Haynes-Shockley experiment with laser

[and661]

Hi!

I am having a bit of a problem comprehending the physics of the Haynes-Shockley experiment; specifically the one where the excess carriers are generated with a laser.

In the ordinary experiment, some electrons or holes are physically injected through a contact, right? Thus, there will be more holes or electrons amongst the excess carriers, and when this pulse (which always moves like the minority carriers, right?) reaches the detector, a voltage change is registered. So far, so good.

But now we use a laser for generating the carriers instead, kicking up electrons into the conduction band and therefore generating just as many excess holes as electrons! And since they all move like the minority carrier, they will pretty much stick together. So no pulse can be registered, since the positive and negative charges cancel out! This makes no sense; can anyone tell me where I am wrong?

EDIT: Sorry, this probably belongs in the Electrical Engineering category. Could an admin please move it there?

 

[eljose79]

Hello,

I can understand your confusion with the Haynes-Shockley experiment. In this experiment, the excess carriers are generated through the use of a laser instead of physically injecting them through a contact. This may seem counterintuitive at first, but it is actually a crucial aspect of the experiment.

When a laser is used to generate excess carriers, they are not all moving in the same direction. Some of the excess carriers will move towards the detector, while others will move away from it. This is due to the random nature of the laser-generated carriers.

So, even though there are equal numbers of excess holes and electrons, they are not all moving in the same direction and therefore do not cancel each other out. This allows for the detection of a pulse when the excess carriers reach the detector.

I hope this helps to clarify your understanding of the Haynes-Shockley experiment. If you have any further questions, please don't hesitate to ask. And don't worry about the category, I'm happy to assist with any questions related to physics.