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Inductor question..

时间:04-08 整理:3721RD 点击:
Dear All:
I have a 1UH Inductor A company and B company.
The working frequency range is 2.2~2.4Mhz,from the
DUT measure result it very closed,but put into the circuit
B company ind return loss degrade 2 dB, I don't know the reason.
Please help me to check the root cause.

Thanks.

One way to roughly determine the quality of an inductor is to compare the out-of-circuit self-resonant frequency (SRF) of the open inductor (do not short the ends together) to the expected operating frequency ..
A good inductor will be self-resonant a minimum of four or more times the operating frequency ..
If both inductors are rated for more or less the same self-resonant-frequency, compare their Q-s ..

Rgds,
IanP

The quality of the inductor depends on the Q-factor. Above friend is right. You first energized the inductor then disconnect supply, you will find a damped sinusoid across the inductor. The more the cycles you get, the better that inductor be. Q=Z/R

I wonder, if the measurement is representative for the actual circuit operation. Without knowing the circuit and
operation conditions, this can't be determined.

It's also inplausible, why you get an equal or even higher Q with factor 1.5 higher Rs. The measurement may be incorrect.

Dear all:
I have check the inductor SRF is 59MHz and in 2.2~2.4MHz Q is around 38 in these two company.
But my question is not how to get inductor Q,
my question is,
Because in 2.2~2.4MHz the Q and Rs are almost the same in these two company,
but why return loss is degrade 2 dB in B company?

Yes, either the Q measurement is not representative for the actual circuit operation or simply inaccurate.

You didn't comment the partly inconsistent measurement results.

Or the circuit has a small bandwidth and must be tuned exactly. A return loss versus frequency plot would reveal this.

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