White LED Learning Curve!

About: Electronics Engineer by trade, general bodger by nature. Always making things, preferably from what-ever bits I have in the garage. You will find more interesting things on my personal web page rather than t...

Needed a Bright Light

I was trying to fix something and I needed a better light to determine one bit of black plastic from another in a confined space ... and what better than a super bright cold white L.E.D (Light Emitting Diode)?

Fortunately, Christmas is almost here again so I bought a cheap 1,50€ garland of 20 L.E.Ds. The instructions said 2 x 1.5 volt "AA" cells which I thought was intriguing as White L.E.Ds need more than 3 volts: On examination, no series resistor is used, just like a 'throwie' (Ref.1) well, o.k., but surely that will not be enough?

So, how does that work then?

The simple answer is that a typical 1.5v battery can give upto 1.6 volts when new and the L.E.Ds have been chosen to have relatively low Vf of 3 volts. This cheap engineering means that individual brightness is varied and reduces relatively quickly and also will not use all the available power to the 0.9 voltage endpoint i.e about a third; at least you can use them for the T.V remote after the L.E.Ds stop working!

But read on...

Step 1: White LED Basics and Operation

Maxim gives an APPLICATION NOTE 3070 explanation of
Standard and White LED Basics and Operation

https://www.maximintegrated.com/en/app-notes/index...

which is a bit confused but this bit is relevant:
"White LEDs - True white-light-emitting LEDs are not available. Such a device is difficult to build because LEDs typically emit one wavelength. White does not appear in the spectrum of colors; instead, perceiving white requires a mixture of wavelengths.

A trick is employed to make white LEDs. Blue-emitting InGaN base material is covered with a converter material that emits yellow light when stimulated by the blue light. The result is a mixture of blue and yellow light that is perceived by the eye as white."

...and goes on to say...

Step 2:

"Figure 7 (above) shows the current-voltage curves from a group of randomly selected white LEDs. Applying a voltage of 3.3V to these LEDs (upper dotted line) produces forward currents in the range 2mA to 5mA, which in turn produces different shades of white color. The Y coordinate in particular changes strongly in this area (Figure 5), resulting in a non-true reproduction of color in the illuminated display. The LEDs also have different light intensities, which produces inhomogeneous illumination. Another problem is the required minimum supply voltage. A voltage well above 3V is necessary for operating the LEDs. Below that level, several LEDs may remain completely dark."
"A Li-Ion battery when fully charged provides an output voltage of 4.2V, which drops to a nominal 3.5V after a short period of operation. That voltage further declines to 3.0V as the battery discharges. If white LEDs are operated directly from the battery as shown in Figure 3, the following problems occur:

At first, when the battery is fully charged, all diodes are illuminated but with different shades of light intensity and color. As battery voltage drops to its nominal level, the light intensities decrease and the differences in white become stronger. The designer must therefore consider the value of battery voltage and diode forward voltage for which the series resistor is calculated. (With a completely discharged battery, some LEDs will be completely dark.)"

Step 3: Out With the DVM!

I took some values from a single sample L.E.D that gave me the following data:

Vf 3.0 3.0 2.96 2.88 2.82 2.75

I(mA) 9.17 7.95 6.24 3.43 1.90 0.76

so you can see that Vf (Forward voltage) settles at 3 volts so problem solved; not so, there is a bit more to this especially if you appreciate that the first three voltage values show a tiny difference of 0.04 volt but the current has changed by 4.24mA - nearly half!

and the battery? A 1.5 volt battery's voltage is not static and is considered exhausted at 0.9 or about 1.35 volts and also varies with current drawn due to it's own internal resistance. (Incidentally, the voltage increased when I made the measurements as the battery became active.)

Here are the voltage measurements from a random battery sample.

Voltage off load (Open circuit) = 1.53v (Digital Voltmeter Impedance)

Voltage with 150R (10mA) = 1.49v i.e 9.93mA

Voltage with 75R (20mA) = 1.44v i.e 19.2mA

Here is what Wiki says:

https://en.wikipedia.org/wiki/Zinc%E2%80%93carbon_...

Step 4: What Have We Learnt?

The internal resistances can be calculated to be:

L.E.D Rf=38ohms

Battery (1.49-1.44)Volts/(9.93-19.2)mA = 5.4ohms Rinternal

A nominal 3 volts forward voltage is required for each L.E.D and the battery supplies 1.53 volts

The internal resistances add: L.E.D Rf=38R plus 2 x Battery at (1.49-1.44)Volts/(9.93-19.2)mA= 5.4Rinternal each so giving 48.8ohms total for two batteries and one L.E.D.

That cannot be enough! (2x 1.53)-3v/48.8ohms is only about 1mA but if the terminal voltage of new batteries is 1.60v it would give (2x 1.60)-3v/48.8ohms = 4mA

even at 1.75, only a quarter of a volt extra, then we now get (2x 1.75)-3v/48.8ohms = 10mA

So what can we do?

Step 5: How Can We Use This Knowledge

Brightness changes with current applied and each L.E.D is slightly different from manufacture. The subject is comprehensively covered by iceng in Ref.2

I started to think of constant current circuits and every time came up against a need for 4 volts or more which made me wonder if this was the reason for sets of three L.E.Ds in the L.E.D chains that we can buy super cheap from China.

I played with the value of series resistor for a 12 volt set of 3 series L.E.Ds i.e 3 volts after 3 x L.E.D Vf of 3 volts and then for a change in voltage of +/- 1 volt to give applied current change.

I will not bother you with the details but 220R for a current of 13.6mA at 12 volts gave the least variation. There may be a lower value fitted in commercial L.E.D chains but that is likely to be a factor of different L.E.Ds being used. 17.3 to 8.2 (delta 11.1mA for 12.8v to 10.8v)

and that knowledge lead me to the following 3 options:

It was obvious that if I was going to use 1.5v batteries then 3 would be required giving a voltage range from 4.8 to a proper 4volt end-point and a resistor of 150ohms causes a current from 12mA to 6.7mA (delta 5.3mA); 120R gives 15mA to 8.3mA (delta 8.7mA). This voltage option would be best served by a constant current circuit.

With what has been learnt is possible to use rechargeables at 1.2v i.e 3.6v without a resistor as the L.E.D Rf is 38R and will limit the current to about 16mA (with,say,delta 7mA) and will simply stop working at 3v!

If I use 4 x 1.5v batteries (because that is a more normal holder size) then the range becomes 6.4v to 5.4v and with 220R making a flatter current range i.e brightness change of 15.4 to 11mA (delta 4.4mA)

The last option appears to be a good balance of brightness, brightness variation and long activity.

I used three L.E.Ds in parallel to get a good coverage and brightness.

Ref.1 https://www.instructables.com/id/LED-Throwies/ Ref.2 https://www.instructables.com/id/LED-Throwies/

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Make it Glow Contest 2018

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