Page 209 - Chapter 3 - Laser/IPL Hair Removal

Chapter 3 – Fundamentals of Laser/IPL Hair Removal 2nd Edition
Inside these electronic devices are usually found various electronic components including capacitors. These can store electrical charge until required. This is how peak powers can exceed average powers in these devices. By storing a certain amount of electrical energy, and then using it in a very short time, it is possible to generate large peak powers, which greatly exceed the average power – but only for a very brief time.
But there are limits on this. In the image above, if the average power indicated, is the maximum power that can be taken from the electrical supply system, then the maximum energy per pulse, pulsewidth and rep rate will all be tied together, by this limit (this can be overcome by use of capacitors in the system).
If you wanted to increase the rep rate (more shots per second) then you would need to lower the maximum energy in each pulse – to allow for more shots with the same average power. Likewise, if you wanted more energy in each individual pulse, then the rep rate would need to be slowed down so that fewer shots were fired per second.
This is why we often see, in diode lasers in particular, a change in one setting (fluence) when another one is changed (rep rate or pulsewidth). These three parameters are all usually tied together, to keep the average power at, or below, its maximum available limit.
Can we see the difference between ‘peak’ power and ‘average’ power?
Yes, we can! Mike made a couple of short videos to demonstrate the difference between these two:
https://youtube.com/shorts/V8ZTKC- ncbo?feature=share
Now let's look at a carbon dioxide laser. In this video I set the laser to a 10 Watt output. We can see it has a much more impressive
This video shows a Q-switched Nd:YAG laser outputting pulses of energy 0.3 Joules in 10 nanosecond pulsewidths. These equate to a peak power of 30 million Watts in each pulse (30 MW).
However, the 'peak' power relates to the individual pulsewidths for each pulse. The 'average' power relates to the total time over which these pulses are delivered. So, if we fire these pulses at 10 Hz (10 shots every second), then the total energy delivered in one second is 10 times 0.3 Joules. The average power is simply the total energy divided by the total time - 3 Joules / 1 second, which is only 3 Watts.
https://youtube.com/shorts/cqea57yzH- 4?feature=share
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