What is a test load?

Can u just measure mA straight across the diode or is that bad?
What other things can you use as a test load? :-?

You can measure a voltage drop across a diode.. a standard silicone rectifier
diode will have a voltage drop of about 0.6 to 0.7 volts... it can NOT be used for
current checking... what you need is a 1 Ohm resistor in series with your load
and a Digital Volt Meter.

You put a Meter lead on each side of that 1 Ohm resistor and put your DVM's
scale to mV (millivolts).  (1 mV = 1mA)

To answer the "what is a test load", in the title.

A test load is a convenient way of checking that your driver is doing exactly what you are expecting, if you connect straight up to your LD and the driver is not working properly you may blow the LD.

4 silicon diodes 1N4001 + 1ohm resistor for a X16 red
6 silicon diodes 1N4001 + 1ohm resistor for an 803T Blu ray

See this thread to find out how to test the driver......
http://www.laserpointerforums.com/forums/YaBB.pl?num=1197651171

Regards rog8811

Rog gave a nice explanation.

Basically a driver that puts out a constant current (not constant voltage) will put out the same current no matter the load, so you can use a test load in place of the laser diode to both set the current AND to prevent damage to the diode from either accidental over turning the pot (which can be easy in small drivers like mine) or if you're testing a driver you personally built or are unsure that is properly functioning so you don't blow your diode if it isn't.

In the case of CCO (constant current output) drivers, any load will do, but having a 1 ohm resistor in series with some rectifier diodes allows you to more accurately simulate the conditions of a particular laser diode AND easily measure the current (by the VOLTAGE across the resistor).

Just measuring current "across" the diode is not going to work and I think you don't fully understand what you're actually measuring.  Current is the "volume" of electrical flow, so if you measure current "across" a diode, you're just shorting or diverting the current flow from the diode to the multimeter, but some current may still flow through the diode depending on the resistance.  You're not actually measuring the current going through the diode, but actually the current going through the meter.  

Putting a resistor of known resistance in series with a load allows you to calculate the current by measuring voltage across while powered and using Ohm's Law (get to know it well in this hobby), Voltage = Current * Resistance, or V = IR.

Rearranged, you get I = V / R, so you see that voltage divided by resistance gives you the current and if you're using a 1 ohm resistor, you can basically ignore the denominator (resistance) so that I = V or you current is equivalent to the voltage you measure across the resistor.  

This is, of course, assuming the resistor is ACTUALLY 1 ohm, but most have a tolerance of 5% or 1%, meaning the resistance may be slightly off, and so your calculation may be off as well, but only by the tolerance of the resistor, so not a huge deal, typically only a few mA for a 1% tolerance.  You could always more accurately measure the resistance, though most meters aren't very accurate at low resistances.

ok thanks guys
lol i found my dads old i ohm resistor its just copper wire raped around a wooden dowel with two aligator clamps one the ends to hook it up to things

I found my dads old 1 ohm resistor its just copper wire wrapped around a wooden dowel

Click to expand...

Wire wound resistor, nothing wrong with that as long as it is the right value....

Regards rog8811