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1156  Forum 2005-2010 (read only) / Interfacing / Re: Non-linear resistance in a digital potentiometer on: January 07, 2011, 03:29:15 pm
The 2.5k to ground is your problem.  It loads the potentiometer (think of 2 resistors in parallel) such that non-linear output occurs.   When you have the pot set at 100% (10k) the output will be 5 volts since the load resistor is across the whole pot and has no effect.  When the pot is set at 50% (5k) the 2.5k in parallel creates a parallel pair which  measures 1.66k.  This is in series with the remaining 5k and the series voltage divider is now 5k +1.66k so the output will be 1.25volts rather than the 2.5volts expected.   The problem occurs at all settings of the pot, other than 0% and 100% and is extreme at higher percentages but has less effect the nearer you get to 0%

To minimise the effect the load resistor should be at least 10 times the value of the pot, say 100k

jack
1157  Forum 2005-2010 (read only) / Interfacing / Re: Non-linear resistance in a digital potentiometer on: January 07, 2011, 04:32:28 am
It sounds like you are "loading" the device with some other component or your coding is non-linear.  Can you provide a diagram of exactly how you have the device connected and a copy of your coding - no-one can critique either your circuit or code unless you supply a copy of them.

jack
1158  Forum 2005-2010 (read only) / Interfacing / Re: scaling and offset voltages. on: January 06, 2011, 06:33:45 am
MK3

Oh dear, you win on that one.  I was so busy looking to get a +5 volt signal I overlooked not being able to get down to zero volts with a single supply line.  However all is not yet lost.

Three choices are open :

a) Using a single power supply (>9 volts) offset the 0 amps output voltage to give an output of say +1.0 volts and the +80 amps to give an output of +5 volts.  This gives a linear range of 4 volts for 80 amps, or 0.05 volts/amp  -  an easy sum to work with and only requiring a simple circuit and one power supply.

b) Use two power supplies to give a +9 and a -9 volt power rail with the centre grounded such that the opamp can give a true 0 to 5 volts output.   This gives a linear output of 0.625volts/amp.  Not so clean maths and requires two sources of power.

c)  Use a single 18 volt power supply with a 9 volt regulator chip to provide a centre voltage (an artificial ground point) of +9 volts.  The +18 now becomes +9 and the true 0 now becomes -9.  All a bit messy but gives the same results as (b) above.

Personally, I'd go for option (a)

jack  
1159  Forum 2005-2010 (read only) / Interfacing / Re: scaling and offset voltages. on: January 05, 2011, 12:50:11 pm
MK3

That's exactly why I suggest powering the opamp from a 9 or 12 volt supply and why I suggested applying half of the 5 volt line (2.5 volts) to act as an offset

By adjusting the offset trimmer the 0volts output can be set and by adjusting the feedback trimmer the output signal (5 volts at 80 amps) can be set.

jack
1160  Forum 2005-2010 (read only) / Interfacing / Re: scaling and offset voltages. on: January 05, 2011, 06:00:57 am
Ok then, what you need is a single supply inverting opamp to run off say 9 to 12 volts.  You need a bias of 2.5 volts onto the +ve input and an overall gain of 2.0.  (If you want a 5 volt output you need a supply somewhat higher than the 5 volts )

Connect the 9 (12)  volt negative terminal to the arduino ground point.

Place a 25k multiturn trimmer between the arduino 5 volt supply (a stabilised 5 volt reference point) and ground.  With the trimmer set at around 50% (2.5 volts) feed this into the +ve terminal of the opamp.

To the output terminal of the opamp connect a 50k multiturn trimmer  and feed this back into the -ve input terminal.   Between the hall effect device output and the opamp -ve terminal connect a 12k resistor.  With the 50K trimmer set at around 24k you will get the 2.0 gain required.

I suggest mutltiturn trimmers so's you can get the calibration set correctly.

To calibrate your hall sensor + opamp setup you will need a variable power supply that can provide a couple of amps output and a suitable high wattage load resistor.

Wind 40 turns of small gauge insulated wire through the sensor core.  Now, if you pass 2 amps through the wire, the sensor will read this as 80 amps (2x40).  Use something like a 12 volt 40watt car headlight bulb as a suitable load resistor since at 2amps on the variable supply you will be dumping something in the order of 20+ watts into the load resistor.

Suggest you google for a suitable single supply opamp.

Happy playing

jack
1161  Forum 2005-2010 (read only) / Interfacing / Re: scaling and offset voltages. on: January 04, 2011, 05:42:47 pm
Firstly, do not use diodes.  These are non linear devices and cannot be used to accurately drop voltage.

Why do you need to scale via an opamp, other than adding complication to what could be an extremely simple circuit.  

With 1000:1 resolution (10bit) over the 5 volt span, an effective range of 2.5 to 5 volts will give you a 500:1 resolution of your 0-80 amps.  Isn't this not good enough for your purpose.   After all, if it was an analogue meter, could you really read 1/500 of scale.  

Added to the argument is the question of what will you use to calibrate your 1000:1 circuit (0.1%).   Most of us work with cheap DVMs which at best are +-1%.

