On Mon, 17 Oct 2011 20:41:06 -0500, "Dave" wrote:


"Phil Hobbs" wrote in message
...
On 10/17/2011 06:53 PM, Dave wrote:
"Phil wrote in message
...
On 10/07/2011 06:36 AM, Dave wrote:
As per my post in SEB, I am trying to use a photo resistor and sunlight
to
turn off a transistor that would otherwise be conducting like crazy,
but
can't quite make the leap of comprehension as to how to actually do
this.
Please see attached schematic parts that show more or less what I am
trying
to do this with. No part number for the photoresistor, sorry. It
drops
from multi-megohms in the dark to single-digit Ohms in the light and
seems
perfectly capable of carrying the 30 to 50 mA current I am working with
at
the voltages indicated. Would really appreciate it if someone could
offer a
hint as to how I should proceed with as few components as possible
(small
circuit board.) I hope that my method of posting the schematic portion
of
what I a working with is not too obtuse. Couldn't figure out any other
way
to do it...

Many thanks...

Dave




You're wasting a gross amount of current in that bias network. How
about
a nice MOSFET?

Say a 2N7002, source to V-, drain to series resistor+LED, LDR plus pot
to
set turn-on level:

(V+)----*----------------*
| |
| R
| R
| R
| R
*---* |
| | ---
| R \ / -----
*- R ----- -----
R |
R | | D
| | |---*
| | |--; 2N7000
*----------J |---*
| | | S
| |
L |
D |
R |
| |
| |
(V-)----*----------------*




Man that was easy. Just had to get the MOSFET. Thank you, Phil Hobbs.

I spent so much time wrestling with that other configuration, and this
just
flew together. And damn, but it works nice. Much appreciated.

---
Geez, I feel kind of miffed since my (Jan's) solution would allow you
to rid yourself of the LDR and its associated circuitry.

With that in mind, why would you find Doctor Phil's circuit, which is
wasteful of power and money by having it drive an unnecessary LDR,
superior to mine?

--
JF