Voltage-to-current and current-to-voltage converters

Why it matters

Industrial plants send measurements as 4–20 mA currents, because a current loop is unaffected by wire resistance and the 4 mA "live zero" reveals a broken wire. That needs a voltage-to-current (V-I) converter in the transmitter. At the other end, and in every photodiode, ionisation chamber or electrochemical sensor, a current must be turned back into a voltage by a current-to-voltage (I-V) converter, also called a transimpedance amplifier. Both are simple op-amp circuits whose accuracy depends on one resistor.

Key ideas

Why current signalling. In a series loop the same current flows everywhere, so lead resistance, contact resistance and moderate voltage drops do not change the reading as long as the transmitter has enough voltage headroom (its compliance). Current loops also pick up less induced voltage noise because the receiving end is a low resistance. The 4 mA offset powers two-wire transmitters and lets 0 mA signal a fault.

V-I converter with floating load. Connect the load as the feedback element of a non-inverting amplifier: input to v₊, load from output to v₋, resistor R₁ from v₋ to ground. The virtual short puts V_in across R₁, so I_L = V_in/R₁, regardless of R_L. Limitation: the load must float (neither end grounded), and the op-amp output must reach V_in + I_L·R_L without saturating.

V-I converter with grounded load (Howland current source). A difference-amplifier-like network with positive and negative feedback. When the four resistors satisfy R₂/R₁ = R₄/R₃ (in the basic form all equal R), the output current into a grounded load is I_L = V_in/R (sign set by which input is driven), independent of R_L. Resistor mismatch makes the output resistance finite, so precision resistors are needed. The improved Howland splits one resistor to raise the output current range.

Transistor-boosted V-I converter. For 4–20 mA loops and larger currents, an op-amp drives the base (or gate) of a transistor whose emitter (source) resistor R_E is inside the feedback loop: V_in appears across R_E, so I = V_in/R_E flows in the collector (drain) and through the load, while the transistor supplies the current and voltage swing. Its β error is removed by using a MOSFET or a Darlington.

I-V converter (transimpedance amplifier). Feed the input current into the virtual-ground node of an inverting op-amp with feedback resistor R_f. All the current flows through R_f, so v_o = −I_in·R_f. The source sees nearly 0 V, which is ideal for photodiodes (photoconductive mode with zero bias gives the most linear and lowest-dark-current response). For small currents choose a FET-input op-amp, because its bias current adds directly to the signal. A small capacitor C_f across R_f is needed to keep the circuit stable against the sensor's capacitance; it also sets the bandwidth f = 1/(2πR_fC_f).

Scaling a 4–20 mA loop. A linear transmitter maps the measurement range onto 4–20 mA (a 16 mA span); a 250 Ω receiver resistor converts it back to 1–5 V.

Formulas

I_L = V_in / R₁ — floating-load V-I converter (A); requires V_in + I_L·R_L ≤ V_sat.

I_L = V_in / R — Howland current source with matched resistors R₂/R₁ = R₄/R₃.

I = V_in / R_E — op-amp plus transistor V-I converter (MOSFET or high-β transistor).

v_o = −I_in · R_f — transimpedance (I-V) converter; R_f in Ω is the transimpedance gain (V/A).

f_c = 1 / (2π·R_f·C_f) — transimpedance bandwidth set by the feedback capacitor.

I = 4 mA + 16 mA × (x − x_min)/(x_max − x_min) — 4–20 mA scaling.

R_loop(max) = (V_supply − V_min,transmitter) / 20 mA — maximum total loop resistance.

Worked examples

Example 1 (standard: floating-load converter and TIA). (a) A floating-load V-I converter has R₁ = 1 kΩ and V_in = 2 V; the op-amp saturates at ±12 V. Find I_L and the largest load. (b) A photodiode delivers 5 µA into a transimpedance amplifier with R_f = 1 MΩ and C_f = 1 pF. Find v_o and the bandwidth.

1. (a) I_L = V_in/R₁ = 2 V/1 kΩ = 2 mA, for any load within range.
2. Output voltage = V_in + I_L·R_L ≤ 12 V → R_L ≤ (12 − 2)/0.002 = 5 kΩ.
3. (b) v_o = −I_in·R_f = −5×10⁻⁶ × 10⁶ = −5 V.
4. f_c = 1/(2πR_fC_f) = 1/(2π × 10⁶ × 10⁻¹²) = 159 kHz.

Answer: (a) I_L = 2 mA, R_L(max) = 5 kΩ; (b) v_o = −5 V, bandwidth ≈ 159 kHz.

Example 2 (GATE level: 4–20 mA loop). A pressure transmitter maps 0–10 bar onto 4–20 mA. The loop has a 24 V supply, a 250 Ω receiver resistor and 20 Ω of resistance in each wire. The transmitter needs at least 10 V across its terminals. (a) What current and receiver voltage correspond to 6.5 bar? (b) Is there enough compliance at 20 mA? (c) What is the maximum total loop resistance?

1. (a) I = 4 + 16 × 6.5/10 = 4 + 10.4 = 14.4 mA. Receiver voltage = 14.4 mA × 250 Ω = 3.6 V.
2. (b) Loop resistance outside the transmitter = 250 + 2 × 20 = 290 Ω. At 20 mA it drops 0.02 × 290 = 5.8 V, leaving 24 − 5.8 = 18.2 V ≥ 10 V. Yes.
3. (c) R_loop(max) = (24 − 10)/0.020 = 700 Ω.

Answer: 14.4 mA and 3.6 V at 6.5 bar; 18.2 V available (enough); maximum loop resistance 700 Ω.

Common mistakes

• Grounding one end of the load in the floating-load converter; it then stops being a current source.
• Forgetting the compliance limit: the output current is constant only while the op-amp or transistor is not saturated.
• Mapping 0–100 % onto 0–20 mA instead of 4–20 mA, or using 20 mA instead of the 16 mA span.
• Dropping the minus sign of the transimpedance amplifier.
• Using a bipolar op-amp with nA bias current to measure pA or nA sensor currents.
• Leaving out the feedback capacitor in a high-gain transimpedance amplifier, causing ringing or oscillation.

For GATE IN

Expect load-current and compliance questions for floating-load and Howland converters, transimpedance output for photodiode currents, and 4–20 mA scaling (current for a given measured value, and the reverse). Process-instrumentation questions often combine a transmitter, a loop and a receiver resistor. Practise the linear scaling equation in both directions.

Quick check

1. A 0–100 °C transmitter outputs 4–20 mA. What current at 25 °C?
2. A TIA has R_f = 100 kΩ and input current 20 µA. Output?
3. In a floating-load converter with R₁ = 2 kΩ and V_in = 1 V, what is I_L for R_L = 1 kΩ and for R_L = 3 kΩ?
4. A 4–20 mA loop reads 2 mA. What does that indicate?

Answers: 1. 8 mA. 2. −2 V. 3. 0.5 mA in both cases. 4. A fault (for example a broken wire or failed transmitter), since valid readings never fall below 4 mA.