LDR, potentiometer and darlington circuit

Design a circuit that will turn on a small incandescent lamp when darkness falls.

It should use a light-dependent resistor (LDR), a potentiometer to set the trigger level, and one or more darlington transistors, and other components as required.

Component values are not required.

(Correct answer attached).

I want to know, in the attached diagram, was an extra transistor used to invert the '0' at the collector (output) of the darlington? As I would have rather have switched places of the LDR and potentiometer, so that when it is dark the darlington goes on, and I wouldnt have to draw an extra transistor.

Why is a resistance needed above the darlington (as it isnt the load)?

Yes, the extra transistor provides an inversion.

And you could switch the pot and LDR location to allow the use of just one darlington transistor (assuming it has enough gain for your requirements).

Oh ok thanks. In the diagram, why is a resistor needed in the darlington, if there is no load (e.g. LED)?

TsAmE said:

Oh ok thanks. In the diagram, why is a resistor needed in the darlington, if there is no load (e.g. LED)?

Click to expand...

hi,
Consider how the 2nd transistor gets its Base current in order to function.

Is this a Darlington circuit, or just two common emitter stages?

mneary said:

Is this a Darlington circuit, or just two common emitter stages?

Click to expand...

The first transistor is a darlington and the second one is a common emitter.

TsAmE said:

The first transistor is a darlington and the second one is a common emitter.

Click to expand...

The D floating near the transistor means darlington? I usually draw a double arrow emitter or two transistors.

mneary said:

The D floating near the transistor means darlington? I usually draw a double arrow emitter or two transistors.

Click to expand...

Yeah it does. Yeah those are alternate ways to draw a darlington.