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1111  Forum 2005-2010 (read only) / Interfacing / Re: Reading RPM from wire coiled round spark plug lead on: December 03, 2010, 08:05:10 am
Hi Mike,
Off topic, but as you mentioned using miniature neons, here's a little story.

Around 50 years ago the cheaper models of vessel echo sounder used a rotating neon light to indicate water depth.  One flash at 12o'clock as the zero reference then as the lamp rotated it would be re-struck to indicate depth.

One of the problems was that the neons were fired with a sharp pulse and would eventually blacken on the side of the glass such that they couldn't be seen.  One of the jobs I used to do, as a youngster, for local fishermen was to replace these neons.

Some varieties wouldn't stike successfully but I found that shining a torch light onto them, made them fire OK.  Grey cells into gear and doing some research, I found that any form of photon "exciting" would do the trick - so I ended up painting a little blob of luminous paint onto the back side of the neons.  In those days it was mildly radio-active !!  

These types of neons were also widely used as high voltage satbilisers, their stricking voltage being around 90 volts.  But that was in the days when 120 volt primary cell radio batteries were available.

jack
1112  Forum 2005-2010 (read only) / Interfacing / Re: Reading RPM from wire coiled round spark plug lead on: December 03, 2010, 07:52:16 am
Silly me, I of course meant "reed switch"
jack
1113  Forum 2005-2010 (read only) / Interfacing / Re: Reading RPM from wire coiled round spark plug lead on: December 03, 2010, 05:54:24 am
Many years ago I built a rev counter using a miniature micro switch to drive the measurement circuit.  You could perhaps try one of those strapped to an HT lead or one of the fuel injector feed wires.  The current flow in the lead creates a magnetic flux that might be strong enough to operate the switch.  Then use the reed switch to feed a digin pin.

And to those who say a reed switch won't stand up to that sort of switching abuse, work out how many miles you get at 2500 RPM, one pulse per 2 engine revs, 60MPH at that RPM.  Belive me it's a lot and the switches are usually rated at >10million operations.

jack
1114  Forum 2005-2010 (read only) / Interfacing / Re: Reading RPM from wire coiled round spark plug lead on: December 03, 2010, 04:23:21 am
In that case buy yourself a commercial rev counter

Arduino project work (or fun) is all about experimentation, trial, error and success - not about "tried and tested" from which you learn zilch

Jack
1115  Forum 2005-2010 (read only) / Interfacing / Re: Reading Temp of Liquid in a High pressure pipe on: December 03, 2010, 04:33:24 am
You'll be above the normal range of thermistors at 300C  Suggest you use a type K thermocouple.  However the output is onlyabout 12.2 millivolts at 300C and you will (may) need cold junction compensation to allow for variable ambient temperature.  Both these problems are resolved by using a thermocouple amplifier module which wil boost the reading to something more sensible for the arduino input and also provide compensation for the ambient temperature.

As to inserting it into the fluid stream, unless your pipework is of large diameter (say > 3") and the fluid flow is truly laminar (very unlikely) I'd avoid the use of any form of insertion device.  As Pluggy says, simply strapping it tightly to the outside of the line and providing line insulation for several inches both upstream and downstream of the measured point will give you the fluid temperature.  There will be a slight time lag since any temperature change in the fluid will take time to pass to the outside wall, but the use of an insertion measurement device (such as a thermowell) would also experience this time delay.

jack
1116  Forum 2005-2010 (read only) / Interfacing / Re: voltage divider 80V on: December 01, 2010, 01:37:54 pm
The LM7805 draws 8ma of quiescent current (see manufacturer's data sheet)  This has to be accounted for in your voltage divider circuit.  The net effect is that the regulator is seeing too low an input voltage to permit it to regulate properly.   Suggest you lower the main dropper resistor to 10k and see where you go from there.  

As mike says, you also need a capacitor in both input and output legs.  Suggest 0.1microfarad on output and 0.22 on input (again taken from data sheet)

jack
1117  Forum 2005-2010 (read only) / Interfacing / Re: voltage divider 80V on: October 25, 2010, 03:13:33 pm
OK I confirm zener is ideal for that purpose but 100k feeder resistor might be a little too high for that, suggest 47k
jack
1118  Forum 2005-2010 (read only) / Interfacing / Re: voltage divider 80V on: October 25, 2010, 02:20:17 pm
You say you want to sense the 80volts.  Do you mean actually measure it or simply detect it is actually there.

If the former then a zener circuit as shown will not do since you will always see 5 volts.

If the latter then you do not need a uno.  A simple relay will do.

jack
1119  Forum 2005-2010 (read only) / Interfacing / Re: voltage divider 80V on: October 25, 2010, 07:48:47 am
Your resistor values are totally screwed up.

Start by letting a reasonable current flow down the chain.  let's say 5ma.  Total power dissipated = v x a = 80 x 5 /1000 = 0.4 watts

R = v / i  =  80 x 1000 / 5  =  16k ohms

Divider ratio = V(r1) / V(r2) = 75 / 5   =  15 : 1

This only gives us a rough guide.  What we now need to do is consider uno safety by permitting an allowance for the voltage to rise above 80.  Let us use +50% so now base calculations on the voltage being 120.  The resistor ratios do not change since ratio is ratio irrespective of values.

