Electronic – Increase 74LS TTL gate’s output current to drive multiple high-power LEDs

For best overall LED efficiency (which is your stated aim) you want high efficiency LEDs plus high efficiency drivers. There are may high power LED modules available but they in many cases use LEDs that are less efficient than the best available. No amount of efficient driving of a very low efficiency LED will make up for its low efficiency.

You mention 0.8A x 40V (which will use multiple LED dies or separate LEDs) = 32 Watts. This input power could be obtained with about 5 to 8 of the top available LEDs and it will often be better to use the best available if top efficiency is wanted.

It will probably be most cost-effective to buy a commercially available LED driver rather than making your own, looking for good efficiency. ebay lists a vast quantity of possible candidates

You also need to specify your energy source. Is this 110 VAC mains, or a 12 Volt battery or?

The standard way of driving LEDs correctly and at maximum efficiency is straightforward (fortunately).

• Determine required LED operating current.

• From data sheets, determine maximum voltage drop that LED will have at desired current.

• Provide a switched mode constant current source that will produce the required current at at least the maximum possible voltage.

• Extra points:

  • Minimise LED temperature with cooling

  • Operating an LED at below its maximum current ratings will result in somewhat increased efficiency - perhaps 10% to 20% more light out per Watt in at 20% of full ratings compared to full power.

  • LED lifetimes increase with decreasing temperature and with decreasing current. While more current will usually result in more heat produced, These effects are independent of each other.

To get an LED driver of essentially any desired rated current. use of an external MOSFET of suitable rating, driven by a controller that sits, will allow whatever current the MOSFET can handle. This can be very large if desired.

For very best efficiencies use "synchronous rectification" - this replaces rectifier diodes with switched MOSFETs and gives lower losses.

Efficiencies around 85% from a low voltage DC source to an LED string will be achievable.Over 90% in selected cases and with care.

Comment on the above and we can provide better information.

First: Two LEDs if operated with identical current, will in practice have different intensities, due to manufacturing tolerances even between LEDs from a single batch. This is not easily perceptible when the LEDs are at high intensity, say the 10 to 20 mA range for your 20 mA LEDs, but the differences become very visible as the LEDs are dimmed.

Now, there are a few things that need improvement in the schematic. For instance:

• The potentiometer at the 6.5 Volt rail: Not needed.
  LEDs are current-driven devices. So long as you supply the right polarity of supply and limit the current through each LED chip with an appropriately calculated resistor in series as you have shown, the supply voltage should be the 6.5 Volt line directly.
  Anything above about 4.5 Volts will work - 3.2 Volts for the green and blue LED's maximum, plus some headroom for the transistor's V_CEsat and for regulation by the resistor.
  .
  Why not use the potentiometer anyway?
  It will need to dissipate power corresponding to the entire current drawn by all 6 LED junctions, so up to 120 mA, times whatever voltage is being dropped across the active part of the potentiometer. That'll require a high power rating pot. Also there will be heat generated on that one part, rather than distributed across several individual current limiting resistors that the schematic already has.

• Why the 6.5 Volt supply? A 5 Volts supply, if available, would reduce wasted power and generated heat at the current limiting resistors.

I am sure others will chime in with other suggestions, but those are a start.