Dark circuit - Dunkelschaltung

Hi!
I built a dark circuit - “Dunkelschaltung” which should light a LED on 9V when it is dark.

I used a BC547 Transistor and some resistors. 1k on the Base and 1x 100k for sensitivity of the circuit.
Everything went fine. When the LDR was covered the LED turned on.

pic1

NOW I wanted to replace the single led with a parralell array, of about 30 leds. (4 Leds serial and those series parallel)
This array is already built and works perfectly with 9V.

I measured voltage between + and the collector of my BC547 (where the LED is connected) … correct 9V. (when LDR is covered - 0 if its in light)

pic2

But if I remove the single LED and replace it with the connection wires of my array nothing happens… the array does not light up even if I use an external power supply instead of the battery.

pic3

What am I doing wrong?

Answers german or english

yours
Martin

If it works with 1 LED, maybe there isn’t enough current to supply many LEDs.

I’m not a electronics guy, so only guessing.

https://www.google.com.au/search?q=dark+activated+switch&client=opera&hs=GF1&tbm=isch&tbo=u&source=univ&sa=X&ved=0ahUKEwjrh5Os9YXQAhWEVZQKHUtWBRIQsAQIHA#tbm=isch&q=dark+activated+led

You are likely running out of gain and/or base drive on the transistor. Assuming r3 is the 100K resistor you aren’t getting a lot of base current flowing. 9v / 100k = around 90 uamps of base current. Assuming a transistor gain of 100 that is about 9ma (or one led worth) of collector current. 9MA will light one led (but if you have 6 of them as shown the same 9ma will flow but likely none of them will light as they are only seeing about 1ma each. You have two choices: reduce the size of the 100k resistor to provide more base current (but that may impact the operation of the cds cell) or add a 2nd transistor with about a 1K resistor for r2 and another 1k resistor in to the base of the second transistor with the leds connected to the collector of the second transistor. That gets you both higher base current drive on the second transistor and another gain of 100 which should provide enough collector current on the second transistor to drive your 6 leds.

Peter Van Epp

Thank you…

I assumed that, but im not so experienced in building circuits.
Can you send me a sketch of that circuit?

yours
Martin

With the note I haven’t tested this, this should work. The position of the ldr changed because there is a logical inversion caused when the second transistor is added. When its dark the ldr will be high resistance, and Q1 will be off. Q2 has around 9 ma flowing in to its base and is thus hard on lighting the leds (you need to watch the collector current rating of Q2 but for 6 leds I expect it will be OK). When there is light the ldr is lowish resistance (10K or so I think) and base current flows in Q1 turning it on (this part works in your current circuit and thus should work here to). That robs the base current from Q2 which shuts off and the leds go out . You could use a higher resistance for r1 to reduce current draw slightly but it isn’t likely to be much against what the leds draw a 10k resistor would reduce the base current to around a milliamp which should still provide enough drive for the leds (if not
reduce the value of r1 til they light correctly).

Peter Van Epp

Thank you verry much!

In simulation I already get much more current to my LEDs than in my original circuit.

Tomorrow I will build it in real and see what will happen!

yours
Martin

You may find you need to adjust the values of the input resistors. For instance R4 may actually need to be much higher depending on the resistance value of the ldr. The objective is to have small to no (although no probably isn’t practical) base current flowing when its dark and enough base current to turn on Q1 when there is light. In any case good luck, and feel free to ask if it doesn’t work.

Peter Van Epp