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Any Jeepers in here? - Page 118 - INGunOwners

Any Jeepers in here? - Page 118 - INGunOwners — Fri, 11 Jan 2013 22:51:30 GMT

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Bi-amping RF-7 II's - Page 3 - Audioholics Home Theater Forums

Bi-amping RF-7 II's - Page 3 - Audioholics Home Theater Forums — Tue, 03 Jan 2012 17:48:09 GMT

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re: Andy's Tightwad Tips #9 - Buy Used

InnerTuber — Wed, 08 Sep 2010 12:25:18 GMT

Sometimes you want a car into which she'll hop into the back seat with you. You could always try some velour throws, but I dunno. You might be pushing your luck!

re: Andy's Tightwad Tips - #8

blsamuel — Thu, 07 Jan 2010 03:09:49 GMT

57 degrees? That's a bit chilly for me. Our youngest daughter would love it. She keeps the vents in her room closed in the winter. 60 doesn't sound quite so cold though I'm hoping to get used to 68 degrees first.

Analog Filters 101 – Part 4

Andy's Blog — Thu, 08 Oct 2009 23:49:50 GMT

Since we are dealing with reactive components we can no longer expect the voltage and current to be in

re: Analog Filters 101 – Part 3

Andy W — Fri, 25 Sep 2009 11:47:29 GMT

The blue voltage curve shows the "output voltage" of the filter, and everywhere EXCEPT at ~500 (resonance frequency) there is a cut. At resonace the voltage on the output is 1V (same as the input voltage), so what you're saying is true, except at resonace, which is a special case that we'll get into later.

re: Analog Filters 101 – Part 3

colterphoto1 — Fri, 25 Sep 2009 10:23:22 GMT

less impedance = more voltage = one more louder,innit? But I thought that passive filter circuits in crossover networks could only CUT the output at certain frequencies. Color me Confused.

Analog Filters 101 – Part 3

Andy's Blog — Fri, 25 Sep 2009 01:07:22 GMT

In today’s blog we will use the same values as last time , R=10Ω, C=10μF, L=10mH, and

re: Analog Filters 101 – Part 2

Andy W — Sat, 19 Sep 2009 16:52:30 GMT

You'll see in the next installment. In these graphs we're looking at impedance not frequency response. Current is inversely proportional to impedance so where impedance goes down current goes up. When current through the voice coil goes up the output from the driver goes up. I'm already working on the next installment.

re: Analog Filters 101 – Part 2

colterphoto1 — Sat, 19 Sep 2009 16:24:17 GMT

ok, thanks (I think) but I"m still thinking-- lower impedance in that portion of the FR curve should mean MORE output at that point, but I KNOW that passive filters in networks can only create DIPS in frequency response. What am I doing wrong?

re: Analog Filters 101 – Part 2

Andy W — Sat, 19 Sep 2009 03:51:22 GMT

Fixed !

re: Analog Filters 101 – Part 2

Andy W — Sat, 19 Sep 2009 02:59:20 GMT

BAH!!! Colter, thanks for making me take a second look at this blog... I should never write so late at night... like I'm doing now... but just by looking I think I need to redo the calculations...

re: Analog Filters 101 – Part 2

Andy W — Sat, 19 Sep 2009 02:30:49 GMT

Exactly. The purple curve (total impedance) being lower in the center of the graph would result in an increase in current where the impedance is lower. More current through the voice coil means more acoustic output. The high impedance at the frequency extremes means lower current and lower acoustic output.

re: Analog Filters 101 – Part 2

colterphoto1 — Fri, 18 Sep 2009 13:30:19 GMT

Perhaps I'm way off, but if the vertical scale is showing resistance then wouldn't a reduction in impedance cause a BUMP in acoustic output at the curve shown in the log diagram? I know I'm reading the graph wrong.

Analog Filters 101 – Part 2

Andy's Blog — Fri, 18 Sep 2009 04:01:04 GMT

Now with pictures! I have been neglecting my blog as of late, so I decided to write at home, and take