A charge amplifier/preamplifier is supposed to have $v_{output}$ proportional to $q(t)$, or proportional to integral $i(t)$. (proof: for example wikipedia : https://en.wikipedia.org/wiki/Charge_amplifier )

I made the computation and concluded differently. Would you know where is my mistake ?

let call $v_{input}$ the voltage at the entrance of the integrator (with feedback capacity $C_f$ and resistance $R_f$) of the charge preamplifier : we have $$v_{output}=-\frac{1}{R C_f} \int v_{input} dt$$

The charge at the entrance is $q(t)=\int i(t) dt$

Let call $C_{detector}$ the capacity of the detector, at the entrance : we have $v_{input}=Q_{detector}(t)/C_{detector}$.

Thus, this makes $$v_{output}(t)=-\frac{1}{R C_f} \int \frac{q(t)}{C_{detector}} dt$$ [this the integrator with operational amplifier]

Thus $v_{output}$ proportional to integral of $q(t)$
Thus $v_{output}$ proportional to integral of integral of $i(t)$

Where is my mistake ?

->It seems that the key is connected to the definition of the condensator, whether we consider $C_{detector}$ or $C_{feedback}$. Is there a link betwee $C_{detector}$ and $C_{feedback}$ ?

it seems that people do $v_s=Q_f/C_f$ but I'm not sure if there is a relationship between $Q_f$ and $Q_{detector}$

thank you


Your first equation is correct for $v_{input}$ making a current through a series resistor $R$ into the feedback input of the opamp. Your second equation is for a capacitance $C_{detector}$ across the opamp inputs and is incorrect because the voltage $v_{input}$ on $C_{detector}$ at time $t$ is not $\frac {q(t)}{C_{detector}}$. Instead, the opamp has worked to keep $$v_{input}=\frac {v_{output}}{Opamp Gain}=\frac {v_{output}}{10^6}\approx 0$$ As negative charge arrives, the opamp transfers the charge from the big capacitance $C_{detector}$ to the smaller capacitance $C_{feedback}$ by making the voltage more positive on the output end of $C_{feedback}$ where the charge will make a larger voltage than it did on $C_{detector}$.

For an infinite gain opamp and no feedback resistor ($R_{feedback}=\infty$) your second equation would be $$v_{output}(t)=\frac {-q(t)}{C_{feedback}}$$

  • $\begingroup$ Thank you. I investigate your nice explanations. $\endgroup$ – Mathieu Krisztian Oct 6 '19 at 14:28
  • $\begingroup$ everything is clear. Thank you Gary for your generous help. I understood now. $\endgroup$ – Mathieu Krisztian Oct 6 '19 at 19:47

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