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Ferrite core diameter effect

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smoothcriminal

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Hi.. two ferrites with same composition (e.g. T50-2 & T68-2) and same no. of turns.. but have different diameter as 0.5inch & 0.68inch.. what effect will be caused by this change?? can this change be compensated by increasing or decreasing no. of turns???
 
Your inductance will increase and so will your Q.
 
Thanx.. actually i am designing an RF receiver front... Required core is T68-2 (Diameter 0.68 inch) with 27 turns.. but i have a T50-2 (Diameter 0.50 inch).... which will require more turns.... i guess to give same inductance... will it in any way effect the required performance after it replaces the original required core???
 
what is the function of the inductor
 
**broken link removed**

this is the circuit I am building L1 & L3 are for tuning purpose (Toroidal Core) and L2 & L4 along with resister prevents from capacitive overloading, isolation you an say (uses Ferrite Core).. this is my understanding of this circuit.!!

now will T68-2 (L1) be replaced by T50-2 (L3) for same results... as i have T50 only.. same material composition but less diameter...??
 
Just doing a quick and dirty calculation, 27 turns on a T50-2 core will give you an inductance of 3.6μH.

If you want to be exact, try using 29 turns on the T50-2, that will give you 4.2μH, (28 turns gives 3.92μH).

In reality, any variation due to the number of turns will probably be masked by the variability in the permeability in the core material.

Some years ago I was getting some strange results with some toroid cores, the published K factor of a T68-6 core is 4.7, I did some measurements and found that mine were showing a K factor of 5.8. Quite a lot different.

JimB
 
Nice link.

I have a spreadsheet for doing my calculations, saves a lot of pain for repetitive calcs.

JimB
 
for difference type core, electrical parameter of core is not same,we select core based on application requirement, for T50 and T68, due to difference dimension, so Al value is not same. to keep same inductance , you can adjust turns.
 
Thanx guys.. and one more thing i have a Ferrite bead, out of my junk box..... I know nothing about it.. only that its color is shinny black and dimension how to exactly find its material like it could be FB 77-6301 or FB 43-6301 both have same physical appearance but different attenuation range .. kindly help me out in finding???
 
Thanx guys.. and one more thing i have a Ferrite bead, out of my junk box..... I know nothing about it.. only that its color is shinny black and dimension how to exactly find its material like it could be FB 77-6301 or FB 43-6301 both have same physical appearance but different attenuation range .. kindly help me out in finding???

Without knowledge of what it actually is, or a good selection of appropriate test equipment, it is just a round black object.

Not very helpfull, but that is as good as it gets.

JimB
 
hmmm............. but i think if i provide signal of different frequencies (through signal generator) at with oscilloscope see above which frequency is it doing attenuation.. so i guess through this way i can characterize it.. Wont't i?????
 
Hi,

If you decrease the cross sectional area of the core then you have to increase the number of turns. For small changes (which is probably all you should allow anyway) less than about 40 percent, it comes out to increasing the number of turns by half the decrease in area. For example, if the area decreased by 10 percent, then you should add 5 percent more turns. So if you had 100 turns to start with, this would mean you would need 105 turns as an approximation, and that should be close enough considering other variables which we have no control over. For a ratio of 0.68 to 0.5, which is 36 percent, would require 18 percent more turns, so if you had 20 turns to start that would mean adding 3 to 4 more turns. The DC resistance will be roughly 20 percent higher so the Q comes down a bit, which only means the selectivity will suffer somewhat if it is used as a tuned circuit. If it is used as a DC rf filter it will just decrease efficiency.

Yeah if you use an inductor in series with some resistive load and the signal at a particular frequency gets attenuated more than with a different inductor (bead on wire) then the first inductor is bigger than the second.
 
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