My Patreon: https://www.patreon.com/TPAI Paypal Donation Link: https://goo.gl/wq74oG Some people are confused how the power dissipation in the dummy load can be calculated when using a PWM rather than a constant voltage. In this case the measured current can simply be multiplied with the output voltage, becausse the Fluke DMM already gives you the MEAN current over time, rather than the size of the current during the on-periods. The power values I stated in the video are correct and I can also back them up without even using the current measurements from the DMM at all. You can simply calculate them using the output voltage, size of the overall resistance of the dummy load and the duty cycle given. The correct equation is the following: (as long as the voltage in the off-period is 0 and a constant value V during the on-period) P_dis = (D* (V^2))/R In the test we had a Duty Cycle of D=0,5 The resistances where 47 Ohm, 23,5 Ohm, 15,66 Ohm, 11,75 Ohm, 9,4 Ohm So the power values you get are: P_dis(R=47 Ohm) = 0.5* ((62V^2) /47 Ohm)= (1922 / 47)W = 40.89 W P_dis(R=23,5 Ohm) = 81,78 W P_dis(R=15,66 Ohm) = 122,73 W P_dis(R=11,75 Ohm) = 163,57 W P_dis(R=9,4 Ohm) = 204,46 W When you simply multiply 62V with the current values measured in the video, you get very similar values. You can find the formula I use on this German website, if you want to know more: http://www.mikrocontroller.net/articl...
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| Science & Technology | Upload TimePublished on 22 Jan 2016 |
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