Some help needed with LED-indicator circuit.

Thread Starter


Joined Jan 30, 2015
As an introduction to analogue electronics I have to design a simple alarm system, starting with a power supply.
I would like to add a small circuit that indicates whether or not a fuse is working correctly. A green LED for a working fuse, a red one for a blown fuse. Now I found a small little circuit, shown in the link below, but i have some questions:

1) Why have they chosen a 5.1V Zenerdiode? And what would be the effect of changing this value? (in general, how doess the Zener funtion in this circuit?)

2) The shown circuit is designed for 230Vac, if i want to adjust this for 12Vac, do i have to anything besides the value of R1?

3) In the circuit the LED's only draw about 2mA. Is there a reason behind this choice? Why not just the more normal 10-20 mA?

Any suggestions for a better circuit are ofcourse always welcome. Preferably with a little bit of explanation since i'm a complete beginner. :)

Thank you,



Joined Feb 17, 2009
Ad1. Zener diode was add to increase the RED LED forward voltage. Without Zener diode RED LED may light even when the fuse is good.
VF for RED diode is around 1.6V and Green LED around 2.2V. This means that the voltage at duo-diode Cathode is 2.2V + 0.7V (D3) = 2.9V smaller than Vsup. So without the Zener diode RED LED will see 2.2V across it and the red diode will lights.
This means that you can use 3 ordinary diodes instead off a Zener diode.
Ad2 reduce R1 to 1K and you can use 1N4148 instead-off 1N4007.
Ad3 Modern LED's will light at the same brightness with much less than 20mA and LED can easily operate at that level of current (5mA) or even smaller currents.