# Why two 1N5401 diodes in the "Auto 12V Bulb Failure Warning" circuit?

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Why two 1N5401 diodes in the "Auto 12V Bulb Failure Warning" circuit?

Assuming that the Light Bulb has a resistance of 4 ohms (36W under 12V):

1) which would the current intensity through diodes D1 and D2 be?

2) which is the internal resistance of each of diodes D1 and D2 under such current intensity?

3) which would the effective tension applied to the Light Bulb be?

1. Diode current = bulb current = slightly less than 3A in  this case

2.  The internal resistance of each diode is about 1V /3A = 0.33Ω

3.  The voltage applied to the light bulb = approx 10V

At rated current, silicon diodes drop approx 1V each--rather than consider the elusive resistance (that changes with load), it is far easer to estimate voltage drop as the silicon diode tends to have a fixed voltage drop.  At light loads, this voltage drop is about 0.65V.

Two series diodes are required in order to obtain sufficient voltage to forward bias the LED inside the opto-coupler.

I can envision a superior circuit that would use a TL080 op amp and a low value series resistor that would drop perhaps 0.1V and apply 11.9V to the bulb thus making it much brighter.  Dim light bulbs bother me...
answered Jan 29, 2017 by (12,800 points)

Simplest way is to put led with resistance ( Depend upon voltage drop for LED)

parallal to fuse, As soon as fuse is off current will pass through LED will indicate blown fuse.
answered Mar 1, 2017 by (240 points)