frogfot
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Estimating effective current
Here it goes, I've powered my chlorate cells from rectifyed transformer current without any smoothening caps. The current was measured by a clamp
multimeter (not a true RMS multimeter).
Now, to know the exact amount of consumed current I wanna calibrate my multimeter by measuring how fast the power source will heat up different
resistances with and without the caps.
The question is, is it possible to use a fat transistor to sample different currents? I thoat to simply change the base resistor to get different
currents and measure the temperature on surface of the transistor..
Maby there's other cheap alternatives to high power resistors.. Using an electrolytical cell as the resistance seems not to be practical at
currents above 3A.. and I have to go up to 22A...
Thanks
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Quibbler
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Measure the voltage dropped across a (very) small (in ohms) resistor, or even a piece of resitance wire (nichrome, constantan).
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IrC
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Most good quality DVM's measure AC current in RMS amperes. Measure the current right from the winding before you rectify it with a good DVM and
you should have the RMS amps. No way is a clampmeter going to be accurate when you are worried about small errors in current in the cell, and you did
not say exactly where in the circuit you were taking the measurement with the clampmeter.
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frogfot
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Huh, I never thoat about measuring on the AC line.. but it would be quite impossible because of the construction of the power supply..
I measure current between the rectifyer and the cell.
Why wouldn't clamp meter be exact if I'll use enough capacitance to fully smooth out the current? Their claimed +/- 3% for DC current is
enough precision for me.
The way I wanned to "calibrate" the clampmeter for the unsmoothened current is following.. didn't want to write this first time because
text will be quite long.. ok..
I'll heat up a variable resistor at different currents (with smoothening cap), record the temp rise and plot temp-time curves. Since these will
be liniar to some extent (y=kx+m), I'll plot a k-current graph..
Same will be done one more time without the smoothening cap.
So in the end I'll have two graphs (where k comes from abovementioned y=kx+m):
k-current
k-current(unsmoothened)
So I'll insert my unsmoothened current values from chlorate cell runs into second graph to get the constant k.. and then insert the k into the
first graph to get the correct (effective) current..
I don't see what could be wrong with this..
The question remains, if it's ok to use a transy to limit the current.. maybe I missed something obvious.. I dunno for example how the transy
will handle the unsmoothened current.. will it evolve the heat that will represent the effective current?
Sorry for boring out everybody.. but I just gotta finish doing this stuff.. 
[Edited on 22-7-2005 by frogfot]
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IrC
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"Why wouldn't clamp meter be exact if I'll use enough capacitance to fully smooth out the current? Their claimed +/- 3% for DC current
is enough precision for me."
It would be ok if this is enough accuracy for you but I thought you initially stated you had no filtering at all. Trying to measure pulsating DC is a
bad idea, the best way is to use a storage scope to capture the waveform, and them do an analysis on the area under the curve to get meaningful
results. If I were doing it I would filter the supply and just measure current and voltage to get a handle on what is going on in the cell
electrically. On the AC measurement I mentioned, this was on the output winding before rectification, not on the input side of the transformer as eddy
current losses would make for much error. I was thinking when I first read your post that no filtering at all was not a good idea. Anytime you start
causing accelerations/decelerations to ions in an electrolyte you invariably end up with other things going on chemically that are unlikely to be what
you were looking for. I firmly believe that a well filtered supply is important in any type of electrochemistry, but that's just me.
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frogfot
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| Quote: | | It would be ok if this is enough accuracy for you but I thought you initially stated you had no filtering at all. |
Um, I did, initially.. but now I wanna calibrate my clampmeter to translate the bad values I initially got, to more accurate values... that's
what I've tried to explain the last posts... 
Those oscilloscopes cost about the same as a true RMS multimeter..
| Quote: | | On the AC measurement I mentioned, this was on the output winding before rectification, not on the input side of the transformer as eddy current
losses would make for much error. |
Yp, that's how I got it. The problem with the power supply is the very short wiers that come from the transformer to the rectifyer.
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bio2
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So just cut the damn wires and put in a shunt made from a piece of wire. They are quite accurate if temperature compensated. 477.8mm of AWG 18 Cu is
1mohm=1mv/A @ 20deg etc.
RMS only applies to AC with a sine wave. Any DC clamp on uses the Hall effect. To estimate RMS w/o a scope or proper meter calculate it from the
average. That's all the RMS meter does anyway.
If you want to increase accuracy with an AC current transformer clamp -on type then use multiple wraps around the jaws then divide by the number of
wires.
Comparing a wire shunt to a direct thru the meter current reading with a Fluke 87 is better than 3%. You can also buy precision shunts or use a 1%
resistor if that's good enough.
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frogfot
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| Quote: |
So just cut the damn wires and put in a shunt made from a piece of wire. They are quite accurate if temperature compensated. 477.8mm of AWG 18 Cu is
1mohm=1mv/A @ 20deg etc.
