Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

LDR, potentiometer and darlington circuit

Status
Not open for further replies.

TsAmE

New Member
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)?
 

Attachments

  • June2009(10).png
    June2009(10).png
    6.3 KB · Views: 561
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)?
 
Oh ok thanks. In the diagram, why is a resistor needed in the darlington, if there is no load (e.g. LED)?

hi,
Consider how the 2nd transistor gets its Base current in order to function.
 
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top