I dont rmember you saying how quickly the spikes repeat. If it many times per second, all you need is an analogue meter, problem solved. MC meters read average current.
If thats no good I'd use a small series R and feed the signal to an opamp integrator.
NT
Might be useful to look at the LEM Hall Effect Current Sensors. This would do isolated current sensing with minimum voltage drop. eg, Radiospares 257-414 is 10V dc out for 200 Ampere-Turns full scale input. Simply pass 20-turns (of insulated flexible wire) through the central hole and it is 10V out for 20A in. The voltage drop caused by the sensor is just the resistance of 20 turns of wire.
There are other HE sensors from LEM which give a precise current ratio. For example one of these could be used to provide a an exact 20A:20mA ratio, with the 0-20mA going sideways into a calorimeter, or to a (digital) integrator, or to a data logger.
It varies from a few hundred Hz to a few 10s of kHz throughout the experiment. The analogue meter is a fine idea, but would require manual number-taking, which I'm trying to avoid (average I doesn't stay constant, and I need to calculate total energy).
That's a great idea, thanks. 257-414 is discontinued, but there are quite a few other suitable ones.
there is, use an opamp integrator. No sampling, along with the errors that introduces.
NT
I use quite a number of the LA55-P (RS 180-7357). 16-turns of ptfe insulated wire to sense 10Adc full scale being pulled from a 347Vdc supply. 15-0-15V supplies, with following opamp to provide the Zero and Gain adjustments.
Calibration can be done off-line, with (say) a 20Vdc source, 2 ohm resistor, and digital ammeter. Once set up it has a repeatability within 0.25% (of fsd).
AFAIK, the sum to be done will be something like.
T*(P1+P2+P3.......Pn) Total Energy = --------------------- Joules. N
P1(etc) are a series of discrete Power calculations, taken at time intervals T/N, where T/N is small compared with T. T/N must also be small compared to the shortest transient P.
For a 5 minute run, T= 300 seconds, and T/N might typically be 100uS. This gives N= 3000000, which is the number of (pairs of?) data points to log.
Now that does seem like a good idea - I could integrate both V and I using opamps, then multiply to get total power - any problem with that thinking?
In general yes there is because the product of the means is not the same as the mean of the product. You need Int[v*i]dt and _not_ Int[v]dt * Int[i]dt.
However in your case v is constant (IIRC) so they are the same, more or less - depending on how good your stabilised supply is and whether voltage drop in the wiring is an issue.
Drift is always a potential problem with analogue integrators though; I'd go with your original data logging idea (provided you can sample fast enough) and then number crunch in the digital domain.
haha, yes, nice one. You think that an analog integrator isn't going to have errors? Especially accumulated over minutes? :)
The data logger is the best idea, just record the I&V waveforms over the period under question then do everything else numerically.
Cheers, Mark.
Doesn't this require the sampling to be at twice the frequency of the shortest spike?
AJH
every method on the planet has errors. Try not to be dumb. We just need to get in below 1%.
Digital sampling is going to be especially problematic with a spikey load, as the digital sampling has to be so fast it maps the spikes out accurately, ie a great number of samples per spike so the spike's shape is followed faithfully throughout. This is necessary to come in at below 1%.
It is difficult to achieve this unless you know the highest possible frequency component of said spikes. In practice, if you dont, you have a mathematical problem to pick the sampling frequency needed to keep total error budget below 1%.
We know the spikes occur at upto 10kHz, but that is the repetition rate, not the highest spike frequency component, which inevitably will be orders of magnitude higher.
Analogue integration has issues like every method, but its relatively easy, simple, and low cost, and coming in at 1% is practical.
There might be a bit more to it.
BTW v and i need to be multiplied before the power product is integrated. Since you need 1% accuracy you will need to measure v, you wont do it just measuring and integrating i.
I always tell people not to ask electronics qusetions in uk.d-i-y, we have a remarkable amount of expertise on diy, building etc, but this can lull some into imagining this is a group with expertise on tronics. While we do have some, most do not, and the truth does not fall out clearly like it does with other topics. Only if you know the answers will you recognise in most cases. sci.electronics.design is the place to go for this stuff.
NT
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