Question:
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 ______
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.
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.
Updated On: Jul 28, 2026
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Correct Answer: 48.8
Solution and Explanation
Step 1: Understand the circuit.
The photodiode current \(I_d\) feeds a transimpedance (current to voltage) amplifier built around an op-amp with feedback resistor \(R_f\). The op-amp forces the whole diode current to flow through \(R_f\), producing an output voltage
\[ V_{out} = I_d R_f \]
This voltage is then digitized by the 10 bit ADC.
Step 2: Recall the ADC resolution limit.
A 10 bit ADC with reference voltage \(V_{Ref}\) divides the input range into \(2^{10} = 1024\) discrete steps. The smallest voltage change it can tell apart from zero is one least significant bit (LSB),
\[ V_{LSB} = \frac{V_{Ref}}{2^{10}} \]
Any current that produces less than this voltage at the amplifier output is indistinguishable from zero on the ADC, so the smallest measurable current is the one that produces exactly \(V_{LSB}\).
Step 3: Compute \(V_{LSB}\).
\[ V_{LSB} = \frac{5\text{ V}}{1024} = 4.8828\times10^{-3}\text{ V} \]
Step 4: Convert this to the smallest measurable current.
\[ I_{d,min} = \frac{V_{LSB}}{R_f} = \frac{4.8828\times10^{-3}}{100\times10^{3}} = 4.8828\times10^{-8}\text{ A} \]
Converting to nano-Amperes,
\[ I_{d,min} = 48.828\text{ nA} \]
Final Answer:
Rounded to one decimal place, \[ \boxed{I_{d,min} = 48.8\text{ nA}} \]
The photodiode current \(I_d\) feeds a transimpedance (current to voltage) amplifier built around an op-amp with feedback resistor \(R_f\). The op-amp forces the whole diode current to flow through \(R_f\), producing an output voltage
\[ V_{out} = I_d R_f \]
This voltage is then digitized by the 10 bit ADC.
Step 2: Recall the ADC resolution limit.
A 10 bit ADC with reference voltage \(V_{Ref}\) divides the input range into \(2^{10} = 1024\) discrete steps. The smallest voltage change it can tell apart from zero is one least significant bit (LSB),
\[ V_{LSB} = \frac{V_{Ref}}{2^{10}} \]
Any current that produces less than this voltage at the amplifier output is indistinguishable from zero on the ADC, so the smallest measurable current is the one that produces exactly \(V_{LSB}\).
Step 3: Compute \(V_{LSB}\).
\[ V_{LSB} = \frac{5\text{ V}}{1024} = 4.8828\times10^{-3}\text{ V} \]
Step 4: Convert this to the smallest measurable current.
\[ I_{d,min} = \frac{V_{LSB}}{R_f} = \frac{4.8828\times10^{-3}}{100\times10^{3}} = 4.8828\times10^{-8}\text{ A} \]
Converting to nano-Amperes,
\[ I_{d,min} = 48.828\text{ nA} \]
Final Answer:
Rounded to one decimal place, \[ \boxed{I_{d,min} = 48.8\text{ nA}} \]
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