# Automatic dark detector: base voltage

Here we have the typical “automatic dark detector” circuit:

Let’s assume LDR=5Kohm with light, and LDR=2Mohm in darkness, the battery is 12V instead of 9V, and the transistor has a forward voltage Vbe=0.7V.

With light, per the voltage divider formula, we have Vbase=0.57V and the transistor is cut off.

In darkness, the formula gives Vbase=11.42V. Since this is much higher than 0.7V. Why does the transistor not burn?

• That's not even a good topology for this. I would not even consider it. But to answer your question, you've forgotten the Thevenin impedance of the divider pair, haven't you?
– jonk
May 21 at 11:02
• @jonk thank you very much for your comment. Yes, definitely I am missing something very basic :) I’m going to research the Thevenin impedance concept: youtu.be/xSRe_4TQbuo?t=656 May 21 at 11:22
• I don't think you included the base-emitter junction's impedance in your voltage divider equations. May 21 at 11:25
• @user_1818839 That's usually pretty small. Ten or twenty Ohms. Or so. The emitter will have some, but usually less than one ohm.
– jonk
May 21 at 11:28
• @jonk oops, edited comment. May 21 at 11:32