3.6V Circuit and 2.8V LED

Question

I'm trying to figure out a good LED + resitor pairing for my circuit. However, I'm provided with a challenge and limited knowledge.

Here's the situation: I'm creating a minimized version of THIS sensor board. The board breaks out THIS sensor, which requires a 3.3V supply voltage. To minimize the size, I took out all voltage regulating components. Now I need to power a white LED with 3.3V and add a current limiting resistor. Getting a sufficient voltage drop is the challenge for which I could use some help.

Now, my solution was to increase the supply voltage to 3.6V instead of 3.3V, which shouldn't be a problem for the color sensor according to the datasheet. I found THIS LED with a supposed forward voltage of 2.8V and current of 20mA. With this situation I CALCULATED that I would need a resistor of at least 47 ohm. Would my suggested solution work or did I miss something??

Answer 1

I found THIS LED with a supposed forward voltage of 2.8V and current of 20mA.

It's worse than you thought!

Figure 1. The ASMT-UWB1-Nxxxx datasheet LED could \$ V_f \$ could vary between 2.8 and 3.6 V at 20 mA.

Figure 2. The blue shaded region represents the possible values of I versus V for your LED. The 47 Ω load-line is superimposed on the chart. Data is for the LTST-C170TBKT but looks similar to OP's LED. Source: Variations in Vf and “binning”.

Depending on the actual value of \$ V_f \$ for your LED the current will be 12.5 mA at lowest \$ V_f \$ to about 1 mA at the highest.

Figure 3. The LED is powered from 3.3 V.

The easiest solution is to disconnect R5 from the 3.3 V supply and power the LED from the unregulated voltage.

Figure 4. The load-line for an 82 Ω resistor on a 5 V supply.

With a 5 V supply and an 82 Ω resistor the current would be 20 mA for the mid-range \$ V_f \$ but could vary by up to 5 mA each way depending on production spread. This is the difficult world of electrical engineering.

If you can increase the supply voltage you can also increase the series resistor which will make it more like a constant current source. The load-line will be less steep and the resultant currents will vary less for a given range of \$ V_f \$.

Answer 2

An alternative is to use a simple current limiter circuit instead of a series resistor. That will minimise the volt drop in cases where current is low. The following is about as simple as you'll get.

Pick Rsense so that it produces about 0.6V at your max current. So for a 20mA LED, choose Rsense = 30Ω. When the current limit is hit, T2 will turn on, drawing the base current away from T1 and therefore inhibiting the current into the load.

Answer 3

The four options you have

1. power the LED separately from a higher voltage.

2. Use 3.6V with the original 3.3V led and a larger resistor.

3. Use 3.3V with the original led and no/a smaller resistor and/or a lower current. You don't have to run it at the 20mA current and associated Vf.

4. Use what you calculated, which is 100% fine.

You could use a constant current led tester set at 20mA to find out the actual forward voltage at Vf and adjust your led calculation based on that.

Answer 4

If you have the room a simple solution is to add a small capacitive booster regulator.

There are many available, but here is one, the LM2750-ADJ from T.I.

They have nicely given you the schematics to drive LEDS in the application examples. (I know you only need one LED, so ignore the extra five shown.)

Notice it regulates the current through the LED, you simply have to chose a value for R1 that sets up your If at 1.23V. It also has a simple logic level shutdown pin you can use to turn the LED on and off.

If you had to drive more than one LED, I'd suggest using one of the fixed output boosters to power all of them, and a traditional transistor/resistor switch for each individually.