Claimed OU circuit of Rosemary Ainslie

[powercat]

Hi MH

Can you please stop all of these riduculous battery voltage before-after tests and spread the word to all of the other experimenters?  You are going to have to develop a strategy for running battery charge-discharge cycle tests to make ENERGY measurements to derive any useful data for all of your experiments.  This will be much more time consuming and much harder to do.

sorry to but in
could this problem be solved by using capacitors instead of batterys
at least when testing for OU ?

cat

[MileHigh]

TK:

I took a glance, very hot links.

Alas, almost all of the people that play with batteries are not even aware that with a few load resistors and a voltmeter you can make some quick measurements to calculate the output impedance of a battery at a given charge state.  That would also be incredibly useful information about the state of health of the battery.  That is the essence of the "battery check" function on a typical multimeter.

Calculating the value of the "invisible resistor" inside the battery is too hard for most of the experimenters unless you give them the exact formulas.

I mentioned this issue to Aaron a few weeks back but I have to assume it flew right over his head.

Ok, sorry for the off-topic stuff.  Back to the real subject:  I want spikes!  Spikes!  Spikes I tell you!  lol

MileHigh

[MileHigh]

Cat:

Great question.  In a nutshell, using a very large capacitor to simulate a battery and then measuring how much time it takes to drop from say 13 volts to 12 volts is a way to give you an accurate power consumption for the device under test.  This will compensate for any arbitrary current consumption waveform from the load and also factor in any power being returned to the capacitor from the device under test.  Some math has to be done to compensate for the dropping voltage.

So, you can get a very accurate power consumption measurement that a standard and possibly even a "true RMS" multimeter would not be able to do.  That might help you in your testing for OU.

MileHigh

P.S.:  There is a way to make the "dropping voltage" problem go away, and I posted what that setup is like on Aaaron's Ainsley thread about two weeks ago.

[TinselKoala]

Hee hee. Yersss, Cat,  and if you should happen to manage to accidentally short-circuit that very large capacitor, you will find out what very low internal resistance really means. Ever drop a wrench across a car battery? Imagine that, times about a tenth of the battery's output impedance.
It's a good idea, just be careful with capacitors, they will eventually bite you and they do bite hard.

But come on, if the energy returned, per cycle, is always and forever measurably less than the energy "lead out" per cycle, there isn't any mystery...no recharge, just slowed discharge.
So the issue dissolves back to the original one: does the slowed discharging battery still produce an anomalously high level of heat? So far, the answer is no.

A question: Does the COP calculated by Rosemary's method actually increase, the longer the test is run?

(And if Kemo-Sabe wantum Spikes, him try 2sk1548 buffalo chip, on a heatsink. Gettum plenty big spikes, much wampum too. )

[allcanadian]

@milehigh
You are correct about the inaccuracies of battery measurement, they will hold their voltage to near the 90% discharged mark as stated and then fall flat on their face. Input/output power measurement to and from the circuit is the easiest way to know where you stand assuming the volts/amps are relative, trying to measure HV/HF output relative to a low voltage DC input is not nearly as easy as it sounds in fact it's a nightmare. As well caps on input/output have many issues ESR, max current ratings, inductance, the caps need to have comparable properties as well as the power charging/discharging them.
AC