THS4631 Datasheet, PDF (11/27 Pages) Texas Instruments – HIGH-VOLTAGE, HIGH SLEW RATE, WIDEBAND FET-INPUT OPERATIONAL AMPLIFIER

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10-kΩ TRANSIMPEDANCE RESPONSES
85
C_S = 18 PF
C_F = 2 PF
80
C_S = 47 PF
C_F = 2.2 PF
75
C_S = 100 PF
C_F = 3 PF
70
V_S = ±15 V
R_L = 1 k
R_F = 10 k
65
10 k
100 k
1M
10 M
1G
f − Frequency − Hz
Figure 35.
100-kΩ TRANSIMPEDANCE RESPONSES
105
C_S = 18 PF
C_F = 0.5 PF
100
C_S = 47 PF
95
C_F = 0.7 PF
C_S = 100 PF
C_F = 1 PF
90
V_S = ±15 V
R_L = 1 k
R_F = 100 k
85
10 k
100 k
1M
10 M
1G
f − Frequency − Hz
Figure 36.
1-MΩ TRANSIMPEDANCE RESPONSES
125
C_S = 18 PF
C_F = 0 PF
120
115
C_S = 47 PF
C_F = 0.2 PF
110
C_S = 100 PF
105
C_F = 0.2 PF
100
V_S = ±15 V
R_L = 1 k
R_F = 1 M
95
10 k
100 k
1M
f − Frequency − Hz
Figure 37.
10 M
MEASURING TRANSIMPEDANCE
BANDWIDTH
While there is no substitute for measuring the per-
formance of a particular circuit under the exact
conditions that are used in the application, the com-
plete system environment often makes measure-
ments harder. For transimpedance circuits, it is diffi-
cult to measure the frequency response with tradition-
al laboratory equipment because the circuit requires a
THS4631
SLOS451A – DECEMBER 2004 – REVISED MARCH 2005
current as an input rather than a voltage. Also, the
capacitance of the current source has a direct effect
on the frequency response. A simple interface circuit
can be used to emulate a capacitive current source
with a network analyzer. With this circuit,
transimpedance bandwidth measurements are simpli-
fied, making amplifier evaluation easier and faster.
Network Analizer
I_O
50 W 50 W
C2
R_S
C1
V_S
ǒ Ǔ ^IO
V_S
(s)
+
2 R_S
1
1
)
C1
C2
(Above the Pole Frequency)
A. The interface network creates a capacitive,
constant current source from a network
analyzer and properly terminates the net-
work analyzer at high frequencies.
Figure 38. Emulating a Capacitive Current Source
With a Network Analyzer
The transconductance transfer function of the
interface circuit is:
s
ǒ Ǔ _IO
V_S
(s)
+
s
2 R_S
)₂
1)^CC12
1
R_S ǒC1)C2Ǔ
(3)
The transfer function contains a zero at dc and a pole
1
at: ^{2 R}S ^{(C1 ) C2)} . The transconductance is constant
1
ǒ Ǔ _at:
2 R_S
1
)
C1
C2
, above the pole frequency, provid-
ing a controllable ac-current source. This circuit also
properly terminates the network analyzer with 50 Ω at
high frequencies. The second requirement for this
current source is to provide the desired output im-
pedance, emulating the output impedance of a
photodiode or other current source. The output im-
pedance of this circuit is given by:
ȧȱȲ _{ǒ Ǔ} ȧȳȴ ^Z_O^(s)
+
C1
C1
)
C2
C2
s
)
2
R_S
1
ǒC1)C2Ǔ
s
s
)
2
1
R_S
C1
(4)
Assuming C1 >> C2, the equation reduces to:
Z_O
[
1
s C2
,
giving
the
appearance
of
a
capacitive
source at a higher frequency.
Capacitor values should be chosen to satisfy two
requirements. First, C2 represents the anticipated
capacitance of the true source. Second C1 is chosen
such that the corner frequency of the
transconductance network is much less than the
transimpedance bandwidth of the circuit. Choosing
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