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russlk

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Posts posted by russlk

  1. I would forget the capacitor and find a solonoid coil, like used in a washing machine water control.  Get an iron rod from the local hardware store that will fit in the coil, an iron or brass washer that will also fit the rod.  When 110VAC is applied to the coil, the washer will be propelled up the rod.  If you still want to use a capacitor, get a 200 volt, 400 microfarad cap and charge it with a bridge rectifier on the 110 volt line.

  2. For safety, keep one hand in your pocket and wear rubber soled shoes or work on an insulated surface. The high voltage won't injure you but you could be injured when you jump back. If you have overheating components, you have to find the cause before replacing parts. Check leakage of filter caps, etc.

  3. You can't use the reactance as a plain number, you have to consider the phase angel. The current leads the voltage by 90 degrees in a capacitor and lags by 90 degrees in an inductor. The reactance of an inductor is given by Xl = 2*PI*f*L
    where PI=3.14162, f= frequency in Hz, L=inductance in henries.
    The reactance of a capacitor is given by Xc = 1/(2*PI*f*C).
    The quantity 2*PI*f is given the symbol ohmega, but since I don't have that symbol on the keyboard, I will use W. The 90 degree phase angle is denoted by j, so now Xl = j*W*L and Xc = 1/(j*W*C).

    The advantage of this nominclature is that it is easy to add series impedances: Z = Xl+R = j*W*L+R
    Parallel impedances is the familiar product over sum: Z = j*W*L*R/(j*W*L+R).
    You can convert a parallel circuit to an equivalent series circuit and vice-versa. I used to be able to practically do that in my head, but it has been many years.

  4. The gain of T1 is nominally RV1/R6, but there is negative feedback thru R2, so the computation is complex. The purpose of R6 is to stabelize the DC current and increase the input resistance of T1. The input resistance is rb+beta*R6. Probably around 4700 ohms. The input capacitor (100nF) is working against the 4700 ohms, which gives an high pass cutoff of 340 Hz. You need a larger input cap to get good bass. Voice frequencies are about 300 Hz and up, so you need low pass starting at 100 hz or lower to have any attenuation at 300 hz.

  5. L2 resonates with C5 and C4 near the center of the TV band. A 75 ohm resistor would probably work as well. L1 is a "peaking coil". It is self resonant near the upper edge of the band and is there to raise the high frequency gain. You cannot put a pot in this circuit because the leads have to be very short and this is a high frequency circuit. The ratio of C5, C4 could be changed, keeping the total constant, to change the gain. FYI: the total is (c5*c4)/(c4+c5).

  6. mF means millifarad, but that is an error, it should be uF. The european style is to use , instead of. so it is 4.7uF.

    #18 wire is rather large and I would not space the turns, use close wound and at least 8 turns.

    The 4 turn inductor is .132uH if you use #18 wire, close wound.

  7. The result of a measurment will depend on how it is done, and the method should relate to the way is is going to be used. If the secondary is to be grounded, the measurement should be done that way. If the transformer includes a faraday shield, the common mode will be nil and hard to measure. I have not had to do this measurement, but I would tie the two primary wires together and drive them relative to ground with a signal generator, measure the signal at the secondary terminals and call the common mode rejection the ratio: sec/pri volts or dB.

  8. The advantage of AC power is that you can transmit at high voltage and low current, for low loss, and transform back to low voltage and high current at the load. But, 15 VAC is not high enough to be efficient, why not send the power as DC? The signal coming back could be on an AC carrier, or if it is logic could be sent directly with Manchester coding.

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