# Charge preamplifier

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 $$v_{output}(t)=\frac {-q(t)}{C_{feedback}}$$