Beginner NPN query


Joined Mar 30, 2015
I am more than a little confused now! I had thought the base was voltage driver, maybe back to the text for me!
Unfortunately, you've picked a circuit that isn't very conducive to learning how a transistor works. The off resistance of the LDR combined with R1 isn't going to give you much base current. If you're trying to operate the LED at 10mA (you need a current limiting resistor as others have already mentioned), you need up to 1mA of base current if you want the transistor to saturate. At the very minimum, you need another transistor for more current gain.

The rule of thumb for voltage dividers is that you want the divider current to be 10 times the load current. There's no way for you to get that much. The transistor will turn on, but it won't saturate.

It would be better if you just used a resistor instead of the LDR and verified that the transistor was amplifying current as it should (by setting a bias point, measuring voltages, and calculating currents).

Something like this:
Note that 5k isn't a standard value and I chose the supply voltage to make it easy to calculate the voltage on the base without using a calculator. Similarly, you can calculate the voltages across R3 and R4 and the currents in them without a calculator.

EDIT: Then you can try this:
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Thread Starter


Joined Oct 28, 2020
Why argue. You can just show it!
Or, confuse me no end!!
With that basic circuit you will not be able to control the 100mA LED current, using a single transistor.
Do you have any more transistors in your stock.?
We could improve the circuit by adding another transistor.
I do, yeah. Would I place them in series to each other? As in, the base of the second transistor attaches to the emmiter of the first? I've not covered this just yet.


Joined Mar 30, 2015
Would I place them in series to each other? As in, the base of the second transistor attaches to the emmiter of the first? I've not covered this just yet.

This is the classical Darlington configuration:
A disadvantage is that the saturation voltage is one diode drop higher than with a single transistor.

This addresses the higher saturation voltage:

EDIT: Note that neither of these circuits is intended to saturate the transistors. I assumed 1.4V for the LED forward voltage (old school value).
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Audioguru again

Joined Oct 21, 2019
The problem is that the saturation spec on the datasheet was not looked at.

EVERY transistor has a saturation spec on its datasheet. Most little transistors including the 2N3904 shows that its saturates with a max Vce loss of 0.2V when its collector current is 10mA and its base current (not its base voltage) is 1mA which is obviously 1/10th its collector current.
Its datasheet also shows that its max Vce saturation loss is 0.3V when its collector current is 50mA and its base current is 5mA (again 1/10th its collector current).

So if you want a 2N3904 to saturate when its collector current is 10mA then give it a base current of 1mA.


Joined May 20, 2015
I do not like dogmas (such as base current 10% of the ratchet in saturation mode). Why oversaturate the transistor. Sometimes 1%, or even less, is enough. Everything depends on the minimum transistor gain.
And look at the resulting voltage drop at the Base Emitter junction. This is to the 0.7 V for emitter junction dogma.

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