Driving an LED with a P-MOSFET or op-amp as a current limiter and no resistor

Question

I have an application where I need to be able to drive an LED without a resistor. Adding a resistor is not an option. I then must be able to blink the LED on or off by shunting to ground.

LED Specs: Vf = 2.2V, I (nominal) = 20mA

Would either of these configurations be viable? For the op amp, I would look for a component with an output short circuit current that's within the range of what an LED would need. For the MOSFET, I would need something in saturation at 5V, that can then have the current lowered by adjusting Vgs.

Is there a better solution or IC on the market that does this already? An LED driver of some kind? The only other condition is that the driver must drive the anode side of the LED.

simulation

Engineering reasons for doing this:

If an LED was accidentally wired without a resistor it will not get destroyed. That's basically it. Convenience. The end user can operate an LED with or without the resistor. Optional.

Another example of why you would do this. Lets say you want to detect whether an enable pin is held to ground or not, without pulling current away from the pull-up.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

A N-channel JFET like BF256 can do this job without any resistor, connecting gate to source (Vgs=0). It is a RF transistor, low power, and there several types.

^{simulate this circuit – Schematic created using CircuitLab}

See the current curves for BF256A, they are not completely flat, but I think it is acceptable for a LED from Vds > 3 V.

Answer 2

Any current source will do, such as this simple design:

^{simulate this circuit – Schematic created using CircuitLab}

You can safely open and close SW1, and current \$I\$ will remain about 20mA. The potential \$V_X\$ at node X will adopt whatever value is necessary to pass 20mA through whatever load is below it, be that an LED, a resistor or a short circuit.

Change R1 to set \$I\$ to whatever (reasonable) current you desire:

$$ R_1 \approx \frac{0.5V}{I} $$

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Alternatively, use a low dropout linear voltage regulator, like the LM1117:

^{simulate this circuit}

R2 is only necessary for currents under about 5mA, necessary to obtain the necessary minimum load current for the LM1117 to regulate properly. For \$I>5mA\$ you may omit R2. Current \$I\$ is determined by R1:

$$ R_1 = \frac{1.25V}{I} $$

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Then there's the good old LM334, designed specifically for this purpose. I include it for completeness, but I admit you can only get 10mA out of this, so it might not be suitable for your needs:

^{simulate this circuit}

$$ R_1 = \frac{0.07}{I} $$

This is somewhat temperature dependent (one of the uses of the LM334 is temperature sensing), but you'll probably not notice much (if any) LED brightness change over typical room temperature variations. On page 8 of the datasheet you can find modifications to cancel this dependency.

Answer 3

You could use a transconductance amp as the current source. MAX426 can supply 20ma, and the output current is proportional to the input voltage. You could also use it as a sink, with the input controlled by your driver.

Answer 4

Adding a resistor is not an option

Yes, it is.

I understand that adding a resistor out at the LED is not an option, because of wiring, physical access, etc. But if you can add a 1- of 2-transistor constant current source, of a FET driver, or an opamp, or whatever at the driving point, then I don't understand why you cannot skip all of those relatively complex solutions in favor or a simple resistor at the same location.

For a 2 V LED operating at 20 mA from a 5 V source, the ideal resistor value is 150 ohms. Power dissipation in the resistor will be 60 mW, so anything rated at or above 100 mW will work.

If the voltage source for the LED is variable, or unknown within a certain range, then a case can be made for a constant-current driver. You do not explicitly state a source voltage, but one statement implies that it is a fixed 5 V source. If so, then a fixed resistor should work nicely.

Answer 5

XY problem. If you can add an active current limiter, you can certainly add a resistor (you’d need several resistors for any sort of a discrete constant-current circuit anyway.)

Probably the most economical is a discrete 2-transistor limiter built from two NPN devices, hacked in a way that allows logic control.

Example (simulate it here):

And I get what you're looking for - the ability to check for stuck-at states, as might apply for functional safety; the current limiter will let you do that.

In a FuSa scenario, your host would monitor the voltage during 'on' and 'off' to detect one of the stuck-at or open-circuit failures. Your host would also also know if a resistor were inserted since there would be an abnormally high voltage at the current source when the driver is 'on'.

Skipping the current limiter is an option too: just use a local load resistor. Since you know that resistance, and what the LED forward drop should be, you'd be able to see not only stuck-at or open-circuit faults, but if an additional resistor were inserted.