What is the voltage drop at a VCSEL / laserdiode

[Bigman]

Hi,

I am designing a constant-current source to power a VCSEL-laserdiode.
The datasheet of that VCSEL can be found here: http://www.optekinc.com/datasheets/OPV330.pdf

I am wondering, about the voltage drop across the VCSEL, when the current-flow through the VCSEL is 7mA.

The datasheet mentions: Forward voltage 2.2V @ 7mA. Is the meaning of those 2.2V a maximum voltage which still "can be APPLIED from the outside"? Or does this mean, when a current of 7mA runs through the VCSEL, the voltage-drop generated by the VCSEL can be up to 2.2V (which would mean, I would need to apply at least 2.2V).

The datasheet also mentions a max. serial resistance of 55 Ohm. Taking this for 7mA, the voltage-drop generated by the VCSEL would be just 7mA*55 Ohm= 0.39V.

2.2V vs. 0.39 ... a significant difference.  So, which voltage drop does really happen accros the VCSEL when 7mA are flowing though it?

Thanks in advance for you comments!

[SeanB]

2.2V is the max forward voltage of the device, using a device with the maximum series resistance. Lowest will be with lowest series resistance, giving a forward drop of just underr 2V. The forward voltage drop will vary both from device to device and with temperature, so you would be best advised to drive it with a constant current source of 7mA, or if you want more light but with a shorter life drive at up to 12mA. Whatever you use it just needs to be able to deliver 7mA while the device will drop between 1V9 and 2V2. Thus just using a resistor will not be very good unless you are using a high supply voltage like 12V, where the change in current over the device voltage change will be small. Using 3V3 for example a simple resistor would have to be selected per device, varying from 150R to 200R. 12V it would need to be between 1440R and 1400R, so a 1k5 with a parallel 22k resistor would work for all diodes with only a small error

[amyk]

"Series Resistance is the slope of the Voltage-Current line from 5 to 8 mA."

In other words, the 20-55 ohms is the differential resistance of the device - and this is an approximation, since the I-V curve is the exponential of a diode.

Or does this mean, when a current of 7mA runs through the VCSEL, the voltage-drop generated by the VCSEL can be up to 2.2V (which would mean, I would need to apply at least 2.2V).

Correct.

[Bigman]

Thanks for your replies. So, it's clear: my constant-current-source does not ONLY have to keep the 7mA constant, BUT also have to "give" AT LEAST 2.2V to the VCSEL.

[Marco]

Your current source has to be able to supply up to 2.2V.

In the end you're always supplying a voltage to it, but with a current source the voltage you are going to supply to it is unknown in this case ... other than it being less than 2.2V.

[awallin]

2.2V vs. 0.39 ... a significant difference.  So, which voltage drop does really happen accros the VCSEL when 7mA are flowing though it?

The voltage drop will change with wavelength. Lowest for e.g. telecom at 1.5um, somewhat higher for NIR at say 980 or 830nm, and highest voltage drop for e.g. blue-ray laser diode at 405nm.
If you want to make a general purpose laser diode drive then allow for up to 6 V voltage drop for 405nm (and maybe UV) diodes.

[Marco]

The resistance is a modeled one and is differential, it's determined from measured forward voltages like this (Vi=8mA - Vi=5mA) / 3mA.

Think of the diode as an ideal one with constant forward voltage with a series resistor, the given resistance is of that series resistor.