Phone-Activated Lamp Controller

Thread Starter

Point_Man

Joined Feb 22, 2004
3
For an assignment, I have to analyze the following circuit:

http://www.geocities.com/spc_point_man/homework.txt

(Click here if the above link does not work)

However, I've been sick the last few classes, and am falling behind. Is there anyone here who can please explain how the current flows through the circuit so I can answer some problems relating to it?

Ex: 1) Determine the current through R3 when the optocoupler is activated.


I am lost and really need help. Any help at all is greatly appreciated!

Thank you in advance! :D
 

Dave

Joined Nov 17, 2003
6,970
:huh: I have no idea, it looks like half a diode! :blink:

Unless its a test point, but I've never seen one drawn like that.
 

Thread Starter

Point_Man

Joined Feb 22, 2004
3
It's called a Shockley Diode... but its not a diode :huh:
It's more like two BJT's connected together with only two external connections available.
In essence, this device acts like two cutoff transistors until the voltage across Anode to Cathode exceeds the breakdown voltage VBRF, at which point minority carrier current flowing through one device into the base of the other device turns it on. The second device now turns the first device on, establishing lots of current to keep both devices turned on.
 

Dave

Joined Nov 17, 2003
6,970
Originally posted by Point_Man@Mar 20 2004, 12:05 AM
It's called a Shockley Diode... but its not a diode :huh:
It's more like two BJT's connected together with only two external connections available.
In essence, this device acts like two cutoff transistors until the voltage across Anode to Cathode exceeds the breakdown voltage VBRF, at which point minority carrier current flowing through one device into the base of the other device turns it on. The second device now turns the first device on, establishing lots of current to keep both devices turned on.
Ah, so its exactly like a BJT, but with an increased breakdown voltage. Strange device <_<
 
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