Problem with reducing the voltage below transistor's minimum

[ozan]

Hello there,

I am newbie here so my question will be accordingly. Please be gentle

I have the following diagram on my breadboard and what I want to achieve here is to turn and off the led with an LDR. I want to place a resistor first where it would only allow voltage pass through to the base just slightly above 0.6v to close the circuit and light up the LED and then place an LDR in series with the resistor to make the circuit light sensitive where it turns on and off according to the daylight. No matter what I did with the resistors available at hand (the biggest I have is 80kohm) I couldn't get the voltage drop below 0.6v. I have also added a diode in series but still the same and I know I am getting something wrong here since I don't know much about electronics. So help me understand it please!

[ozan]

Okay guys I got it. I have placed the ldr in parallel with the transistor and I guess this is what you call voltage divider? See the attached picture pls.

[Gyro]

@quince, an LDR is a light dependent resistor.

[ledtester]

Okay guys I got it. I have placed the ldr in parallel with the transistor and I guess this is what you call voltage divider? See the attached picture pls.

It's being used in a voltage divider and the circuit is normally drawn like this to clearly show the divider:



Putting the LDR on the high side of the divider will invert the behavior of the circuit, eg:

[Kim Christensen]

Okay guys I got it. I have placed the ldr in parallel with the transistor and I guess this is what you call voltage divider? See the attached picture pls.

I see that you're powering this with 3.7V so I assume that the circuit is battery powered?
Just want to point out that your circuit will draw approximately 3mA when the transistor/LED is turned off by the LDR. (3.7V across the 1200 resistor)