loop compensation and transfer function for this circuits

You can break the loop at any point so you can pick the point that makes doing the math easiest.

You may need to add a small resistor in the base leg of the transistor.

The small signal model of the BJT and the Laser can be used but you need to step through the whole current range. 3 or 5 steps is likely to be enough. You can often simplify the problem by looking at the frequencies that matter and removing poles that won't matter.

The BJT circuit can be modeled without the op-amp. Chances are it will have its poles way above the corner frequency of the whole system. If you drop those poles, the BJT will be simpler. The base impedance of the BJT is too low in this case to let you assume a simple transconductance but you can load the op-amp resistively and be close enough.

"You may need to add a small resistor in the base leg of the transistor. "

Yup, that would be a good idea. What's the current gain of the BJT, does the opamp have enough output current to drive it? What is the turn on voltage of the laser diode? It might be something like 2.0 to

2.5 volts, in which case dropping the 0.6 Volts over the BJT might be too much a FET might be better here.

Are you interested in fast pulses?

George

Or use the LoopGain function from my website and get it exactly right ;-)

...Jim Thompson

--
| James E.Thompson, P.E.                           |    mens     |
| Analog Innovations, Inc.                         |     et      |
| Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
| Phoenix, Arizona  85048    Skype: Contacts Only  |             |
| Voice:(480)460-2350  Fax: Available upon request |  Brass Rat  |
| E-mail Icon at http://www.analog-innovations.com |    1962     |
             
 I love to cook with wine     Sometimes I even put it in the food

IIRC: The OPA350 limits him to short of 40MHz bandwidth and has a strong output.