[AMRadio] AM Audio |
John E. Coleman (ARS WA5BXO)
wa5bxo2005 at pctechref.com
Sat Aug 20 14:44:02 EDT 2005
Whoa, Bacon, that ASCII thing doesn't look like a triangle to me. But I think this link will so what you wanted to show. http://wa5bxo.shacknet.nu/QuickBridge/Quick%20Bridge%20test.GIF John -----Original Message----- From: amradio-bounces at mailman.qth.net [mailto:amradio-bounces at mailman.qth.net] On Behalf Of Bob Bruhns Sent: Saturday, August 20, 2005 11:00 AM To: Discussion of AM Radio Subject: Re: [AMRadio] AM Audio Cool! Here is a crude Ascii-art diagram. Heh heh, yeah, it's supposed to be a triangle. The base of the triangle is the input voltage, the left side is the voltage across the resistor, and the right side is the voltage across the inductor. This is the situation when the resistance equals the inductor impedance. The angle between the V(R) and V(L) is 90 degrees, and you would calculate the voltage compared to V(in) as V(L) = V(R) = V(in) * sin(45 deg) = V(in) * 0.7071 because the angle of V(R) and V(L) compared to V(in) is 45 degrees. /\ / 90 \ V(R) / \ V(L) / \ / \ /__45____________________45__\ V(in) Bacon, WA3WDR (modern artist!) ----- Original Message ----- From: "Geoff" <w5omr at satx.rr.com> To: "Discussion of AM Radio" <amradio at mailman.qth.net> Sent: Saturday, August 20, 2005 11:23 AM Subject: Re: [AMRadio] AM Audio > Bob Bruhns wrote: > > > <>Yes, that will work. But remember, you want > > equal voltages across the resistor and the inductor. > > Surprisingly, this will not be 50% of the applied > > voltage, but it will be about 71% of the applied > > voltage. This adds up to about 1.414 times the > > applied voltage! You should confirm this by > > measuring both voltages directly. > > > > This happens because the impedance of the > > inductor is reactive, and the current is equal, so > > the phases ofthe voltages differ by 90 degrees. > > This phase shift causes a partial cancellation when > > the voltages are added, so 71 + 71 = 100 instead >> of 142. It's interesting to observe. > > > Actually, Bacon, John's entire message was thus... > you two agree. and I present it here, so that others > may learn from this experience, as well. > > ________________ > > If all you need is a measurement of inductance, the easiest thing is to series up a variable resistor with the choke and apply a low voltage from a filament XFMR (6 - 20 volts). Adjust the resistor so that the voltage measured across the choke is equal to the voltage across the resistor. If for instance a 10 volt source is used, you will measure about 7 volts across the resistor as well as across the choke. > > You are probably asking why it is not 5 Volts across each and the answer to that is because of the phase difference of 45 degrees when XL = R. The sine of 45 degrees is .707 and .707 X 10 Volts is about 7 volts. But the main thing is what ever the voltage is, by equalizing the voltage across the resistor and choke results in the creation of a resistance that is equal to XL of the choke at 60 CPS. > > Measure the resistance of the variable resistor after equalizing the voltage > drops and you have the XL of the choke at 60 cps. > > L = XL / 2 * pi * F > > L = XL / 6.28 * 60 > > L = XL / 376.8 > > Divide the resistance by 376.8 and you have the inductance. > > ________________________________ > > Thanks for the response. From all the active posters in here, > I admire and respect the intellect for the electronic abilities > and talents mostly of WA3WDR, K4KYV, WA3VJB, > KD5OEI and of course, WA5BXO, and not necessarily > in that order. > > -- > 73 = Best Regards, > -Geoff/W5OMR > > > _______________________________________________________ _______ > AMRadio mailing list > Home: http://mailman.qth.net/mailman/listinfo/amradio > Help: http://mailman.qth.net/mmfaq.html > Post: mailto:AMRadio at mailman.qth.net > ______________________________________________________________ AMRadio mailing list Home: http://mailman.qth.net/mailman/listinfo/amradio Help: http://mailman.qth.net/mmfaq.html Post: mailto:AMRadio at mailman.qth.net
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