# Is there constant current flow in this circuit?

I am studying basic circuits, and have a very elementary question.

When the photoresistor is exposed to light in the following circuit, is there a constant drain on the power supply?

If this were used in a battery powered nightlight, the battery would eventually drain even if the nightlight were never in the dark?

Am I thinking correctly?

• No. It will not draw the same current when the LED is off Vs on. It will draw significantly more power when the photoresistor is in the dark. You are correct that it will drain the battery when the LED is off, but it will draw less than 30 microAmps. In that situation, two AA batteries in series (2000mAh) could hold this circuit in standby for more than 7 years. Sep 28, 2020 at 19:33
• Thank you, that answers my question! How did you calculate 30 microAmps
– Ben
Sep 28, 2020 at 20:44
• 3.3Volts/100K Ohms= 33microAmps. It will be less than that because there will still be resistance from the photodiode in series with the 100k Ohm resistor. Sep 28, 2020 at 22:09
• Many LEDs will not light when the battery is 3V and is brand new because some LEDs need 3.6V. The 330 ohms resistor also needs additional voltage. If you use a 2V red LED, a 3V battery and the transistor saturates with a 0.4V drop then the LED current will be only (3V - 2V - 0.4V)/330 ohms = 1.8mA which is VERY dim. For a bright 20mA of current with the 2V red LED then the battery must be (20mA x 330 ohms) + 2v = 0.4V= 9V. then the transistor base current must be 1mA and the base resistor must be (9V - 0.7V)/1mA= 8300 ohms. Sep 29, 2020 at 0:10
• The 100K is rather large. It will restrict the transistor base current to about 25uA. Even with a transistor with beta = 100 that's only 2.5mA of LED current. Sep 29, 2020 at 2:57

Yes, you are thinking correctly. It will draw some minute current, even when 'off'.

The quiescent current would be that current going through the base bias resistors in the 'off' state. It would be very small when 'off', and while the bias current would drop slightly when 'on', the total current would be more with it on; (I know you knew that part!).

That would probably be a common cadmium sulfide cell (photo-cell), which I recall are about 100KΩ to 500KΩ in the dark, depending on it's specs; and a few kohms or so when exposed to moderate light. So it would be 100kΩ plus a few kΩs drawing current when exposed (transistor is off). When 'on', it would draw roughly this: take the battery voltage, reduce that by .3V for the transistor if saturated* (saturated drop) and by whatever the LED will drop when lit (V-forward). a red LED may drop about 1.2V IIRC. That gives you 1.5V left over across the resistor. So divide 1.5V remaining by 330Ω and you get about 5mA* of current to light the LED. The 'off' quiescent current would be about 3V ÷ ~100kΩ = 30µA. Very small as ambitiose_sed_ineptum aptly pointed out above!

*EDIT: td127 spotted a big error I made - Thanks td127! The 2N3904 will not be fully turned on by this choice of bias resistor values. As td127 correctly commented earlier, (and I incorrectly contradicted him/her!) - the 2N3904 could have a gain of 100 or even less. If you use a different transistor with adequate gain to saturate (fully turn on) at 30µA, then you should get about 5mA to the LED using the 330Ω current limiting resistor. You could also decrease the 100kΩ bias resistor (to provide enough base current to saturate a 2N3904), but that will also increase the 'standby' current some. That will also affect the 'turn-on' point/sensitivity.

I think the comments above already pretty much answered it all, but I put it in a formal answer anyway, with (excess) details, for other readers to understand all aspects of current flow in this circuit, which the question eluded to...

• td127 pointed out the error I made. (I even incorrectly contradicted his/her correct calculations earlier). Thanks td127, you have been very gracious! Sep 30, 2020 at 15:33
• Thanks for taking the time to explain so clearly! Much appreciated.
– Ben
Sep 30, 2020 at 19:09