"o pere o"  wrote in message 
news:j6ml9s$ql9$1_at_dont-email.me...
> On 10/07/2011 12:10 PM, Dave wrote:
>> Hate to be obtuse, but its been nearly 40 years since I was in school, 
>> and I
>> am now out of practice.  Any idea as to how I could use a photoresistor 
>> (and
>> sunlight) to turn off a transistor that would otherwise be conducting 
>> like
>> crazy?  Please see my post in ABSE for the particulars of what I am 
>> working
>> with.  Many thanks for any help...
>>
>> Dave
>>
>>
>
> Without looking at your posts, to turn off a transistor with sunlight you 
> first have to be able to get a "low" voltage when sunlight is present and 
> "high" voltage when dark.
> Since an LDR exhibits a low resistance in sunlight, this voltage divider 
> does exactly this:
>
> Vcc
> |
> R
> |
> o----- Vout = PhR/(PhR+R)*Vcc
> |
> PhR
> |
> GND
>
> When dark, PhR is very high (PhR>>R) and the circuit is almost equivalent 
> to
>
> a)  Vcc-----R-----x
>
> In sunlight, PhR is low (PhR<
> b)  Vcc*Phr/R-----Phr----x
>
> Now, your design goal is that the transistor turns on when connected to 
> circuit a and turns off when connected to the circuit b. Specific values 
> depend on the values of PhR. A typical LDR may exhibit 1K or less in the 
> sun and 1M when dark. Then, R between 10K and 100K should work ok.
>
> Vcc       Vcc
> |          |
> |         load
> R          |
> |          C
> o----x--- B
> |          E
> PhR        |
> |          |
> GND       GND
>
> Pere

Why, *thank you* Pere.  Between your post and that of Phil Hobbs, I think 
I'm set.  Many, many thanks.

Dave

>