dark activated led with 2.4V power supply

[kalaman]

I use two Ni-Mh battery. So my power supply is 2.4V. The load is a 2V led. What do you advise for this voltage limit ? what kind of circuit correct for this supply ?

 

[audioguru]

If the max allowed current for the 2.0V LED is 30mA then Ohm's Law says that a current-limiting resistor of 13.33333 ohms will cause it. I don't know what is the max allowed current for your LEDs.

You must limit the current to an LED, not the voltage. One million volts will light an LED for a very long time if the current is limited to its current rating or less.
The two Ni-MH cells will have a voltage of 2.8V when they are fresh from the charger.

 

[kalaman]

thank you for this information but i just want to make a circuit which is dark activated ldr circuit for one led with only two battery. also the activation must be precise.

 

[audioguru]

Try this circuit. Change the value of the 100k resistor in series with the LDR to change its sensitivity.

 

[kalaman]

Thank you very much. I will try this.

Best regards.

 

[kalaman]

hi it's working. But it is not like my wishing. the led is lighting up slowly.

 

[audioguru]

My solar garden lights also have an unsensitive LDR and a couple of transistors. They also light slowly.

 

[ronsimpson]

Connect a resistor 220k from 2N3904 base to 2N3906 collector. That will add positive feedback. The LED should be on or off but not dim. Try different values of resistor.

 

[Hero999]

Correct, it will introduce some hysteresis into this circuit so the LED will turn on at one threshold and turn off at another.

 

[audioguru]

The collector voltage of the PNP is always much higher than the voltage at the base of the NPN. Therefore there won't be any positive feedback, the LED will just be turned on all the time.
It might work if you carefully make a voltage divider to reduce the 2V to 2.3V at the collector of the PNP down to about 0.55V to 0.65v at the base of the NPN.

 

[Hero999]

I see what you mean.

I would use an adapted version of this circuit.

 

[kalaman]

The collector voltage of the PNP is always much higher than the voltage at the base of the NPN. Therefore there won't be any positive feedback, the LED will just be turned on all the time.
It might work if you carefully make a voltage divider to reduce the 2V to 2.3V at the collector of the PNP down to about 0.55V to 0.65v at the base of the NPN.

Sorry i'm late for answering. The feedback resistor made me happy. That is great.

@Hero999
I have just tried your circuit. that is perfect too.

Those spend 2uA standby current. Thank you so much again and again.

Best regards.