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Everything posted by Hero999

  1. You need a permanent 12V power source for the relay. Your description is a bit vague. A microcontroller should be able to do this.
  2. It should work. Be careful with the PCB layout which is important for stability and efficiency. The data sheet contains an example PCB so stick to it.
  3. It should be easy to add a timer to delay the pulse.
  4. JFETs are symmetrical - the source and drain connections are interchangeable. http://en.wikipedia.org/wiki/JFET I was talking about a constant current diode which is just a JFET with the gate connected to the source, often via a resistor. http://docs-europe.electrocomponents.com/webdocs/0027/0900766b80027bdf.pdf
  5. No it won't and can't violate Ohm's law. The only exception is if it had a minimum load current which means you need a load connected, in order to regulate properly, otherwise the output voltage would be higher than specified or be unstable but 350mA should be enough to satisfy that.
  6. No, look up Ohm's law. Hint: the voltage output is constant, the current is not & depends on the load.
  7. It doesn't work like that because as you say, the current needs to flow. You can get constant current diodes which pass a fixed current, regardless of the voltage - it's just a J-FET with the gate connected to the source.
  8. That should work. You may need to connect a small capacitor (say 100nF) from the feedback to 0V for stability.
  9. You could be able to connect two capacitors to the output of a single supply, each via a diode so each will discharge separately but you need to make sure the power supply is powerful enough to drive two coils.
  10. Can't you charge each capacitor via a separate diode? Please post a schematic.
  11. Each has its own advantages. Bifilar is good because it ensures each winding has exactly the same characteristics but there will be a greater capacitance between each winding. Centre tapped is easier to wind but the winding won't be quite symmetrical. If you need to charge two capacitors, why not simply connect them in parallel?
  12. You connected it up backwards so that was going to happen. It's a negative regulator, thus its input voltage should be negative but you gave it postive 12V and fried it.
  13. Comparator, a few resistors and a relay .
  14. I had to Google is because I've never heard of it before. Are you talking about a card skimmer to steal credit card information? No it's illegal, we don't help you with that.
  15. The thread is 6 months old so why should you expect a response? Please stop resurrecting old threads.
  16. What's the maximum current output from the PLC output? How about simply changing the fan for a 24V version? It seems silly to have another 12V power supply for the motor. A solid state relay may work but you need one with a fast turn on/off time <0.25ms.
  17. Some of those transistors are NPN and the others PNP so are not interchangeable. You need to makse sure the transistor can handle the current surge when the motor starts. The base resistor depends on the control voltage i.e. what's connected to the base. To ensure good saturation, the base current should be 1/10 the collector current but you'll probably be fine with 1/20.
  18. I think you mean reverse voltage. The reverse current is typically [email protected] and the maximum rated voltage. The higher voltage parts (>=600V) have a lower junction capacitance, 8pF as opposed to 15pf which makes no difference in most applications. unless you're using it as a varactor in a voltage controlled filter or voltage controlled oscillator.
  19. Yes you'll burn a starter motor out by running it continuously.
  20. The voltage are what you should expect - nothing wrong there.
  21. No never remove the fuse from the meter. The meter's insulation is only 1kVDC, if you try to measure 100kV, an arc will jump through the case which will both damage it and you'll get a nasty shock.
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