EL5100IS Intersil, EL5100IS Datasheet - Page 12

• Manufacturer Part Number: EL5100IS
• Description: IC OP AMP HS V-FEEDBACK 8-SOIC
• Manufacturer: Intersil

Specifications of EL5100IS

• Amplifier Type: Voltage Feedback
• Number Of Circuits: 1
• Slew Rate: 2200 V/µs
• -3db Bandwidth: 200MHz
• Current - Input Bias: 2µA
• Voltage - Input Offset: 1000µV
• Current - Supply: 2.5mA
• Current - Output / Channel: 100mA
• Voltage - Supply, Single/dual (±): 3.3 V ~ 12 V, ±1.65 V ~ 6 V
• Operating Temperature: -40°C ~ 85°C
• Mounting Type: Surface Mount
• Package / Case: 8-SOIC (3.9mm Width)
• Lead Free Status / RoHS Status: Contains lead / RoHS non-compliant
• Output Type: -
• Gain Bandwidth Product: -

C1
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:
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!
F
=
1n
2
π
C2
1
L
×
C
Z
0.1µF
F RES
C3
22µF
FIGURE 31.
FIGURE 32.
=
12
F
C1
1n
L1
<
Ci
C2
Li
L2
EL5100, EL5101, EL5300
0.1µF
C3
22µF
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.)
R3 = 0
R3 = 3
The final power supply bypass circuit will look:
Z
C1
1n
R10
C12
F RES
FIGURE 34.
FIGURE 33.
3R3
33nF
C11
F
22µF
C1
Vs+
C3
R3
L3
May 3, 2007
FN7330.3
0.1µF
2 to 3Ω