Please see
I have a unit with 4 LED diodes (HFBR fiber optic connectors). These are today connected to Vcc= 5 V and each driven at their recommended drive current, 60 mA trough use of a 48 Ohm resistor in series with each LED.
In order to pass safety testing for our product (this is a medical device) we have to prove that our system is safe to use even in a single fault condition For this case, the LED diode is considered to present a hazard (class II laser) to the user if it's drive current exceed 100 mA.
A single fault condition is considered as one of the components (resistor, diode, regulator) fails and are either shortcutted or cut. You only have to calculate with one component fail at a time, not that several components fails. For my circuit either a resistor or diode is shorted, or the regulator is shorted giving Vcc = 9V and the goal is to avoid that the drive current for the LED diode exceeds 100mA.
I have made the circuit referred to in the beginning but I'm not satisfied with the solution. It solves the problem but is not much elegant, and a small deviation in any of the values, especially the Zener diode, would easily lead to a drive current >100 mA. It also have to be used for each of the 4 LED diodes which introduces many additional components.
Do anyone have a suggestion for a better solution?
Thanks for any help!
Stian