Hi, I am not so familiar with circuit design.

Now, we use a variable gain amplifier, AD603, and we found the page 14 in the spec.

When the BW is varied, the value of capacitor connected to the FDBK pin is different.

Do you know how the capacitor value to be determined? What's the purpose for this capacitor?

As I know, the BW should be relevant with only the feedback resistor or it's relevant with this capacitor as well?

Our PCB is about to gerber out so we need to confirm this question.

Thanks in advance.

Hi Hun-Ching.

The resistor will dictate the gain and the Gain will dictate the bandwidth. The purpose of the capacitor is to extend the bandwidth of the part at higher gains, this is due to the pole added by the capacitor. On figure 6, 7 and 8, as the Gain goes up you will notice that the bandwidth lowers and the peaking goes down. Figures 33 and 34 shows AD603 at different gains.

In figure 33 it is 0 dB to 40 dB, the computation goes this way:

Total Gain = 20log( [(6.44k+694)/20] + 1) + (0 to -42.14 dB)

Total Gain = 41.31 dB + ( 0 to -42.14 dB)

Total Gain = -0.83 dB to 41.31 dB but in figure 36, a worst-case gain error of 1.07dB is expected

Total Gain = 0 dB to 40 dB.

In figure 34

Total Gain = 20log( [((6.44k//2.15k)+694)/20] + 1) + (0 to -42.14 dB)

Total Gain = 51.07 dB + ( 0 to -42.14 dB)

Total Gain = 8.93 dB to 51.07

Total Gain = 10 dB to 50 dB.

So at higher gains, a roll-off in the bandwidth is expected. Adding a capacitor on the feedback pin could extend the pin by introducing poles at higher frequency(peaking), extending the bandwidth.

Regards.