Calculations still to be based on a chain current of 5ma and because the uno has a high input impedance we shall "choose" to ignore it.

V(r2) = say 7.5 volts (This allows your 80 volt supply to rise to 120volts without damaging the uno)

R2 = V(r2) / I  =  7.5 * 1000 / 5   =  1.5k ohms

R1 = 15 * V(r2) / I =  15 * 1.5k ohms  =  22.5k ohms

Verification calculation :  V(r1) = I * R1  =  5 * 22.5  =  112.5v

V(R1) + V(R2) = 7.5 + 112.5  =  120 volts (which was where we started from)

Power dissipation in R1 = V(R1) * I =  112.5 * 5 / 1000 =  0.562 watts (Note that this figure is higher than the original 0.4 since we are now basing calculation on 120 volts)

Power dissipation in R2 = V(R2) * I = 7.5 * 5 / 1000 = 0.0375 watts

So R1 should be rated at say 1 watt and R2 rated at 0.25 watt

jack
1120  Forum 2005-2010 (read only) / Interfacing / Re: voltage divider 80V on: October 25, 2010, 07:31:14 am
To get the power for your uno you cannot (well you can but it'b totally stupid unless all you had was a bunch of expensive high power resistors) use a voltage divider because of the heat dissipated in the dropper resistor and/or the voltage regulator.  What you need is a switch-mode power supply that can happily accomodate your 80 volt input and will output the required 9 volts.  Because of the way switch mode units operate the input power will be about 120% or less than the output power.  So your 9 volts at up to 500ma equates to 4.5 watts so your switch mode PSU will only pull say 5 watts out of your 80 volt supply, with very little heat generated.

You can of course use a simple dropper system for your measuring circuit because it is delivering practically zero power.  However, for accuracy of measurement, you need to take into consideration the input impedance of the uno since it will form part of the divider chain.

jack
1121  Forum 2005-2010 (read only) / Interfacing / Re: Connect latching relay on: October 24, 2010, 04:00:23 am
More information needed as to what you are trying to do.

Until now you have answered your own question.  Powering one coil triggers relay, which then latches, and powering the other releases it.

jack
1122  Forum 2005-2010 (read only) / Interfacing / Re: how to connect CURRENT TRANSDUCER LEM HTFS 200-P on: November 30, 2010, 04:10:32 pm
I would suggest you are worse off with the ACS sensor for several reasons :
a) They require direct connection to the current flow and hence present a danger if working with high voltage circuits. In effect they defeat your first requirement "to be non-invasive"
b) The specified error is 2%
c) You cannot multi-feed the current to reduce error when measuring low currents
jack
1123  Forum 2005-2010 (read only) / Interfacing / Re: how to connect CURRENT TRANSDUCER LEM HTFS 200-P on: November 30, 2010, 03:09:35 pm
The accuracy is defined by the manufacturer's data sheet as +/- 1% of IPn.  For a 200amp unit this means an accuracy specification (or typical error) of +/- 2 amps.  Contrary to what might be intuitive, measured error is not prorata over the device range.  With 0 amps of current flowing the output could read + or - 2 amps and still meet the manufacturer's specified accuracy !!

So, if you are measuring a maximum of 10 amps the possible error could be +/- 2 amps.  It could read anywhere between 8 amps to 12 amps.  This is a measurement error of +/- 20%.  For obvious reasons the lower your measured current the larger the percentage of error.

By increasing the number of core passes, say 5 times, your 10 amps is measured as 50 amps and the error tolerance is still +/- 2 amps which represents a +/- 4% error.  Not good but very much better than +/- 20%

Therefore the larger the number of turns you can pass through the core, within the limits of the device range, the better the measurement accuracy and the larger the signal change for your measured current.

jack
1124  Forum 2005-2010 (read only) / Interfacing / Re: how to connect CURRENT TRANSDUCER LEM HTFS 200-P on: November 30, 2010, 01:19:10 pm
You can measure on either live or neutral line.  Bear in mind that only one of them should pass through the "core".  If you feed both through, the currents (which are flowing in opposite directions) will cancel each other out.  

Also be aware that if you have bought a 200amp unit and you want to measure only 60 amps you can increase the sensitivity of the system by feeding the wire three times through the "core".   You do this by feeding the wire into the core then once it leaves the sensor's core, loop it back around the sensor and feed it back in etc.  This will triple the output slope since the currents effectively add and the core interprets this as 180 amps.  I used this technique to calibrate a 200 amp unit by winding 100 turns of fine wire through the core and passed 2 amps through it.  100 feeds @ 2amps = 200amps.

jack
1125  Forum 2005-2010 (read only) / Interfacing / Re: increase sensor to arduino sensitivity on: November 26, 2010, 02:51:50 pm
Expanding to focus on a smaller section of the system span does not increase accuracy, on the contrary it reduces accuracy.  This is because system accuracy (% error) is always quoted with respect to full range.  By way of example let us say the system accuracy is quoted as +/- 1.5%.  If you only look at an expanded 20% of system span then the accuracy of this expanded view is still +/- 1.5% of full range but actually +/- 7.5% of the reduced span.
jack

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