You set your software to set 2.5 volts  as 0 amps  and 5 volts as +80 amps, with an effective resolution of around 160ma.  No calibration required !

jack

1162  Forum 2005-2010 (read only) / Interfacing / Re: Waterproof ultrasonics on: January 07, 2011, 01:19:25 pm
Who's thread is this ?  It seems to have been high-jacked
1163  Forum 2005-2010 (read only) / Interfacing / Re: Waterproof ultrasonics on: January 07, 2011, 12:57:34 pm
Transducers are usually fairly narrow beam so I doubt anything with more than 20 degrees of separation between a pair of transducers would give a reliable range and direction detection and even if you got a double echo the reliability of the calculation of direction might be questionable.

An option might be to build a system for rotating the transducer head assembly, rather like a radar scanner.  This would then rotate the transmission and echo beam through 360 degrees so you could "look" and "see" in all directions with only a single head.   By feeding your circuit with information of rotational position you get both distance and direction of any obstructions.

All the transmission and receiver electronics could be housed within the rotational member with only the necessary power supply running through slip rings.  The "coded" data generated by your electronics could be transmitted wirelessly via a low power RF system into the main control system so avoiding the necessity to transmit data via a slip-ring assembly.

jack
1164  Forum 2005-2010 (read only) / Interfacing / Re: Waterproof ultrasonics on: January 03, 2011, 10:03:16 am
Alternatively you could try bonding a piezo sounder module to a piece of glass or acrylic and seal the back surface with silicon or similar type agent

jack
1165  Forum 2005-2010 (read only) / Interfacing / Re: Waterproof ultrasonics on: January 03, 2011, 10:00:02 am
Try a marine supplier or yacht chandler, they should provide exactly what you are looking for - ie ultrasonic transducers that are guaranteed totally waterproof (they are used in echo location depth sounders).   I am unsure of the frequency, I think it is around 22kHz that they use which is not the "normal" 40kHz that air ultrasonic devices operate on; the lower frequency being required to achieve adequate transmission through water.

jack
1166  Forum 2005-2010 (read only) / Interfacing / Re: Outputting 12v and -12v capable of 5A on: January 06, 2011, 01:04:14 pm
Are you "simply" trying to superimpose an RS232 signal onto a motor control circuit - if so you might not really need to meet the power switching demands you are asking for; or do you really need to meet the power switching rates you specified

If you only want to superimpose a signal then this can be relatively easily accomplished using techniques that are typically used to transmit digital control signals over the AC mains power system

jack
1167  Forum 2005-2010 (read only) / Interfacing / Re: Building a stamping press - need urgent help on: January 04, 2011, 08:23:24 am
Without wishing to be guilty of submitting to pointless safety issues, have you considered how you will prevent someone accidentally putting their hand or fingers under the power driven press when they operate the foot pedal.   Industrial presses have many in-built safety features such as limit switches, photo or proximity sensors, including high integrity processors, to prevent such types of accidents.  They also generally require both hands to be activating separate switches to permit the foot pedal to operate.  (If the hands are busy doing something they cannot be under the press tool)  You might also need to consider how you prevent the arduino from accidentally activating the press without the foot pedal being operated (such can occur due to electrical "noise")

jack
1168  Forum 2005-2010 (read only) / Interfacing / Re: real dumb question on: December 30, 2010, 06:29:59 am
If you have nothing attached to pin 7 then it is "floating" and susceptible to ANY input signal, from whatever source.  Your body develops a static charge simply because parts of you are moving about (rubbing insulators together).  The arduino circuit is also "floating" since you presumably have not grounded it to any metallic safety earth.  In effect there are potential differences between your body and the arduino circuit.  When you touch the input terminal (7) your body voltage excites the arduino input so causing the LED to light, but touching also effectively discharges you with respect to the arduino so the LED goes out.  When you next let go, you recharge yourself and the cycle repeats.
1169  Forum 2005-2010 (read only) / Interfacing / Re: real dumb question on: December 29, 2010, 07:12:37 am
If you are shortening +5, pin 7 and ground together you are effectively killing the supply to the board.  Shortening +5 to ground forces the +5 to 0volts which may amage your onboard regulator.  (Probably won't damage it due to its inbuilt protection but is still a highly unreccomended action).

The resistor from 7 to ground is to stop the point "floating about" and biases towards ground (Lo).  When you close the switch to the +5 line you force pin 7 to go "Hi" (as in high, not hello)

There will be a flow of current via the resistor from +5 to ground but this is only 5millamps which is perfectly acceptable.

No questions are dumb - unless you ask the same one more than two times.  First one because you don't know and want to learn, second one because you forgot you'd already asked it, and third because , by then, you have proven that you can't be bothered.

jack
1170  Forum 2005-2010 (read only) / Interfacing / Re: Transistor behaviour on: December 30, 2010, 12:43:30 pm
I am having difficulty comprehending why you wish to consider "transistor theory" with one of the wires disconnected.  Such a situation is irrelevant to the operation of a transistor (unless you are considering IR detection, UV detection or temperature measurement)

The circuit you show, with all wires connected is a classic emitter follower circuit, where the output (emitter voltage) follows the base input voltage and control of the LED is determined by the current flowing into the base.  The collector should be connected to a permanent +5 supply.

If you are switching both emitter and collector (separately or together) they are in fact acting as inputs to an AND gate (the transistor) since both need energising to get an output and considering you are using arduino outputs to act as these inputs it's all very much a waste of technology.

jack
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