LED activated by LDR

[HEMOglobina]

Let's say I have a circuit like this:

http://img504.imageshack.us/i/ldr.png/

When I shine the LDR, the LED instead of having only 2 states (on or
off), it lights up kind of proportonally to the amount of light I put
on the LDR (from weak to fully lit). I know this is how the transistor
works, but What can I do to make the LED either light up fully or not?
I know I could use a relay (to act as a switch), but I'm pretty sure
that would be overkill. Is there another component or circuit
configuration that I could use?
As you might have guessed I'm pretty newbie in electronics.

Thank you for your help,
HEMOglobina

 

[krw]

On Sat, 3 Oct 2009 09:44:51 -0700 (PDT), HEMOglobina
<cassiano@gmail.com> wrote:

Let's say I have a circuit like this:

http://img504.imageshack.us/i/ldr.png/

When I shine the LDR, the LED instead of having only 2 states (on or
off), it lights up kind of proportonally to the amount of light I put
on the LDR (from weak to fully lit). I know this is how the transistor
works, but What can I do to make the LED either light up fully or not?
I know I could use a relay (to act as a switch), but I'm pretty sure
that would be overkill. Is there another component or circuit
configuration that I could use?

You could, more or less, duplicate the circuit to make it a
differential amp, or comparator.


+9V +9V
| |
.-. .-.
| | | |
| | | |
+9V '-' '-'
| ^ | +
.-/ | |
|/| V -> |
/ | - | +9V
/'-' | + |
| |/ \| .-.
+---------| |--------| |
| |> <| | |
| | | '-'
.-. +-------+ |
| | | Gnd
| | .-.
'-' | |
| | |
Gnd '-'
|
Gnd

As you might have guessed I'm pretty newbie in electronics.

That's why this group is named sci.electronics.basics.

 

[Tim Wescott]

On Sat, 03 Oct 2009 09:58:33 -0700, larwe wrote:

On Oct 3, 12:44 pm, HEMOglobina <cassi...@gmail.com> wrote:

When I shine the LDR, the LED instead of having only 2 states (on or
off), it lights up kind of proportonally to the amount of light I put

You could use a second transistor to increase the gain of the circuit,
or add a hysteresis resistor so that when the transistor switches on, it
pulls itself further on.

You'll need the second transistor if you want to give the thing positive
feedback.

You'll use less circuit board area if you just use a comparator, with
suitable hysteresis to make sure things stay on. They can be had with
enough capacity to drive an LED.

You may even be able to use a tiny logic schmitt trigger -- it's easy to
find one that'll pull 5mA, that's visible indoors with an old round red
through-hole LED, and downright painful with a new high-intensity 0603
surface-mount LED.

--
www.wescottdesign.com

 

[John Fields]

On Sat, 3 Oct 2009 09:44:51 -0700 (PDT), HEMOglobina
<cassiano@gmail.com> wrote:

Let's say I have a circuit like this:

http://img504.imageshack.us/i/ldr.png/

When I shine the LDR, the LED instead of having only 2 states (on or
off), it lights up kind of proportonally to the amount of light I put
on the LDR (from weak to fully lit). I know this is how the transistor
works, but What can I do to make the LED either light up fully or not?
I know I could use a relay (to act as a switch), but I'm pretty sure
that would be overkill. Is there another component or circuit
configuration that I could use?

---
Yes, use a comparator: (View in Courier)


+9V>---------+----+----------+------------+
| | | |
| | | [3600]
[LDR] | | |
| | +---|--[1M]--+ |
| | | | | |
| | | | | [LED]
+----|------+--|+\ | |K
| | | >------+---+
| [POT]<------|-/
| | LM393 | +------+
[20K] | U1A | | |
| | +-+-|+\ |
| | | | >-+
GND>---------+----+----------+---|-/U1B
SPARE

With the LDR in ambient light, adjust the pot so the LED goes out and
then crank it a little bit more in the same direction, maybe 10 degrees
or so.

