Hello,

I have some questions on parts of the AD745 datasheet including some basics.

I am referring to the most recent datasheet of the AD745 which can be found here:

http://www.analog.com/media/en/technical-documentation/data-sheets/AD745.pdf

*[Minor] On page 8 lower left corner it says: "Figure 5 shows that these two circuits have an identical frequency response and the same noise performance (provided that C_s/C_f = R_1/R_2)."*But, Fig. 5 shows the circuit of a charge amplifier circuit. As can be deduced from the caption of Fig. 7 it is Fig. 7 that should be referre to.*On page 8 lower left corner it says: "One feature of the first circuit is that a “T”network is used to increase the effective resistance of R_B and improve the low frequency cutoff point by the same factor.*Could somebody please clarify "effective Resistance", the resistor R_B with or without "*" (because I think it is confused in the text) and how the T network has an impact on the resistance of R_B since R_B is on the non inverting input and there is virtual ground in between.*On page 8 on the right side it says: "The graph of Figure 8 shows how to select an R B large enough to minimize this resistor’s contribution to overall circuit noise."*As far as I understand the voltage noise (Johnson-Nyquist) increases with rising resistance. Thus the overall contribution to voltage noise increases (added root-sum-squared). If modeled as a parallel noise current source the "R" in the denominator should be square-rooted (equivalent current noise). All noise currents cause voltage noise when passing a resistor so minimizing the current noise is advantageous. In the discussed circuit decreasing the current noise of the resistor to a point where the current noise of the amplifier is dominant is usefull. Still I do not understand the advantage in the noise balance sheet when voltage noise increases by sqrt(4kTR).*On page 11 on the left column are calculations of the bandwidth.*When I inset the given values I get different results, for example C_s = 300 pF, R_f = 100 kOhm and f_c = 20 MHz yields a f_B = 325 kHz, not 360 kHz. Plus if I insert the values of R_f and C_s into the given formula for C_L I again get different values -> C_L is approximately given as 1/(2*pi*R_f*C_s) -> how can C_L be still 4.5 pF if you change R_f from 100 kOhm to 360 kOhm. Probably I misunderstand something.- I am looking for a low noise charge amplifier for signals up to 120 kHz. From what I found the AD745 is my best choice, especially combining a low voltage AND current noise with a gain bandwidth product of 20 MHz. Maybe any other suggestions?

Thank you very much for answers and explanations.

Hi, Sebastian.

Thank you for using ADI products and sorry for the very late response. As for your questions,

1. Thank you for noticing this error. The caption should refer to Figure 7 and not Figure 5. We'll get this fixed.

2. I'm also confused about this sentence. I think the effective resistance here is the equivalent resistance of the T-network which is Rth= Rs +R1//R2. I'm just not sure on this. I'll consult the product owner for this question.

3. You are correct that the equation of the current noise should be sqrt(4kT/R). We'll also get this fixed. Regarding Figure 8, I'll get back to you on that.

4. The newest alternative part of AD745 is ADA4637-1. You can look at its specifications here - http://www.analog.com/media/en/technical-documentation/data-sheets/ADA4627-1_4637-1.pdf

Regards,

Anna