Comparing effects of using pulsed and steady state power to illuminate a high brightness LED

I was having a discussion here with a colleague about the merits of powering a high brightness LED (HBLED) using pulsed power versus using steady state DC power.

My opinion was: “Basically, amperes in proportionally equates to light flux out, so you will get about the same amount of illumination whether it is pulsed or DC.”

His argument was: “Because the pulses will be brighter, it’s possible the effective illumination that’s perceived will be brighter. Things appear to be continuous when discrete fixed images are updated at rates above thirty times a second, and that should apply to the pulsed illumination as well!”

I countered: “It will look the same and, if anything, will be less efficient when pulsed!”

So instead of continuing our debate we ran a quick experiment. I happened to have some HBLEDs so I hooked one up to an N6781A DC source measure module housed in an N6705B DC Power Analyzer sitting at my desk, shown in Figure 1. The N6781A has excellent current sourcing characteristics regardless whether it is DC or a dynamic waveform, making it a good choice for this experiment.

Figure 1: Powering up an HBLED

First we powered it up with a steady state DC current of 100 mA. At this level the HBLED had a forward voltage drop of 2.994 V and resulting power of 0.2994 W, as seen in Figure 2, captured using the companion 14585A control and analysis software.

Figure 2: Resulting HBLED voltage and power when powered with 100 mA steady state DC current

We then set the N6781A to deliver a pulsed current of 200 mA with a 50% duty cycle, so that its average current was 100 mA. The results were again captured using the 14585A software, as shown in Figure 3.

Figure 3: Resulting HBLED voltage and power when powered with 200 mA 50% DC pulsed current

Switching back and forth between steady state DC and pulsed currents, my colleague agreed, the brightness appeared to be comparable (just as I had expected!).  But something more interesting to note is the average current, voltage, and power. These values were obtained as shown in Figure 3 by placing the measurement markers over an integral number of waveform cycles. The average current was 100 mA, as expected. Note however that the average voltage is lower, at 2.7 V, while the average power is higher, at 0.3127 W! At first the lower average voltage together with higher average power would seem to be a contradiction. How can that be?

First, in case you did not notice, the product of the RMS voltage and RMS current are 0.3897 W which clearly does not match our average power value displayed. What, another contradiction? Why is that? Multiplying RMS voltage and RMS current will give you the average power for a linear resistive load but not for a non-linear load like a HBLED. The average power needs to be determined by taking an overall average of the power over time computed on a point-by-point basis, which is how it is done within the 14585A software as well as within our power products that digitize the voltage and current over time. Second, the average voltage is lower because it drops down towards zero during periods of zero current. However it is greater during the periods when 200 mA is being sourced through the HBLED and these are the times where power is being consumed.

So here, by using pulsed current, our losses ended up being 4.4% greater when powered by the comparable steady state current. These losses are mainly incurred as a result of greater resistive drop losses in the HBLED occurring at the higher current level.

There is supposed to be one benefit however of using pulsed power when powering HBLEDs. At different steady state DC current levels there is some shift in their output light spectrum. Using pulsed current provides dimming control while maintaining a constant light spectrum. This prevents minor color shifts at different illumination levels. Although I would probably never notice it!

Powerlifting Agilent style!

I have been working out at a gym including lifting weights since the early 1980’s. We have a small gym here in our office building that I use a few times per week. The other day, while doing incline bench presses, my mind was wandering and I began to wonder how much power it took for me to lift the barbell and weights.

I could put the barbell and weights on a battery operated lift we have here in the office and instead of the battery, use one of our power supplies to power the lift and measure the power while operating the lift. I also wanted to calculate how much power would be required. I admit that I had to take out my old physics book to refresh my memory on how to convert weight moved through a distance to watts, but this turned out to be pretty simple: the power is just the force (weight in newtons) times the velocity. Here is the justification:

Force is mass times acceleration. F = mass*acceleration = kg-m/s^2 = newton = N which is weight when the acceleration is due to gravity (weight = mass*gravity).

Work (energy) is force (weight) applied over distance. Work = F*distance = N-m = joule = J.

Power is work per unit of time. Power = J/s = watt = W.

So power in watts = W = J/s = N-m/s = kg-m/s^2-m/s = mass * acceleration * velocity = kg*gravity*velocity = weight*velocity (gravity = 9.8 m/s^2).

During my investigation, I did go off on a tangent for a short time looking at why we talk about measuring weight in kilograms even though kilograms are units for mass and not weight. It would be proper to measure weight in newtons, not in kilograms, but that’s a different story!

So when I lift 205 lbs (93 kg) a distance of 15 inches (0.38 m) in 1.5 seconds, I use 231 watts of power to do so (mass*gravity*velocity = 93 kg * 9.8 m/s^2 * 0.38 m/1.5 s). As I mentioned above, I wanted to see if I could measure something similar with a power supply connected to a battery operated lift by using our power supplies in place of the 24 V batteries. Here is what I found:

I did a baseline power measurement of just the lift lifting some wooden pallets needed to support the barbell I was about to put on the lift. I used 2 Agilent N7972A (40 V, 50 A, 2kW) power supplies connected in parallel (I needed the extra current capacity) and set to 24 V along with our 14585A Control and Analysis Software to capture the power over time. I could then add weight and measure the incremental power required to lift the added weight.

I found that the lift itself consumes 1502 W as my baseline measurement. Then I added a 288 lb (130.6 kg) battery compartment along with 295 lbs (133.8 kg) of barbell weighs for an added 583 lbs (264.4 kg). Again, I measured the power consumed by the lift while it moved the weights vertically and found it to be 1638 W. Lifting the incremental 264.4 kg consumed an additional 136 W. Let’s see if this makes sense with a calculation. The lift moved 4.5 inches vertically in 2.2 seconds which equals 0.052 m/s. The calculated power is then 264.4 kg * 9.8 m/s^2 * 0.052 m/s = 134.7 W. That’s very close to the measured 136 W!!

It is no surprise that the laws of physics work as expected here and that our power supplies can provide insight into those laws. Agilent has added new meaning to the term “powerlifting”!