Laser Parking Aid Schematic

[f4cepl4nt]

Lately my parents have been having a hard time parking their car in the garage because of visual obstructions. I have a circuit here that will use a 5v power supply (probably plugged into the wall, batteries wouldn't last long enough)

--d1 is a laser pointer
--r1 dissipates .5V so that d1 runs on 4.5v, and at 200mA (hopefully thats the right value, correct me if I'm wrong) so it should be 2.5 ohms right?

The problem is, I'm a little bit confused about using not gates. d2 should be a photodiode, so that when d1 shines on it, r2 (which would be a buzzer or some indicator that the light beam has been broken) doesn't do anything. r3 limits the current when r2 doesn't have current through it, and r2 limits current when r3 has no current through it, am I right? r2 May be just a buzzer, or a buzzer and another resistor if the buzzer doesn't run on 5v.

Hopefully I've explained my problem, just let me know if I'm doing things right. Oh yeah, both r3 and r2 should be 25 ohms right? (5V / .2A)

Here's a link to my schematic:

[audioguru]

--d1 is a laser pointer
--r1 dissipates .5V so that d1 runs on 4.5v, and at 200mA (hopefully thats the right value, correct me if I'm wrong) so it should be 2.5 ohms right?

Wrong! If a laser diode had a voltage of 4.5V and a whopping current of 200mA then it would dissipate 0.9W and melt. If it could be kept cool enough, and sent out only 30% of its power then its laser power would be 0.3W and it could burn holes through things.

Please don't just guess, look it up.
Digikey lists 92 laser diodes. The cheap ones for pointers operate at only 7mA. Their absolute max current is only 20mA and their typical forward voltage is only 2.2V. They typically have about 38 ohms in series internally and the high internal resistance of a three button cell battery limits the current to 7mA.
You were going to blow-up your laser diode!

d2 should be a photodiode

Look on the web about info about photo-diodes. If they are reverse-biased then their output current of uA must be amplified. If they are forward-biased with an extremely low current then their output voltage of mV must be amplified. Your circuit doesn't have bias for the photo-diode and it doesn't have an amplifier.

Oh yeah, both r3 and r2 should be 25 ohms right? (5V / .2A)

No. 200mA through a photo-diode will blow it up. The input of a NOT gate isn't grounded so no current or very little current will flow through the photo-diode anyway. Your photo-diode doesn't have a bias voltage, doesn't have a load and doesn't have an amplifier so it won't work.

Also, Look on the web for info about NOT gates. I don't know of a NOT gate that is strong enough for an output current of a whopping 200mA.