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Science Forum Index » Electronics - Basics Forum » A single fault safe design for a LED
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| Guest |
 Posted: Fri Jan 26, 2007 6:21 pm |
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Please see http://anders.sagevik.com/beregninger.jpg for circuit
details.
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 |
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