# LDR / photo resistor - increasing resistance with increasing light???

Discussion in 'Electronic Basics' started by mike, Oct 19, 2004.

1. ### mikeGuest

Is there such a thing as a LDR that increases resistance with
increasing light?

Or, is there a simple circuit that will invert the resistance of a
common LDR?

Disclaimer: I have very little experience with electronic/circuits.

Any help is appreciated. Thanks!
m

2. ### Terry PinnellGuest

What is the objective? If you want to control a circuit using an LDR,
you would normally arrange the LDR with a normal resistor (or pot or
preset) as a voltage divider pair. You'd then use the output voltage
from that as input to a subsequent (analog or digital) circuit
section. So, simply reversing the position of the LDR and the other
resistor would reverse the operation of the circuit.

As an example, see the first section of my garden lamp circuit here:
http://www.terrypin.dial.pipex.com/Images/GardenLamp.gif

In the unlikely event that I wanted my lights to go on at dawn rather
than dusk, I could achieve that by swapping the positions of the LDR
and the resistors.

3. ### John PopelishGuest

I don't know of any.
This is fairly weary in some cases.
Tell us all the details you can think of about what you are trying to
accomplish. This will save us a lot of useless questions.

4. ### mikeGuest

wow, thanks for the response. here is exactly what i want to do:

i live in a basement w/o windows, and i want to rig inside light so it
coincides with the outside light. i was figuring i could run an LDR
placed outside, to a dimmable ballast in my basement apt.

i have an electronic dimming ballast, made by the now defunct JRS
Technology (#C232120RS501). it has a dimming control by way of a 0 to
10v class 2 circuit, 0.5mA output. it is 0-10v out, and by connecting
a 0-100k ohm potentiometer to this circuit i can dim the connected
flourescent bulbs no problem-- the greater the resistance, the greater
the V, the greater the resulting brightness.

however, what i want to do is connect an LDR in such a way that a
greater amount light hitting the LDR makes the bulbs brighter. this
means that in order to get brighter bulbs, i need a greater voltage
accross this circuit, which means i need greater resistance connected
to this circuit (i think, right?)

i have an LDR (from a radioshack multi-pack) connected in series (with
a 0-100k trimmer/potentiometer) to this 0-10v circuit, and it does
what i want it to do, only in reverse. that is, as lighting
conditions on the LDR get brighter, the dimmer makes the bulbs get
dimmer.

this makes sense since an LDR decreases resistance as more light hits
it. what i need is increasing resistance with more light. or a
circuit that simulates this.

if this looks familiar, i posted 6 weeks ago before i had any idea
what i was talking about (thanks to those who offered suggestions):

6. ### rayjkingGuest

Mike,

If you add a cad photocell in series with a 10k resistor that is connected
to a low current ( 10ma or more ) 10/12 volt dc power supply the signal you
wish will appear across the 10k resistor. The better circuit would
substitute the 10k for a 100k pot and a series 470 ohm ( safety resistor )

Ray

7. ### John PopelishGuest

You can make a voltage divider with a light dependent resistor in
series with a fixed resistor that has the LDR in either the pull up or
pull down side of the divider. If you have a 10 volt supply available
(a wall wart, perhaps) or if this voltage is available from the
ballast, you can experiment with the LDR and fixed resistor replacing
the potentiometer. Varying the value of the fixed resistor shifts the
light level that produces about half light output.

8. ### CFoley1064Guest

Subject: Re: LDR / photo resistor - increasing resistance with increasing
Hi, Mike. I'm hearing you say you tried the RS photoresistor, but it operated
in a bass-ackwards fashion. You want resistance to increase as light
increases.

You might want to try something like this (view in fixed font or M\$ notepad):

VCC
+
|
| VCC
.-. 1/2 LM358 +
| | VCC |
1K| | |\| |
'-' .---|-\ ___ |/
| | | >-|___|--|2N3904
.----o-----|---|+/ 100 |>
| | | |/| |
| | | GND | ___ 1 H11F1 6
~~.-. .-. '-----------------o-|___|---o--. .---o--->
~~| | | | 150 ohms | |
LDR| |1K| | V ~~ ||-+
'-' '-' - ~~ || To Ballast
| | 2 | ||-+
| | .---------o--' | 4
| | | '---o--->
| | |
V | |
- | ===
| | GND
=== ===
GND GND .------------.
| |
1N4001 | |\ |
'---|-\ |
| >-----'
.---|+/
| |/
|
===
GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

There are a couple of components here you'll have to purchase, and I'm afraid
they're special order at RS. You can try mouser.com for better prices and good
service for the hobbyist. You also might want to pick up a 9VDC wall wart for
the Vcc in the diagram.

