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| | #1 |
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Hi, The attached is the circuit which i've been asking for many questions about it. The red circled part, previously there is a 2.7nF capacitor parallel with it, to pick up the square wave electromagnetic field at around 9.6kHz. If the capacitor is removed, what will i get? Can the inductor pick up the electromagnetic field from any wire which constant DC flow insite the wire? How about AC of squarewave? Another question is, is there any way to increase or reduce the value of inductor? Like capacitor and resistor, we series or parallel them to get the value we want. What happen if i series a few inductor together, and parallel them together? Thanks | |
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If you series inductors, you increase the inductance. If you parallel you decrease them (opposite of capacitors, same as resistors). (I think): You can only change inductance by using inductors (hence the name). You can't change it by using resistors and capacitors to get the same effect (well you can, but it doesn't look like it in this circuit to me). You can use a few inductors of different values to get an electrically equivelant inductor. It looks like removing the capacitr in parallel with the inductor will change one of the time constants of the circuit (which represents the period of of the wave that will be picked up. I think the time constant is square_root(L*C). If C is zero like you have made it by removing the capacitor, it will only pick up DC? I am not quite sure about anything I said in this paragraph with regards to the circuit's function without analyzing it with equations. Last edited by dknguyen; 15th June 2006 at 03:27 AM. | |
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| | #3 |
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In your revised circuit, the inductor by itself isnt tuned, so will pickup magnetic fields at many frequencies. It will have a big problem with mains hum. The coil is a very high impedance at high frequencies and the input of the opamp is also a very high impedance, so the wire connecting the coil to the opamp is an antenna for picking up interference.
__________________ Uncle $crooge | |
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| | #4 |
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What is main hum?
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| | #5 | |
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__________________ I thought what I'd do was I'd pretend I was one of those deaf-mutes. | ||
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| | #6 |
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The supply current frequency to the house?? If i want to detect DC current, what should i do? Can this circuit to be modified to get what i want? | |
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| | #7 | ||
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Quote:
__________________ Gods own Country Incredible !ndia www.flickr.com/photos/_akg/ "Give a man a fish, and he will eat for a day. Teach that man to fish, and he will eat for a lifetime." | |||
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| | #8 | |
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Previously, i use that circuit (with capacitor) to detect the output of the 555 timer.. Everything works well.. Now (just for extra knowledge), i want to know whether this circuit can be modified to detect DC current or not but not squarewave or AC current. | ||
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| | #9 |
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The loop of wire around your room is the primary winding of a transformer. The coil in your receiver is the secondary winding of the transformer. Transformers work only on AC, not DC.
__________________ Uncle $crooge | |
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| | #10 | |
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So, is there any way to detect the DC? Or it is impossible? | ||
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| | #11 | |
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| | #12 |
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I have never seen an inductive-pickup-based device used for measuring DC current. Inductive pickup responds only to changes in the magnetic flux around a wire, which a steady DC current doesn't have. To make it work you would presumably have to integrate the response over time to know the DC value; it would be very prone to drift, unsuitable for long-term measurements, I would expect. I have, however, seen it done with hall effect sensors, as nigel mentioned. Hall sensors CAN measure the DC magnetic field. The january 2006 issue of EPE Magazine has a very simple project that does this, it outputs a simple voltage and is designed to be plugged into a DMM for reading the output.
__________________ EEgeek.net Last edited by evandude; 16th June 2006 at 04:58 PM. | |
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