Does using maximum power transfer theory on an LED circuit make any sense? - Page 1
Forum section: electronics / general

author: tylerssims
author_rank: Contributor
timestamp: April 28, 2025, 11:01:04 pm
content: I truly cannot understand what my professor is asking. I don't think he understands how maximum power transfer theory works.In this project, we are supposed to make a simple circuit with a small resistor network and a potentiometer. The goal is to design the circuit so that the potentiometer can be used to control the LED intensity over its full range. So we are simply selecting resistor values to limit the current to the LED while supplying sufficient voltage. Simple.But then we are supposed to treat the resistor network as the source resistance and the LED as the load resistance, and apply maximum power transfer theory to simplify the circuit to a single resistor. This seems totally backward from what I understand about the theory. The entire point of the resistor network is to limit current, so why on earth would we want to maximize power? We might as well just remove all of the resistors at that point since the LED will be fried either way.From what I understand about the theory, it is typical to treat the source resistance as fixed and optimize the load resistance to be as close to that as possible, like when doing impedance matching for antenna design.Thoughts? Thanks!

author: hans
author_rank: Super Contributor
timestamp: April 29, 2025, 07:06:11 am
content: Yes that does sound very much back to front.The maximum power transfer states that you need (conjugate) impedance between your source and load. Then both will distribute 50% of the supply voltage, and you can show mathematically that then maximum power is delivered to the load. However, this is only true for resistive loads. Diodes are not resistors.I guess what prof wants to see is use R1 and R2 to create a "4V supply" from 6V, and then "impedance match" using R2,R5 and R3 to the LED "impedance" at its maximum current (e.g. 2V 20mA => 100R), because running the LED at any lower setting means you deliberately want to transfer less power.But if I were to design this circuit for real, I would leave R2,R5,R6 out, choose R1=(6-2)/20m=200R, and then put a 10k (logarithmic) potentiometer in there to dim the LED "full range" (whatever that means for the "zero" LED current, as it will never be actually zero with a resistive dimmer). Wastes a lot less power too, because in order to rely on R1/R2 divider to create a stable 4V supply (e.g. it doesn't shift around when the load varies), it needs to be sufficiently low impedance. Good reason to use active voltage regulators...

author: tom66
author_rank: Super Contributor
timestamp: April 29, 2025, 09:12:54 am
content: Maximum power transfer theory in this case I think refers to having the source and load impedances matched where maximum power is dissipated in the load (and equal power is dissipated in the source impedance). In this case you can treat the LED as being an impedance although it is of course non-linear and quite temperature dependent. Whether the LED survives this situation is another matter! You could set the source impedance up so that the maximum current is something like 20mA which will be safe for most 5mm LEDs. I do agree it is a strange circuit and an even stranger way to teach the concept.

author: nctnico
author_rank: Super Contributor
timestamp: April 29, 2025, 09:32:32 am
content: The way I read it is that the goal is to drive the LED from the circuit where R3 can be between 0 and R. So with R3 as zero Ohms, the current through the LED should not exceed the limits. The hint to use 'maximum power transfer' could be confusing but probably hints towards using formula's to calculate resistive pi-attenuator circuits. I'd probably attack the circuit by going from Vforward at the maximum LED current. That way the voltage across R5 is known. From there you can create some equations to calculate the voltages at the nodes or the current through the resistors.@tom66: it depends on the type of LED whether the forward voltage is temperature dependant. Elektor has published several audio amplifiers in which they use a red LED instead of a diode to create a current mirror which is less sensitive to temperature changes.