THS4031, THS4032

100-MHz LOW-NOISE HIGH-SPEED AMPLIFIERS

SLOS224C – JULY 1999 – REVISED APRIL 2000

23

POST OFFICE BOX 655303

APPLICATION INFORMATION

optimizing frequency response

Internal frequency compensation of the THS403x was selected to provide very wide bandwidth performance

and still maintain a very low noise floor. In order to meet these performance requirements, the THS403x must

have a minimum gain of 2 (–1). Because everything is referred to the noninverting terminal of an operational

amplifier, the noise gain in a G = –1 configuration is the same as a G = 2 configuration.

One of the keys to maintaining a smooth frequency response, and hence, a stable pulse response, is to pay

particular attention to the inverting terminal. Any stray capacitance at this node causes peaking in the frequency

response (see Figure 46 and Figure 47). Two things can be done to help minimize this effect. The first is to simply

remove any ground planes under the inverting terminal of the amplifier, including the trace that connects to this

terminal. Additionally, the length of this trace should be minimized. The capacitance at this node causes a lag

in the voltage being fed back due to the charging and discharging of the stray capacitance. If this lag becomes

too long, the amplifier will not be able to correctly keep the noninverting terminal voltage at the same potential

as the inverting terminal’s voltage. Peaking and possible oscillations will then occur if this happens.

Figure 46

_

+

150

Ω

VO

VI

50

Ω

Ci–

300

Ω

7

6

2

0

100 k

1 M

10 M

8

9

f – Frequency – Hz

OUTPUT AMPLITUDE

vs

FREQUENCY

10

100 M

500 M

5

4

3

1

Ci– = 10 pF

No Ci–

(Stray C Only)

VCC = ± 15 V

Gain = 2

RF = 300 Ω

RL = 150 Ω

VO(PP) = 0.4 V

300

Ω

Figure 47

_

+

360

Ω

150

Ω

VO

VI

56

Ω

Ci–

360

Ω

1

0

–4

–6

100 k

1 M

10 M

2

3

f – Frequency – Hz

OUTPUT AMPLITUDE

vs

FREQUENCY

4

100 M

500 M

–1

–2

–3

–5

Ci–= 10 pF

No Ci–

(Stray C Only)

VCC = ± 15 V

Gain = –1

RF = 360 Ω

RL = 150 Ω

VO(PP) = 0.4 V

small-signal response. This is why in Figure 30, a feedback resistor of 3.9 k

Ω with a gain resistor of 1 kΩ only

shows a small peaking in the frequency response. The parallel resistance is only 800

Ω. This value, in

conjunction with a very small stray capacitance test PCB, forms a zero on the edge of the amplifier’s natural

frequency response. To eliminate this peaking, all that needs to be done is to reduce the feedback and gain

resistances. One other way to compensate for this stray capacitance is to add a small capacitor in parallel with

the feedback resistor. This helps to neutralize the effects of the stray capacitance. To keep this peaking out of

the operating range, the stray capacitance and resistor value’s time constant must be kept low. But, as can be

seen in Figures 26 – 29, a value too low starts to reduce the bandwidth of the amplifier. Table 1 shows some

recommended feedback resistors to be used with the THS403x.