Good job guys, and thanks for sharing your results.

I tried your method in PSpice, and it comes very close to the actual summed power in the transistor and LED, but it is lower by about 1.5%.

It is certainly good to know that the measured Pototal is LOWER than the actual. I'd like to do an analysis to show that it is indeed the case IF we also include the power spent on the 1k resistor (our measuring method unchanged).

Pototal = P1k + Pbe + Pce + Pled.

P1k = power on the 1k resistor = I1k * V1k,

Pbe = power on the be knot of the transistor = Ibe * Vbe,

Pce = power on the ce knot of the transistor = Ice * Vbe,

Pled = power on the LED = Iled*Vled.

On the input side it is relatively simple:

Iin = Icsr1 = Ibat, Vin = Vbat - Vcsr1, so Pitotal = Iin*Vin = Icsr1 * V1

The measurement is exact (Note: Vin=V1, Iin=Icsr1=Vcsr1=V2, with Rcsr1=1).

So I will concentrate on the analysis of the output side, and consider the ON and OFF state of the transistor separately.

When the transistor is OFF, the analysis of the output is simple: Iin = Iled, Pototal = Pled = Vled * Iin. Thus our measurement is exact in this case.

Let's turn to the ON state:

Note that Ie = Ibe + Ice, I1k = Ibe, Iin=Ibat=Iscr1=Ibe+Ice+Iled, and Vled = Vce.

As the current Ice is climbing up, so the coil connected to the collector maintains a voltage drop, we have Vin > Vce. On the other hand, the time interval for Ibe to rise to its working level is tiny, so most of the time, Vbe+V1k = Vin + Vinduced (ignoring the resistance in the coil), where Vinduced is the induced voltage because the current in the other coil is increasing.

P1k + Pbe = I1k * V1k + Ibe * Vbe

= I1k * (V1k + Vbe)

> Ibe * Vin

> Ibe * Vce.

From which we can proceed as follows:

Pototal = P1k + Pbe + Pce + Pled

> Ibe*Vce + Ice * Vce + Iled * Vled

= (Ibe + Ice + Iled) * Vce

= Iin * Vce

Note that Iin*Vce is what we measured as Pototal (to see this, Iin = Iscr1 = Ibat, Vce = Vled = V3), so it is indeed less than the actual Pototal.

On the other hand, the measured Pototal should be very close, as Ibe is relatively small to Ice on the average. The bigger the beta, the better the approximation. The gap can be roughly estimated by integrating Ibe * (Vin - Vce) during the ON state of the transistor (ignoring Vinduced), now (Vin - Vce) is the voltage drop over the coil connected to the collector, Ice * (Vin-Vce) is the power stored on that coil, which will be released through the LED later. However, Ice is increasing almost linearly from zero up to (beta * Ibe) [Ibe is relatively constant, so is Vin-Vce, during the ON state], therefore MEAN[beta * Ibe * (Vin - Vce)] during the ON state = 2 * MEAN[ Ice * (Vin-Vce) ] during the ON state. The stored energy MEAN[ Ice * (Vin-Vce) ] during the ON state is released later on the LED, which can be estimated by Pled * ( 1 - Vin/Vled) during the OFF state of transistor. From our previous tests, Pled ~ Pototal, therefore the gap is estimated to be 2 * Pototal * (1 - Vin/Vled) / beta, which is about 200*(1-Vin/Vled)/beta percent of the Pototal. If beta = 50, Vin = 1V, Vled=2V, the gap is then estimated to be 2%. Which is close to your 1.5%, and that 2% also included the 1k resistor. If assume Vinduced = Vin-Vce, as the coils have the same # of windings, the gap becomes 400*(1-Vin/Vled)/beta; in this case if beta=100, Vin=1, Vled=2, the gap should be 2%.

In conclusion: the simplified method of measurement gives a good

**underestimate** of efficiency n.

@PhysicsProf:

It all Looks exciting -- look forward to your verified results. And your encouraging words are appreciated. I will do my best to join you [might take some time, I need to get a scope and learn how to use it].

@feynman:

Thank you for your encouragement, and it is great that you could join the force here.

cheers, lanenal