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Sorry for my stubborness, but IMO this would add up several incertainities(eng) giving much bigger error than 3%.. For example all the heat loss that
happens after the current meter.. in the rectifyer (probably at least 20W lost) and in all the wires.
The reason I wanned to estimate the current with measurement of heat evolution (as I explained in second post) is because all the possible errors will
be incorporated in the calculation. Although I'll make lots of measurements, the calculations will be very simple..
I abandoned the idea with using a transistor as the current limiter.. gonna find a simple resistance wire that'll heat up water or oil (where
temp will be sampled)...
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bio2
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....... gonna find a simple resistance wire that'll heat up water or oil (where temp will be sampled)...........
That's one way to do it. 1% or better resistors can be paralleled and would be more accurate than plain wire. If you have a 3.5 digit meter they
can be sorted out. Try the .47ohm 10 watt ones as they are readily available.
If measuring Watts this way (Calorie meter) the uncertanties may be worse trying to achieve perfect thermal insulation using uncalibrated thermometers
etc. This is not as easy as it seems!
To measure DC Watts at the load simply place the current and voltage measuring device at the load using the minimum resistance if shunt or multiple
wraps on the calibrated clamp on.(calibrate against the thru the meter reading with one wrap only) most DVM's measure 10-20A directly.
Calculate the Watts using an average of several samples only after everything is
at operating temp. Line fluctuations are taken into account thru the xfmr turns ratio but good to take some AC input voltages and currents as well for
comparison at different loads.
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frogfot
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Posted while I wasn't logged in and all text dissapeared.. here we go again 
| Quote: |
That's one way to do it. 1% or better resistors can be paralleled and would be more accurate than plain wire. If you have a 3.5 digit meter they
can be sorted out. Try the .47ohm 10 watt ones as they are readily available.
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Actually, I dont have to know the resistance exactly. Only have to know the current and the change in the temp. Decided to use handmade steel wier
coils, cheaper that way.
| Quote: |
If measuring Watts this way (Calorie meter) the uncertanties may be worse trying to achieve perfect thermal insulation using uncalibrated thermometers
etc. This is not as easy as it seems!
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It's also not necessary to know the total amount of evolved heat. I'll only compare the heat evolved with smoothed and unsmoothed current.
| Quote: |
Line fluctuations are taken into account thru the xfmr turns ratio but good to take some AC input voltages and currents as well for comparison at
different loads
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Umm, whats xfmr?
Anyway, I did some measurements as described in my second post.. had two Al rails with 4 slots for different steel wier coils. The coils were immerced
into 300 ml water in a beaker. The beaker was inserted into another to protect from air fluctuations during the day.
After each measurement, the water was changed by new, from a big reservoar with constant temp...
Measuring buttloads of time, I got a bunch of linear temp-time curves (22-50*C) that gave following two curves:
http://img.photobucket.com/albums/v113/frogfot/truermscurve....
So, one should insert the "bad" value (from my multimeter) of unsmoothened current into the "pink curve", get the k value and
insert it into the "deep blue curve" to get the real current.
Well, to anyone who's gonna calibrate their multimeter this way.. this is a pain to do and this consumes lots of time.. Better to use smoothing
caps from the beginning or borrow a true RMS multimeter or an oscilloscope..
[Edited on 29-7-2005 by frogfot]
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bio2
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..........Actually, I dont have to know the resistance exactly. Only have to know the current and the change in the temp. Decided to use handmade
steel wier coils, cheaper that way................
I suppose then you only care about the relative change. The current is derived from the resistance Volts/Resistance=Amps!
For the current range in your chart a battery shunt found at an RV or solar store is ideal. It is a very accurate resistance in milliOhms. Also a car
amp meter can be scavenged for a shunt but sometimes they are remote
BTW xfmr is abbreviation for transformer.
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froot
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Very interesting question.
Personally I would measure from the AC input to the power supply.
First leave the optput of the power supply disconnected. Take a reading of the input
Call this 'ie' for energising current.
Now you can work out output current (before the rectifier) rather easily:
io=(ii-ie)xRatio
io = output current
ii = input current
The accuracy of this depends on many things, all depends on how deep you want to dig. ie..
PSU internal impedance
Linearity of transformer efficiency
Rectifier losses.
Loading the PSU with known values will also help working out linearity and other sneeky losses.
[Edited on 29-7-2005 by froot]
We salute the improvement of the human genome by honoring those who remove themselves from it.
Of necessity, this honor is generally bestowed posthumously. - www.darwinawards.com |
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bio2
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.......Now you can work out output current (before the rectifier) rather easily: ...........
Yes but not very meaningfull unless you have the "ideal" transformer.
In the "real" world it is done exactly as I outlined.
Enjoy your textbooks, I did. 
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