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I urge you to reconsider Professor.When dealing with DC power sources, heavy averaging of both the battery voltage and current signals is the most reliable way to measure input power. You simply multiply the two DMM values together (taking the CSR value into account), and the result is an accurate net average INPUT power measurement..99
As noted earlier, I am trying to find caps that do not leak so fast, or at all. Any ideas on this would be helpful.
Professor,Under ideal conditions, the scope method is accurate.What I am suggesting is this; if the scope and DMM methods do not agree, one of them must be wrong. DC power sources have a power factor of 1.0, therefore heavily averaging the current and voltage measurements is not only the best way to measure the INPUT power, but it is the easiest and most accessible..99
Hey, I appreciate your comments Xee2 and Nul-points. [...]Meanwhile, a quick test today: Four 10K uF caps, to run the sj1 circuit. By measuring the volts before and after 30 seconds on the caps, I can calculate input power easily.delta-E = 1/2 C(Vi**2 - Vf**2) , Pin = deltaE / delta-T , 30 seconds. C = 40mF.Start, Vinitial = 1.385V , Vfinal = 1.255VSo delta-E = 6.8 mJoules. and Pin = 6.8/o.5min = 13.6 mW, pls check my math.in reasonable agreement with the Tek-scope measurement under similar conditions, 10 mW (see reply #1 for the data, Pin on the left).[...]
hi Steveni think you have an incorrect method for calculating Einyou should calculate the Energy stored in C for each voltage, start & endTHEN subtract to get total hence:-40mF1.385V => 38.37mJ1.225V => 30.01mJ Ein 8.36mJPin = 8.36/0.5 = 16.72mWanother potential issue to note - the cap value can be +/- 10-20%when doing these calcs, it's wise to measure the cap valueof course, for a 'ball-park' calculation it's not necessary! hope this helpsnpPS i admire your other 'extra-mural' work , investigating & providing low-cost solar cooking solutions for developing countries!http://docsfreelunch.blogspot.com
delta-E = 1/2 C(Vi**2 - Vf**2) , Pin = deltaE / delta-T , 30 seconds. C = 40mF.Start, Vinitial = 1.385V , Vfinal = 1.255VSo delta-E = 6.8 mJoules. and Pin = 6.8/o.5min = 13.6 mW, pls check my math.
[...]therefore over 30 seconds = 6.864/30 = 0.2288 mWnote > watt = Joule/sec What did I do wrong?
[...] a Joule is a Watt-second - i see you've divided Ein by units of minutes;[...]Steven, apologies - your Ein calc method is good, but Pin needs units of seconds, not minutes]Pin = 6.87/30 = 0.23mW[...]np[...]