# A single fault safe design for a LED

G

#### [email protected]

Jan 1, 1970
0
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

Do anyone have a suggestion for a better solution?

Thanks for any help!

Stian

D

#### DJ Delorie

Jan 1, 1970
0
Why not two current limiting FETs in series?

E

#### Eeyore

Jan 1, 1970
0
DJ said:
Why not two current limiting FETs in series?

Highish voltage drop.

You could use 2 constant current sources though.

Graham

T

#### Terry Given

Jan 1, 1970
0
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

Do anyone have a suggestion for a better solution?

Thanks for any help!

Stian

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

S

#### Stef Mientki

Jan 1, 1970
0
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. Q1: NIRS ?

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.
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.
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.
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

G

#### [email protected]

Jan 1, 1970
0
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.

Yup, your right, but limiting the current to 100mA would keep the outout power within the specified limit (stated in datasheet)

BS2: at these voltage it's impossible that a resistor will change in a
short-circuit.
but if you insist ;-)

You can discuss this with the people at the IEC commities
A1: use 3 resistors in series
A2: lower the current to 50 mA and use 2 resistors in series
A3: use LEDs with feedback

Thank's, I'll look into this but I can't lower the current much below
60mA because of the cable lenght of the fibre optic cable.

G

#### [email protected]

Jan 1, 1970
0
(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- Hide quoted text -- Show quoted text -

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!

D
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