Building Test Load - Help

You guys can correct me if I'm wrong..

Here's the explanation on how to select # of diodes based on the diode breakdown voltage:

I noticed that if you look at a datasheet of the laser diode, it will tell you operational voltage. This is the voltage value is the target value you want to build up your diodes in series. For example if your IR LD is 2.1V Operation Voltage, you want to try to get 3 diodes (with breakdown voltage of 0.7V) to get to that value of 2.1V. Another example, if you have breakdown voltage of 0.5V diodes, then you would need 4 of them.

If you don't have the datasheet then the rule of thumb is 3 diodes for IR, 4 diodes for red, and 6 diodes for BR. This assumes the breakdown voltage of the diodes are 0.7. Do your math accordingly.

The 1 ohm resistor in series is for measuring the current going through the path. Voltage = Current x Resistance.. and since you use Resistance = 1 ohm, the voltage measured across the resistor would be the current (+/- % of error of resistor value)

You can assume wit 99% of possibilities that all silicon rectifier diodes have 0,7 / 0,75 V dropout ..... cause this is the typical dropout of a silicon junction (to be more precise, 0,6V at absolute minimum load, this value increase a bit with some current increasing ..... with our typical currents, is 0,7 / 0,75)

Germanium diodes have 0,20 / 0,22 V dropout, typically, but is so difficult to find germanium diodes that take more than 40 / 50 mA, that it don't worth the effort (and also if you find them, probably you end paying them a little fortune )

So, just go for approximation ..... 4 diodes are ok for anything red / IR (you can also use just 3 for IR, is not so important if the LD is an 1,8V FV) ..... 6 for BR like PHR or 4X ones ..... better 7 for 6X / 8X LDs ..... in any case, if the total dropout of the dummy load is also a bit more than the one of the LD, it's OK ..... is worse if the dropout of the dummy load is less than the one of the LD.

As example, if you use a LD with 5V FV, and a dummy load with the dropout of 5,2 (7 diodes), is ok, so if the driver have, as example, a maximum output voltage of 5,5 / 5,6 V (like flexdrive), you still can set correctly the current ..... now, suppose to have a LD with a FV of 6V (8X or more) ..... with a dummy load that uses 6 or 7 diodes, you still can set the current with the driver, cause the dropout is under the maximum voltage output of the flexdrive, but when you connect the LD in place of the dummy load, the driver find at the output something that require more voltage than its maximum, for keep the same current, and start to work in bad conditions (can just give less current, or go in overload, depend from the configuration, anyway is always not the good mode of working for it)