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Everything posted by EdwardM

  1. Hi Dan doesn't sound too difficult, essentially it's just an alarm clock (4, 7 segment LED displays or LCD display plus some means of setting time and an alarm time). Thereafter it's just a dimmer circuit, fired from fully off to fully on for a programmable time (another 2 switches). I'm sure there are many on this forum who could design one, maybe even me, if and when I get the time. Nor should it be horribly expensive! (Take a look at the 'giveaway' clocks available and imagine the cost of adding dimming....zilch + zilch = Best Ed
  2. Hi don't know if it will help but I'd consider using 3 of the parallel port lines to clock 3, 4-bit serial/parallel converters, each of which would be attached to a motor driver chip. This will leave you a minimum of 5 parallel port lines available for your obstacle circuit (which could also be serialised in the other direction). The software requirements shouldn't be too onerous as the stepper motors require on/off times to be in the millisecond range whilst the parallel port is capable of considerably higher speeds. Ed Actually, you only need one parallel port line for serial data if
  3. Hi u from what I understand, a magnetic sensor actually senses current, therefore if your closed loop circuit is driven by a voltage then it should be possible to detect that current using a combination of X, Y and Z coils to determine an exact position, it depends also on how far away you expect to detect the wire, is it a few cm? or 30-40m? Hope that helps Ed You could also try a Google search on magnetic sensors
  4. Hi Imcondor welcome. As far as I remember the LM353 has been superseded by the lower power LF353, which is pin for pin compatible. Best of Luck Ed
  5. Hi Staigen don't think so... I shifted the input gradually from 2.5v to 5v and got a gradual change of output from 0v to 5v which was the original requirement, this led me to believe that my original equation might be correct. Do let us know why you think that it's not linear. As you rightly say, "when the input 2.5v the output saturate at 0v" It is a rail to rail opamp with a single supply of 5v so that's what I'd expect to happen. And, "When the input is 5volt, the output saturate at 5 volt" ... Yes, also. "And the scale between is not linear", please explain - there are no reactive co
  6. Hi wk I now realize I was being a bit lazy about the circuit I described, I used an online opamp calculator to provide the starting values but they aren't quite correct. What I have done is obtained some AD8601 rail-rail single supply opamps and did a quick build using a 500K pot as the feedback element and a 50K pot to derive the reference. The reference was set to about 4.75v and the feedback set (by eye) at about 1:1, and lo and behold it did work. Put in 2.5v and you get 0v out, put in 5v and you get 5v out. Unfortunately, I don't have the time to produce a graph to prove that the t
  7. Hi Chaps R1 and R2 provide an offset voltage of about +2.5v (your minimum input), this means you need something more positive than that at the non-inverting input to make the output voltage rise above 0v. The value of all the resistors is almost immaterial provided that they maintain the same ratio. With a bit of good luck and a following wind, it ought to work. Best of Luck Ed PS If you are interested in the theory behind this, I'd suggest you read chapter 4 of OP AMPS for EVERYONE it's freely available on the web.
  8. Hi Staigen I know what you mean, this site is much more personal than any of the others... Seems to be often on monday evenings UK time that I can't reach E-lab, maybe Mixos doesn't get out of jail until tuesdays? ;D Ed
  9. Hi anybody here, + or - 0.5% usually means that range of inaccuracy of full scale, there is also a residual potential error of + or - 1 digit. Is that any clearer? Ed
  10. Hi wk with this sort of problem you have to start with the transfer function. In this case you are transferring an input of between 2.5 and 5v into an output of 0 to 5v and it should be linear, ie Vout = m*Vin+C. (The equation for a straight line). So m will equal the slope of the line (5-0)/(5-2.5) = 2 and C from the same formula = -5 after plugging-in the value of m. Vin Vo=2*Vin -5 2.5 0 3.0 1 3.5 2 4.0 3 4.5 4 5.0 5 You then need an Opamp circuit to achieve this transfer f
  11. EdwardM


    Hi mettula Test for open circuit (which a capacitor should be) using high Ohms range, there may be an initial 'kick' of the meter display but it should become infinity. Test for short circuit, anything lower than infinity Ohms indicates leakage and the worst case being a short circuit (0 Ohms). Test for good: One way is to connect a battery or power supply to the capacitor via a high value resistor and at the same time measure the voltage across the capacitor. From a knowledge of the supply voltage and resistance of the resistor you can make a reasonably accurate guess at the capacitors va
  12. Hi Gogo you could do worse than start where I did, after reading lots of contradictory material I bought a book called Easy Step'n, it's not high tech but down to earth practical advice on the care and feeding of stepper motors and how to get started with them and I highly recommend it. Best of Luck Ed http://www.melabs.com/products/books/stbook.htm
  13. Hi Adam, the short answer is a qualified yes. 1000kSPS = 1MSPS = 1 million samples per second thus 1 sample per microsecond. Have a look at some ADC datasheets (Maxim, Analog Devices and others) to get some more detail. Best of Luck Ed
  14. One of the best resources on the net for the HD44780 is http://ouwehand.net/~peter/lcd/lcd.shtml Best of Luck Ed
  15. EdwardM


    Hi All you could start your websearch at Analog Devices http://www.analog.com/en/cat/0,2878,760,00.html Best of Luck Ed
  16. Hi gogo I believe so, I think it is to prevent moisture interfering with components and the PC board and altering their electrical characteristics. Ed
  17. Thanks MP the conformal coating is a good idea and should enable the transition from prototype to 'production'. The reason for posting this idea was mainly to pass on the results of a lot of frustration when designs didn't work, whether mine or someone elses'. Many is the hour I've spent debugging a circuit only to find that I'd inadvertently joined A to B instead of B to C and if you are not sure whether the design or wiring is correct then inevitably getting the thing working seems to take forever. So, if you can take one variable out of the equation - Life Gets Easier Cheers Ed
  18. Hi For some years now I've used a simple way of protoyping on 0.1in perforated boards which usually guarantees first time working. I simply make an outline of each IC and pin name used as well as power supply terminals in a drawing program and then print them on paper. Each outline is then stuck down on the perforated board using paper glue. It's then very easy to follow your circuit diagram to join up each node of the circuit and it's a very useful aid in debugging. My favourite program is Visio, I'm sure there are many others, where you set up a 0.1in printing grid for each drawing and
  19. Hi Ritmo2k I've tried to answer what I think is your question with a diagram... +5v means a positive supply relative to some common terminal and likewise -5v means a negative supply relative to some common terminal. Hope that helps Ed
  20. EdwardM


    Hi shiks Whilst I have no 'hands-on' experience of jammers, from what I've read, this is a big subject... If you are talking of radio reception jamming, this means making the reception of radio signals unintelligible or impossible in a particular region. At least two and many other ways to do this, 1. Is to use the original transmission and retransmit with a time delay or several time delays with antennas beamed to the place you want to cover, making the reception tedious in the extreme. 2. Is to use the transmission frequency and beam 'white noise' at much higher power than the orig
  21. ....and another thing The 50K resistor is most cheaply produced by connecting two 100K resistors in parallel ;D Ed
  22. Hi cynicmonster, First off, interesting handle. ;D Makes me assume you've not had one before - You have to ask, what do I need it for? If your interests are in (relatively) low frequency analogue/audio then almost anything modern will do. If you do a lot of digital work, even at low frequency, then think of looking at a scope with 5 to 10 times better bandwidth than the highest frequency you normally work with. Also, if mainly digital, then storage facilities can be extremely useful. Unfortunately, at this stage, the options start to mushroom together with the price. Myself, I work
  23. Just because you're paranoid, doesn't mean the whole world ISN'T out to get you Ed
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