated current, but by what safety factor, 2x, 5x, 10x ? If it's not in the data sheet, one way is to contact the manufacturer, another is to test a motor, with the expectation of destroying it. After demagnetisation, the Kt will be lower, zero or even reversed, a quick measurement of output torque will detect this.

Heating is a bit more tractable, but still needs some guesswork. The limitation is the temperature of the windings, as the insulation is good only up to a certain temperature. Fortunately it's easy to estimate the temperature of the windings by measuring their resistance, copper has a relatively large 10% change in resistance for every 25C increase in temperature. Unfortunately, the specs probably do not tell you what the limiting temperature on the windings is. Best is to play safe and assume (say) 80C for the max temperature, \$t_{max}\$

The method is to measure the approximate \$I^2t_{max}\$ of the windings. Measure the starting temperature \$t_{initial}\$. Apply a 10 second pulse (short enough to be roughly adiabatic, long enough to be useful) of rated current, and estimate the temperature rise from the change in resistance. The \$I^2t\$ you have just applied, \$I^2t_{test}\$, is 10*Irated*Irated. The maximum \$I^2t\$ you can apply safely, \$I^2t_{max}\$, is $$I^2t_{max}=\frac{t_{max}-t_{initial}}{ temperature rise}I^2t_{test}$$

The maximum you can apply on thermal grounds is any current I, as long as the application time is less than $$\frac{I^2t_{max}}{I^2}$$ in a single pulse, starting with a cold motor. If you apply multiple pulses, then you must use less current, to keep the temperature down.