# AD745 datasheet and low noise charge amplifier

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:

1. [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.
2. 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.
3. 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).
4. 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.
5. 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.

Parents
• 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.

Regards,

Anna

• 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.