-2

:)

Circuit diagram

So. How do I correctly calculate the power consumption of this circuit? :)
I don't have an oscilloscope or an DMM that measures I. My DMM is an Extech EX330, which is quite good for it's price.

Here are the measurements:
It is drawing 9.21 mA DC and 6.48 mA AC in series with the battery.
It shows 1.364 VDC and 0.010 VAC in parallel with the battery.
Output on the LED is 2.572 VDC and 0.011 VAC with no measurable frequency. (with my DMM at least)
All measurements are taken with the same configuration and voltage

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Please note that I put 20.000 µF electrolytic, a bit metal film and a tiny ceramic capacitor in parallel with the LED (for no particular reason. just more smoothing) while measuring the above stated data points!
Also, I switched both 3-pin TO22 dual schottky-diode packages, of which I bridged pin 1 and 3 to put diodes in parallel, (shown as weird schottky-diode things in the schematic) with regular 1N4007 diodes, because they work more efficient for some weird reason. (Reverse leakage current in the schottky packages maybe?? BTW, does it get worse when two are in parallel?)

Sorry I couldn't update the schematic, but thank you for your help! :)

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Why is it different from the other question, you ask? My two questions of the same kind were ignored in the other thread. Also it is totally different question. So keep it, please.

Distelzombie
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  • You need to measure the phase between voltage and current too, so beast case scenario is that you can estimate the total apparent power. Most multimeters won’t measure 3.5 kHz current. – winny Mar 27 '18 at 20:14
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    This is really just a continuation of your original, unanswered question https://electronics.stackexchange.com/questions/364647/why-does-my-dmm-show-ac-and-dc-current-going-in-my-circuit-joule-thief Rather than re-posting you should improve that question, especially as it already has context this one lacks. – Chris Stratton Mar 27 '18 at 20:50
  • No one answered the question there. I asked two times. It was ignored. I thought it is frowned upon to ask a second question in one thread. Therefore I made a new one that is totally different from the other one. – Distelzombie Mar 28 '18 at 05:01
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    @Distelzombie what you should do instead is to edit appropriately your original question, removing all the obsolete parts and clarify schematic and problem as best as you can – clabacchio Mar 28 '18 at 12:48
  • Ok, you know what? Forget it. I spend enough time trying to comply to your weird rules. This is obviously the wrong place to ask a simple, short question. How can anyone even consider this to be a duplicate - let alone an "exact copy"?! WTF?! It is a different question!! I only have bad experience trying to ask two question in the same Thread in any forum. I started to get the same vibe here since one question (this) got ignored two times. And now this ridiculous assertion that this an "exact copy"! Do you even know what that means, guys?! illogical, arrogant, unfriendly and stiff and starchy! – Distelzombie Mar 28 '18 at 15:05
  • You happy now? :) btw i am quite a bit irritable due to medicinal side effects atm. Sorry – Distelzombie Mar 29 '18 at 11:38

1 Answers1

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The power consumption from a DC power supply is \$V_{DC}\times I_{DC}\$. Taking into account the AC part of the current is nonesense if the supply is DC.

Why is this so you might ask.

Power is volts x amps at its most basic and if the current taken from a supply is a sinewave wholly superimposed on a DC value then multiplying \$V_{DC}\$ by the AC content of the sine wave produces a power waveform that is positive in one half cycle and negative in the other. The average power would therefore be zero.

Andy aka
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  • Hi Andy, I want to apologise. I had posted a comment saying since there can be no phase relationship when voltage is DC Power is \$ V_{DC} \times I_{rms} \$. Thanks to your post I have just recalculated from first principles. You are correct, my bad. +1. – Warren Hill Mar 27 '18 at 21:40
  • Ok, my DMM is relatively cheap and doesn't measure I. So I guess I can simply use the DC current reading? – Distelzombie Mar 28 '18 at 05:02
  • Yes, if your DC range is sluggish enough not to have its mean value affected by AC components then it should be ok. – Andy aka Mar 28 '18 at 08:01
  • @Distelzombie That means "if it displays the average reading over a second (or so) instead of constantly flickering between changing readings." – user253751 Mar 28 '18 at 10:27
  • Could you post this answer in the "original" thread? :D I just changed the original to include all of this information and this question, actually making it an exact copy imo, but meh. If they want it that way... – Distelzombie Mar 29 '18 at 11:41
  • @Distelzombie done! – Andy aka Mar 30 '18 at 07:32