The Howland current pump, shown in Figure 1a, is a circuit that accepts an input voltage *v _{I}*, converts it to an output current

*i*

_{O}=*Av*

*with*

_{I},*A*as the transconductance gain, and pumps

*i*to a load LD, regardless of the voltage

_{O}*v*

_{L}*developed by the load itself. To see how it works, label it as in Figure 1b, and apply Kirchoff's Current Law and Ohm’s Law.*

_{ }

**Figure 1.** (a) The Howland pump. (b) Properly labeling the circuit for its analysis.

**Figure 1.**(a) The Howland pump. (b) Properly labeling the circuit for its analysis.

*Equation 1*

The op-amp, together with *R _{3} *and

*R*, forms a non-inverting amplifier with respect to

_{4}*v*, thus giving

_{L}*Equation 2*

Substituting *v _{A}* into Equation 1 and collecting, we put

*i*into the insightful form

_{O}*Equation 3*

where *A* is the transconductance gain, in A/V,

*Equation 4*

and where *R _{o} *is the output resistance presented by the circuit to the load,

*Equation 5*

To make *i _{O} *independent of

*v*

_{L}*we must impose*

_{ }*R*→ ∞, or the balanced-bridge condition.

_{o}*Equation 6*

Take a look at the example in Figure 2 and observe, row-by-row, how the op-amp adjusts *i _{2}*, via

*v*, so as to ensure the same current

_{A}*i*regardless of

_{O}*v*.

_{L}

**Figure 2.** (a) A 2 mA current source, and (b) its inner workings for different values of *vL** (voltages in volts, currents in milliamps; a negative current value means that current flows in the direction opposite to the arrow). *

**Figure 2.**(a) A 2 mA current source, and (b) its inner workings for different values of

With the polarity of *V _{REF}* as shown, the pump sources

*i*to the load. Inverting the polarity of

_{O}*V*will cause the pump to sink

_{REF}*i*from the load. Note that for the pump to work properly

_{O}*v*must always be confined within the linear range of op-amp operation. If the op-amp is driven into saturation, the pump will cease to operate properly.

_{A}

### The Effect of Resistance Mismatches

A practical bridge is likely to be unbalanced because of resistance tolerances, so *R _{o}* is likely to be less than infinity. Denoting the tolerances of the resistances in use by

*p*, we note that the denominator

*D*of Equation 5 is maximized when

*R*and

_{2}*R*are maximized and

_{3}*R*and

_{1}*R*are minimized. For

_{4}*p*<< 1, we write

Here we have incorporated the relationship of Equation 6, applied approximation

and ignored quadratic terms in *p*. Substituting into Equation 5 gives

*Equation 7*

As an example, using 1% (p = 0.01) resistances in Figure 2a can lower *R _{o}* from ∞ to as little as 1,000/(4×0.01) = 25 kΩ, thus making

*i*depend upon

_{O}*v*, by Equation 3. If the bridge is unbalanced in the opposite direction of above, then the worst-case condition for

_{L}*R*is –25 kΩ. So, depending on the mismatch,

_{o}*R*may lie anywhere from +25 kΩ to ∞ to –25 kΩ.

_{o}

**Figure 3.** (a) Using a potentiometer R_{p} to balance the resistive bridge. (b) Calibration set up.

**Figure 3.**(a) Using a potentiometer R

_{p}to balance the resistive bridge. (b) Calibration set up.

For improved performance, we must either use lower-tolerance resistances or balance the bridge using a potentiometer *R _{p}*, as in Figure 3a. To calibrate the circuit, ground the input as in Figure 3b and use an ammeter A. First, flip the switch to ground, and if necessary, zero the op-amp’s input offset voltage until the ammeter reads zero. Then flip the switch to a known voltage, such as 5V, and adjust

*R*until the ammeter reads again zero. By imposing that

_{p}*i*with

_{O}*v*= 5 V be equal to

_{L}*i*with

_{O}*v*= 0 V, we are making

_{L}*i*independent of

_{O}

*v*, in effect driving

_{L}*R*to infinity, by Equation 3.

_{o}

### The Effect of Op-Amp Nonidealities

#### Common-Mode Rejection Ratio

A practical op-amp is sensitive to its common-mode input voltage, a feature that is modeled with a small internal offset voltage in series with the noninverting input. In the case of the Howland pump, this offset voltage can be expressed as *v _{L}*/CMRR, where CMRR is the common-mode rejection ratio as reported in the op-amp’s datasheet. With reference to Figure 4a, we note that Equation 1 still holds, but Equation 2 changes to

Substituting into Equation 1, solving for* i _{O}*, and putting

*i*in the form of Equation 3 gives

_{O}*Equation 8*

As an example, using an op-amp with CMRR = 60 dB (=1000) in the above example will lower* R _{o}* from ∞ to (10

^{3}||10

^{3})×1000 = 500 kΩ. With an arrangement of the type of Figure 3b, we can use the potentiometer to compensate for the cumulative effect of bridge imbalance as well as non-infinite CMRR.

#### Open-Loop Gain

So far we have assumed the op-amp to have infinite open-loop gain. The gain *a* of a practical op-amp is finite, so let us now see how this affects circuit behavior.

**Figure 4.** Circuits to investigate the effect of (a) non-infinite common-mode rejection ratio and (b) non-infinite open-loop gain.

**Figure 4.**Circuits to investigate the effect of (a) non-infinite common-mode rejection ratio and (b) non-infinite open-loop gain.

With reference to Figure 4b, we now have

Solving for *v _{A}*, substituting into Equation 1, solving for

*i*, and putting i

_{O}_{O}in the form of Equation 3 gives

*Equation 9*

As an example, using an op-amp with a DC gain of 100 dB (=100,000 V/V) will lower *R _{o}* from ∞ to (10

^{3}||10

^{3})×(1 + 100,000/2) ≅ 25 MΩ. With an arrangement of the type of Figure 3b, we can use the potentiometer to compensate for the cumulative effect of bridge imbalance, non-infinite CMRR, and non-infinite open-loop DC gain, and raise

*R*as close as possible to ∞.

_{o}However, as we increase the frequency of operation, the gain a rolls off with frequency, leading to a progressive deterioration of *R _{o}*. For example, if an op-amp with a DC gain of 100 dB has a gain-bandwidth product of 1 MHz, its open-loop gain vs. frequency (assuming a single-pole response) will look like this:

**Figure 5.** *Single-pole** frequency response of a 1 MHz op-amp with a DC open-loop gain of 100 dB.*

**Figure 5.**

Thus, the gain *a* drops to 60 dB (=1000 V/V) at 1 kHz, and the value of *R _{o}* will drop to 500×(1 + 1000/2) ≅ 250 kΩ. At 10 kHz

*R*drops to 500×(1 + 100/2) ≅ 25 kΩ, and so on.

_{o}

#### Further Reading

A Comprehensive Study of the Howland Current Pump (PDF): an application note published by Texas Instruments.

3 CommentsLoginMyla Iza2019-05-22For reducing the mismatch, how precise does my ammeter have to be? Because setting mine to 40u, current seems to be zero. Still, there’s a lot of current variation as the voltage changes.

Sergio Franco2019-05-22Are you sure your op amp does not saturate or even oscillate as toy vary vL?

Myla Iza2019-05-22It does not saturate. I have a square wave as the input. Changing the amplitude of the square wave, the output changes accordingly and everything seems fine. But when I change the load (220, 1k, 10k), the output current changes considerably.