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Cabwood's Achievements


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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 cut off current flow becomes a smaller air gap than anywhere else in the circuit, and the coil's energy will be dumped into that gap. In the case of a transistor switching off, as the transistor's conductance drops, the coil responds by increasing its voltage in an attempt to maintain current flow - several hundred volts is not uncommon. Being a lower impedance than any local air gap, that poor transistor will suffer the energy dump instead of surrounding air. Few transistors will tolerate this abuse even once. As Audioguru states, the coil's resistance has little influence. The rule is simple: whatever current the coil was conducting when switched on, it will continue to conduct immediately after it is switched off, and any conduction path, however resistive, will do.
  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/resistor pair. Connect it only across the galvanometer's coil connections.
  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 the emitter for feedback. From this point of view, its behaviour can be described starting with a common emitter setup, introducing a resistance in the emitter to diminish and linearise the response to base voltage fluctuations. I agree with Zeppelin in this sense, that classical transistor theory teaches this, and it is not wrong. I teach this too. But here (on this forum) I enjoy some license for deviation from the classical approach. Since the subject is inherent negative feedback in transistor amplifiers, I found it helpful to treat circuit B as a common collector system (an emitter follower), thus: Given that for circuit B to work, the collector voltage must always be greater than the emitter, and the emitter voltage will still follow the base. Thus this configuration is still an emitter follower at heart. The introduction of a collector resistance capitalises on this behaviour. It merely develops the emitter current (resulting from the buffered base voltage appearing at the emitter) into a voltage which is proportionally larger (in most cases) than the emitter voltage. Which approach is best? I think flexibity is more useful than any rigid textbook approach, but as Zeppelin notes, sometimes flexibility comes at the cost of technical correctness. He is right. Circuit B is a common emitter configuation. I would ask this though - take a phase spilitter, where RE and RC are identical, and outputs are derived from both emitter and collector. Is this a common emitter or common collector circuit?
  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 operates with a fairly constant voltage between base an emitter - about 0.7V. If the base input voltage rises, then the voltage difference between base and emitter increases. The transistor becomes more conductive, increasing current flow from the collector down through the emitter and resistor. More current through the resistor means a greater voltage across it, meaning the emmitter voltage also rises. When the base voltage falls, the base-emitter voltage difference is reduced, causing a drop in the transistor's conduction. So emitter current drops, and therefore also the voltage across the resistor. In this way, the emitter voltage (output) rises and falls with the base voltage (input) in order to maintain a constant 0.7V between base and emitter - hence the name 'emitter follower'. Even though we have placed no explicit feedback loop in this circuit, the very nature of the transistor and it's configuration in this circuit exhibits negative feedback, to ensure that the enmitter voltage follows the base. This is called "inherent negative feedback". If we modify the emitter follower slightly by adding a resistor at the collector (circuit B), we can reduce the amount of feedback inherent in the circuit to something less than 100%. The voltage gain is thus controlable (-R1/R2), but still there is no visible negative feedback path! The feedback present in this circuit is still inherent, and due to the emitter resistor. In the absence of an emitter resistor, there is no inherent negative feedback present, and so we must provide it explicitly (circuit C). Resistor R1 does this. The voltage gain is then controlable (-R1/R2), and the feedback greatly linearises the amplifier's response. Sometimes we do not need any feedback at all, especially if we are switching something on or off. In switching applications we are not concerned with linearity, and we want as much gain as possible, and so we omit the emitter resistor, and provide no explicit feedback path. That's what's happening in circuit D. Operational amplifiers (except in switching applications) are always connected with external components to provide feedback that determines the response of the system. Transistors, though, can be connected in such a way that negative feedback occurs even though no feedback path is provided explicitly.
  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 - otherwise by the time the second stage is ready to fire, the caps will already have lost a lot of charge.
  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 doesn't work because he's connected them all together. That is sure to break the preamp because of the fake ground. My design is to address the single supply issue, remove any ambiguity regarding ground points, and provide 1M input impedance, all with a single op-amp. Perfection, as you state, would require another op-amp to buffer the output from the volume pot.
  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) Input impedance 1M, 2) Ground should be common to all sections. In my schematic I've shifted the volume control to after the op-amp, and I've decoupled the 4.5V mid-point with a capacitor in the feedback loop. This frees me to offset the input by +4.5V (so the amp can operate from a single supply) and create the 1M input impedance with only two resistors. Put this preamp as the first stage, then the equaliser.
  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 through the capacitor will be 90 degrees out of phase with the resulting sinusoidal voltage developed across it. This phase shift means that a simple value of resistance is not enough to completely describe the behaviour of a capacitor. Instead we use a complex number, which is really a vector encapsulating both 'phase angle' and 'magnitude' information. This complex number is called impedance. It is analagous to resistance, in that it describes the relationship between current through and voltage across some device, but it includes phase information as well as magnitude. You can think of resistance R as being a complex number (R + 0j) with an angle of zero and magnitude of R. In other words, a resistance is an impedance with no imaginary part. Thus a resistance is an impedance through which current remains in phase with and in direct proportion to voltage across it. Ohm's law can be applied using impedances instead of resistance. If impedance is represented by Z, then V = IZ. Just like resistances, impedances connected in series have a combined impedance of Z = Z1 + Z2 + ... + Zn Parallel impedances are treated as you would treat parallel resistances: 1 / Z = 1 / Z1 + 1 / Z2 + ... + 1 / Zn One more complication: the impedance of a reactive component (like a capacitor) varies with frequency. For a given frequency f (Hz), the impedance of a capacitance C (Farads) is Zc = -j / (2 Pi f C). Note that this a purely imaginary value! That's describing the fact that the current in a capacitor is always 90 degrees out of phase with the voltage across it. Better brush up on your complex arithmetic. It will be worth it in the end.
  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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