Then when you "shine the LDR" the LED should come on.

The 3600 ohm current limiting resistor is based on a high-efficiency 2mA
LED.

If you want to sink more current you can get about 10mA out of one
comparator, but it goes out of saturation very quickly after that, which
will keep the LED from lighting up.

Alternatively, you could use your transistor to drive the LED, like
this:

+9V>---+----+----------+-------+------+
| | | | |
| | | [10K] [R]
[LDR] | | | |
| | +---|----+ | |A
| | | | | | [LED]
| | | | [1M] | |
+----|------+--|+\ | | C
| | | >--+--+----B
| [POT]<------|-/ E
| | LM393 | +------+ |
[20K] | U1A | | | |
| | +-+-|+\ | |
| | | | >-+-----+
GND>---+----+----------+---|-/U1B
SPARE


JF

 

[John Fields]

On Sat, 03 Oct 2009 13:35:37 -0500, John Fields
<jfields@austininstruments.com> wrote:

Oops... The LDR and the 20k are backwards.

They should be like this:

+9V>---------+----+----------+------------+
| | | |
| | | [3600]
[20k] | | |
| | +---|--[1M]--+ |
| | | | | |
| | | | | [LED]
+----|------+--|+\ | |K
| | | >------+---+
| [POT]<------|-/
| | LM393 | +------+
[LDR] | U1A | | |
| | +-+-|+\ |
| | | | >-+
GND>---------+----+----------+---|-/U1B
SPARE

For the transistor output circuit they're OK the way they're shown.

JF

 

[larwe]

On Oct 3, 12:44 pm, HEMOglobina <cassi...@gmail.com> wrote:

When I shine the LDR, the LED instead of having only 2 states (on or
off), it lights up kind of proportonally to the amount of light I put

You could use a second transistor to increase the gain of the circuit,
or add a hysteresis resistor so that when the transistor switches on,
it pulls itself further on.

 

[Jamie]

HEMOglobina wrote:

Let's say I have a circuit like this:

http://img504.imageshack.us/i/ldr.png/

When I shine the LDR, the LED instead of having only 2 states (on or
off), it lights up kind of proportonally to the amount of light I put
on the LDR (from weak to fully lit). I know this is how the transistor
works, but What can I do to make the LED either light up fully or not?
I know I could use a relay (to act as a switch), but I'm pretty sure
that would be overkill. Is there another component or circuit
configuration that I could use?
As you might have guessed I'm pretty newbie in electronics.

Thank you for your help,
HEMOglobina
You need a hysteresis circuit..

that would be a comparator type circuit with a (+)
feed back to slightly offset the trigger point of the
(-) input when switched.

 

[default]

On Sat, 3 Oct 2009 09:44:51 -0700 (PDT), HEMOglobina
<cassiano@gmail.com> wrote:

Let's say I have a circuit like this:

http://img504.imageshack.us/i/ldr.png/

When I shine the LDR, the LED instead of having only 2 states (on or
off), it lights up kind of proportonally to the amount of light I put
on the LDR (from weak to fully lit). I know this is how the transistor
works, but What can I do to make the LED either light up fully or not?
I know I could use a relay (to act as a switch), but I'm pretty sure
that would be overkill. Is there another component or circuit
configuration that I could use?
As you might have guessed I'm pretty newbie in electronics.

Thank you for your help,
HEMOglobina

I got there without trying.

I made a linear on at dusk off at dawn circuit. I figured the LEDs
wouldn't matter too much - constant current device powering constant
voltage drop device - what could go wrong?

Three transistors - two for constant current and two for light sensing
(using the pass transistor for both functions).

Turns out the simple current load of 100 milliamps was dropping my
supply voltage by 2 tenths of a volt. Instant hysteresis! Once it
came on, the lower supply voltage turned in on harder, It had to get
really light to turn it off again.

Synergy . . .

--