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Does the use of BC547 and 2N3904 transistor interchangeable in a Dark sensing circuit

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Newbie level 1
Jul 21, 2015
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I have made dark sensing circuit using one LDR, 390 ohm and 100 kilo ohm resistor, 2N3904 transistor, white LED and 9V battery?
But the problem is that, when I turn on the switch in the presence of light or in its absence the LED is glowing everytime? Why is my circuit not sensing light or working as expected? Give me suggestions.

My circuit diagram is attached with it. IMG_6389[1].JPG

Using the transistor as a saturated switch means knowing the Ic:Ib ratio. In this case it is limited by the R ratios. 100k/330R
Reverse the LDR and R and use a Darlington and adjust R for threshold, so that LDR conduicts current to drop 1.4V across R at required threshold.


is this a 100k resistor?

If yes: then the max. voltage across this resistor about 8.5V. Making a max current of 8.5V / 100kOhms = 85uA.
Now the bjt has a gain of about 100. This means max. expectable LED current is 85uA * 100 = 8.5mA

the LDR should be specified. Therefore your question can´t be answered.
There are many types around. What type is it?


You should consider using a photodiode instead of the 100k. When you disconnect the LDR, then the bjt acts like a current amplifier.
The more brightness, the more current through the photodiode, the more base current, the more LED current. It is limited to about 25mA LED current.

If you want a sharp ON/OFF output, then consider using a comparator (with hysteresis) or a schmitt trigger circuit).


I have two LDR's from solar garden lights. In daylight (not sunlight) they measure about 24k ohms which is not low enough to completely turn off the transistor, unless it is very cold (the circuit is also a thermometer).

The datasheet for the 2N3904 shows that its current gain is barely enough to use a resistor as high as 100k in that circuit so even more light would be needed to turn off the transistor. A BC547C has much more current gain and will work better.
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