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Voltage-To-Current Converters

Published On:
May 31, 2016
Last Updated:
Jul 15, 2026

Voltage-to-current converters are circuits which convert an input voltage into an output current. The output current is usually proportional to the input voltage, over some suitable range.

Many of these circuit designs use op-amps. Please see the Op-Amp page for more general information on operational amplifiers.

Howland Current Source

The Howland current source is an op-amp based constant current sink/source. The basic Howland current source schematic is shown below.

A schematic of the basic Howland current source.

How It Works

First, we’ll annotate the schematic with some labels:

The basic Howland current source schematic, annotated with labels.

Appling Kirchhoff’s Current Law (KCL) and Ohm’s Law to the circuit, we get:

IL=I1+I2=VIVLR1+VAVLR2\begin{align*} I_L &= I_1 + I_2 \\ &= \frac{V_I - V_L}{R_1} + \frac{V_A - V_L}{R_2} \end{align*}

We also can form a relationship between the voltage at the op-amps non-inverting input VLV_L and the op-amps output voltage VAV_A. We can see that it forms a non-inverting amplifier with a gain of 1+R4/R31 + R_4 / R_3:

VA=VL(1+R4R3)V_A = V_L \left(1 + \frac{R_4}{R_3}\right)

Now we substitute the expression for VAV_A directly into the KCL equation and simplify. Notice that in the R2R_2 term the VLV_L contributions cancel, leaving just VLR4R3V_L \frac{R_4}{R_3}, after which we group the VLV_L terms:

IL=VIVLR1+VAVLR2=VIVLR1+VL(1+R4R3)VLR2=VIVLR1+VLR4R3R2=VIR1VLR1+VLR4R2R3=VIR1VL(1R1R4R2R3)\begin{align*} I_L &= \frac{V_I - V_L}{R_1} + \frac{V_A - V_L}{R_2} \\ &= \frac{V_I - V_L}{R_1} + \frac{V_L \left(1 + \frac{R_4}{R_3}\right) - V_L}{R_2} \\ &= \frac{V_I - V_L}{R_1} + \frac{V_L \frac{R_4}{R_3}}{R_2} \\ &= \frac{V_I}{R_1} - \frac{V_L}{R_1} + \frac{V_L R_4}{R_2 R_3} \\ &= \frac{V_I}{R_1} - V_L \left(\frac{1}{R_1} - \frac{R_4}{R_2 R_3}\right) \end{align*}

This separates the output current into a driven term and a load-dependent term:

IL=AVIVLRoI_L = A V_I - \frac{V_L}{R_o}

where A=1R1A = \frac{1}{R_1} is the transconductance gain and RoR_o is the output resistance the circuit presents to the load:

Ro=(1R1R4R2R3)1=R1R2R3R2R3R1R4R_o = \left(\frac{1}{R_1} - \frac{R_4}{R_2 R_3}\right)^{-1} = \frac{R_1 R_2 R_3}{R_2 R_3 - R_1 R_4}

The trick here is to realize we can make the load current independent of the load voltage VLV_L by driving RoR_o to infinity (this drives the term VLRo0-\frac{V_L}{R_o} \to 0). This requires balancing the bridge so that the denominator R2R3R1R4R_2 R_3 - R_1 R_4 goes to zero.

The easy way to do this is to make R1=R3R_1 = R_3 and R2=R4R_2 = R_4, which satisfies R2R3=R1R4R_2 R_3 = R_1 R_4 automatically:

R1=R3,R2=R4R_1 = R_3, \quad R_2 = R_4

Setting The Output Current

With the bridge balanced, the load-dependent term drops out and the output current depends only on the input voltage and R1R_1:

IL=AVI=VIR1I_L = A V_I = \frac{V_I}{R_1}

So there are two knobs for setting the current. R1R_1 fixes the transconductance gain A=1R1A = \frac{1}{R_1} (the amps-per-volt), and the input voltage VIV_I then sets the operating point along that slope. Note that R2R_2 (and its match R4R_4) does not appear in this expression — once the bridge is balanced it affects the output compliance and impedance, but not the current.

For example, to source 20mA20\unit{mA} from a 10V10\unit{V} input, we need:

R1=VIIL=10V20mA=500ΩR_1 = \frac{V_I}{I_L} = \frac{10\unit{V}}{20\unit{mA}} = 500\unit{\Omega}

and, to keep the bridge balanced, R3=R1=500ΩR_3 = R_1 = 500\unit{\Omega}. Keep in mind that the op-amp output VAV_A must stay within its linear range — if the load voltage VLV_L climbs high enough to saturate the op-amp, the source can no longer maintain the set current.

The Howland current source is commonly used to build a 4-20mA current-loop transmitter.

Weaknesses

One weaknesses of the basic Howland current source is that it is sensitive to resistor tolerances in the balanced bridge. If the resistors are not well matched, the output current will depend on the load voltage VLV_L and the output resistance RoR_o will be finite. This can be mitigated by using precision resistors or trimming the resistors to balance the bridge.1

History

The “Basic Howland Current Pump” was invented by Prof. Bradford Howland in 1962 when working for MIT. It was first published in January 1964 in the “Lightning Empiricist”, Volume 12, Number 1.1 2

The Howland current source as published in the Lightning Empiricist, Volume 12, Number 1, January 1964.2

Op-amp Current Sink

An op-amp, a MOSFET and a sense resistor make a simple and accurate voltage-controlled current sink. Like the Howland source above it converts an input voltage into a proportional output current, but here the load is referenced to the positive supply rail (making it a current sink rather than a ground-referenced source).

