Parametric EQ Math with gyrators - Page 1
Forum section: electronics / beginners

author: iampoor
author_rank: Frequent Contributor
timestamp: November 07, 2013, 07:49:54 am
content: Hi!Ive been attempting to figure out the calculations for these eq's for awhile now. I know I am missing something plainly obvious, so I could use a guide in the right direction.I am trying to determine the frequency ranges for the eq circuits in the 4th figure down.http://www.geofex.com/article_folders/eqs/paramet.htmI want to find the values of C1/C2 and the respective eq range." the inductance is equal to the value of C2 times the 470 ohm resistor times the series resistance of the 51K and the setting of the 1M pot. Or more prosaicly, Leff = R1*R2*C, where R1=470, R2= series resistance, and C = C2's value."SO for the example quote, this would equal 470 * 51000 * 0.02 = 479400Moving on"Doing a little math, we can come up with some values of the frequency range for an LC section. The L value isLmin = 470*51K C2 and Lmax = 470 1.05M*C2, orLmin/Lmax = 51K/1.05M = 0.048 or about a 20 to one range."Im confuse how C2 relates to this example. I understand the range, but why is C2 not included? It doesnt appear that we are solving for C2.Now here is where I get really confused. Im not sure what the value of L is. I presumed it was the value above (479400) If I enter C2 then get...."The center frequency of an LC filter is F0=1/(2*pi*L*C),"1/(2*3.14*479400*0.02)=1.6607808593013028493685046860593e-5What am I missing? I feel really dumb for not understanding this!!!

author: ElectroIrradiator
author_rank: Frequent Contributor
timestamp: November 07, 2013, 08:52:33 am
content: Quote from: iampoor on November 07, 2013, 07:49:54 amC2 is 0.02 uF, 20nF, which is 0.00000002 F, so the product is 470 * 51000 * 2E-8 = 0.479

author: iampoor
author_rank: Frequent Contributor
timestamp: November 07, 2013, 10:17:34 pm
content: Quote from: ElectroIrradiator on November 07, 2013, 08:52:33 amAh, I knew there was an obvious mistake somewhere! Thank you so much!!!Now I know Im missing something else however.."The center frequency of an LC filter is F0=1/(2*pi*L*C)"So, Ive tried a few options and none of them seem correct. Im missing how to factor C1 in as well.F0=1/(2*3.14*23970000 * 0.02)F0=1/(2*3.14*0.048*0.02)I know neither of these is right...I know this is wrong, but isnt there a different value for L in this equation?I know Im missing so many obvious things right now, thank you for your patience. :-)

author: ElectroIrradiator
author_rank: Frequent Contributor
timestamp: November 07, 2013, 11:17:28 pm
content: That article doesn't seem to explain things very well, and there are lots of typographical errors and a haphazard lack of component labels. However you really need to get on top of your orders of magnitude.*) The trick is that he uses the virtual inductor circuit, the gyrator, to make a variable inductor for a resonant LC filter.*) The 'inductor' simulated by a gyrator circuit has a size Leff= R1 * R2 * C, where C can be anything. In his first filter section he uses 20 nF (2E-08 F), R1 is 470 ohm and R2 is the variable resistance through which we adjust the 'inductor'. It is the sum of the two series connected resistors in each gyrator, like the 51K + 1M. Say this is adjusted to 100K in total. This gives Leff= 470 * 100000 * 2E-08 = 0.94 Hy (Henry).*) The resonant frequency of any LC circuit is given by Fres= 1/(2 * Pi * sqrt[L * C]) for any L and C. Note the mistake in the article.If we use the values for the first gyrator, we have L = 0.94 Hy and C = C1 = 0.15 uF = 15e-8 F. So Fres= 1 / (2 * 3.14 * sqrt[0.94 * 15e-08]) = 423 Hz.Basically the article is shock full of typographical errors. Don't believe everything you read on the net. Double check it against other sources.