LMH6642, LMH6643, LMH6644
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SNOS966P –MAY 2001–REVISED MARCH 2013
No matter how low an Rf is selected, there is a need for Cf in order to stabilize the circuit. The reason for this is
that the Op Amp input capacitance and Q1 equivalent collector capacitance together (CIN) will cause additional
phase shift to the signal fed back to the inverting node. Cf will function as a zero in the feedback path counter-
acting the effect of the CIN and acting to stabilized the circuit. By proper selection of Cf such that the Op Amp
open loop gain is equal to the inverse of the feedback factor at that frequency, the response is optimized with a
theoretical 45° phase margin.
CF =
SQRT (CIN)/(2p ∂ GBWP ∂ RF)
ꢀ
(1)
where GBWP is the Gain Bandwidth Product of the Op Amp
Optimized as such, the I-V converter will have a theoretical pole, fp, at:
fP =
SQRT GBWP/(2pRF ∂ CIN)
(2)
With Op Amp input capacitance of 3pF and an estimate for Q1 output capacitance of about 3pF as well, CIN
=
6pF. From the typical performance plots, LMH6642/6643 family GBWP is approximately 57MHz. Therefore, with
Rf = 1k, from Equation 1 and Equation 2 above.
Cf = ∼4.1pF and fp = 39MHz
C
f
5pF
Photodiode
Equivalent
Circuit
V
bias
R
f
1kW
V
+5V
=
R
CC
bias
C
C1
100nF
Q1
2N3904
-1mA
PP
-
V
out
d
R2
R
d
Photodiode
1.8kW
10
-
200pF
x
R5
510W
+
ı100kW
I
d
R11
D1
1N4148
910
W
R10
1kW
R3
1kW
+5V
Figure 63. Single Supply Photodiode I-V Converter
For this example, optimum Cf was empirically determined to be around 5pF. This time domain response is shown
in Figure 64 below showing about 9ns rise/fall times, corresponding to about 39MHz for fp. The overall supply
current from the +5V supply is around 5mA with no load.
200 mV/DIV
20 ns/DIV
Figure 64. Converter Step Response (1VPP, 20 ns/DIV)
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