Current to Voltage Converter

A current-to-voltage converter produces an output voltage proportional to an input current. Photodiode amplifiers, DAC output stages, and current-output sensors all feed into a current-to-voltage converter to produce a readable voltage signal.

Core Concept

The current-to-voltage converter is also called a transresistance amplifier because its gain has units of ohms (V/A). The input current source connects directly to the inverting input of the op-amp. The non-inverting input is at ground. Because of virtual ground, the inverting input is also at 0V, so the input current flows entirely through the feedback resistor Rf.

The voltage across Rf is Iin * Rf, and since the inverting input is at 0V, the output equals -Iin * Rf. The input impedance of this converter is nearly zero (the virtual ground sinks all current at near-zero voltage), which is exactly what current sources need. A real current source requires a low-impedance load to maintain its current accurately.

Photodiode preamplifiers use this topology with Rf in the range 100 kΩ to 10 MΩ. The BPW34 photodiode generates about 0.5 µA per lux under 850 nm illumination. With Rf = 1 MΩ, this gives 0.5V output per lux, making it suitable for light measurement. A feedback capacitor Cf (1 pF to 10 pF) in parallel with Rf is often added to limit bandwidth and reduce noise.

Key Equations

Output voltage:

Vout = -Iin * Rf

Transresistance gain:

Rm = Vout / Iin = -Rf (in ohms)

Input impedance (closed loop):

Rin = Rf / (1 + Aol) ≈ Rf / Aol (very low, typically milliohms)

Bandwidth when Cf is added in parallel with Rf:

f_-3dB = 1 / (2 * pi * Rf * Cf)

For photodiode applications, the noise equivalent power (NEP) is also important, but for GATE, the key formula is Vout = -Iin * Rf.

Given:
  Photodiode current Iin = 2 µA = 2 × 10^-6 A
  Feedback resistor Rf = 500 kΩ = 500,000 Ω
  TL071 op-amp, Aol = 200,000

Why this formula:
  Virtual ground forces Iin through Rf.
  Vout = -Iin * Rf

Formula:
  Vout = -Iin * Rf

Substitution:
  Vout = -2 × 10^-6 × 500,000

Calculation:
  = -2 × 10^-6 × 5 × 10^5
  = -2 × 5 × 10^(-6+5)
  = -10 × 10^-1
  = -1.0 V

  Input impedance:
  Rin = Rf / Aol = 500,000 / 200,000 = 2.5 Ω

Final Answer:
  Vout = -1.0 V
  Input impedance = 2.5 Ω (very low, ideal for current source loading)

Exam Tip: GATE asks for the input impedance of the I-V converter. It is NOT zero and NOT Rf. It is Rf/(1+Aol) ≈ Rf/Aol. For Rf = 1 MΩ and Aol = 100,000, Rin = 10 Ω. Also, Vout = -IinRf always has a negative sign because the current enters the inverting input. Students often write a positive output. The sign is negative because current flows into the virtual ground node, which is the inverting input.

Key Properties

• Vout = -Iin * Rf. Transresistance gain Rm = -Rf in ohms.
• Input impedance is very low: Rin = Rf/(1+Aol) ≈ Rf/Aol.
• Virtual ground at inverting input ensures all input current flows through Rf.
• BPW34 photodiode with Rf = 1 MΩ gives approximately 0.5V/lux.
• Adding Cf in parallel with Rf limits bandwidth to 1/(2piRfCf) and reduces wideband noise.
• TL071 is standard; for very low currents (femtoamperes), the LMC6001 (5 fA input bias current) is used.
• Used in photodiode preamplifiers, DAC output buffers, and electrometer circuits.

Quick Revision

• Also called transresistance amplifier.
• Vout = -Iin * Rf. Negative sign always present.
• Rf is the only component setting gain.
• Input impedance = Rf/Aol (very low).
• Cf in parallel with Rf sets bandwidth = 1/(2piRf*Cf).
• Input current source sees near-zero load impedance.
• Photodiode, DAC, ionisation chamber outputs all use this converter.
• Exam trap: Writing Vout = +Iin * Rf (positive sign). The output is always negative for a positive input current entering the inverting terminal. This is the most common sign error in GATE numerical problems on this circuit.