OPA656 datasheet(11/35 Pages) TI1

11 / 35 page

10 k

100 k

1 M

10 M

1 G

f − Frequency − Hz

VS = ±15 V

RL = 1 k

RF = 10 k

CS = 100 PF

CF = 3 PF

CS = 18 PF

CF = 2 PF

CS = 47 PF

CF = 2.2 PF

50 W

Network Analizer

(s) +

2 R

1 )

(Above the Pole Frequency)

(s) +

2 R

s )

2 R

S C1)C2

100

105

10 k

100 k

1 M

10 M

1 G

f − Frequency − Hz

VS = ±15 V

RL = 1 k

RF = 100 k

CS = 47 PF

CF = 0.7 PF

CS = 100 PF

CF = 1 PF

CS = 18 PF

CF = 0.5 PF

2 R

S (C1 ) C2)

2 R

1 )

100

110

120

10 k

100 k

1 M

10 M

f − Frequency − Hz

125

VS = ±15 V

RL = 1 k

RF = 1 M

115

105

CS = 18 PF

CF = 0 PF

CS = 47 PF

CF = 0.2 PF

CS = 100 PF

CF = 0.2 PF

O(s) +

) C2

)

2 R

S C1)C2

s s

)

2 R

O [

s C2

THS4631

www.ti.com

SLOS451B

– DECEMBER 2004 – REVISED AUGUST 2011

Table 1. Transimpedance Performance Summary

is difficult to measure the frequency response with

for Various Configurations (continued)

traditional laboratory equipment because the circuit

requires a current as an input rather than a voltage.

10-k

Ω TRANSIMPEDANCE RESPONSES

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, trans- impedance bandwidth measurements

are simplified, making amplifier evaluation easier and

faster.

Figure 35.

The interface network creates a capacitive,

100-k

Ω TRANSIMPEDANCE RESPONSES

constant current source from a network

analyzer

and

properly

terminates

the

network analyzer at high frequencies.

Figure 38. Emulating a Capacitive Current Source

With a Network Analyzer

The

transconductance

transfer

function

the

interface circuit is:

(3)

The transfer function contains a zero at dc and a pole

Figure 36.

1-M

Ω TRANSIMPEDANCE RESPONSES

at:

. The transconductance is constant

at:

above

the

pole

frequency,

providing 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

impedance, emulating the output impedance of a

photodiode or other current source. The output

impedance of this circuit is given by:

Figure 37.

(4)

Assuming C1

>> C2, the equation reduces to:

MEASURING TRANSIMPEDANCE

BANDWIDTH

, giving the appearance of a capacitive

While there is no substitute for measuring the

source at a higher frequency.

performance of a particular circuit under the exact

conditions that are used in the application, the

Capacitor values should be chosen to satisfy two

complete

system

environment

often

makes

requirements. First, C2 represents the anticipated

measurements harder. For transimpedance circuits, it

capacitance of the true source. Second C1 is chosen

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