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參數(shù)資料
| 型號(hào)： | EL5300 |
| 廠商： | Intersil Corporation |
| 英文描述： | 200MHz Slew Enhanced VFA |
| 中文描述： | 200MHz的回轉(zhuǎn)增強(qiáng)揮發(fā)性脂肪酸 |
| 文件頁(yè)數(shù)： | 11/11頁(yè) |
| 文件大?。?/td> | 333K |
| 代理商： | EL5300 |
11
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FN7330.1
The reason for that is the very small but not zero value serial
inductance of the capacitor.
The capacitor will behave as a capacitor up to its resonance
frequency, above the resonance frequency it will behave as
an inductor.
Just 1nHy inductance serial with 1nF capacitance will have
serial resonance at:
1
F
×
π
C = 1nF, L = 1nHy, F = 159 MHz
And an other 1nHy is very easy to get together with the
inductance of traces on the pcb, and therefore you could
encounter resonances from ca 50MHz and above anywhere.
So if the amplifier has a bandwidth of a few hundred MHz,
the proper power supply by-pass could become a serious if
not difficult task.
Intuitively, you would use capacitors value 0.1μF parallel
with a few μF tantalum, and to cure the effect of it’s serial
resonance put a smaller one parallel to it.
The result will surprise to you, because you will get even
something worse than without the small capacitor.
What is happening there Just look what we get:
Above its serial resonance C2* the ideal capacitance of C2 is
a short, the Tantalum capacitor for high frequencies is not
effective, the left over is C1 capacitor and L1 + L2 inductors,
we get a parallel tank circuit, which is at it’s resonance a high
impedance path and do not carry any high frequency
current, it does not work as bypass at all!
The impedance of a parallel tank circuit at resonance is
dependent from it’s Q. High Q high impedance.
The Q of a parallel tank circuit could be reduced by
bypassing it with a resistor, or adding a resistor in serial to
one of the reactive components. Since the bypassing would
short the DC supply we do have to go to add resistor in serial
to the reactive component, we will ad a resistor serial with
the inductor. (See Figure 33.)
The final power supply bypass circuit will look:
FIGURE 31.
Ci
Li
Z
F
F RES
I
C
C
L
2
=
C1
1n
C3
22
μ
F
C2
0.1
μ
F
L2
L1
C1
1n
C3
22
μ
F
C2
0.1
μ
F
=
<
FIGURE 32.
C3
0.1
μ
F
L3
C1
Z
F
F RES
R3 = 3
R3 = 0
R3
2 to 3
FIGURE 33.
C1
1n
C11
22
μ
F
R10
3R3
C12
33nF
Vs+
FIGURE 34.
EL5100, EL5101, EL5300
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