About the gain switching circuit of instrumentation AMP such as AD8553

Hi UBOTI 

Thank you for reaching out. To further understand the problem and be able to make the proper recommendations, we would like to know more about your applications and design parameters.

Here are several items for your confirmation:

• What application is the part being used? Example: ADC driver, AFE for Pressure, Temp, Strain gauges, Weight sensing, etc..
• I see four ( 4) resistor configs for Figures 1 and 2, but you mentioned only the x900, 200x, and 50x Analog Gain. Is there a 4^th Gain you want to use?
• What are your desired values of the supplies? Vs+ (Pin3), Vs- (pin8), Vref (pin7)
• What are the expected inputs on both VINP and VINN terminals? I’m guessing these will be low since you plan to use higher Gains.

 Regarding your inquiries, please check my answers below:

QQ21: I want to switch the gain (900x, 200x, and 50x)
I would like to switch the resistor using an analog SW, but is there a better configuration?

ArSol: Your Gain circuits Fig1 and Fig2 are fine and will be functional. I suggest using the latter (Figure 2) to reduce layout branches from the switch to the resistor to the AD8853, eliminating additional parasitics that will cause the Gain errors.

 Q2：I would like to build a filter with a cutoff frequency of 1KHz.
Datasheet Figure 2. What is the Gain vs. Frequency filter?
Is it the filter created with R2C2 in Figure 31?

ArSol: Figure 2 This plot shows your filter's BW, cutoff (breakpoint), and Slope (-20dB Roll-off) at different Gains.  

R1 and R2 is the factor that dictates your Analog Gain, and the additional C2 is used to limit the switching noise present in the output. The suggested approach is to identify first your R1 and R2 values ( where: R2 > R1) then compute C2 by this equation: C2 = 1/(1400 × 2 × π × R2).  An additional R3C3 filter of 1.4kHz on the output is required for BW=10Hz. These two filters produce an overall BW = 1Khz.

 Thanks

Thank you for answering.
I will answer your questions.
Used with strain gauges. We are planning to increase the gain by about 400 times (it is possible without it).
The power supply is planned to be a single 5V power supply.

I plan to use it in conjunction with a wireless module, and I want a signal bandwidth of up to 200Hz, so I'm considering a filter that cuts off at around 2KHz, which is 10 times the bandwidth.
Determine the ratio of R1 and R2,
Is it correct to decide C2 by C2 = 1/(2K × 2 × π × R2)?

Thank you for answering.
I will answer your questions.
Used with strain gauges. We are planning to increase the gain by about 400 times (it is possible without it).
The power supply is planned to be a single 5V power supply.

I plan to use it in conjunction with a wireless module, and I want a signal bandwidth of up to 200Hz, so I'm considering a filter that cuts off at around 2KHz, which is 10 times the bandwidth.
Determine the ratio of R1 and R2,
Is it correct to decide C2 by C2 = 1/(2K × 2 × π × R2)?

Hi UBOTI 

You are right; use this equation to identify the values for C2: (C2 = 1/(1400 × 2 × π × R2). Please note that before finding values for C2, identify first your intended Gain. Then, finding R1 and R2 values by the equation G = 2 × (R2/R1). Where R2>R1.

Thanks,