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Use a current divider placed after the fuse.
A current divider is very simple. You set the 2mA curent flowing into a resistor R1 and LED, and 8mA into a resistor R2 of a smaller value.
In this way, both current paths maintain a common node of 3V potential difference, which is needed to drive the LED in series with R1 because LED usually requires 2.2V for Red, 1.9V for Green or Yellow. Do not use Blue LED because you need 2.9 to 3.4V at least.
R1 is used to provide current-limiting resistance to LED and part of the current divider parameter.
10mA = [(3V-2.2V)/R1] + [3V/R2] = 2mA + 8mA
Thanks for help. The part that I don't get is that if I use this current devider after the fuse like you said then when the fuse blows there won't be 3.3V on the R1 to drive the LED. Maybe I am seeing it wrong.
if you put the LED in series with Fuse, the LED will be off after fuse blown.
if you put the LED in parallel with fuse, the circuit will have some current after fuse blown and it is not appropriate.
So, the complete solution is that you put a currect sense resistor in series with diode. if the fuse is close (not blown) some voltage will be present on sense resistor and after fuse blown, the voltage across the sense resistor will be zero. you can use this voltage to turn on/off a LED.
That's precisely what I want to achieve. If the fuse hasn't blown, the LED still lights up telling your visually that the fuse is OK! If the fuse blow, the LED will not light up anymore because no current will flow into the current-divider.
I hope you understand my point, and I think the previous two forumers understood my point from the way they tried to explain to you as well.
Thank you, Except I need the exact opposite. I need the LED to be OFF when the FUSE is NOT BLOWN. I need the LED to turn on only after the FUSE BLOWS.
I don't want to consume extra current when the circuit is running. I only want to turn on the LED when circuit is not running to show that there is a problem.
Use a PNP transistor and connect its Emitter before the fuse (+3V) .. connect LED with a small resistor between its Collector and 0V .. connect 2-resistor voltage divider between the other side of the fuse and 0V and connect the middle point of the voltage divider to the base of transistor ..
Select resistor values in such a way that if the fuse is OK the voltage at transistors base will keep the transistor OFF, whereas if the fuse is blown the other resistor connected between 0V and the Base will opent transistor and turn the LED ON ..
For example, you can have 1kΩ connected between output and the Base and 10kΩ connected between the Base and 0V ..
you need the res/led across the fuse but for 10ma it will not work because the led will supply the circuit power because now it becomes the source of 10ma. you need now a voltage comparator and some deviders to trigger the comparator and provide the led power on. need more help ask
u can try putting a relay after the fuse , where the other two terminals are NC and one is connected to VDD and the other to res.+LED , so when current pass throught the relay the two terminals are opened and the LED is OFF , when fuse is blown >>NO current throught relay and the terminals will connect and LED is ON,
u can also use a switch IC instead of the relay "i think ther was one called 4066"