Hello,

For a school project i had to design a power supply for an alarm system. In this power supply, there had to be a fuse and a fuse indicator. This indicator indicates whether the fuse is still in tact or not (using 2 leds). I found a circuit online that does just dat, but i don't really understand how it works. i guess the zener diode acts as a voltage limiter, but what do al the other diodes do and why does the second led not need a zener diode for limiting the voltage? Could somebody help me with this? Thank you!

 

if you look at LED2 (GRN), D3, R1 and D4 it's basically a power ON indicator. Diodes protect against reverse polarity, F1 has to be good.

In the other case, the voltage has to be greater than ~8.2 V for LED2 (RED) to light. there's 2 diode drops, one LED drop and a 5.1 V zener diode drop to overcome. In practice, the LED gets say 8.2V peak less than the ON condition.

The usual technique was a Neon lamp and resistor across the fuse. This was perfect because the nean lamp needed about 80V to fire, so the circuit was indeed open. The LED/RESISTOR across a fuse works, but it limits the current to a few mA,

 

Was that schematic originally for 230V or did you put that voltage there? Wasn't it originally for a lower DC voltage?
You maybe breaking one of the TOS(terms of service) with a circuit like that.

That said, if it was a low volt DC circuit, when the fuse is good, the zener keeps the green led on and the red one off. When the fuse blows, there would no longer be a voltage on the green led, so the zener would then be allowed to conduct, turning the red one on. The 120k resistor is what limits the current through what ever led is on at the time, not the zener. The zener just acts like a simple electronic "switch" not a current limit.

kiss types faster than me.

 

Was that schematic originally for 230V or did you put that voltage there? Wasn't it originally for a lower DC voltage?
You maybe breaking one of the TOS(terms of service) with a circuit like that.

That said, if it was a low volt DC circuit, when the fuse is good, the zener keeps the green led on and the red one off. When the fuse blows, there would no longer be a voltage on the green led, so the zener would then be allowed to conduct, turning the red one on. The 120k resistor is what limits the current through what ever led is on at the time, not the zener. The zener just acts like a simple electronic "switch" not a current limit.

kiss types faster than me.

I use a 230 to 12v transformer as an input voltage, also i used a 1k ohm resistor instead of a 120k ohms resistor. It al works, i just wonder WHY...

 

I understand how a fuse indicator might be useful but what is wrong with a simple ON/OFF indicator? If the fuse is blown the ON/OFF indicator will remain OFF.

I'm not sure if anyone can correct me here but I seem to remember someone telling me that there was a fuse which existed with a built in indicator. I heard that these were discontinued as the indicator was causing flashovers/explosions.

 

I understand how a fuse indicator might be useful but what is wrong with a simple ON/OFF indicator? If the fuse is blown the ON/OFF indicator will remain OFF.

I'm not sure if anyone can correct me here but I seem to remember someone telling me that there was a fuse which existed with a built in indicator. I heard that these were discontinued as the indicator was causing flashovers/explosions.

Well that was one of the things that we had to include in the power supply. Offcourse we could have not done this, but it's a school project so we are supposed to be learning something, not just do the thing that is the easiest.

 

Since both the red and green legs have diodes in them, D4 is redundant and can be deleted.

When the fuse is intact, there are two parallel circuit paths from L to the top of R1 and through to N:
Path 1: F1 (a very small resistance), D3, Green
Path 2: D1, D2, Red

Whichever path has the lower voltage drop across its components before an LED comes on will be the path that lights up. Let's assume that these are normal little LED chips, with an average forward voltage (Vf) of approx. 2.0 V. D1 and D3 have a known Vf of 0.6 V an low currents. For all practical purposes in this circuit, F1 is a dead short.

Path 1: 0.0 V + 0.6 V + 2.0 V = 2.6 V
Path 2: 0.6 V + 5.1 V + 2.0 V = 7.7 V

So as the input sinewave voltage increases from zero, the green LED will start to come on at around 2.6 V and the red LED will start to come on at around 8.2 V. BUT, once the green LED comes on, it clamps the maximum voltage across *both* paths to 2.6 V because both paths are in direct parallel. Because path 2 never reaches 7.7 V and above, the red LED never comes on.

When the fuse blows, it is an open circuit and path 1 no longer is connected to path 2. Now the voltage across path 2 can increase to 8.2 V and light the red LED.

ak

 

MOD NOTE: Moved to Homework Help.

Possible ToS violation being discussed.

 

Possible ToS violation being discussed.

See post #4:

I use a 230 to 12v transformer as an input voltage,

 

That makes it moot for this thread. The general discussion still continues.

 

I'm not sure if anyone can correct me here but I seem to remember someone telling me that there was a fuse which existed with a built in indicator.

Some old fuses for VW bugs had them. The fuses were a funny ceramic cylinder that had a pointed end on both ends. The fuse element couldn't be seen so had to be tested with a voltmeter or powered test light. The after market came out with indicator fuses. This would have been in the late 1960's or early 70's.

 

You can buy fancy fuses with visual indicators and electrical switching.

https://www.galco.com/buy/Daito/P413

 

I seem to remember someone telling me that there was a fuse which existed with a built in indicator.

Telecom fuses have a spring that is held in place by the fuse element. When the fuse blows, the spring releases and pushes up a little indicator pin.

https://www.fusesunlimited.com/products/alarmindicatingfuse

ak