The following schematic shows such a device which can control between 01A0-1\unit{A} through the load (shown as RloadR_{load}):

Schematic of a basic op-amp based current sink.

You set the desired load current by providing a voltage to VinV_{in}. This voltage typically comes from a resistor divider (fixed current), potentiometer (manually variable current) or DAC (digitally variable current). The load current is given by the simple Ohm’s law equation:

Iload=VinRsense\begin{align*} I_{load} = \frac{V_{in}}{R_{sense}} \end{align*}

The circuit works like this:

  1. Desired voltage to set current is provided to VinV_{in} which is applied to the positive input of the op-amp, Vop+V_{op+}.
  2. The op-amp will then drive its output high in an attempt to bring its VopV_{op-} to the same voltage.
  3. As the op-amp raises the voltage on its output, this is connected to the gate of the MOSFET, which will begin to turn it on.
  4. As the MOSFET turns on, current begins to flow through the load and sense resistor, RsenseR_{sense}.
  5. The op-amp will keep turning the MOSFET on until the voltage drop across RsenseR_{sense} is equal to VinV_{in}, meaning VopV_{op-} is the same as Vop+V_{op+}.
  6. This voltage drop will occur when we have the desired amount of current flowing through it, leading to the equation Iload=VinRsenseI_{load} = \frac{V_{in}}{R_{sense}}.

Things to note:

  • The op-amp is powered here with a slightly negative voltage rail on its VSSV_{SS} pin. This is so that the op-amp remains operational when you set it at low current levels. At low current levels, the voltages at Vop+V_{op+} and VopV_{op-} are very close to zero. Even rail-to-rail op-amps can have trouble performing well if the negative voltage rail was at 0V0\unit{V}.
  • The power dissipation through the MOSFET and sense resistor has to be considered. The sense resistor is easy, just make sure it can handle the power given by P=I2RP = I^2 R at the maximum current. The MOSFET power dissipation will depend on the load current and voltage drop across it. The MOSFET is used in its active region --- the region where it is not fully on nor fully off. The MOSFET will drop the remaining voltage from the voltage source provided to the load, once the load voltage drop and sense resistor voltage drop has been subtracted. Use the equation P=VIP = VI to determine the power dissipation in the MOSFET.
  • The gate capacitance of the MOSFET can load the op-amp output to the point that it introduces enough phase lag to cause the circuit to go unstable. See below to recommended compensation circuitry to add to the basic schematic to make the design more stable.

Current Sinking Accuracy

The accuracy of the current sink primarily depends on three aspects:

  • The input offset voltage of the op-amp.
  • The accuracy of the DAC (or other voltage source) providing the voltage to VinV_{in}.
  • The tolerance of the current-sense resistor.

MOSFET Gate Capacitance Compensation

The gate capacitance of the MOSFET can load the op-amp output to the point that it introduces enough phase lag to cause the circuit to go unstable. Compensation circuitry can be added as shown in the below circuit to limit the phase lag and prevent the circuit from becoming unstable.

A schematic showing gate capacitance compensation circuitry on a op-amp based current sink using the LT14923.

See this Electronics Stack Exchange thread for more information on op-amp stability with capacitive loads.4

Pre-built ICs

XTR111

The XTR111 is a current-loop transmitter by Texas Instruments. It can drive grounded loads. It requires both an external BJT and MOSFET. As of May 2016, this IC was US$1.28 (100).

XTR117

The XTR117 is a current-loop transmitter by Texas Instruments. It cannot drive grounded loads. It only requires an external MOSFET. As of May 2016, this IC was US$1.77 (100).

DigiKey stocks a number of pre-built voltage-to-current converter ICs. First, navigate to the following component sub-section on www.digikey.com:

Product Index -> Integrated Circuits (ICs) -> Interface - Sensor and Detector Interfaces.

Then select the filtering options:

  • Type: Current Transmitter
  • Input Type: Voltage

Footnotes

  1. Texas Instruments (2013, Apr). AN-1515 A Comprehensive Study of the Howland Current Pump [application note]. Retrieved 2026-07-14, from https://www.ti.com/lit/an/snoa474a/snoa474a.pdf. 2

  2. George A. Philbrick Researches (1964, Jan). The Lightning Empiricist, Volume 12, Number 1 [periodical]. Retrieved 2026-07-14, from http://www.philbrickarchive.org/1964-1_v12_no1_the_lightning_empiricist.htm. 2

  3. Linear Technology. LT1492/LT1493: 5MHz, 3V/µs, Low Power Single Supply, Dual and Quad Precision Op Amps [datasheet]. Retrieved 2020-12-25, from https://www.analog.com/media/en/technical-documentation/data-sheets/14923f.pdf.

  4. Electronics Stack Exchange (2013, May 16). Stability problem in unity gain opamp [forum post]. Retrieved 2026-06-08, from https://electronics.stackexchange.com/questions/69506/stability-problem-in-unity-gain-opamp.