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  1. An inductor, such as a relay coil, opposes changes in current. That means if an inductor is conducting one amp, and the source of that current is suddenly removed (like a switch opening or transistor switching off), the inductor will still conduct one amp. The question is " through what?" If there's nothing for the coil to conduct its current through except air, then air will do just fine. In order to pass current through air, the coil will develop hundreds or thousands of volts so that the air ionises and provides a conductive path, causing a spark. Usually a mechanical switch opening to cu
  2. One application of interpolation in engineering is as a challenging subject for a academic research assignment.
  3. What would make this REALLY easy - can you use a 24V supply instead of 12V? Or, really, any supply higher than 12V?
  4. If you can measure the resistance of the coil for me, I'll figure something out. This is no ordinary gauge, methinks, it's an interesting one.
  5. Daisy, I believe you are connecting the capacitor across the whole gauge (resistor and galvanometer together), effectively putting it in parallel with the power supply, which is why it doesn't slow the rise to 12, but does slow the drop to zero. What Audioguru correctly suggested was to connect the capacitor (100uF should make a significant difference), across the galvanometer itself. Only the galvanometer part. Inside the gauge's case there should be a resistor in series with the galvanometer's coil. Do not connect the capacitor across this resistor. Do not connect it across the galvanometer
  6. You need a resistor in series with the LED. 220 Ohms will do. Without this modification, your circuit will kill the LED when the switch is closed. With the modification, your circuit will light the LED when the switch is closed. Positive signal? Controls a negative signal? What does that mean?
  7. Zeppelin pointed out to me in a nice and discrete PM: That's a good point, and he asked me to clarify it. In my explanation I suggested that circuit B is a small modification to circuit A, insinuating that B is also a common collector setup, with a slight change. But, as Zeppelin says, the modification isn't really that trivial. It's quite a leap from A to B, with the two circuits operating in fundamentally different manners. Given that the output is derived from the collector (in contrast to circuit A), circuit B is technically a common emitter configuration with a resistance present in
  8. The negative feedback in many transistor amplifier circuit is not always apparent. Take circuit A, for instance. It's an emitter follower configuration, which has a voltage gain of +1. There is 100% negative feedback present in this amplifier, as I will explain. The transistor will conduct from collector to emitter when the base voltage is 0.7V or so above the emitter voltage. Increasing this base-emitter difference even slightly will cause the transistor to conduct heavily, and decreasing it only a tiny amount will cause the transistor to completely block current. Therefore, this transistor
  9. You'll need to wire those capacitors in parallel. Two 1000uF caps in parallel have a combined capacitance of 2000uF. A 1000uF cap's voltage will drop 3V (from 12V to 9V, or 15 to 12, for example) in 1 millisecond, if it is supplying 3A. Is this long enough to ignite the rocket? A 2000uF cap (or two 1000uF caps in parallel) supplying 3A will drop by 3V after 2 milliseconds (twice as long), and so on. Use 'low ESR' caps to maximise the current available in these short bursts. Capacitors discharge on their own, even with no load, so you'll need to charge them immediately prior to launch - oth
  10. Audioguru, I considered the pot's effect, and decided (subjectively) that the equaliser wouldn't suffer much from the 5k output impedance of the volume control. You are right though, there would be attenuation, and shift of band freqencies. I haven't figured out how much. I made an assumption - that the supply was a single 9V battery. The original design would be OK if supplied from dual rails. I assume that there is only a single supply because of the 0V centre point being derived from 2 resistors. The ground symbol is visible in all three circuits, and my guess is that Jesus's circuit does
  11. In the preamp section, 4.5V is derived from the potential divider of two 4k7 resistors. On the diagram, this is connected to ground. You must not connect that 4.5V point to the ground connectors of the other sections, because for those other sections ground is the negative supply rail. In other words, the ground symbol in your preamp schematic shouldn't be there. The preamp in this design has an ouput offset of 4.5V (assuming a supply of 9V). That will play merry hell with the 386, so you must couple with a capacitor from the preamp to the next stage. The specs for your preamp should be: 1)
  12. For a resistor the relationship between voltage across it and the current through it is very simple - the current is always proportional to the voltage (V = IR). We say that an alternating current through a resistance will develop an alternating voltage across it which is in phase with the current, or that there is zero phase difference. For a capacitor or inductor though, the relationship is more 'complex' (see what I did there?). For example, the current through a capacitor is proportional to the rate of change of the voltage across it. This means that a sinusoidal alternating current throu
  13. Not only. I am talking principally about phase shift from input voltage to output voltage.
  14. I don't where I went with that. Apologies for the rambling. It sure helped me cement a few ideas though.
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