A single fault safe design for a LED

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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

Why not two current limiting FETs in series?

for a start, reduce the value of the 5.1V zener, so it is conducting (slightly) at Vcc = 5V. Then when Vcc goes to 9V, the LED current wont change much at all. make sure the zener is OK with the current & power.

change the input resistors so two are in series. When one shorts, the zener current gets higher but the LED current doesnt change much.

you will need a suitably sized zener - possibly even a shunt regulator. A paralleled pair of zeners gets around the single failure issue.

If you have to drive many strings of LEDs, then have a single clamp circuit. it may even be easier to have a simple series-pass regulator, or some form of crowbar circuit to deal with the regulator failure issue.

a crude series-pass regulator that is normally driven into saturation (so it drops very little voltage) is effectively failure-proof, as you only consider single component failures. If the series-pass shorts, then the 5V regulator can be assumed to be working, and vice-versa. It also shifts the losses into a transistor, which is easier to heatsink.

the failure of R2 can be resolved by using a current sink and a series resistor. Set the current sink to the desired LED current, and select the resistor to give the desired current when the current sink saturates to say 0.2V. Then if the current sink shorts, R2 keeps the LED current almost constant; conversely if R2 is shorted the current source does likewise; you could use a simple voltage divider from the zener for the base drive of an npn current sink.

even easier is to use more resistors in series, further reducing the effect of one shorting, although what you have at the moment meets the requirements. Four 1206 SMT resistors doesnt cost very much....

Cheers Terry

Q1: NIRS ?

BS1: the relation between current and optical output power is very variant, and I guess there's only a limit on the optical output power not on the current.

BS2: at these voltage it's impossible that a resistor will change in a short-circuit. but if you insist ;-) A1: use 3 resistors in series A2: lower the current to 50 mA and use 2 resistors in series A3: use LEDs with feedback

cheers, Stef Mientki

BS1: the relation between current and optical output power is very variant,

current.

within the specified limit (stated in datasheet)

BS2: at these voltage it's impossible that a resistor will change in a short-circuit.

You can discuss this with the people at the IEC commities :)

Thank's, I'll look into this but I can't lower the current much below

Thank you, I'll really look into this and consider your suggestions. Keeping the amount of components, PCB space and cost as low as possible is also an issue that I have to take into consideration so I'm greatful that you share several solutions and thoughs on this one!