Testing Current Load

[Xplorer877]

Ok so I'm implementing home made drivers and want a means to test for voltage and current across a laser diode emulating load. I have put together a quick and easy little circuit for blu-ray and red diode leads.

The diodes are 1N4001 rectifier diodes. To those who have experience with laser emulating loads, is this a good setup? How exactly does a series of standard diodes and a resistor emulate a Laser diode?

When the switch is closed it will bypass three diodes for testing red diodes. Otherwise it's for blu-ray diodes.

What do ya think?

-Tony

 

[phoenix3200]

I haven't looked at the I-V characteristics of the BR or red diodes, but you should take a look at the test board from StoneTek's website.  The I-V characteristics of the 1N4001 and the diodes he uses are different, but it shouldn't matter in the long run.  I would use 6 diodes for the BR and 4 diodes for the red if I was you (200mA ~ 0.8V with the 1N4001), and minimize the resistance of the test resistor to avoid putting improper emphasis on the resistor and to reduce the power output P = I^2 R.

Just make sure you're using the test load for a constant-current driver.  Besides the fact that constant voltage dividers aren't good for diodes, you're using the test load to measure current, and to do a very rough approximation of the voltage.

Edit:  I completely overlooked the pic.  It looks good, other than you should bridge only two diodes, not three.

Edit2: Do NOT flip the switch while the circuit is on/there are attached charged capacitors.

 

[rkcstr]

Yeah, the circuit essentially the same (without one of the non-essential parts) as my Test Load (that Stonetek sells).

Most people, for standard 1N400x diodes, use 4 for red to get appx. 3V on load, but I use 3 because of the diodes I use, at the currents typical for red (and the added resistor voltage), will be around 3V. While 6 of them will be more in-line with a blu-ray diode at their usual currents. They obviously won't be exact, but good enough.