Hi Les and welcome to Electronics Point

Your calculations in point 1 are good. You have calculated the voltage that must be dropped by the resistor, and divided that by the current in amps, to come up with a value of 43 ohms for the resistor.

But the data sheet says that the forward voltage V

_{F} is 3.1V ±10%. So I'm not sure where you got the 3.5V from. I think you need to re-calculate the resistor value based on the nominal laser forward voltage of 3.1V. In that case the calculation is:

R = V / I

= (5V - 3.1V) / 0.035A

= 1.9 / 0.035

= 54.2857Ω

Closest preferred value is 56Ω.

Using a 56Ω resistor gives a nominal current (assuming V

_{F} = 3.1V) of 34 mA. Slightly less than the recommended 35 mA.

As Gryd3 explained above, using a simple resistor as a current limiter is not a very exact way to get the current you want. And with LEDs (and I'm assuming this laser module contains a type of LED internally), the current is the important factor - it determines the brightness, and it must be kept below the specfied maximum, which is 40 mA in this case, to prevent damage.

Using a resistor, the actual current flow is determined by the resistor value (which will be pretty accurate if you use a resistor with 1% tolerance - don't use 5% resistors for this), and the voltage that is dropped across the resistor. This voltage depends on two factors: the forward voltage of the laser, V

_{F}, which is specified as 3.1V ±10% which is quite a wide range: 2.79~3.41V, and the power supply voltage, which you've already said is only 4.56V!

Just taking into account the V

_{F} range of 2.79~3.41V and assuming a perfect 5V supply and no error in the resistor value, the voltage remaining across the resistor could be anywhere from (5V - 3.41V) to (5V - 2.79V) which is a range of 1.59~2.21V. Using Ohm's Law, I = V / R, this corresponds to a current range of 28.4~39.5 mA, which is about ±18% compared to the expected current of 34 mA we calculated for a 56Ω resistor. That's quite a large range! Remember that brightness is roughly proportional to current.

Since your "5V" power supply rail is only 4.56V (this sounds wrong to me, but I'll use it anyway), the lowest possible LED current is even less than 28.4 mA. It's actually (4.56V - 3.41V) / 56Ω which is 20.5 mA. That's about half the maximum current I calculated for a supply voltage of 5V and the minimum V

_{F} earlier. So you would have a brightness range of almost 2:1 depending on the supply voltage and V

_{F}. That's probably not acceptable.

The proper way to drive an LED when the current is critical is to use a cicuit called a current regulator or constant current source (or sink). This circuit replaces the resistor with an active component such as a transistor (or two transistors, depending on the circuit) that regulates the current directly, so that the current is almost completely independent of the supply voltage and V

_{F} - at least, as long as they're within the design range of the circuit.

A current regulator is very valuable when the operating current (35 mA) is close to the maximum allowable current (40 mA) because it means you can reliably pass the desired current through the LED without being in danger of exceeding the maximum rated current and risking damage to the LED. Three paragraphs ago I calculated the current range using a 5V supply and a 56Ω limiting resistor as 28.4~39.5 mA. That upper limit is very close to the maximum rated current of that laser LED.

A current regulator also provides a simple way to turn the LED ON and OFF using a logic-level output from a microcontroller or other circuit. These outputs usually can't directly supply the 35 mA required by the LED.

There are various ways of building a current regulator. Here's one of the simpler circuits (captured from

http://e-project4u.blogspot.co.nz/p/101-200-transistor-circuits.html):

The control signal at the left enables the circuit when high, and disables it when low. The resistor labelled R sets the LED current; the formula for the relationship is shown in the diagram. Two LEDs are shown in the collector circuit but you can use any number, as long as there's enough voltage left across the transistor for the circuit to operate. It needs at least 1.0V but more is better for best accuracy. 1.5V is good.

In your case with a 4.56V supply and a worst case V

_{F} of 3.41V, only 1.15V is left across the current regulator. This is probably workable with the design above. I would reduce the 10k resistor to something like 2k2, and use a high gain transistor in the right hand position (e.g. BC548C or BC337-40).