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  1. BobK

    PWM channel to control LED's

    You do not need a low pass filter or an opamp to do what you want. All you need is a transistor switch which would increase the current available while still using the PWM controller. A mosfet would be best. What are the specs of the LED's? Given that I could tell you how to wire it. Bob
  2. BobK

    LED array getting hot

    With a 181 (unlikely, did you mean 180) ohm resistor it is not likely they would get hot. Are you sure it is not a n 18 Ohm? Also what color are the LEDs? Red LEDs have a voltage drop of about 2V, while blue or white have a voltage drop of about 3.6. I would guess they are the latter if...
  3. BobK

    How to boost signal voltage level to make it audible

    How did you do that, if not with an amplifier? It would be audible but not very loud is my guess. I calcuate that it is 50mW. Bob
  4. BobK

    clamping circuits.......mechansim

    If the diode and capacitor are ideal, the capacitor would come to the same voltage of the AC input instantly (drawing infinite current), so, since it spent no time doing this, the load would always be drawing current. With a real diode and capacitor, the diode will drop a certain voltage...
  5. BobK

    How to boost signal voltage level to make it audible

    It's called an audio amplifier. LM386 is the classic IC audio amplifier. Bob
  6. BobK

    how to charge a 12V, 44Ah car battery with trickle charger ?

    It would take 44 / 1.5 = 29 hours to fully charge an empty battery if it were 100% efficient. I believe charging is somewhere around 60% efficient, so, about 29 / .6 = 48 hours. 2 hoiurs ain't gonna do it. Good luck. Bob
  7. BobK

    Electronics & computer Science Question

    If you are looking for the equivalent of an app store, I don't think it exists for hardware. Bob
  8. BobK

    RF Rain Detector Circuit Help

    The radio modules are not what you want. They would require a microcontroller on each side to operate and are serious overkill for what you are doing. I am not sure where to look for a simple on / off radio transmitter and receiver that is what you need. Bob
  9. BobK

    Oscilloscope advice

    In my 50+ years as an electronics hobbyist I have never used a curve tracer. Why do you think you need one? Bob Edited to add: But an oscilliscope is essential!
  10. BobK

    electronic pollution ?

    No CRT compares favorably with even the worst LCD you could buy today. You are fooling yourself. Bob
  11. BobK

    4066 and audio

    An audio signal is typcially going to go above and below ground. So the positive part of the signal may be < 5V, but the negative part is below 0, putting it out of range. The biasing brings the signal up to the midpoint (2.5V) when the signal is 0, so it allows the signal to go either...
  12. BobK

    Using a LED chaser circuit 110V, no transformer.

    The CD4017B is not capable of driving 1 LED, let alone 8. Why would you want to do this transformerless anyway? It is much safer (both for you and your components) to use a low voltage DC power supply. And transformerless supplies are usually used when only a small amout of current is...
  13. BobK

    Parallel Resistor Calculator Thing...

    Nice job, thanks! Bob
  14. BobK

    HD TV screen broken

    Even if you could, the part would most likely cost more than a new TV. Bob
  15. BobK

    Question About Transistor

    A silicon controlled rectifier is not a transistor, it is a different sort of device. Google it. Bob
  16. BobK

    Good SOund

    Fleetwood Mac before or after they became a Pop band? Bob
  17. BobK

    best volume control

    If V1 and V2 are ganged, i.e. a double POT that is turned together, then I think you already have what you are asking for. The V1/V2 control will control the overall volume and the bass control will effect only the bass. Does not not perform that way? Bob
  18. BobK

    help with Joule thief please

    So what did you change that finally made it work? Bob
  19. BobK

    resistor question

    First, find the current through the resistor: V = I * R (12 - 0.4) = I * 1000 11.6 / 1000 = I So the current through the resistor is 11.6 mA or 0.0116 A Now the power dissapated in a resistor is: P = I * I * R P = 0.0116 * 0.0116 * 1000 P = 0.13456 The power is less...
  20. BobK

    Opto-isolator basic question

    Thanks, I didn't know that. File away for future use! Bob
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