The two components you need are an LM358 (dual op amp) and a H11F1 (an opto
analog output FET). The distinguishing characteristic of the H11F1 is that, as
current through the LED increases, the resistance of the FET decreases. You
can use that to advantage with your CdS photoresistor by seting up a voltage
divider, and using the CdS to shunt the lower resistor. That voltage is picked
up by the op amp, and used to drive the LED.

Notice that, as ambient light increases, the voltage decreses, so the current
througfh the LED decreases, causing the resistance the ballast sees as
increasing. Also note that this assumes your CdS LDR has less than a couple
hundred ohms "ON" resistance. Since there's such a wide variety in the RS CdS
parts, you'll probably have to tweak this circuit for your part to make it
work.

I hope this is of value, even though it's a little on the complex side. If you
need more help, feel free to email. Please put "LDR" in the heading to avoid
having the spam filter dump you.

Good luck
Chris

9. ### Rich GriseGuest

I seem to remember this from all those months ago, and it seems like
someone had suggested another LDR inside, with a comparator/difference
amp, to make the inside lights _really_ track the outside. Then, as
long as you get it scaled right, you won't have to worry about polarity,
or even nonlinearity.

Of course, on cloudy days, it could be a little dim. )-;

Have Fun!
Rich

10. ### Robert MonsenGuest

But eclipses and airplanes would be fun.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

11. ### mikeGuest

ha! actually it *would* be great if the lights dimmed a little when a

you guys may think i'm a little crazy for wanting my lights this way--
ok, i may be crazy, but for entirely different reasons.

here's a little more background, hopefully to put things in better
perspective: i have regular lights/lamps in the room, apart from this
project. for this project, dimmable flourescents will be mounted in a
"light box". this light box will actually be an old window i'll be
hanging on the wall, with a reflective box framed behind it (open to
the glass in front). the lights will be mounted behind the framing of
the window, the plan being that you won't see the bulbs, but only the
light. think "hangable" windows, with real light shining through. i'm
using flourescents because the natural-sunlight models are much closer
(in color) to real sunlight.

now back to the gritty. i've been reading up on voltage dividers and
op amps, but they only seem to apply when you want to change or divy
up the voltage of a known voltage source. the dimming control of the
ballast i'm using has 2 wires out for a 0-10V, .5mA, class 2(DC)
circuit. this is a voltage source built into the ballast. i can dim
the lights by adding a 100k trimmer to these 2 wires. i was measuring
today and found response on the lights came from a resistance between
15k and 75k ohm (the higher the resistance, the brighter the light).
this didn't make sense to me because when I did my equations (V=IR) I
thought response should have been when R was between 0-20k ohm. i
figure i didin't consider dynamics in the ballast or internal
resistance or one of a million other concepts i've never heard of.

the LDR, hooked in series with a trimmer on these 2 control wires,
makes the flourescents go dimmer as light hitting the LDR gets
brighter-- i need it the other way around.

i really like the idea of having an inside LDR and an outside LDR
balance each other out, because (1) i don't think i'd ever get a
linear response otherwise and (2) this will eliminate problems with
the resistance of the outside LDR changing when it get's wicked cold
out. (i understand other inside lighting may interfere, but creative
placement of the LDR should fix that.)

BUT... it there a way to do this not with voltage, but with
resistance in to the ballast control? from your suggestions maybe the
ballast 0-10v controls hooked up to one side of an op amp, a similar
but steady vcc hooked up to the other side, both sides with LDRs in
series and whatever necessary trimmers/resistors-- but how would this
raise or lower the voltage of the ballast control wires?

sorry, i'm a little dense. but i'm slowly learning

12. ### CFoley1064Guest

Subject: Re: LDR / photo resistor - increasing resistance with increasing
Again, take a look at the H11F1. Its output is a bilateral FET which is seen
by your ballast as a plain old resistor, as long as the voltage applied across
it is less than 30V or so. Depending on the amount of LED current, it runs
from 300 Megohms down to a couple of hundred ohms -- ideal for your
application. The op amp circuit I provided is definitely a fiirst cut -- it
can definitely be improved. With a little more effort and a trimmer with an RC
filter instead of a straight voltage divider to set the current servo point,
you should have a good solution. But that shows the way.

Good luck
Chris

13. ### mikeGuest

thanks for the help chris-- i have the H11F1 on order. i'll let you
know how it goes.