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If I understand you correctly, you propose having the CDS and R1 in series, from base of Q1 to the negative terminal of the battery. This won't work. The Darlington beta will be so high that the LED will never turn off.hjl4 said:Your CDS is doing nothing.
Tell me what do you think it is doing in this circuit?
Answer.. NOTHING.
Q2 Emiter, has constant access to Positive and the Base of Q1 has its negative. Your circuit is always On as long as the base of Q1 is negative, So your CDS should be in series from Base of Q1 to R1. And R1 should not be directly connected to Base of Q1.
Make Sense?????
Awwww! You spoiled it. :cry:Ron H said:The only thing wrong was the value of R1, although the value of R2 (100 ohms) is going to allow about 50 or 60 ma to flow through the LED. A safer current level for most LEDs is about 10 - 20 ma. I would make R2=300 ohms or greater.
You sure are a sadistic SOB! Sorry I spoiled your fun. :wink:Awwww! You spoiled it.
I wanted him to smoke his LED and learn about having too much current the hard way. 60mA continuously through an LED would look spectacular (for about 5 seconds)!
There is only one battery. The emitters are connected to the positive terminal. There is no way the base can be more positive than the emitters.electronist said:Im confused. Why would the ckt not work. If the resistance of the CDS falls to 3k as stated in the diagram wont the base be at a positive potential hence causing the darlington pair to become inactive.
If you look at both the above circuits they form voltage divider working on the same principle. We may need to simply change the value from 4.7K to 10K to provide a higher tolerance