The output signal (current \(I_d\)) of a reversed biased (with a voltage \(V_b\)) photodiode, on which light is incident, is fed to an amplifier (see figure). The output voltage is digitized by a 10 bit Analogue to Digital convertor (ADC) which has a reference voltage of 5 V. The smallest current which can be measured by the circuit in nano-Amperes (rounded off to one decimal place) is ______
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Hint:
The transimpedance stage gives \(V_{out} = I_dR_f\). Find the ADC least significant bit, \(V_{Ref}/2^{10}\), then divide by \(R_f\) to get the smallest resolvable current.
Step 1: Find the full scale current the ADC can register.
Since $V_{out} = I_dR_f$ from the transimpedance stage, the largest current the ADC can fully register corresponds to its full reference voltage:
\[ I_{d,max} = \frac{V_{Ref}}{R_f} = \frac{5\text{ V}}{100\times10^3\ \Omega} = 5\times10^{-5}\text{ A} = 50000\text{ nA} \]
Step 2: Split this range into the ADC steps.
A 10 bit converter divides its full input range into $2^{10} = 1024$ equal current steps, since $V_{out}$ and $I_d$ are directly proportional through the fixed resistor $R_f$. So each digital step, the smallest change the ADC can register, corresponds to a current of
\[ I_{d,min} = \frac{I_{d,max}}{1024} = \frac{50000\text{ nA}}{1024} \]
Step 3: Compute the value.
\[ I_{d,min} = 48.828\text{ nA} \]
This is exactly the same number as scaling the reference voltage directly by $R_f$ and the bit count, just reached by normalizing the full scale current instead of the full scale voltage.
Final Answer:
Rounded to one decimal place,
\[ \boxed{I_{d,min} = 48.8\text{ nA}} \]
